Here we will state and prove the Spectral Theorem for Bounded, Self-Adjoint Operators. This theorem sits at the core of much of AQFT, and extensive use of this theorem will be made in subsequent work.

Here, we generally follow the clear, straightforward presentation of Quantum Theory for Mathematicians.

A note on conventions. Throughout this post, results that are standard and whose proofs lie outside the scope of the development—for instance the Heine–Borel Theorem, the Riesz Representation Theorem, the Stone–Weierstrass Theorem, and the Bounded Convergence Theorem—are stated in full but not proven. They are marked as such by the absence of an accompanying Proof. Every other result stated here is proven in full.

Spectral Theorem: Bounded Self-Adjoint Operators

In this section we will state and prove the Spectral Theorem for bounded, self-adjoint operators. However, we must introduce “extensive machinery” before we are able to state and prove the theorem. To that end we begin by examining some properties of bounded operators.

Elementary Properties of Bounded Operators

In this section we will introduce and prove some relatively “elementary” properties of bounded operators that will be of use when proving the Spectral Theorem for bounded, self-adjoint operators. We begin by introducing some notation

Definition (Bounded Operator Notation) We notate the set of operators on a separable, complex Hilbert space \(\mathbf{H}\) that are bounded with respect to the operator norm as \(\mathcal{B}(\mathbf{H})\).

Before proceeding we record two elementary continuity properties of the Hilbert space norm and inner product. Both are used repeatedly throughout this post, and so we state them once here rather than re-deriving them at each point of use. Their proofs rest on two standard inequalities, which we state first.

Proposition (Cauchy–Schwarz Inequality) If \(V\) is a space with an inner product, then for all \(\phi, \psi \in V\), we have the Cauchy–Schwarz inequality

\[\lvert \left< \phi, \psi \right> \rvert^2 \le \left< \phi, \phi \right> \left< \psi, \psi \right>.\]

Equivalently, writing \(\left\| \phi \right\| \equiv \sqrt{\left< \phi, \phi \right>}\) for the norm induced by the inner product, one has

\[\lvert \left< \phi, \psi \right> \rvert \le \left\| \phi \right\| \left\| \psi \right\|.\]

Proposition (Reverse Triangle Inequality) Let \(V\) be a normed space. Then for all \(\phi, \psi \in V\)

\[\left| \left\| \phi \right\| - \left\| \psi \right\| \right| \le \left\| \phi - \psi \right\|.\]

With these stated, we turn to the continuity properties themselves.

Proposition (Continuity of the Norm and Inner Product) Let \(\mathbf{H}\) be a Hilbert space and let \(\{ \psi_n \}_{n \in \mathbb{N}}\) be a sequence in \(\mathbf{H}\) converging to \(\psi \in \mathbf{H}\). Then

1. (Continuity of the norm) \(\left\| \psi \right\| = \lim\limits_{n \rightarrow \infty} \left\| \psi_n \right\|\).
2. (Continuity of the inner product) For any fixed \(\phi \in \mathbf{H}\), \(\left< \psi, \phi \right> = \lim\limits_{n \rightarrow \infty} \left< \psi_n, \phi \right>\) and \(\left< \phi, \psi \right> = \lim\limits_{n \rightarrow \infty} \left< \phi, \psi_n \right>\).

Proof Part 1: By the Reverse Triangle Inequality, for any \(n \in \mathbb{N}\)

As \(\psi_n \rightarrow \psi\) the righthand side tends to \(0\), and hence \(\left\| \psi_n \right\| \rightarrow \left\| \psi \right\|\), the desired Part 1 result.

Part 2: By the Cauchy–Schwarz Inequality, for any \(n \in \mathbb{N}\)

where we have used the additivity of the inner product in its first argument. As \(\psi_n \rightarrow \psi\) and \(\left\| \phi \right\|\) is a fixed finite real number, the righthand side tends to \(0\), giving the first claim. An identical argument, using additivity in the second argument, gives the second claim. This is the desired Part 2 result.

Combining Part 1 and Part 2 gives the desired result.\(\blacksquare\)

With that stated, we now give an “elementary” lemma that proves \(\mathcal{B}(\mathbf{H})\) is a Banach space

Lemma (Bounded Operators form a Banach Space) \(\mathcal{B}(\mathbf{H})\) forms a Banach space under the operator norm.

Proof By definition a Banach space is a normed vector space that is complete with respect to the distance function associated to its norm. Hence, we must prove that \(\mathcal{B}(\mathbf{H})\) is a normed vector space that is complete with respect to the distance function associated to its norm.

The norm we place on \(\mathcal{B}(\mathbf{H})\) is the operator norm. The operator norm is indeed a norm. Hence, \(\mathcal{B}(\mathbf{H})\) is normed.

Next we must prove that \(\mathcal{B}(\mathbf{H})\) is a vector space. Consider \(A, B \in \mathcal{B}(\mathbf{H})\) as well as \(\alpha, \beta \in \mathbb{C}\). As the operator norm is a norm we have

where the final step follows from the fact that \(A, B \in \mathcal{B}(\mathbf{H})\) and thus \(\|A\|\) and \(\|B\|\) are bounded with respect to the operator norm. This implies that \(\alpha A + \beta B\) is bounded and thus a member of \(\mathcal{B}(\mathbf{H})\). This in turn implies that \(\mathcal{B}(\mathbf{H})\) is a vector space.

Finally we must prove that \(\mathcal{B}(\mathbf{H})\) is complete with respect to the distance function associated to the operator norm.

Consider a Cauchy sequence \(\{A_i\}_{i \in \mathbb{N}}\) in \(\mathcal{B}(\mathbf{H})\). The definition of Cauchy sequence implies that for any \(\epsilon > 0\) there exists an \(N \in \mathbb{N}\) such that for all \(i,j \ge N\) we have

Fix an arbitrary \(\psi \in \mathbf{H}\). With this fixed \(\psi \in \mathbf{H}\) consider the sequence \(\{A_i\psi\}_{i \in \mathbb{N}}\) in \(\mathbf{H}\). For any \(i,j \in \mathbb{N}\) we have

where the first line follows from linearity and the second line follows from the operator norm definition. As \(\{A_i\}_{i \in \mathbb{N}}\) is a Cauchy sequence, for any \(\epsilon > 0\) there exists an \(N \in \mathbb{N}\) such that for all \(i,j \ge N\) we have

Thus for such \(\epsilon\), \(i\), and \(j\) we have

This implies that \(\{A_i\psi\}_{i \in \mathbb{N}}\) is a Cauchy sequence in \(\mathbf{H}\).

In particular for the case \(\|\psi\| \neq 0\) we can choose the \(\epsilon\) to be \(\epsilon = \epsilon' / \|\psi\|\), where \(\epsilon' > 0\) is arbitrary. Using this the previous equation takes the form

proving that for an arbitrary \(\epsilon' > 0\) there exists an \(N \in \mathbb{N}\) such that for all \(i,j \ge N\) we have

In other words for \(\|\psi\| \neq 0\) the sequence \(\{A_i\psi\}_{i \in \mathbb{N}}\) is a Cauchy sequence in \(\mathbf{H}\).

In the case \(\|\psi\| = 0\) the sequence \(\{A_i\psi\}_{i \in \mathbb{N}}\) consists only of zeros. Hence, it’s trivially a Cauchy sequence.

As \(\mathbf{H}\) is complete, this Cauchy sequence \(\{A_i\psi\}_{i \in \mathbb{N}}\) converges to an element of \(\mathbf{H}\). With this knowledge, we can define a map \(A : \mathbf{H} \rightarrow \mathbf{H}\) pointwise as follows

As we have just proven, the fact that \(\{A_i\psi\}_{i \in \mathbb{N}}\) is a Cauchy sequence implies this map is well-defined.

Linearity of \(A\) follows from the fact that each \(\{A_i\}_{i \in \mathbb{N}}\) is linear. In more detail, given \(\psi,\phi \in \mathbf{H}\) and \(\alpha,\beta \in \mathbb{C}\) we have

proving \(A\) is linear.

Next let us prove that \(\{A_i\psi\}_{i \in \mathbb{N}}\) converges to \(A\) in the operator norm.

As we previously established, for a fixed \(\psi \in \mathbf{H}\) and any \(i,j \in \mathbb{N}\) we have

If this fixed \(\psi\) is such that \(\|\psi\| = 1\), then for any \(\epsilon > 0\) there exists an \(N \in \mathbb{N}\) such that for all \(i,j \ge N\) we have

Fixing an \(i \ge N\), recall that \(A\) was defined pointwise by \(A\psi \equiv \lim\limits_{j \rightarrow \infty} A_j\psi\), so that the sequence \(\{ A_i\psi - A_j\psi \}_{j \in \mathbb{N}}\) converges in \(\mathbf{H}\) to \(A_i\psi - A\psi\). Hence, taking the limit as \(j \rightarrow \infty\) we have, as a result of Part 1 of Proposition (Continuity of the Norm and Inner Product) and our previous result

As \(\epsilon\) is independent of \(\psi\), this limit holds uniformly for all \(\psi \in \mathbf{H}\) that satisfy \(\|\psi\| = 1\). Thus we can take the supremum over such \(\psi\) to obtain

This proves that \(\{A_i\}_{i \in \mathbb{N}}\) converges to \(A\) in the operator norm.

Finally let us prove that \(A\) is bounded. For fixed \(i \ge N\) we have

where the second line follows from the definition of a norm, the third from the fact that \(\{A_i\}_{i \in \mathbb{N}}\) converges to \(A\) in the operator norm, and the final from the fact that \(A_i\) is an element of \(\mathcal{B}(\mathbf{H})\) and is thus bounded. So we conclude that \(\|A\| < \infty\) and thus \(A\) is bounded.

So in conclusion every Cauchy sequence \(\{A_i\}_{i \in \mathbb{N}}\) in \(\mathcal{B}(\mathbf{H})\) converges in the operator norm to an operator \(A\) that is linear and bounded, and thus \(A\) is an element of \(\mathcal{B}(\mathbf{H})\). This implies that \(\mathcal{B}(\mathbf{H})\) is complete with respect to the distance function associated to the operator norm, the desired result.

So in summary we have proven that \(\mathcal{B}(\mathbf{H})\) is a normed vector space that is complete with respect to the operator norm. Thus \(\mathcal{B}(\mathbf{H})\) is a Banach space, the desired result. \(\blacksquare\)

With this first, elementary result out of the way, our next step is the introduction of several definitions required by the Spectral Theorem. The first of these is the notion of “bounded inverse”

Definition (Bounded Inverse) The bounded inverse of \(A \in \mathcal{B}(\mathbf{H})\) is an element \(B \in \mathcal{B}(\mathbf{H})\) such that \(AB = BA = \mathbf{1}\), where \(\mathbf{1} \in \mathcal{B}(\mathbf{H})\) is the multiplicative identity element.

This is required to define the “spectrum” of an operator, which is required by much of what follows

Definition (Resolvent and Spectrum) For \(A \in \mathcal{B}(\mathbf{H})\), the resolvent set of \(A\), denoted as \(\rho(A)\), is the set of all \(\lambda \in \mathbb{C}\) such that the operator \((A - \lambda \mathbf{1})\) has a bounded inverse. The spectrum of \(A\), denoted by \(\sigma(A)\), is the complement of \(A\)’s resolvent set \(\rho(A)\) in \(\mathbb{C}\). For \(\lambda\) in the resolvent set of \(A\) the bounded inverse of \((A - \lambda \mathbf{1})\), which we notate as \((A - \lambda \mathbf{1})^{-1}\), is called the resolvent of \(A\) at \(\lambda\).

Spectral Theorem for Bounded Self-Adjoint Operators

In this section we will actually be able to state the Spectral Theorem. However, we will only be able to do so after introducing “substantial machinery” related to “projection-valued measures”.

Projection-Valued Measures

“Projection-valued measures” are “core” to the Spectral Theorem. Basically, they generalize the notion of a measure. A “projection-valued measure”, instead of taking on positive, real-values as a standard measure does, takes on “bounded orthogonal projection” values. Formally, we define this by first introducing the notion of a “bounded orthogonal projection”

Definition (Orthogonal Projection) A bounded orthogonal projection, sometimes shortened to orthogonal projection or simply projection, is an element \(P \in \mathcal{B}(\mathbf{H})\) such that \(P^2 = P\) and \(P^* = P\).

The notion of a bounded orthogonal projection can then be employed to define a “projection-valued measure”

Definition (Projection-Valued Measure) Let \(X\) be a set and \(\Omega(X)\) a \(\sigma\)-algebra on \(X\). A map \(\mu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) is called a projection-valued measure if the following properties are satisfied:

1. For each \(E \in \Omega(X)\), it follows that \(\mu(E)\) is a bounded orthogonal projection.
2. \(\mu(\emptyset) = 0\), where \(\emptyset \in \Omega(X)\) is the empty set, and \(\mu(X) = \mathbf{1}\), where \(\mathbf{1}\) is the multiplicative identity element.

3. If \(E_1\), \(E_2\), \(E_3\)… in \(\Omega(X)\) are pairwise disjoint, then for all \(v \in \mathbf{H}\), we have

   \[\mu \left( \bigcup_{j = 1}^{\infty} E_j \right) v = \sum_{j = 1}^{\infty} \mu(E_j)v,\]

   where the convergence of the sum is in the norm topology on \(\mathbf{H}\).

4. For all \(E_1, E_2 \in \Omega(X)\), we have \(\mu(E_1 \cap E_2) = \mu(E_1) \mu(E_2)\).

Two projection-valued measures \(\mu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) and \(\nu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) are said to be equivalent if \(\mu(E) = \nu(E)\) for all \(E \in \Omega(X)\).

Now, we can associate a positive, real-valued measure \(\mu_\psi\) to a projection-valued measure \(\mu\) and any \(\psi \in \mathbf{H}\) as follows:

Theorem (Projection-Valued Measure’s Associated Measure) Given a projection-valued measure \(\mu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) and any \(\psi \in \mathbf{H}\) the map \(\mu_\psi\) defined by

\[\begin{align} \mu_\psi : \Omega(X) &\longrightarrow \mathbb{R} \\ E &\longmapsto \mu_\psi(E) \equiv \left< \psi, \mu(E) \psi \right> \end{align}\]

defines a positive, real-valued measure \(\mu_\psi\) on \(\Omega(X)\).

Proof To prove that \(\mu_\psi\) defines a positive, real-valued measure on \(X\) with \(\sigma\)-algebra \(\Omega(X)\) we must prove

• Empty set is of measure zero - \(\mu_\psi(\emptyset) = 0\), where \(\emptyset\) is the empty set.
• Non-negativity - For all \(E \in \Omega(X)\), it follows that \(\mu_\psi(E) \ge 0\).
• Countable additivity - For pairwise disjoint \(E_1\), \(E_2\), \(E_3\)… in \(\Omega(X)\)

Let us first prove the empty set is of measure zero, i.e. \(\mu_\psi(\emptyset) = 0\). The definition of \(\mu_\psi\) along with the definition of the projection-valued measure \(\mu\) imply

Hence, the empty set is of measure zero \(\mu_\psi(\emptyset) = 0\) as desired.

Next let us prove non-negativity, i.e. for all \(E \in \Omega(X)\), it follows that \(\mu_\psi(E) \ge 0\). The definition of \(\mu_\psi\), the projection-valued measure \(\mu\), and of an orthogonal projection imply

where the final step follows from the definition of an inner product. Thus, \(\mu_\psi(E) \ge 0\) as desired.

Finally let us prove countable additivity. Let \(E_1\), \(E_2\), \(E_3\)… in \(\Omega(X)\) be pairwise disjoint. The definition of \(\mu_\psi\) along with the definition of the projection-valued measure \(\mu\) imply

where the third equality follows from the definition of an inner product. This implies

which is the desired result.

Together these imply that \(\mu_\psi\) defines a positive, real-valued measure \(\mu_\psi\) on \(\Omega(X)\). \(\blacksquare\)

Projection-valued measures give rise to a type of integration known as “operator-valued integration”. The primary properties of “operator-valued integration” are described by the following theorem

Theorem (Operator-Valued Integration) Let \(\Omega(X)\) be a \(\sigma\)-algebra on a set \(X\) and let \(\mu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) be a projection-valued measure. Then there exists a unique linear map, denoted by

\[f \longmapsto \int_X f \, d\mu,\]

from the space of bounded, measurable, complex-valued functions on \(X\) into \(\mathcal{B}(\mathbf{H})\) such that

\[\left< \psi, \left( \int_X f \, d\mu \right) \psi \right> = \int_X f d\mu_\psi,\]

for all \(f\) and \(\psi \in \mathbf{H}\), where \(\mu_\psi\) is the positive real-valued measure of Theorem (Projection-Valued Measure’s Associated Measure) and \(\left< \cdot, \cdot \right>\) is the Hilbert space inner product on \(\mathbf{H}\). This unique linear map has the following additional properties

1. For all \(E \in \Omega(X)\), we have

   \[\int_X 1_E \, d\mu = \mu(E),\]

   where \(1_E\) is the indicator function of \(E\). In particular, the integral of the constant function \(1\) is the multiplicative identity \(\mathbf{1}\).

2. For all bounded, measurable, complex-valued functions \(f\) on \(X\), we have

   \[\left\| \, \int_X f \, d\mu \, \right\| \le \sup\limits_{\lambda \in X} \left| f(\lambda) \right|,\]

   where \(\| \cdot \|\) is the operator norm and \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\).

3. Integration is multiplicative: For all bounded, measurable, complex-valued functions \(f\) and \(g\) on \(X\), we have

   \[\int_X fg \, d\mu = \left( \int_X f \, d\mu \right) \left( \int_X g \, d\mu \right).\]

4. For all bounded, measurable, complex-valued functions \(f\) on \(X\), we have

   \[\int_X \overline{f} \, d\mu = \left( \int_X f \, d\mu \right)^*,\]

   where \(\overline{f}\) is the complex conjugate of \(f\) and the superscript \(*\) denotes the adjoint on \(\mathcal{B}(\mathbf{H})\) arising from the Hilbert space inner product. In particular, if \(f\) is real-valued, then \(f = \overline{f}\) and

   \[\left( \int_X f \, d\mu \right) = \left( \int_X \overline{f} \, d\mu \right) = \left( \int_X f \, d\mu \right)^*\]

   is self-adjoint.

Proof

To streamline the proof of this theorem, we will introduce a few new terms

Definition ((Bounded) Sesquilinear Form) A sesquilinear form on a Hilbert space \(\mathbf{H}\) is a map \(L : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) that is conjugate linear in the first factor and linear in the second factor. A sesquilinear form \(L\) is a bounded sesquilinear form if there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi, \psi \in \mathbf{H}\)

\[|L(\phi, \psi)| \le C \|\phi\| \, \|\psi\|,\]

where \(\mid\cdot\mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

Definition ((Bounded) Quadratic Form) A quadratic form on a Hilbert space \(\mathbf{H}\) is a map \(Q : \mathbf{H} \rightarrow \mathbb{C}\) with the following properties:

1. \(Q(\lambda\psi) = \mid\lambda\mid^2 Q(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\).

2. The map \(L : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

   \[\begin{align} L(\phi, \psi) &\equiv \frac{1}{2} \left[ Q(\phi + \psi) - Q(\phi) - Q(\psi) \right] \\ &-\frac{i}{2} \left[ Q(\phi + i\psi) - Q(\phi) - Q(i\psi) \right] \end{align}\]

   is a sesquilinear form on \(\mathbf{H}\).

A quadratic form \(Q\) is a bounded quadratic form if there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

\[|Q(\phi)| \le C \|\phi\|^2,\]

where \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

These will now let us begin the proof of Theorem (Operator-Valued Integration)

By hypothesis we have a projection-valued measure \(\mu\). Consider any bounded, measurable, complex-valued function \(f\) on \(X\) and any \(\psi \in \mathbf{H}\). With this, let us define a map \(Q_f : \mathbf{H} \rightarrow \mathbb{C}\) by

where \(\mu_\psi\) is the positive real-valued measure of Theorem (Projection-Valued Measure’s Associated Measure). It turns out that such a \(Q_f(\psi)\) is a bounded quadratic form which we now prove

Lemma Let \(\Omega(X)\) be a \(\sigma\)-algebra on a set \(X\) and let \(\mu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) be a projection-valued measure. For any bounded, measurable, complex-valued function \(f\) on \(X\) and any \(\psi \in \mathbf{H}\) the map \(Q_f : \mathbf{H} \rightarrow \mathbb{C}\) defined by

\[Q_f(\psi) \equiv \int_X f \, d\mu_\psi,\]

where \(\mu_\psi\) is the positive real-valued measure of Theorem (Projection-Valued Measure’s Associated Measure), is a bounded quadratic form.

Proof To prove this result we will first prove the result for indicator functions, then for simple functions, and finally for bounded, measurable, complex-valued functions.

Let us start this proof by proving it is true for the case of indicator functions. Consider an arbitrary \(E \in \Omega(X)\) and its indicator function \(1_E\). In this case the definition of \(Q_{1_E}\), standard properties of integration, and the definition of \(\mu_\psi\) implies

So, \(Q_{1_E}(\psi) = \left< \psi, \mu(E) \psi \right>\).

To prove that this \(Q_{1_E}\) is a bounded quadratic form we must prove that

1. \(Q_{1_E}(\lambda\psi) = \mid\lambda\mid^2 Q_{1_E}(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\).

2. The map \(L_{1_E} : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

   \[\begin{align} L_{1_E}(\phi, \psi) &\equiv \frac{1}{2} \left[ Q_{1_E}(\phi + \psi) - Q_{1_E}(\phi) - Q_{1_E}(\psi) \right] \\ &-\frac{i}{2} \left[ Q_{1_E}(\phi + i\psi) - Q_{1_E}(\phi) - Q_{1_E}(i\psi) \right] \end{align}\]

   is a sesquilinear form on \(\mathbf{H}\).

3. There exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

   \[|Q_{1_E}(\phi)| \le C \|\phi\|^2,\]

   where \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

Let us first prove that \(Q_{1_E}(\lambda\psi) = \mid\lambda\mid^2 Q_{1_E}(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\). As a result of our previous derivation, the definition of an inner product, and the definition of the norm on \(\mathbb{C}\) one has

which proves \(Q_{1_E}(\lambda\psi) = \mid\lambda\mid^2 Q_{1_E}(\psi)\), the desired result.

Next let us prove the map \(L_{1_E} : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined above is a sesquilinear form on \(\mathbf{H}\). As a result of our previous derivation and the definition of \(L_{1_E}\) we have

for all \(\phi, \psi \in \mathbf{H}\). Using the definition of an inner product, this simplifies as follows

which implies

The definition of an inner product then implies that \(L_{1_E}(\phi, \psi)\) is conjugate linear in the first factor and linear in the second factor. Thus \(L_{1_E}(\phi, \psi)\) is a sesquilinear form, the desired result. The results up until this point prove that \(Q_{1_E}\) is a quadratic form.

Finally, to prove that \(Q_{1_E}\) isn’t only a quadratic form but is a bounded quadratic form we must prove that there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

Using the results of our previous derivation, the definition of an orthogonal projection, and standard properties of an inner product and its associated norm we have

where the final inequality follows from the fact that \(\mu(E)\) is an orthogonal projection. This proves that

which implies that the constant required to prove that \(Q_{1_E}\) is a bounded quadratic form is simply \(1\). This concludes the proof of the indicator function result, \(Q_{1_E}\) is a bounded quadratic form for any indicator function \(1_E\).

Next we will prove that any simple function \(s\), i.e. any finite linear combination of indicator functions

where \(\alpha_i \in \mathbb{C}\) and \(E_i \in \Omega(X)\) are pairwise disjoint, also results in a bounded quadratic form \(Q_s\).

In this case, following a logic similar to the indicator function case, we have

So in summary

To prove that such a \(Q_s\) is a bounded quadratic form we must prove the same three results.

First we must prove that \(Q_s(\lambda\psi) = \mid\lambda\mid^2 Q_s(\psi)\). This follows from our indicator function result

giving the desired result \(Q_s(\lambda\psi) = \mid\lambda\mid^2 Q_s(\psi)\).

Next we must prove the map \(L_s : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

is a sesquilinear form on \(\mathbf{H}\). Basically this result follows from linearity and our indicator function result.

As

we have

From our indicator function result we know that each \(L_{1_{E_i}}(\phi, \psi)\) is conjugate linear in the first factor and linear in the second factor. Hence, \(L_s(\phi, \psi)\) is conjugate linear in the first factor and linear in the second factor. Thus \(L_s(\phi, \psi)\) is a sesquilinear form on \(\mathbf{H}\), the desired result.

Finally, we must prove that there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

This again follows from linearity and our indicator function result. We have

This gives the desired result

for

This completes the proof that \(Q_s\) is a bounded quadratic form for any simple function \(s\).

Next we will prove that for any bounded, measurable, complex-valued function \(f\) the map \(Q_f : \mathbf{H} \rightarrow \mathbb{C}\) defined by

is a bounded quadratic form. This proof relies upon our previous simple function result along with the Complex-Valued Simple Approximation Theorem

Theorem (Complex-Valued Simple Approximation Theorem) Let \(X\) be a measurable space, i.e. a set equipped with a \(\sigma\)-algebra \(\Omega(X)\). Given any bounded, measurable, complex-valued function \(f\) on \(X\), there exists a sequence of complex-valued simple functions \(\{s_i\}_{i \in \mathbb{N}}\) on \(X\)—i.e. functions of the form \(s_i = \sum_{j = 1}^{n_i} \alpha_{ij} 1_{E_{ij}}\) with \(\alpha_{ij} \in \mathbb{C}\) and \(E_{ij} \in \Omega(X)\) pairwise disjoint—such that \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to \(f\) on \(X\).

To wit we must first prove that \(Q_f(\lambda\psi) = \mid\lambda\mid^2 Q_f(\psi)\). This follows from our simple function result and the Complex-Valued Simple Approximation Theorem. One has

where the second equality follows from the Complex-Valued Simple Approximation Theorem, the third from the fact that \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to \(f\) and thus the limit can be pulled out of the integral, the fourth from the definition of \(Q_{s_i}\), the fifth from our simple function result, and the final as before from uniform convergence and the Complex-Valued Simple Approximation Theorem. So in summary we have proven that

which is the first desired result.

A similar argument implies that \(L_f\) defined by

is a sesquilinear form on \(\mathbf{H}\). In particular, using the same argument as above but for \(\lambda = 1\) we have

Hence,

This implies

which—as a result of our simple function proof that \(L_{s_i}(\phi, \psi)\) is conjugate linear in the first factor and linear in the second factor—implies that \(L_f(\phi, \psi)\) is conjugate linear in the first factor and linear in the second factor, and thus a sesquilinear form on \(\mathbf{H}\).

Finally, we must prove that there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

for all \(\phi \in \mathbf{H}\). It is to this we now turn.

The definition of \(Q_f\), triangle inequality for integrals, and the fact that \(f\) is bounded imply

However, the definitions of \(\mu_\phi\) and \(\mu\) imply

where the final step uses the definition of the Hilbert space norm in terms of the Hilbert space inner product.

The previous two results together imply

which, if we make the identification

is nothing more than the statement that there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

for all \(\phi \in \mathbf{H}\), the desired result.

This concludes our proof that for any bounded, measurable, complex-valued function \(f\) on the set \(X\) with \(\sigma\)-algebra \(\Omega(X)\) the map \(Q_f\) is a bounded quadratic form. \(\blacksquare\)

Our next step in the larger proof is establishing several propositions we will have need of later in our argument. To wit let us first prove the proposition (Proposition A.61 of Hall)

Proposition If \(Q\) is a quadratic form on \(\mathbf{H}\) and \(L\) is the associated sesquilinear form

\[\begin{align} L(\phi, \psi) &\equiv \frac{1}{2} \left[ Q(\phi + \psi) - Q(\phi) - Q(\psi) \right] \\ &-\frac{i}{2} \left[ Q(\phi + i\psi) - Q(\phi) - Q(i\psi) \right], \end{align}\]

then we have the following results

1. For all \(\psi \in \mathbf{H}\), we have \(Q(\psi) = L(\psi, \psi)\).
2. If \(Q\) is bounded, then \(L\) is bounded.

3. If \(Q(\psi)\) belongs to \(\mathbb{R}\) for all \(\psi \in \mathbf{H}\), then \(L\) is conjugate symmetric, that is

   \[L(\phi, \psi) = \overline{L(\psi, \phi)}\]

   for all \(\phi, \psi \in \mathbf{H}\)

Proof Let us first prove for all \(\psi \in \mathbf{H}\), we have \(Q(\psi) = L(\psi, \psi)\).

The definition of a quadratic form along with standard properties of the norm \(\mid \cdot \mid\) imply

the desired result \(L(\psi, \psi) = Q(\psi)\).

Let us next prove that if \(Q\) is bounded, then \(L\) is bounded.

In this case by hypothesis \(Q\) is bounded. Hence, there exists a \(C\) in \(\mathbb{R}\) such that

for all \(\psi \in \mathbf{H}\).

Consider for the moment \(\phi, \psi \in \mathbf{H}\) such that \(\|\phi\| = 1\) and \(\|\psi\| = 1\). The definition of a norm then implies

which implies \(\|\phi + \psi\| \le 2\) as well as

which implies \(\|\phi + i\psi\| \le 2\).

This, along with the definition of a norm and the fact that \(Q\) is bounded, implies

Now for arbitrary \(\phi, \psi \in \mathbf{H}\) that need not satisfy \(\|\phi\| = 1\) and \(\|\psi\| = 1\), we can obviously always find unit vectors \(\widehat{\phi}\) and \(\widehat{\psi}\) in \(\mathbf{H}\) such that

As the definition of a quadratic form implies that \(L\) is sesquilinear, we thus have for these arbitrary \(\phi, \psi \in \mathbf{H}\)

where the inequality follows from our previous result. Thus we have proven the second desired result

that \(L\) is bounded.

Finally, we will prove that if \(Q(\psi)\) is valued in \(\mathbb{R}\) for all \(\psi \in \mathbf{H}\), then \(L\) is conjugate symmetric.

In this case, by hypothesis \(Q(\psi)\) is valued in \(\mathbb{R}\) for all \(\psi \in \mathbf{H}\). This implies that for arbitrary \(\phi, \psi \in \mathbf{H}\)

If we define \(M(\phi, \psi) \equiv \text{Re} \left[ L(\phi, \psi) \right]\), then we have shown

Now let us derive several basic properties of \(M\).

First one can note that by inspection of

that \(M(\phi, \psi)\) is symmetric, i.e. \(M(\phi, \psi) = M(\psi, \phi)\).

Second, as \(M(\phi, \psi) \equiv \text{Re} \left[ L(\phi, \psi) \right]\) and \(L\) is a sesquilinear form, and thus conjugate linear in the first factor and linear in the second factor, it follows that \(M\) is linear in both factors, i.e. real-bilinear.

Finally, as \(Q\) is a quadratic form, and thus satisfies \(Q(\lambda\psi) = \mid\lambda\mid^2 Q(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\), one has

which implies \(M(i\phi, i\psi) = M(\phi, \psi)\).

Together these basic properties of \(M\) imply

which implies \(M(\phi, i\psi) = -M(\psi, i\phi)\).

Now noting that

we have

where the second to last step follows from our previous result \(M(\phi, i\psi) = -M(\psi, i\phi)\) and the last from the symmetry of \(M\). As \(M\) is real this derivation can be continued as

proving that \(L(\phi, \psi) = \overline{L(\psi, \phi)}\), the desired result, i.e. if \(Q\) is real, then \(L\) is conjugate symmetric.\(\blacksquare\)

The next in the set of “helper” propositions that we will prove is the proposition (Proposition A.63 of Hall)

Proposition If \(Q\) is a bounded quadratic form on \(\mathbf{H}\), there is a unique \(A \in \mathcal{B}(\mathbf{H})\) such that \(Q(\psi) = \left< \psi, A\psi \right>\) for all \(\psi \in \mathbf{H}\). If \(Q(\psi)\) belongs to \(\mathbb{R}\) for all \(\psi \in \mathbf{H}\), then the operator \(A\) is self-adjoint.

Proof By hypothesis \(Q\) is a bounded quadratic form. Hence, as a result of the Proposition we just proved, the sesquilinear form associated to \(Q\) is bounded. This implies that there exists a constant \(C\) in \(\mathbb{R}\) such that

for all \(\phi, \psi \in \mathbf{H}\).

Hence, for any fixed \(\phi \in \mathbf{H}\), the operator norm of the linear map \(\psi \mapsto L(\phi, \psi)\) is bounded, with operator norm at most \(C \|\phi\|\).

Explicitly, \(\psi \mapsto L(\phi, \psi)\) is linear as a result of \(Q\) being a quadratic form, which implies \(L\) is a sesquilinear form, which in turn implies that \(\psi \mapsto L(\phi, \psi)\) is linear in \(\psi\).

Furthermore, the definition of operator norm implies

The fact that \(L\) is bounded as a sesquilinear form implies

for our fixed \(\phi\) and any \(\psi \in \mathbf{H}\). So for any \(\psi\) such that \(\|\psi\| = 1\)

Hence, these together imply

the desired result.

Now as one will recall the Riesz Theorem (Theorem A.52 (Riesz Theorem) of Hall) states

Theorem (Riesz Theorem) If \(\xi : \mathbf{H} \rightarrow \mathbb{C}\) is a bounded linear functional on the Hilbert space \(\mathbf{H}\), then there exists a unique \(\chi \in \mathbf{H}\) such that

\[\xi(\psi) = \left< \chi, \psi \right>\]

for all \(\psi \in \mathbf{H}\). Furthermore, the operator norm of \(\xi\) as a bounded linear functional is equal to the norm of \(\chi\) as an element of \(\mathbf{H}\).

As a result of the Riesz Theorem, for any fixed \(\phi\) there exists a \(\chi\) in \(\mathbf{H}\) such that \(L(\phi, \psi) = \left< \chi, \psi \right>\). In addition, the “operator norm conclusion” of the Riesz Theorem and our finding that the operator norm of \(\psi \mapsto L(\phi, \psi)\) is bounded for any fixed \(\phi\) imply that

As this is true for any \(\phi\), we can use it to define a map \(B : \mathbf{H} \rightarrow \mathbf{H}\) by \(B\phi \equiv \chi\). It turns out that \(B\) is linear and bounded relative to the operator norm.

Boundedness of \(B\) relative to the operator norm follows from

along with \(B\phi = \chi\). These imply

This is simply the statement that \(B\) is bounded in the operator norm.

Linearity of \(B\) follows from explicit calculation. For \(\phi_1, \phi_2 \in \mathbf{H}\) we have by definition

for any \(\psi \in \mathbf{H}\). Similarly, for any \(\alpha_1, \alpha_2 \in \mathbb{C}\) we have by definition

As \(L\) is a sesquilinear form and thus conjugate-linear in its first argument

Hence, for any \(\psi \in \mathbf{H}\)

where the final equality uses the fact that the inner product is conjugate-linear in its first argument. As this is true for any \(\psi \in \mathbf{H}\) it implies

which is none other than the statement of linearity.

It turns out that the unique operator \(A \in \mathcal{B}(\mathbf{H})\) of the proposition, i.e. the operator satisfying \(Q(\psi) = \left< \psi, A \psi \right>\), is given by \(A \equiv B^*\). Let us prove that this is the case.

As a result of the last Proposition we proved,

for all \(\psi \in \mathbf{H}\). The definition of \(B\) implies that

Together these imply

which implies \(Q(\psi) = \left< \psi, B^*\psi \right>\). This in turn implies that \(A \equiv B^*\) is indeed the appropriate definition.

Uniqueness of \(A \equiv B^*\) follows from the uniqueness of the Riesz Theorem. Explicitly, as a result of the Riesz Theorem, for any fixed \(\phi\) there exists a unique \(\chi\) in \(\mathbf{H}\) such that \(L(\phi, \psi) = \left< \chi, \psi \right>\). We then defined the map \(B\) by \(B\phi \equiv \chi\) and the map \(A\) by \(A \equiv B^*\). As \(\chi\) is unique, any other possible \(B'\) one could choose would have to satisfy \(B'\phi = \chi\) too. Hence, \((B - B')\phi = 0\) for all \(\phi \in \mathbf{H}\). This then implies that \(B - B'\) is the zero operator, and thus \(B = B'\), i.e. \(B\) and thus \(A \equiv B^*\) is unique.

Finally we must prove that if \(Q(\psi)\) belongs to \(\mathbb{R}\) for all \(\psi \in \mathbf{H}\), then the operator \(A\) is self-adjoint.

Assuming that \(Q(\psi)\) belongs to \(\mathbb{R}\) for all \(\psi \in \mathbf{H}\), the last Proposition we proved implies that \(L\) is conjugate symmetric,

for all \(\phi, \psi \in \mathbf{H}\). This along with our definition of \(B\) and \(A \equiv B^*\) imply

for all \(\phi, \psi \in \mathbf{H}\), where the final step used the definition of an inner product. This implies

for all \(\phi, \psi \in \mathbf{H}\), which is simply the statement that \(A\) is self-adjoint.\(\blacksquare\)

With these two “helper” propositions proven, we can now join the main thread of the Theorem (Operator-Valued Integration).

As one will recall we had established that for any bounded, measurable, complex-valued function \(f\) the map \(Q_f : \mathbf{H} \rightarrow \mathbb{C}\) defined by

is a bounded quadratic form. Hence, Proposition implies that there is a unique, bounded operator \(A_f \in \mathcal{B}(\mathbf{H})\) such that

for all \(\psi \in \mathbf{H}\). We then define the operator-valued integral of \(f\) as follows

By construction it is a map from the space of bounded, measurable, complex-valued functions to \(\mathcal{B}(\mathbf{H})\), as required.

Property 0: Tracing definitions it is obvious that this satisfies the required property

of an operator-valued integral. Explicitly, the definition of the operator-valued integral along with the definition of \(Q_f\) imply

giving

the desired result.

Property 1: Next we must prove that for all \(E \in \Omega(X)\), we have

where \(1_E\) is the indicator function of \(E\).

Consider then the case \(f = 1_E\). The definition of \(Q_{1_E}\) and that of the measure \(\mu_\psi\) imply

So, \(Q_{1_E}(\psi) = \left< \psi, \mu(E) \psi \right>\).

As we proved, Proposition implies that there is a unique, bounded operator \(A_{1_E} \in \mathcal{B}(\mathbf{H})\) such that

for all \(\psi \in \mathbf{H}\). Hence, \(A_{1_E} = \mu(E)\). Thus the definition of the operator-valued integral implies

which is the desired result.

Property 2: The next result we must prove is that for all bounded, measurable, complex-valued functions \(f\) on \(X\), we have

where \(\| \cdot \|\) is the operator norm and \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\).

To prove this we will first prove a “utility” lemma that will aid our argument.

Lemma Let \(X\) be a set with \(\sigma\)-algebra \(\Omega(X)\), and let \(\mu : \Omega(X) \rightarrow \mathcal{B}(\mathbf{H})\) be a projection-valued measure. If \(E_1, E_2, \ldots, E_n \in \Omega(X)\) are a finite set of elements that are pairwise disjoint and satisfy

\[X = \bigcup\limits_{i = 1}^n E_i,\]

then for any \(\psi \in \mathbf{H}\)

1. The vectors \(\mu(E_1)\psi, \mu(E_2)\psi, \ldots, \mu(E_n)\psi\) in \(\mathbf{H}\) are pairwise orthogonal.

2. The norm \(\|\psi\|\) of \(\psi\) can be written as follows

   \[\|\psi\|^2 = \sum_{i = 1}^n \|\mu(E_i)\psi\|^2.\]

Proof Let us first prove that the vectors \(\mu(E_1)\psi, \mu(E_2)\psi, \ldots, \mu(E_n)\psi\) in \(\mathbf{H}\) are pairwise orthogonal.

Let \(i \neq j\) be indices of the elements \(E_1, E_2, \ldots, E_n\). The projection-valued measure definition along with the definitions of inner product and orthogonal projection imply

proving that \(\left< \mu(E_i)\psi, \mu(E_j)\psi \right> = 0\), the desired pairwise orthogonality.

Next let us prove that the norm \(\|\psi\|\) of \(\psi\) can be written as follows

The definition of a projection-valued measure implies

which implies

As we just proved, these summands are pairwise orthogonal; hence

giving

the desired result and completing the proof of our “utility” lemma.\(\blacksquare\)

One more “utility” lemma we will require is

Lemma Let \(\mathcal{B}(\mathbf{H})\) be the set of operators on a separable, complex Hilbert space \(\mathbf{H}\) that are bounded with respect to the operator norm as \(\mathcal{B}(\mathbf{H})\). For any \(A\) in \(\mathcal{B}(\mathbf{H})\) we can write the operator norm \(\|A\|\) of \(A\) as follows

\[\|A\| = \sup\limits_{\|\chi\| = \|\psi\| = 1} \lvert \left< \chi, A \psi \right> \rvert\]

where \(\chi, \psi \in \mathbf{H}\).

Proof Let us begin this proof by deriving an alternative means of writing the norm of an element \(\psi \in \mathbf{H}\) as

First let us prove this is true for the case \(\psi = 0\).

In this case the definition of a norm implies \(\|\psi\| = 0\). Similarly, for all \(\chi \in \mathbf{H}\) we have \(\lvert \left< \chi, 0 \right> \rvert = 0\). Thus we have the trivial equality in this case: both the lefthand and righthand sides of the desired equation are zero.

Now we can safely assume that \(\psi \neq 0\).

Consider an arbitrary \(\psi \in \mathbf{H}\) and an arbitrary \(\chi \in \mathbf{H}\) such that \(\|\chi\| = 1\). The Cauchy–Schwarz Inequality implies that

Hence, we have

Alternatively, as \(\psi \neq 0\) we can consider a particular \(\chi_0\) of the form

Obviously \(\|\chi_0\| = 1\). With this form for \(\chi_0\) the definitions of norm and inner product imply

Hence, the previous bound we derived

is actually attained for this specific \(\chi_0\). Thus it must be the case that

Now at this point we have proven that

Together these imply the desired result

Now, the definition of the operator norm implies

Our recent result allows us to re-write the norm \(\|A\psi\|\) as

Hence,

This is the final desired result of this lemma.\(\blacksquare\)

With this lemma proven, we can once again continue on with our main argument.

Consider, as in the lemma, a finite set of elements \(E_1, E_2, \ldots, E_n \in \Omega(X)\) that are pairwise disjoint and satisfy

If we are given in addition a set of complex numbers \(c_1, c_2, \ldots, c_n \in \mathbb{C}\), we can define a simple function \(s\) by

As we have already proven,

Linearity of the map

then implies that

Hence, for any \(\phi, \psi \in \mathbf{H}\) we have

where we have used our previous equation for \(A_s\), the definition of an inner product, definition of a projection-valued measure, and the definition of an orthogonal projection to conclude

Now using the definition of a norm and applying Cauchy–Schwarz twice, first to each summand and then across the sum (viewing \(\|\mu(E_i) \phi\|\) and \(\|\mu(E_i) \psi\|\) as vectors in \(\mathbb{R}^n\)), one obtains

where in the final step we employed the second result of Lemma. So in summary

for all \(\phi, \psi \in \mathbf{H}\).

As a result of Lemma we can write the operator norm of \(A_s\) as follows

Hence, our result implies

Obviously

Hence, we have proven the desired result

for our simple function \(s\). What remains to do is to generalize this to a bounded, measurable, complex-valued function \(f\).

This generalization relies upon the Theorem (Complex-Valued Simple Approximation Theorem). This theorem implies that a sequence of complex-valued simple functions \(\{s_i\}_{i \in \mathbb{N}}\) on \(X\) exists such that \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to \(f\) on \(X\).

Now for any two simple functions \(s_i\) and \(s_j\) in this sequence, their difference \(s_i - s_j\) is also a simple function. Furthermore, linearity of the map

implies

and thus

However, the result we just proved for simple functions implies

This in turn implies

However, the definition of a norm implies

Hence, we can continue our derivation

However, as the sequence \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to \(f\), for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(k \ge N\)

This along with our previous derivation allows us to conclude that for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i, j \ge N\) we have

This is nothing more than the statement that \(\{A_{s_i}\}_{i \in \mathbb{N}}\) is a Cauchy sequence.

By construction each \(A_{s_i}\) is an element of \(\mathcal{B}(\mathbf{H})\). Hence, \(\{A_{s_i}\}_{i \in \mathbb{N}}\) is a sequence in \(\mathcal{B}(\mathbf{H})\). As we proved in Lemma (Bounded Operators form a Banach Space) \(\mathcal{B}(\mathbf{H})\) is a Banach space. So, in particular, \(\mathcal{B}(\mathbf{H})\) is complete. Thus there exists an operator \(A_s\) in \(\mathcal{B}(\mathbf{H})\) that is the limit of the sequence \(\{A_{s_i}\}_{i \in \mathbb{N}}\) relative to the operator norm on \(\mathcal{B}(\mathbf{H})\).

As the sequence \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to \(f\), for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(k \ge N\)

Hence, using the definition of a norm, we can conclude that for all \(k \ge N\)

This implies

We can also create a similar derivation switching the roles of \(s_k\) and \(f\) as follows

This implies

The last two conclusions imply

Now tying the last results together

In other words

As

this is almost the desired result

We simply need to identify \(A_f\) with \(A_s\) and we will have completed the proof. It is to this we now turn.

We previously proved

for any \(\psi \in \mathbf{H}\). So in particular as

for any \(s_i\) of our sequence, our result implies

Using this along with the fact that \(\{A_{s_i}\}_{i \in \mathbb{N}}\) converges uniformly to \(A_s\) in the operator norm

where the fact that the convergence is uniform allows us to bring the limit under the integral. Hence,

Now previously we proved that there is a unique linear map

such that

As this map is unique and

it follows that

and thus the desired relation

is proven.

Property 3: Next we must prove that integration is multiplicative. In other words for all bounded, measurable, complex-valued functions \(f\) and \(g\) on \(X\), we have

As in other proofs, we will first prove this result for indicator functions, then simple functions, then finally for bounded, measurable, complex-valued functions.

Let us begin with indicator functions. Consider \(E_1, E_2 \in \Omega(X)\). Property 1, which we have already proved, along with the projection-valued measure definition imply

which gives the desired result

for indicator functions.

Let us next prove the result for simple functions. This result follows from the indicator function result and linearity.

To wit, consider two simple functions \(s_1\) and \(s_2\)

where \(\alpha_i, \beta_j \in \mathbb{C}\) and \(E_i, F_j \in \Omega(X)\). Linearity implies

Hence, using linearity and our previous indicator function result we have

proving

the desired simple function result.

Let us next prove the result for bounded, measurable, complex-valued functions.

As one will recall Theorem (Complex-Valued Simple Approximation Theorem) implies that there exist sequences of complex-valued simple functions \(\{s_i\}_{i \in \mathbb{N}}\) and \(\{r_j\}_{j \in \mathbb{N}}\) on \(X\) such that \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to a bounded, measurable, complex-valued function \(f\) on \(X\) and similarly \(\{r_j\}_{j \in \mathbb{N}}\) to \(g\).

This along with linearity of operator-valued integration implies for any \(s_i\)

where in the final step we employed our Property 2 result. As \(\{s_i\}_{i \in \mathbb{N}}\) converges uniformly to \(f\), this implies that for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i \ge N\) one has

in other words the operator-valued integral of \(s_i\) converges to the operator-valued integral of \(f\). One can establish using similar logic that the operator-valued integral of \(r_i\) converges to the operator-valued integral of \(g\).

Similarly, uniform convergence along with linearity imply for any \(s_i\) and \(r_j\)

where in the final step we employed our Property 2 result. Looking at this result and our previous similar result, one concludes that if we can prove that given any \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(i,j \ge N\) one has

then we can conclude that the operator-valued integral of \(s_ir_j\) converges to the operator-valued integral of \(fg\).

One can prove this desired convergence as follows. Consider any \(s_i\) and \(r_j\). One has

As \(f\) is bounded and \(r_j\) is a simple function, their suprema are finite numbers. This allows us to continue this derivation as follows

Now as \(g\) is bounded there exists a real constant \(M_g\) such that

As \(r_j \rightarrow g\) uniformly, for \(\epsilon = 1\) there exists an integer \(N_1\) such that for all \(j \ge N_1\) and all \(\lambda \in X\) one has

The definition of a norm and our previous results then imply that for all \(j \ge N_1\) and all \(\lambda \in X\) one has

Taking the supremum while still requiring \(j \ge N_1\) gives

Noting that the \(r_j\) are simple functions and thus bounded we can define the real constant \(C\) by

then it obviously follows that

In other words there is a bound \(C\) on the \(r_j\) that holds uniformly for all \(\lambda \in X\) and all \(j\).

Now as \(s_i\) converges uniformly to \(f\), for any \(\epsilon > 0\) there exists an \(N\) such that for all \(i \ge N\) one has

Similarly, as \(r_j\) converges uniformly to \(g\), for this same \(\epsilon > 0\) there exists an \(M\) such that for all \(j \ge M\) one has

This implies that for all \(i,j \ge \max(N,M)\) we have

where the third line follows from our previous result

This implies that for any \(\epsilon > 0\) there exists a natural number \(L\) such that for all \(i,j \ge L\) we have

This along with our previous result

implies that for any \(\epsilon > 0\) there exists a natural number \(L\) such that for all \(i,j \ge L\) we have

In particular, this holds when \(j = i\): for any \(\epsilon > 0\) there exists a natural number \(L\) such that for all \(i \ge L\) we have

In other words, the single-indexed sequence of operator-valued integrals of \(s_i r_i\) converges, in the operator norm, to the operator-valued integral of \(fg\)

Now recall from our simple function result that for every \(i\), as \(s_i\) and \(r_i\) are both simple functions,

exactly, with no limit involved. Combining this exact identity with the convergence just established gives

as \(i \rightarrow \infty\).

On the other hand, as we established earlier in this proof,

in the operator norm. Operator multiplication is jointly continuous with respect to the operator norm: for any \(X_i \rightarrow X\) and \(Y_i \rightarrow Y\) in \(\mathcal{B}(\mathbf{H})\), submultiplicativity gives

where we have used that \(\{ \|X_i\| \}_{i \in \mathbb{N}}\) is bounded, being a convergent sequence of real numbers. Applying this with

gives

as \(i \rightarrow \infty\).

We have thus shown that the single sequence

converges, in the operator norm, to both

and

As limits in \(\mathcal{B}(\mathbf{H})\) with respect to the operator norm are unique, this implies

the desired result of Property 3, integration is multiplicative for bounded, measurable, complex-valued functions \(f\) and \(g\).

Property 4: Finally we must prove that for all bounded, measurable, complex-valued functions \(f\) on \(X\), we have

where \(\overline{f}\) is the complex conjugate of \(f\) and the superscript \(*\) denotes the adjoint on \(\mathcal{B}(\mathbf{H})\) arising from the Hilbert space inner product.

Let us start by considering the case in which \(f\) is real. The definition of \(Q_f\) states

for all \(\psi \in \mathbf{H}\). As \(f\) is real, this implies that \(Q_f(\psi)\) is real. As \(Q_f(\psi)\) is real, the Proposition we previously proved along with the definition of \(A_f\)

imply that \(A_f\) is self-adjoint. As \(A_f\) is self-adjoint, it implies that

is also self-adjoint.

Consider now the case in which \(f\) is complex. This implies that \(f\) can be written as

where both \(f_1\) and \(f_2\) are real. Linearity then implies

where when taking the adjoint we have also used our previous result that \(A_{f_i} = A_{f_i}^*\) when \(f_i\) is real. This proves the desired result

for any bounded, measurable, complex-valued function \(f\). This also completes the proof of Theorem (Operator-Valued Integration).\(\blacksquare\)

The Spectral Theorem

Now we are finally in the position to state the spectral theorem for bounded operators.

Theorem (Spectral Theorem for Bounded, Self-Adjoint Operators) If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint, then there exists a unique projection-valued measure \(\mu^A\) on the Borel \(\sigma\)-algebra of \(\sigma(A)\), the spectrum of \(A\), with values in orthogonal projections on \(\mathbf{H}\) such that

\[\int_{\sigma(A)} \lambda \, d\mu^A(\lambda) = A.\]

Proof To facilitate the proof of this theorem, we first introduce the useful notion of “Functional Calculus”.

Definition (Functional Calculus) If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint and \(f : \sigma(A) \rightarrow \mathbb{C}\) is a bounded measurable function on the spectrum \(\sigma(A)\) of \(A\), functional calculus defines an operator \(f(A)\) by

\[f(A) \equiv \int_{\sigma(A)} f(\lambda) \, d\mu^A(\lambda),\]

where \(\mu^A\) is the unique projection-valued measure of Theorem (Spectral Theorem for Bounded, Self-Adjoint Operators) associated to \(A\).

With functional calculus defined, we can now outline the proof of Theorem (Spectral Theorem for Bounded, Self-Adjoint Operators). This proof consists of two main stages.

Stage 1: In the first stage any self-adjoint \(A \in \mathcal{B}(\mathbf{H})\) is used to construct a “continuous functional calculus” that associates to each continuous function \(f\) on \(\sigma(A)\) an operator \(f(A)\).

This association is such that for any natural number \(m\) the function \(f(\lambda) = \lambda^m\) is associated with the operator \(f(A)=A^m\). The full “continuous functional calculus” is then constructed by approximating arbitrary continuous functions \(f\) on \(\sigma(A)\) by polynomials.

The Stone–Weierstrass Theorem for Real Numbers implies that polynomials are dense in the space of continuous functions on \(\sigma(A)\). Hence, for any continuous function \(f\) on \(\sigma(A)\) there exists a sequence of polynomials \(\{p_i\}_{i \in \mathbb{N}}\) that converge uniformly to \(f\) on \(\sigma(A)\). The final step of stage 1 then proves that the sequence of operators \(\{p_i(A)\}_{i \in \mathbb{N}}\) converge to an operator denoted as \(f(A)\).

Stage 2: The second stage of the proof shows that for a continuous function \(f\) on \(\sigma(A)\) the operator \(f(A)\) of the first stage can be represented as integration against a projection-valued measure. This amounts to an operator-valued version of the Riesz Representation Theorem from measure theory.

Stage 1: The Continuous Functional Calculus

We begin this stage of the proof with “utility” lemmas and propositions that we will have need of later in this stage.

Lemma (Bounded Operator Product is Submultiplicative) Let \(A, B \in \mathcal{B}(\mathbf{H})\), then the operator product is submultiplicative

\[\|AB\| \le \|A\| \, \|B\|.\]

Proof Consider arbitrary \(A\) and \(B\) in \(\mathcal{B}(\mathbf{H})\) and an arbitrary element \(\psi\) in \(\mathbf{H}\) such that \(\|\psi\| = 1\).

Let us first consider the case in which \(B\psi \neq 0\). As \(B\psi \neq 0\) it follows that \(\|B\psi\| \neq 0\). Hence,

Obviously,

has norm \(1\). Furthermore, by definition

Hence, as \(\|\psi\| = 1\) the previous derivation can proceed as follows

In other words for \(\|\psi\| = 1\) such that \(B\psi \neq 0\)

Taking the supremum of the lefthand side this gives

which as

is almost the desired equation \(\|AB\| \le \|A\| \,\|B\|\). We just need to prove it holds for \(B\psi = 0\).

If \(\|\psi\| = 1\) and \(B\psi = 0\), then \(AB\psi = 0\), and thus \(\|AB\psi\| = 0\). Hence, the supremum of \(\|AB\psi\|\) over such \(\psi\) is \(0\). As the supremum over such \(\psi\) is zero, this supremum is always less than or equal to \(\|B\| \, \|A\|\). Hence, the bound above

is also satisfied for \(\psi\) that satisfy \(B\psi = 0\).

Hence, we have proven that

which as a result of the definition of operator norm implies

the desired result, operator multiplication in \(\mathcal{B}(\mathbf{H})\) is submultiplicative.\(\blacksquare\)

The next “utility” lemma we must prove is the following

Lemma Suppose \(X \in \mathcal{B}(\mathbf{H})\) satisfies \(\|X\| < 1\), where \(\|X\|\) is the operator norm of \(X\). Then the operator \(I - X\) has a bounded inverse \((I - X)^{-1}\) in \(\mathcal{B}(\mathbf{H})\); and this bounded inverse is given by the following series

\[(I - X)^{-1} = \mathbf{1} + X + X^2 + X^3 + \cdots\]

that is convergent in \(\mathcal{B}(\mathbf{H})\) with respect to the operator norm.

Proof As a result of Lemma (Bounded Operator Product is Submultiplicative) the product of operators \(A,B \in \mathcal{B}(\mathbf{H})\) is submultiplicative,

thus for an arbitrary natural number \(m\) one has

In other words \(\|X^m\| \le \|X\|^m\).

Now the hypothesis \(\|X\| < 1\) implies that the geometric series \(\{\|X\|^m\}_{m \in \mathbb{N}}\) converges

Furthermore, as \(\|X^m\| \le \|X\|^m\) this further implies the series \(\{\|X^m\|\}_{m \in \mathbb{N}}\) also converges

Explicitly this follows from the definition of norm, \(\|X^m\| \le \|X\|^m\), and convergence of the series \(\{\|X\|^m\}_{m \in \mathbb{N}}\) together implying

Finally, recalling the fact established in Lemma (Bounded Operators form a Banach Space) that \(\mathcal{B}(\mathbf{H})\) is a Banach space with respect to the operator norm, one can use the Proposition

Proposition If \(V\) is a Banach space, then absolute convergence implies convergence in \(V\). That is, if \(\{\psi_i\}_{i \in \mathbb{N}}\) is a sequence in \(V\) and

\[\sum\limits_{i \in \mathbb{N}} \|\psi_i\| < \infty,\]

then

\[\sum\limits_{i \in \mathbb{N}} \psi_i\]

converges in \(V\).

to conclude that the series \(\{X^m\}_{m \in \mathbb{N}}\) converges in \(\mathcal{B}(\mathbf{H})\) with respect to the operator norm. In other words the series

of the hypothesis converges in \(\mathcal{B}(\mathbf{H})\), one of the desired results.

Now if we consider this convergent series of the hypothesis

and multiply by \((\mathbf{1} - X)\) we find

proving the right inverse is given by

A similar computation also proves

establishing that the left inverse is given by the same expression

Together these imply the final desired result.\(\blacksquare\)

Proposition For all \(A \in \mathcal{B}(\mathbf{H})\), the following results hold.

1. The spectrum \(\sigma(A)\) of \(A\) is a closed, bounded, and non-empty subset of \(\mathbb{C}\).
2. If \(\lvert \lambda \rvert > \|A\|\), where \(\|A\|\) is the operator norm of \(A\), then \(\lambda\) is in the resolvent set of \(A\).

Proof For any non-zero \(\lambda \in \mathbb{C}\) and \(A\) from our hypothesis, consider the operator

If \(\lvert \lambda \rvert > \|A\|\), then \(\|A / \lambda\| < 1\) and as a result of the Lemma we just proved \((\mathbf{1} - A / \lambda)\) is invertible. As a result of the previous equation and \(\lambda \neq 0\), this then implies that \(A - \lambda \mathbf{1}\) is also invertible, with inverse given by

Hence, \(\lambda\) is in the resolvent set of \(A\).

This establishes that if \(\lvert \lambda \rvert > \|A\|\), then \(\lambda\) is in the resolvent set of \(A\), which is the desired result of Part 2 of the proposition.

Furthermore, as \(\lvert \lambda \rvert > \|A\|\) implies that \(\lambda\) is in the resolvent set of \(A\) and the spectrum \(\sigma(A)\) is the complement of the resolvent set in \(\mathbb{C}\), this also proves that the spectrum \(\sigma(A)\) of \(A\) is bounded. This is one of the desired results of Part 1 of the proposition.

Suppose now that \(\lambda_0 \in \mathbb{C}\) is in the resolvent set of \(A\). For any other \(\lambda \in \mathbb{C}\) one has

where in the second line we have used the fact that \(\lambda_0\) is in the resolvent set of \(A\) to factor out \((A - \lambda_0 \mathbf{1})\).

Looking at the previous equation along with the Lemma we just proved, we can conclude that if

then both factors on the righthand side of the previous equation are invertible, and thus \(A - \lambda \mathbf{1}\) would also be invertible. This implies that the resolvent set of \(A\) is open and thus the spectrum \(\sigma(A)\) of \(A\), the complement of the resolvent set in \(\mathbb{C}\), is closed. This is another one of the desired results of Part 1 of the proposition.

The final result that needs to be proven in the proposition is that the spectrum \(\sigma(A)\) is non-empty, which is in Part 1 of the proposition.

Continuing on with the previous calculation where \(\lambda_0\) is in the resolvent set of \(A\) but now under the additional assumption that

we see that the righthand side of

is invertible. This gives

where the second equality follows from the Lemma we just proved.

This implies that in the neighborhood of any point \(\lambda_0\) in the resolvent set of \(A\) the resolvent \((A - \lambda \mathbf{1})^{-1}\) can be expressed by this locally convergent series in powers of \((\lambda - \lambda_0)\) with coefficients of these powers being elements of \(\mathcal{B}(\mathbf{H})\).

Hence, for any \(\phi, \psi \in \mathbf{H}\) the map

can be expressed as a locally convergent power series with coefficients in \(\mathbb{C}\). In other words it is an analytic function on the resolvent set of \(A\) which, as we have proven, is open. Thus, as a result of the Analytic Equivalence Theorem

Theorem (Analytic Equivalence Theorem) Let \(U\) be an open subset of \(\mathbb{C}\) and let \(f: U \rightarrow \mathbb{C}\) be a function. Then \(f\) is analytic on \(U\) if and only if it is holomorphic on \(U\).

this function is holomorphic on the resolvent set of \(A\).

Now, as we proved earlier, for \(\lambda\) in the resolvent set of \(A\) we can also write \((A - \lambda \mathbf{1})^{-1}\) as

This manner of writing \((A - \lambda \mathbf{1})^{-1}\) makes it clear that \(\|(A - \lambda \mathbf{1})^{-1}\|\) tends to zero as \(\lvert \lambda \rvert\) tends to infinity. Hence, the righthand side of our other expression for \(\| (A - \lambda \mathbf{1})^{-1} \|\)

also tends to zero as \(\lvert \lambda \rvert\) tends to infinity. And thus so does the holomorphic function

in the same limit.

Now let us assume the spectrum \(\sigma(A)\) of \(A\) is the empty set. Then the resolvent set of \(A\), the complement of \(\sigma(A)\) in \(\mathbb{C}\), would be all of \(\mathbb{C}\). This would then imply that the holomorphic function \(\lambda \mapsto \left< \phi, (A - \lambda \mathbf{1})^{-1} \psi \right>\) is holomorphic on all of \(\mathbb{C}\). In other words it is an entire function.

Hence, by evaluating the entire function \(\lambda \mapsto \left< \phi, (A - \lambda \mathbf{1})^{-1} \psi \right>\) on ever larger closed disks \(\overline{B}\), the Maximum Modulus Principle

Theorem (Maximum Modulus Principle) Let \(B\) be a bounded, non-empty, connected open subset of \(\mathbb{C}\). Let \(\overline{B}\) be the closure of \(B\). Suppose \(f : \overline{B} \rightarrow \mathbb{C}\) is a continuous function that is holomorphic on \(B\). Then \(\lvert f(z) \rvert\) attains its maximum at some point on the boundary of \(B\).

implies that the maximum of \(\lvert \left< \phi, (A - \lambda \mathbf{1})^{-1} \psi \right> \rvert\) is zero. As a result of the definition of a norm, this in turn implies that \(\left< \phi, (A - \lambda \mathbf{1})^{-1} \psi \right>\) is identically zero.

As this is true for any \(\phi, \psi \in \mathbf{H}\), it implies that \((A - \lambda \mathbf{1})^{-1}\) has operator norm zero and is thus the zero operator. However, we know that \((A - \lambda \mathbf{1})(A - \lambda \mathbf{1})^{-1} = \mathbf{1}\). Thus, \((A - \lambda \mathbf{1})^{-1}\) can not be the zero operator, and we have arrived at a contradiction.

Hence, our assumption that the spectrum \(\sigma(A)\) of \(A\) is the empty set is false. The spectrum \(\sigma(A)\) is non-empty. This is the final desired result of Part 1.\(\blacksquare\)

Another proposition we will have need of is

Proposition For any \(A \in \mathcal{B}(\mathbf{H})\), we have

\[\left( \text{Range}(A) \right)^\perp = \text{Ker}(A^*)\]

where \(\text{Ker}(A^*)\) is the kernel of \(A^*\), \(\text{Range}(A)\) is the range of \(A\), and \(\left( \text{Range}(A) \right)^\perp \equiv \{ \psi \in \mathbf{H} : \left< \psi, A\phi \right> = 0 \text{ for all } \phi \in \mathbf{H} \}\).

Proof Assume that \(\psi \in \left( \text{Range}(A) \right)^\perp\). Hence, for all \(\phi \in \mathbf{H}\)

As this is true for all \(\phi \in \mathbf{H}\), it implies that \(A^*\psi = 0\) and thus \(\psi \in \text{Ker}(A^*)\). Hence, we have proven that \(\left( \text{Range}(A) \right)^\perp \subseteq \text{Ker}(A^*)\).

Assume now that \(\psi \in \text{Ker}(A^*)\). Hence, \(A^*\psi = 0\). Thus for all \(\phi \in \mathbf{H}\)

As this is true for all \(\phi \in \mathbf{H}\), it implies that \(\psi \in \left( \text{Range}(A) \right)^\perp\). Hence, we have proven that \(\text{Ker}(A^*) \subseteq \left( \text{Range}(A) \right)^\perp\).

As we have proven that \(\left( \text{Range}(A) \right)^\perp \subseteq \text{Ker}(A^*)\) and that \(\text{Ker}(A^*) \subseteq \left( \text{Range}(A) \right)^\perp\) it follows that

which is the desired result.\(\blacksquare\)

Another result we will require is

Lemma If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint, then for all \(a,b \in \mathbb{R}\) and associated \(\lambda \equiv a + ib\) in \(\mathbb{C}\), we have

\[\left< (A - \lambda \mathbf{1}) \psi, (A - \lambda \mathbf{1}) \psi \right> \ge b^2 \left< \psi, \psi \right>\]

for all \(\psi \in \mathbf{H}\).

Proof The definition of \(\lambda\) in terms of \(a,b \in \mathbb{R}\) along with the definition of an inner product imply that for any \(\psi \in \mathbf{H}\)

Now by hypothesis \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint and \(a \in \mathbb{R}\), hence \((A - a\mathbf{1})\) is self-adjoint. Thus

Thus the second and third summands of the righthand side of our previous equation cancel leaving us with

The definition of inner product implies

Hence, the previous equation implies

which is the desired result.\(\blacksquare\)

Now we move onto the result

Proposition If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint, then the spectrum \(\sigma(A)\) of \(A\) is in \(\mathbb{R}\).

Proof A statement obviously equivalent to that of this proposition is the following: For \(a,b \in \mathbb{R}\) and \(\lambda \equiv a + ib\) it follows that if \(b \neq 0\), then \(\lambda\) is in the resolvent set of \(A\). We will prove this equivalent statement.

Now the Lemma we just proved implies that

for arbitrary \(\psi \in \mathbf{H}\). By hypothesis \(b \neq 0\); this along with the previous inequality and the inner product definition imply that if \(\psi \neq 0\), then \((A - \lambda \mathbf{1}) \psi \ne 0\). In other words \((A - \lambda \mathbf{1})\) is one-to-one.

As \(\overline{\lambda}\) also has a non-zero imaginary part the previous argument also applies to \((A - \overline{\lambda} \mathbf{1})\) and thus \((A - \overline{\lambda} \mathbf{1})\) is also one-to-one.

Recall that as a result of the Proposition we just proved we have

where the second equality follows from the hypothesis that \(A\) is self-adjoint. As \((A - \overline{\lambda} \mathbf{1})\) is one-to-one, \(\text{Ker}(A - \overline{\lambda} \mathbf{1})\) consists of the zero vector \(\psi = 0\). Hence, the previous equation implies \(\left( \text{Range}(A - \lambda \mathbf{1}) \right)^\perp\) consists of the zero vector. This in turn implies that \(\text{Range}(A - \lambda \mathbf{1})\) is dense in \(\mathbf{H}\).

Next we will prove that \(\text{Range}(A - \lambda \mathbf{1})\) is not only dense in \(\mathbf{H}\) but is in fact all of \(\mathbf{H}\).

Consider any \(\phi \in \mathbf{H}\). As \(\text{Range}(A - \lambda \mathbf{1})\) is dense in \(\mathbf{H}\), there exists a sequence \(\{ \phi_i \equiv (A - \lambda \mathbf{1}) \psi_i \}_{i \in \mathbb{N}}\) in \(\text{Range}(A - \lambda \mathbf{1})\) such that \(\{ \phi_i \}_{i \in \mathbb{N}}\) converges to \(\phi\).

The Lemma we just proved implies that for any natural numbers \(i\) and \(j\) we have

As a result of the fact that \(\{ \phi_i \}_{i \in \mathbb{N}}\) converges to \(\phi\) it follows that \(\{ \phi_i \}_{i \in \mathbb{N}}\) is a Cauchy sequence. As \(\{ \phi_i \}_{i \in \mathbb{N}}\) is a Cauchy sequence \(\{ (A - \lambda \mathbf{1}) \psi_i \}_{i \in \mathbb{N}}\) is a Cauchy sequence.

Hence, for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i,j \ge N\) one has

As a result of the previous inequality involving \(b^2\) this, along with the definition of the norm on \(\mathbf{H}\), implies that for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i,j \ge N\) one has

This in turn implies that for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i,j \ge N\) one has

In other words \(\{ \psi_i \}_{i \in \mathbb{N}}\) is a Cauchy sequence. Furthermore, as \(\{ \psi_i \}_{i \in \mathbb{N}}\) is a Cauchy sequence and \(\mathbf{H}\) is complete, there exists a \(\psi \in \mathbf{H}\) such that \(\{ \psi_i \}_{i \in \mathbb{N}}\) converges to \(\psi\).

Now as \(A \in \mathcal{B}(\mathbf{H})\) it is a bounded linear operator from the normed space \(\mathbf{H}\) to the normed space \(\mathbf{H}\). Thus, as a result of the standard proposition Bounded Operators are Continuous

Proposition (Bounded Operators are Continuous) A linear operator between normed spaces is bounded if and only if it is continuous.

\(A\) is continuous.

As \(A\) is continuous our definitions imply

As \(\phi \in \mathbf{H}\) was arbitrary, this implies that an arbitrary \(\phi \in \mathbf{H}\) is in \(\text{Range}(A - \lambda \mathbf{1})\), and thus \(\text{Range}(A - \lambda \mathbf{1})\) is all of \(\mathbf{H}\), the desired result.

Hence, we have proven that \((A - \lambda \mathbf{1})\) is one-to-one and onto. There is one remaining result required to show that \(\lambda\) is in the resolvent set of \(A\). It remains to prove that \((A - \lambda \mathbf{1})^{-1}\) is bounded.

As we just proved, for an arbitrary \(\phi \in \mathbf{H}\) there exists a \(\psi \in \mathbf{H}\) such that \((A - \lambda \mathbf{1}) \psi = \phi\). In this case Lemma and the definition of norm on \(\mathbf{H}\) imply

which proves that \((A - \lambda \mathbf{1})^{-1}\) is bounded.

Hence, we have proven that for any \(a,b \in \mathbb{R}\) with \(b \neq 0\) and \(\lambda\) defined by \(\lambda \equiv a + ib\), it follows that \(\lambda\) is in the resolvent set of a self-adjoint \(A\). This is then equivalent to the statement that if \(A\) is self-adjoint, then the spectrum \(\sigma(A)\) of \(A\) is in \(\mathbb{R}\).\(\blacksquare\)

Before proceeding, let us consolidate several structural facts about the spectrum \(\sigma(A)\) of a self-adjoint operator \(A\) into a single lemma. These facts are used repeatedly throughout the remainder of this post—in particular as the standing hypothesis “\(\sigma(A)\) is a compact metric measurable space” required by several later results—and collecting them here allows us to cite them once rather than re-deriving them at each point of use.

However, in consolidating these facts we will have need of the Heine–Borel Theorem which states

Theorem (Heine–Borel Theorem) For any positive, natural number \(n\), any subset of \(\mathbb{C}^n\) is compact if and only if it is closed and bounded.

with that stated let us begin the consolidation.

Lemma (The Spectrum is a Compact Metric Measurable Space) Let \(A \in \mathcal{B}(\mathbf{H})\) be self-adjoint. Then the spectrum \(\sigma(A)\) of \(A\) satisfies the following.

1. \(\sigma(A)\) is a non-empty subset of \(\mathbb{R} \subset \mathbb{C}\).
2. \(\sigma(A)\) is compact.
3. \(\sigma(A)\) is a metric space under the metric \(d(z_1, z_2) \equiv \lvert z_1 - z_2 \rvert\) inherited from \(\mathbb{C}\).
4. \(\sigma(A)\) is a measurable space when equipped with its Borel \(\sigma\)-algebra, i.e. the smallest \(\sigma\)-algebra containing the open sets of \(\sigma(A)\) in the topology induced by the metric of (3).

In particular, \(\sigma(A)\) is a non-empty, compact metric measurable space. Throughout the remainder of this post, whenever \(\sigma(A)\) is described as measurable, or a subset of \(\sigma(A)\) is described as measurable, it is with respect to this Borel \(\sigma\)-algebra. Consequently, on \(\sigma(A)\) the terms “measurable” and “Borel-measurable” are synonymous, both for sets and for functions.

Proof Part 1: As \(A\) is self-adjoint, Proposition implies that \(\sigma(A)\) is a subset of \(\mathbb{R}\), and \(\mathbb{R}\) is a subset of \(\mathbb{C}\). Furthermore, part 1 of Proposition implies that \(\sigma(A)\) is non-empty. This is the desired Part 1 result.

Part 2: Part 1 of Proposition implies that \(\sigma(A)\) is a closed and bounded subset of \(\mathbb{C}\). The Heine–Borel Theorem then implies that \(\sigma(A)\) is compact, the desired Part 2 result.

Part 3: As the norm \(\lvert \cdot \rvert\) on \(\mathbb{C}\) defines a metric

on \(\mathbb{C}\), and as the restriction of a metric to a subset is again a metric on that subset, the subset \(\sigma(A)\) of \(\mathbb{C}\) is a metric space by way of this inherited metric. This is the desired Part 3 result.

Part 4: By Part 3 the set \(\sigma(A)\) carries a metric, and hence a topology generated by the open balls of that metric. The Borel \(\sigma\)-algebra of \(\sigma(A)\) is by definition the smallest \(\sigma\)-algebra containing the open sets of this topology, and a set equipped with a \(\sigma\)-algebra is by definition a measurable space. Hence \(\sigma(A)\), equipped with its Borel \(\sigma\)-algebra, is a measurable space, the desired Part 4 result.

Combining Part 1 through Part 4, \(\sigma(A)\) is a non-empty, compact metric measurable space, the desired result.\(\blacksquare\)

Definition (Spectral Radius) For any \(A \in \mathcal{B}(\mathbf{H})\) the spectral radius \(R(A)\) of \(A\) is defined by

\[R(A) \equiv \sup\limits_{\lambda \in \sigma(A)} |\lambda|.\]

Note that as a result of Proposition, \(\sigma(A)\) is a closed, bounded, and non-empty subset \(\mathbb{C}\). Hence, \(R(A)\) is a finite real number.

The first property that one can easily ascertain of the spectral radius is the following corollary:

Corollary If \(A \in \mathcal{B}(\mathbf{H})\) then the spectral radius \(R(A)\) of \(A\) is less than or equal to the operator norm \(\|A\|\) of \(A\),

\[R(A) \le \|A\|\]

Proof Proposition Part 2 implies that if \(\lvert \lambda \rvert > \|A\|\), then \(\lambda\) is in the resolvent set of \(A\). The contrapositive of this statement is: if \(\lambda\) is not in the resolvent set of \(A\), then \(\lvert \lambda \rvert \le \|A\|\).

By definition the spectrum \(\sigma(A)\) of \(A\) is the complement of the resolvent set of \(A\) in \(\mathbb{C}\). Hence, our contrapositive is equivalent to the statement: if \(\lambda\) is in the spectrum \(\sigma(A)\) of \(A\), then \(\lvert \lambda \rvert \le \|A\|\).

Hence, for all \(\lambda \in \sigma(A)\) it follows that \(\lvert \lambda \rvert \le \|A\|\). This then implies

the desired result.\(\blacksquare\)

The next “utility” proposition we will require details properties of the operator norm on \(\mathcal{B}(\mathbf{H})\).

Proposition For any \(A \in \mathcal{B}(\mathbf{H})\) the operator norm satisfies

\[\|A\| = \|A^*\|\]

along with

\[\|A^*A\| = \|A\|^2.\]

In particular if \(A\) is self-adjoint, it satisfies \(\|A^2\| = \|A\|^2\).

Proof Let us begin the proof of the first result \(\|A\| = \|A^*\|\) by noting that as a result of our previous Lemma we can write the operator norm of any \(A \in \mathcal{B}(\mathbf{H})\) as

This also allows us to write the operator norm of \(A^*\) as

where we have used the definition of \(A^*\) as well as the definition of norm and inner product. This gives us

the first desired result.

Now let us begin the proof of the second desired result \(\|A^*A\| = \|A\|^2\) by recalling that as a result of Lemma (Bounded Operator Product is Submultiplicative) operator multiplication in \(\mathcal{B}(\mathbf{H})\) is submultiplicative,

for any \(A,B \in \mathcal{B}(\mathbf{H})\). Hence, for an arbitrary \(A \in \mathcal{B}(\mathbf{H})\) we have

where the equality uses the first result \(\|A^*\| = \|A\|\) proved in this proposition.

Using the alternative means of expressing the operator norm derived in Lemma one has

where we used the definition of \(A^*\), standard supremum properties, the norm definition, and the operator norm definition. So in summary this implies

However, we have already proven

Together these imply

the final desired result.\(\blacksquare\)

As a consequence of the first result of the Proposition just proven, we record the following continuity property of the adjoint, which we will have need of later.

Proposition (Continuity of the Adjoint) Let \(\mathbf{H}\) be a Hilbert space and let \(\{ B_n \}_{n \in \mathbb{N}}\) be a sequence in \(\mathcal{B}(\mathbf{H})\) converging in the operator norm to \(B \in \mathcal{B}(\mathbf{H})\). Then \(\{ B_n^* \}_{n \in \mathbb{N}}\) converges in the operator norm to \(B^*\), i.e.

\[B^* = \lim\limits_{n \rightarrow \infty} B_n^*.\]

Proof Let us first prove that the adjoint is additive, i.e. that \((C - D)^* = C^* - D^*\) for any \(C, D \in \mathcal{B}(\mathbf{H})\). For any \(\chi, \psi \in \mathbf{H}\) one has

where the first and third lines follow from the definition of the adjoint and the second and fourth from the additivity of the inner product. As this holds for all \(\chi \in \mathbf{H}\), it follows that \((C - D)^* \psi = (C^* - D^*)\psi\) for all \(\psi \in \mathbf{H}\), and hence \((C - D)^* = C^* - D^*\).

Now, as a result of the first result of Proposition, the operator norm satisfies \(\|C\| = \|C^*\|\) for any \(C \in \mathcal{B}(\mathbf{H})\). Combining this with the additivity just proven gives, for any \(n \in \mathbb{N}\),

As \(B_n \rightarrow B\) in the operator norm the righthand side tends to \(0\), and hence \(B_n^* \rightarrow B^*\) in the operator norm, the desired result.\(\blacksquare\)

Lemma If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint, then the operator norm \(\|A\|\) of \(A\) is equal to the spectral radius \(R(A)\) of \(A\),

\[\|A\| = R(A)\]

Proof Before diving in to the details, let’s present an outline of the proof. At core, the proof consists of 4 steps:

1. Prove that if \(\lvert \lambda \rvert > \|A\|\), then \((A - \lambda \mathbf{1})^{-1}\) can be expressed as a series convergent in the operator norm topology.
2. Prove that if \(\lvert \lambda \rvert \le \|A\|\), then this series doesn’t converge in the operator norm topology.
3. Prove that if \(\lvert \lambda \rvert > R(A)\), then this series converges in the operator norm topology.
4. Conclude that a contradiction arises if \(R(A) < \|A\|\), and thus \(R(A) = \|A\|\).

Let us first (1) prove that if \(\lvert \lambda \rvert > \|A\|\), then \((A - \lambda \mathbf{1})^{-1}\) can be expressed as a series convergent in the operator norm topology.

This first result was established in the proof of Proposition. There we established that for \(\lvert \lambda \rvert > \|A\|\), the following series is convergent in the operator norm topology

and in fact is equivalent to \((A - \lambda \mathbf{1})^{-1}\)

This is the first desired result.

Next let us (2) prove that if \(\lvert \lambda \rvert \le \|A\|\), then this series doesn’t converge in the operator norm topology.

The first step in this proof is to derive a useful identity expressing \(\|A^{2^n}\|\) in terms of \(\|A\|^{2^n}\). For self-adjoint \(A\) Proposition states

For natural number \(n\) iterating this identity gives

This identity will be of use when we prove our series doesn’t converge in the operator norm topology. In particular we will prove this using the Nth-Term Test

Lemma (Nth-Term Test) Let \(\{a_i\}_{i \in \mathbb{N}}\) be a series in a normed vector space. If

\[\lim\limits_{i \rightarrow \infty} \| a_i \| \neq 0,\]

then this series does not converge.

Consider the limit

By hypothesis \(\lvert \lambda \rvert \le \|A\|\). Hence

This implies that our derivation continues as follows

Hence, the Nth-Term Test implies that the series does not converge.

With that we have proven the desired result: if \(\lvert \lambda \rvert \le \|A\|\), then our series doesn’t converge in the operator norm topology.

Now let us (3) prove that if \(\lvert \lambda \rvert > R(A)\), then this series converges in the operator norm topology.

Recall that in the proof of Proposition we showed that if \(\lambda_0\) is in the resolvent set of \(A\) and \(\lambda \in \mathbb{C}\) satisfies

then \((A - \lambda \mathbf{1})^{-1}\) exists in \(\mathcal{B}(\mathbf{H})\) and is expressible as the following locally convergent power series in \((\lambda - \lambda_0)\)

that converges in the operator norm topology. Furthermore, as \((A - \lambda \mathbf{1})^{-1}\) exists in \(\mathcal{B}(\mathbf{H})\) this implies that \(\lambda\) is also in the resolvent set of \(A\).

Consider now any element \(\xi\) in the dual space of \(\mathcal{B}(\mathbf{H})\). Using \(\xi\) and the above convergent series, the map

can be expressed as a locally convergent power series with coefficients in \(\mathbb{C}\). Hence, it is an analytic function on the resolvent set of \(A\), which as shown in the proof of Proposition is open. Hence, the Analytic Equivalence Theorem implies that this function is holomorphic on the resolvent set of \(A\).

Now, as mentioned in Definition (Spectral Radius), the spectral radius

of \(A\) is a finite real number. Explicitly, Proposition implies that \(\sigma(A)\) is a closed, bounded, and non-empty subset \(\mathbb{C}\). Hence, \(R(A)\) is a finite real number.

The spectrum \(\sigma(A)\) of \(A\) is defined as the complement of the resolvent set of \(A\) in \(\mathbb{C}\). Hence, all \(\lambda \in \mathbb{C}\) such that \(\lvert \lambda \rvert > R(A)\) are in the resolvent set of \(A\). Thus, the function

is holomorphic on the (unbounded) open annulus \(R(A) < \lvert \lambda \rvert\).

Now recall that Laurent’s Theorem states

Theorem (Laurent’s Theorem) Any function holomorphic on an open annulus in \(\mathbb{C}\) can be expanded uniquely as a Laurent series on that open annulus.

Hence, the function

can be expanded uniquely into a convergent Laurent series on the (unbounded) open annulus \(R(A) < \lvert \lambda \rvert\).

Now in Part 1 of this proof we established that if \(\lvert \lambda \rvert > \|A\|\), then \((A - \lambda \mathbf{1})^{-1}\) can be expressed as the series

that is convergent in the operator norm topology. Acting on this series with \(\xi\) we obtain the convergent Laurent series

on the (unbounded) open annulus \(\lvert \lambda \rvert > \|A\|\).

Hence, we have two convergent Laurent series expansions of \((A - \lambda \mathbf{1})^{-1}\) on the (unbounded) open annulus \(\max(R(A), \|A\|) < \lvert \lambda \rvert\). The first resulting from the initial application of Laurent’s Theorem and the second from the result from Part 1 of this proof. The uniqueness of Laurent’s Theorem implies that these convergent Laurent series expansions must be identical. Hence, we can write the convergent Laurent series expansion in both cases as

which converges for \(R(A) < \lvert \lambda \rvert\). This completes the proof of Part 3.

Next let us (4) conclude that a contradiction arises if \(R(A) < \|A\|\), and thus \(R(A) = \|A\|\).

Recall we already established in Corollary that \(R(A) \le \|A\|\). We will now establish that \(R(A) = \|A\|\) using proof by contradiction.

Let us for the moment assume that \(R(A) < \|A\|\). Then it is possible to select a \(\lambda\) such that \(R(A) < \lvert \lambda \rvert < \|A\|\). Let us fix this \(\lambda\) for the remainder of the proof.

The unique Laurent series from Part 3, evaluated at this fixed \(\lambda\), converges. This implies that all of its summands are bounded. In other words, for each \(\xi\) in the dual space of \(\mathcal{B}(\mathbf{H})\) there exists a \(C_\xi \in \mathbb{R}\)—which may depend on \(\xi\) and on our now-fixed \(\lambda\)—such that for any natural number \(m\)

Now, as we established in Lemma (Bounded Operators form a Banach Space), \(\mathcal{B}(\mathbf{H})\) forms a Banach space under the operator norm. Recalling the Theorem on Completeness of the Dual

Theorem (Theorem on Completeness of the Dual) If \(V\) is a Banach space, then its dual \(V^*\) is also a Banach space.

we can conclude that the dual \(\mathcal{B}(\mathbf{H})^*\) of \(\mathcal{B}(\mathbf{H})\) is also a Banach space.

As both \(\mathcal{B}(\mathbf{H})\) and \(\mathcal{B}(\mathbf{H})^*\) are Banach spaces and we have a set of bounded summands we can apply the Principle of Uniform Boundedness

Theorem (Principle of Uniform Boundedness) Suppose \(\{ T_m \}\) is any family of bounded linear maps from a Banach space \(V_1\) to a normed space \(V_2\). Suppose that for each \(\xi \in V_1\), there is a real constant \(C_\xi\) such that

\[\| T_m \xi \| \le C_\xi\]

for all \(m\). Then there exists a real constant \(C\) such that for all \(m\)

\[\|T_m\| \le C,\]

where \(\|T_m\|\) is the operator norm of \(T_m\).

by identifying \(V_1\) with \(\mathcal{B}(\mathbf{H})^*\), \(V_2\) with \(\mathbb{C}\), the operators \(\{ T_m \}\) with the operators

evaluated at our fixed \(\lambda\), and the bounds we derived previously

with those in this theorem. Doing so we find that there exists a real number \(C\)—which may depend on our fixed \(\lambda\)—such that for all natural numbers \(m\)

where here the operator norm is used.

Now, as one will recall, in Part 2 of this proof we established that for any natural number \(n\)

Applying this to the bound we just derived, at our fixed \(\lambda\), gives

Now, as \(R(A) < \lvert \lambda \rvert < \|A\|\), we have

Hence, it is possible to select an \(n\) large enough to violate the above inequality

So it can not be the case that \(R(A) < \|A\|\). As we know \(R(A) \le \|A\|\), the only option left is \(R(A) = \|A\|\), the desired result.\(\blacksquare\)

The next step in this Stage 1: The Continuous Functional Calculus is to understand how the spectrum \(\sigma(A)\) of an operator \(A \in \mathcal{B}(\mathbf{H})\) is related to the spectrum \(\sigma(p(A))\) of a polynomial \(p(A)\) in \(A\). The relation between \(\sigma(A)\) and \(\sigma(p(A))\) is “straightforward” and described by the Spectral Mapping Theorem.

However, to prove the Spectral Mapping Theorem we will first have to prove this utility lemma

Lemma If \(A,B \in \mathcal{B}(\mathbf{H})\) commute and \(A\) is not invertible, then \(AB\) is not invertible.

Proof We will prove this result using proof by contradiction. We will assume that \((AB)\) is invertible, then derive a contradiction.

Assuming \((AB)\) is invertible implies that there exists a \((AB)^{-1}\) such that

This implies that

This is simply the statement that \(A\) has a right inverse \((B(AB)^{-1})\).

Similarly, under the assumption that \((AB)\) has an inverse, \((AB)^{-1}\) exists and satisfies

As \(A\) commutes with \(B\) this implies

This is simply the statement that \(A\) has a left inverse \(((AB)^{-1}B)\).

Now if we let the right inverse \((B(AB)^{-1})\) act on the right of \(((AB)^{-1}B)A = \mathbf{1}\), we have

However, \((B(AB)^{-1})\) is the right inverse of \(A\); so

The last two equations then imply

In other words the left inverse \(((AB)^{-1}B)\) and the right inverse \((B(AB)^{-1})\) agree and there is a single unique inverse of \(A^{-1}\).

However, by hypothesis \(A\) is not invertible. Thus our assumption that \(AB\) is invertible is false, and \(AB\) is not invertible. This is the desired result.\(\blacksquare\)

With this lemma complete we may now move on to the Spectral Mapping Theorem.

Lemma (Spectral Mapping Theorem) For all \(A\) in \(\mathcal{B}(\mathbf{H})\) and any polynomial \(p(\lambda)\) of degree \(m\) on the spectrum \(\sigma(A)\) of \(A\)

\[p(\lambda) = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda^2 + \cdots + \alpha_{m - 1} \lambda^{m - 1} + \alpha_m \lambda^m\]

with \(\mathbb{C}\) valued coefficients \(\alpha_i\), let us define a map \(p \mapsto p(A)\) from such polynomials to elements of \(\mathcal{B}(\mathbf{H})\) by

\[p \longmapsto p(A) \equiv \alpha_0 \mathbf{1} + \alpha_1 A + \alpha_2 A^2 + \cdots + \alpha_{m - 1} A^{m - 1} + \alpha_m A^m.\]

Then the spectrum \(\sigma(p(A))\) of \(p(A)\) is given by

\[\sigma(p(A)) = \{ p(\lambda) : \lambda \in \sigma(A) \}.\]

Note that in an abuse of notation this is often written as \(\sigma(p(A)) = p(\sigma(A))\), despite the fact that \(p(\sigma(A))\) is ill-defined.

Proof Before getting into the details, let us present an outline of the 4-step proof.

1. Prove that the desired result

   \[\sigma(p(A)) = \{ p(\lambda) : \lambda \in \sigma(A) \}\]

   holds for a constant polynomial \(p(\lambda) = \alpha_0\).

2. Prove that

   \[\{ p(\lambda) : \lambda \in \sigma(A) \} \subseteq \sigma(p(A))\]

   for a generic polynomial \(p\) of positive degree.

3. Prove that

   \[\sigma(p(A)) \subseteq \{ p(\lambda) : \lambda \in \sigma(A) \}\]

   for a generic polynomial \(p\) of positive degree.

4. Conclude that

   \[\sigma(p(A)) = \{ p(\lambda) : \lambda \in \sigma(A) \}\]

   as a result of the proofs of Parts 1-3.

Let us begin by (1) proving that the desired result

holds for a constant polynomial \(p(\lambda) = \alpha_0\).

The resolvent set of \(p(A) = \alpha_0 \mathbf{1}\) is defined as the set of \(\lambda \in \mathbb{C}\) such that \(\alpha_0 \mathbf{1} - \lambda \mathbf{1}\) has a bounded inverse in \(\mathcal{B}(\mathbf{H})\). Obviously the resolvent set of \(p(A)\) in this case is \((\mathbb{C} - \alpha_0)\), the set of all elements in \(\mathbb{C}\) not equal to \(\alpha_0\). The spectrum \(\sigma(p(A))\) of \(p(A)\) is defined as the complement of the resolvent set of \(p(A)\) in \(\mathbb{C}\). Hence, \(\sigma(p(A)) = \{ \alpha_0 \}\).

On the other hand, by definition the spectrum \(\sigma(A)\) is some subset of \(\mathbb{C}\). In addition, for any \(\lambda\) in \(\mathbb{C}\) we have \(p(\lambda) = \alpha_0\). Hence,

Thus we have proven that if \(p\) is a constant polynomial \(p(\lambda) = \alpha_0\), then

which implies \(\sigma(p(A)) = \{ p(\lambda) : \lambda \in \sigma(A) \}\), the first desired result.

Next let us (2) prove that

for a generic polynomial \(p\) of positive degree.

Consider an arbitrary \(\lambda\) in the spectrum \(\sigma(A)\) of \(A\). Linearity implies

However,

This allows us to pull out a \((A - \lambda \mathbf{1})\) factor from each summand in our expression for \(p(A) - p(\lambda) \mathbf{1}\) giving

where \(q(A)\) is a \(\lambda\) dependent polynomial.

Now as \(\lambda\) is in \(\sigma(A)\), by definition \((A - \lambda \mathbf{1})\) is not invertible in \(\mathcal{B}(\mathbf{H})\). Furthermore, by construction \((A - \lambda \mathbf{1})\) commutes with \(q(A)\). Hence, as a result of Lemma the lefthand side \((A - \lambda \mathbf{1}) q(A)\) of the previous equation isn’t invertible. Hence, \(p(A) - p(\lambda) \mathbf{1}\) isn’t invertible, and thus \(p(\lambda)\) is an element of the spectrum \(\sigma(p(A))\) of \(p(A)\). In other words

the second desired result.

Next let us (3) prove that

for a generic polynomial \(p\) of positive degree.

Consider an arbitrary \(\gamma\) in the spectrum \(\sigma(p(A))\) of \(p(A)\). As a result of the Fundamental Theorem of Algebra

Theorem (Fundamental Theorem of Algebra) The field of complex numbers is algebraically closed.

we can factor the polynomial \(p(z) - \gamma\) as a function of \(z\) as follows

where \(c, b_i\in \mathbb{C}\). Thus, as \(A\) commutes with itself and \(\mathbf{1}\), we also have

Now as \(\gamma \in \sigma(p(A))\) it follows that \(p(A) - \gamma \mathbf{1}\) is not invertible. Thus there must exist some \(j\) such that \((A - b_j \mathbf{1})\) is not invertible. If no such \(j\) existed, then all the terms on the righthand side of this equation would be invertible which would imply that \(p(A) - \gamma \mathbf{1}\) is invertible, which we already know is not the case.

Now as \((A - b_j \mathbf{1})\) is not invertible, this implies that \(b_j\) is an element in the spectrum \(\sigma(A)\) of \(A\). However, from our equation for \(p(z)\) we know that \(p(z)\) evaluated at this \(b_j\) satisfies

This implies that

which is none other than the statement that an arbitrary \(\gamma\) in the spectrum \(\sigma(p(A))\) of \(p(A)\) is of the form \(p(\lambda)\) for a \(\lambda\) in \(\sigma(A)\). This is exactly the third desired result

for a generic polynomial \(p(\lambda)\) of positive degree.

Finally we (4) conclude that

as a result of the proofs of Parts 1-3.

Part 1 establishes this result for a constant polynomial. Part 2 establishes that

for a generic polynomial of positive degree while Part 3 establishes that

also for a generic polynomial of positive degree. So Part 2 and Part 3 imply that

for a generic polynomial of positive degree, while Part 1 establishes the result for a polynomial of degree zero. This establishes the desired result for a generic polynomial of arbitrary finite degree.\(\blacksquare\)

The last step in Stage 1: The Continuous Functional Calculus is to generalize the map of the Spectral Mapping Theorem

taking complex-valued polynomials on \(\sigma(A)\) to elements of \(\mathcal{B}(\mathbf{H})\), to take real-valued, continuous functions \(f\) on \(\sigma(A)\) to elements \(f(A)\) of \(\mathcal{B}(\mathbf{H})\).

We will do so using the Stone–Weierstrass Theorem for Real Numbers to prove that the set of polynomials on \(\sigma(A)\) is dense in \(C^0(\sigma(A); \mathbb{R})\)—the space of continuous, real-valued functions on \(\sigma(A)\). Then we will use this fact along with the Bounded Linear Transformation Theorem to extend the Spectral Mapping Theorem map \(p \mapsto p(A)\) to all of \(C^0(\sigma(A); \mathbb{R})\)

Let’s get started.

Proposition Let \(A \in \mathcal{B}(\mathbf{H})\) be self-adjoint. Then there exists a unique bounded linear map from \(C^0(\sigma(A); \mathbb{R})\)—the space of continuous, real-valued functions on the spectrum \(\sigma(A)\) of \(A\)—to \(\mathcal{B}(\mathbf{H})\)

\[f \longmapsto f(A)\]

such that when restricted to real-valued polynomials on \(\sigma(A)\) it agrees with the restriction of the Spectral Mapping Theorem map \(p \mapsto p(A)\). This map \(f \mapsto f(A)\) is called the (real-valued) functional calculus for \(A\).

Proof Consider a real-valued polynomial on \(\sigma(A)\) the spectrum of \(A\). The map

of the Spectral Mapping Theorem results in

with \(\mathbb{R}\) valued coefficients \(c_i\). If \(A\) is self-adjoint, then \(p(A)\) is also self-adjoint. Explicitly,

where we employed the fact that \(A\) is self-adjoint, the definition of the involution \(B \mapsto B^*\), and the fact that the \(c_i\) are real-valued.

Now we can apply the Spectral Mapping Theorem to this real-valued polynomial to conclude that

Furthermore, as \(p(A)\) is self-adjoint, we can apply the Lemma to conclude that

The definition of spectral radius, however, implies that

Putting this all together we conclude that

proving that

which is simply the statement that the map \(p \mapsto p(A)\) is isometric.

Explicitly, the map \(p \mapsto p(A)\) from the set of real-valued polynomials on \(\sigma(A)\) equipped with the supremum norm into \(\mathcal{B}(\mathbf{H})\) equipped with the operator norm, is isometric. This map is also linear; for real-valued polynomials \(p\) and \(q\) we have \((p + q) \mapsto (p + q)(A) = p(A) + q(A)\).

Before proceeding, we pause to address a subtlety in the definition of this map. The map \(p \mapsto p(A)\) has been described as a map on the set of real-valued polynomials on \(\sigma(A)\), i.e. on polynomial functions with domain \(\sigma(A)\), whereas the operator

is computed from the coefficients \(c_0, c_1, \ldots, c_m\). These are not the same datum: distinct coefficient tuples can determine the same function on \(\sigma(A)\). For instance, if \(\sigma(A) = \{ 0, 1 \}\), then the polynomials \(p(\lambda) = \lambda\) and \(q(\lambda) = \lambda^2\) agree at every point of \(\sigma(A)\) while having different coefficients. For the map \(p \mapsto p(A)\) to be well-defined on polynomial functions we must therefore verify that whenever two real-valued polynomials \(p\) and \(q\) agree at every point of \(\sigma(A)\), the operators \(p(A)\) and \(q(A)\) coincide.

This follows from the isometry just established. Let \(p\) and \(q\) be real-valued polynomials with \(p(\lambda) = q(\lambda)\) for all \(\lambda \in \sigma(A)\). Then \(p - q\) is a real-valued polynomial satisfying \((p - q)(\lambda) = 0\) for all \(\lambda \in \sigma(A)\). Applying the isometry to \(p - q\) and using the linearity of the map noted above gives

As \(\| \cdot \|\) is a norm on \(\mathcal{B}(\mathbf{H})\), the fact that \(\left\| p(A) - q(A) \right\| = 0\) implies \(p(A) - q(A)\) is the zero operator, i.e.

Hence, the operator \(p(A)\) depends only on the function \(p\) restricted to \(\sigma(A)\), and not on the particular choice of coefficients representing it. Thus the map \(p \mapsto p(A)\) is well-defined as a map on the set of real-valued polynomial functions on \(\sigma(A)\).

Now in preparation for the application of the Stone–Weierstrass Theorem for Real Numbers let us examine explicitly some of the properties of the objects we are currently considering.

As one will recall, Lemma (The Spectrum is a Compact Metric Measurable Space) established that the spectrum \(\sigma(A)\) of \(A\) is a non-empty, compact metric measurable space.

In addition, any element \(p\) in the algebra of real-valued polynomials on \(\sigma(A)\) is continuous. Hence, the algebra of real-valued polynomials on \(\sigma(A)\) is an algebra in \(C^0(\sigma(A); \mathbb{R})\), the space of continuous, real-valued functions on \(\sigma(A)\).

Obviously the algebra of real-valued polynomials on \(\sigma(A)\) contains the constant functions

In addition this algebra separates-points

Definition (Separates Points) Let \(X\) be a compact metric space and let \(\mathcal{A}\) be an algebra in \(C^0(X; \mathbb{R})\), the space of continuous, real-valued functions on \(X\). The algebra \(\mathcal{A}\) is said to separate points if for any \(x,y \in X\) such that \(x \neq y\) there exists an \(f \in \mathcal{A}\) such that \(f(x) \neq f(y)\).

Explicitly, let \(x\) and \(y\) be any elements in \(\sigma(A)\) such that \(x \neq y\). Note that Proposition along with the hypothesis that \(A\) is self-adjoint, imply that \(x,y \in \sigma(A) \subset \mathbb{R}\). Hence, the polynomial

on \(\sigma(A)\) is a real-valued polynomial. Evaluating \(p_s\) at \(x\) and \(y\) gives

where the final \(\neq\) follows from the fact that \(x \neq y\). Hence, the real-valued polynomials on \(\sigma(A)\) separate points.

With all of this in-hand we can apply the Stone–Weierstrass Theorem for Real Numbers

Theorem (Stone–Weierstrass for Real Numbers) Let \(X\) be a compact metric space and let \(\mathcal{A}\) be an algebra in \(C^0(X; \mathbb{R})\), the space of continuous, real-valued functions on \(X\). If \(\mathcal{A}\) contains the constant functions and separates points, then \(\mathcal{A}\) is dense in \(C^0(X; \mathbb{R})\) with respect to the supremum norm.

to the current situation. Identifying \(X\) with \(\sigma(A)\) and \(\mathcal{A}\) with the real-valued polynomials on \(\sigma(A)\), the Stone–Weierstrass Theorem for Real Numbers allows us to conclude that real-valued polynomials on \(\sigma(A)\) are dense in \(C^0(\sigma(A); \mathbb{R})\).

Next we must prepare for the application of the Bounded Linear Transformation Theorem. To do so, we must establish some relatively straightforward properties of objects we are currently considering.

Consider \(C^0(\sigma(A); \mathbb{R})\), the space of continuous, real-valued functions on \(\sigma(A)\). As \(\sigma(A)\) is compact the Boundedness Theorem

Theorem (Boundedness Theorem) A continuous real-valued function on a non-empty compact set \(C\) is bounded on \(C\).

implies that any element of \(C^0(\sigma(A); \mathbb{R})\) is bounded. Hence, the supremum norm

on \(C^0(\sigma(A); \mathbb{R})\) is finite. Furthermore, as a result of the fact that \(\lvert \cdot \rvert\) defines a norm on \(\mathbb{R}\), this supremum norm defines a norm on \(C^0(\sigma(A); \mathbb{R})\). In other words \(C^0(\sigma(A); \mathbb{R})\) is a normed space.

Recall, as proven in the Lemma (Bounded Operators form a Banach Space), \(\mathcal{B}(\mathbf{H})\) is a Banach space under the operator norm.

Also recall we proved that the map \(p \mapsto p(A)\) from the real-valued polynomials on \(\sigma(A)\) into \(\mathcal{B}(\mathbf{H})\) is linear and isometric satisfying

This implies

which is none other than the statement that the linear map \(p \mapsto p(A)\) is bounded using constant \(1\).

With all of this in hand we can apply the Bounded Linear Transformation Theorem

Theorem (Bounded Linear Transformation Theorem) Let \(V_1\) be a normed space and \(V_2\) a Banach space. Suppose \(W\) is a dense subspace of \(V_1\) and \(T: W \rightarrow V_2\) is a bounded linear map. Then there exists a unique bounded linear map \(\widetilde{T}: V_1 \rightarrow V_2\) such that \(\widetilde{T}|_W = T\). Furthermore, the norm of \(\widetilde{T}\) equals the norm of \(T\).

to the current situation. Identifying \(V_1\) with \(C^0(\sigma(A); \mathbb{R})\), \(V_2\) with \(\mathcal{B}(\mathbf{H})\), \(W\) the real-valued polynomials on \(\sigma(A)\), and \(T\) with our map \(p \mapsto p(A)\) allows us to conclude that there exists a unique, bounded, linear map

from \(C^0(\sigma(A); \mathbb{R})\) to \(\mathcal{B}(\mathbf{H})\) that when restricted to real-valued polynomials agrees with our map \(p \mapsto p(A)\) and that has the same norm of our map \(p \mapsto p(A)\).

This map \(f \mapsto f(A)\) is the desired result of this proposition and is known as the (real-valued) functional calculus for \(A\).\(\blacksquare\)

As a final step in Stage 1: The Continuous Functional Calculus, we will derive some basic properties of the (real-valued) functional calculus of a self-adjoint operator \(A\) in \(\mathcal{B}(\mathbf{H})\). These properties require the following definition

Definition (Non-Negative Bounded Operator) An operator \(A \in \mathcal{B}(\mathbf{H})\) is called a non-negative bounded operator if

\[0 \le \left< \psi, A\psi \right>\]

for all \(\psi \in \mathbf{H}\).

as well as the following lemma

Lemma If \(A \in \mathcal{B}(\mathbf{H})\) is invertible in \(\mathcal{B}(\mathbf{H})\), then there exists an \(\epsilon > 0\) such that for all \(B \in \mathcal{B}(\mathbf{H})\) that satisfy

\[\|B - A\| < \epsilon\]

\(B\) is also invertible in \(\mathcal{B}(\mathbf{H})\).

Proof We start this proof by noting that as \(A\) is by hypothesis invertible we have

where the final equation follows from factoring out a common \(A\) at the expense of introducing \(A^{-1}\).

Now, as a result of Lemma, if

then \((\mathbf{1} - A^{-1}(A - B))\) is invertible in \(\mathcal{B}(\mathbf{H})\). As \(A\) is also invertible in \(\mathcal{B}(\mathbf{H})\), this condition implies that as \(B\) is given by

it is invertible in \(\mathcal{B}(\mathbf{H})\).

As a result of Lemma operator multiplication in \(\mathcal{B}(\mathbf{H})\) is submultiplicative. This along with the norm definition imply

Now if we define the \(\epsilon\) of the hypothesis by

then if

it follows that

and thus

However, from submultiplicativity we already established that

Hence, we have

This is none other than the condition

which we previously found is the condition required for \(B\) to be invertible.\(\blacksquare\)

The properties of the (real-valued) functional calculus are captured in the following proposition

Proposition If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint, the (real-valued) functional calculus for \(A\), mapping \(C^0(\sigma(A); \mathbb{R})\) into \(\mathcal{B}(\mathbf{H})\), has the following properties

1. Multiplicativity: For all \(f,g \in C^0(\sigma(A); \mathbb{R})\), we have

   \[(fg)(A) = f(A)g(A),\]

   where \((fg)\) denotes the pointwise product of \(f\) and \(g\), i.e. \((fg)(\lambda) \equiv f(\lambda)g(\lambda)\).

2. Self-adjointness: For any \(f \in C^0(\sigma(A); \mathbb{R})\), the operator \(f(A)\) is self-adjoint.
3. Non-negativity: For any \(f \in C^0(\sigma(A); \mathbb{R})\) such that \(f\) is non-negative, it follows that \(f(A)\) is a non-negative bounded operator.

4. Norm and spectrum properties: For any \(f \in C^0(\sigma(A); \mathbb{R})\), we have

   \[\|f(A)\| = \sup\limits_{\lambda \in \sigma(A)} \lvert f(\lambda) \rvert,\]

   where \(\|f(A)\|\) is the operator norm of \(f(A)\) and \(\sigma(A)\) is the spectrum of \(A\), and

   \[\sigma(f(A)) = \{ f(\lambda) : \lambda \in \sigma(A) \},\]

   where \(\sigma(f(A))\) is the spectrum of \(f(A)\).

Proof Part 1: Let us begin with Part 1 multiplicativity and prove that for all \(f,g \in C^0(\sigma(A); \mathbb{R})\), we have

where \((fg)\) denotes the pointwise product of \(f\) and \(g\), i.e. \((fg)(\lambda) \equiv f(\lambda)g(\lambda)\).

As a result of the proof of Proposition we know the real-valued polynomials on the spectrum \(\sigma(A)\) of \(A\) are dense in \(C^0(\sigma(A); \mathbb{R})\) with respect to the supremum norm. Hence, there exists a sequence \(\{ s_i \}_{i \in \mathbb{N}}\) in the set of real-valued polynomials on \(\sigma(A)\) such that \(s_i \rightarrow f\) uniformly. Similarly, there exists a sequence \(\{ r_j \}_{j \in \mathbb{N}}\) in the set of real-valued polynomials on \(\sigma(A)\) such that \(r_j \rightarrow g\) uniformly.

Recall that by Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a non-empty, compact metric measurable space. Thus as a result of the Boundedness Theorem \(f\), \(g\), and all the \(s_i\) and \(r_j\) are bounded.

This setup will now allow us to prove that \(s_ir_i \rightarrow fg\) uniformly.

To wit, consider any \(s_i\) and \(r_j\). For any \(\lambda \in \sigma(A)\) the norm definition implies

As \(f\) and \(r_j\) are bounded, their suprema are finite numbers. Hence, we can continue this derivation as follows

Now as \(g\) is bounded there exists a real constant \(M_g\) such that

As \(r_j \rightarrow g\) uniformly, for \(\epsilon = 1\) there exists an integer \(N_1\) such that for all \(j \ge N_1\) and all \(\lambda \in \sigma(A)\) one has

The norm definition and our previous results then imply for all \(j \ge N_1\) and all \(\lambda \in \sigma(A)\)

Taking the supremum while still requiring \(j \ge N_1\) results in

Noting the \(r_j\) are bounded we can define the real constant \(C\) by

then it is obviously the case that

That is to say there is a bound \(C\) on the \(r_j\) that holds uniformly for all \(\lambda \in \sigma(A)\) and all \(j\).

Now as \(s_i\) converges uniformly to \(f\), for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i \ge N\) one has

Similarly, as \(r_j\) converges uniformly to \(g\) and \(f\) is bounded, for this same \(\epsilon > 0\) there exists an \(M\) such that for all \(j \ge M\) one has

This implies that for all \(i,j \ge \max(N,M)\) we have

where the third line follows from our previous result

This implies that for any \(\epsilon > 0\) there exists a natural number \(L\) such that for all \(i,j \ge L\) we have

As this applies for all \(i,j \ge L\) it implies in particular if \(j = i\) and \(i \ge L\). So, for any \(\epsilon > 0\) there exists a natural number \(L\) such that for all \(i \ge L\) we have

for all \(\lambda \in \sigma(A)\). In other words \(s_ir_i \rightarrow fg\) uniformly.

Hence, we have

where the first equality follows from our previous derivation, the second equality from Proposition proving that for real-valued polynomials \(p\) the Spectral Mapping Theorem map \(p \rightarrow p(A)\) is isometric, the third equality from \((s_ir_i)(\lambda) \equiv s_i(\lambda)r_i(\lambda)\), and the fourth equality follows from Proposition which proved \(s_i(A)r_i(A) \rightarrow f(A)g(A)\) relative to the operator norm.

In summary this proves the desired Part 1 multiplicativity result, \((fg)(A) = f(A) g(A)\).

Part 2: Next let us prove Part 2 self-adjointness, proving for any \(f \in C^0(\sigma(A); \mathbb{R})\), the operator \(f(A)\) is self-adjoint.

As mentioned in Part 1 there exists a sequence \(\{ s_i \}_{i \in \mathbb{N}}\) in the set of real-valued polynomials on \(\sigma(A)\) such that \(s_i \rightarrow f\) uniformly.

Furthermore, as any \(s_i\) is a real-valued polynomial, the map \(s_i \mapsto s_i(A)\) of the Lemma (Spectral Mapping Theorem) gives

with \(\mathbb{R}\) valued coefficients \(c_i\). As \(A\) is self-adjoint, it obviously follows that \(s_i(A)\) is also self-adjoint. Explicitly,

This then implies, using Proposition (Continuity of the Adjoint) to exchange the adjoint with the limit,

proving that \(f(A)^* = f(A)\) and thus that \(f(A)\) is self-adjoint, the desired Part 2 self-adjointness result.

Part 3: Next let us prove Part 3 non-negativity, proving that for any \(f \in C^0(\sigma(A); \mathbb{R})\) such that \(f\) is non-negative, it follows that \(f(A)\) is a non-negative bounded operator.

If \(f \in C^0(\sigma(A); \mathbb{R})\) is non-negative, then there exists a continuous function \(g\) in \(C^0(\sigma(A); \mathbb{R})\) such that \(g = \sqrt{f}\). This follows from the fact that \(f\) is by hypothesis continuous and the square root function

is continuous. Hence, as a result of the Composition Theorem

Theorem (Composition Theorem) If a function \(f\) is continuous at \(c\) and a function \(h\) is continuous at \(f(c)\), then the composition \(h \circ f\) is continuous at \(c\).

we know \(h \circ f = \sqrt{f}\) is continuous on \(\sigma(A)\) and thus an element of \(C^0(\sigma(A); \mathbb{R})\).

With \(g \equiv \sqrt{f}\) it follows that \(f = g^2\). Applying the result of Part 1 we have \(f(A) = g(A)g(A)\). Applying the result of Part 2 we know that \(g(A)\) is self-adjoint. Hence, for any \(\psi \in \mathbf{H}\) we have

where the final inequality follows from the definition of an inner product. This implies that for any \(\psi \in \mathbf{H}\) we have

As \(f(A)\) is bounded as a result of Proposition, this is none other than the statement that \(f(A)\) is a non-negative bounded operator, the desired result of Part 3.

Part 4: Finally let us prove Part 4 norm and spectrum properties, proving that for any \(f \in C^0(\sigma(A); \mathbb{R})\), we have

where \(\|f(A)\|\) is the operator norm of \(f(A)\) and \(\sigma(A)\) is the spectrum of \(A\), and

where \(\sigma(f(A))\) is the spectrum of \(f(A)\).

Let us first prove that

As mentioned in Part 1 there exists a sequence \(\{ s_i \}_{i \in \mathbb{N}}\) in the set of real-valued polynomials on \(\sigma(A)\) such that \(s_i \rightarrow f\) uniformly.

As proven in Proposition for a real-valued polynomials \(p\) on \(\sigma(A)\) the Spectral Mapping Theorem map \(p \rightarrow p(A)\) is isometric

This implies

where the first equality follows from the fact that \(s_i \rightarrow f\), the second from the fact that this convergence is uniform, the third equation from the fact that \(p \rightarrow p(A)\) is isometric, and the final from Proposition which proved \(s_i(A) \rightarrow f(A)\) relative to the operator norm. So with this we have proven the first, desired result

of Part 4.

Next let us prove the remaining result of Part 4

where \(\sigma(f(A))\) is the spectrum of \(f(A)\) and \(\sigma(A)\) is the spectrum of \(A\).

We will prove this by first proving that \(\sigma(f(A) \subseteq \{ f(\lambda) : \lambda \in \sigma(A) \}\). We will then prove \(\{ f(\lambda) : \lambda \in \sigma(A) \} \subseteq \sigma(f(A)\). Together this will entail the desired result

We start by proving \(\sigma(f(A) \subseteq \{ f(\lambda) : \lambda \in \sigma(A) \}\) by proving its contrapositive. In other words we will prove that any element of \(\mathbb{C}\) not in \(\{ f(\lambda) : \lambda \in \sigma(A) \}\) is not in \(\sigma(f(A)\).

To that end consider an arbitrary element \(\lambda_0 \in \mathbb{C}\) such that \(\lambda_0\) is not in \(\{ f(\lambda) : \lambda \in \sigma(A) \}\). In other words, there exists no \(\mu \in \sigma(A)\) such that \(f(\mu) = \lambda_0\).

Such a \(\lambda_0\) could lie in the complement of \(\mathbb{R}\) in \(\mathbb{C}\) or it could lie in \(\mathbb{R}\) in \(\mathbb{C}\). We will deal with these two cases separately.

Assume first that \(\lambda_0\) is in the complement of \(\mathbb{R}\) in \(\mathbb{C}\). This in particular implies that the imaginary component of \(\lambda_0\) is non-zero.

Now, as a result of Part 2 \(f(A)\) is self-adjoint. As \(f(A)\) is self-adjoint, Proposition implies that \(\sigma(f(A))\) is a subset of \(\mathbb{R}\) in \(\mathbb{C}\).

However, as the imaginary component of \(\lambda_0\) is non-zero this implies that \(\lambda_0\) is not in \(\sigma(f(A))\).

This proves the contrapositive of \(\sigma(f(A) \subseteq \{ f(\lambda) : \lambda \in \sigma(A) \}\) holds when \(\lambda_0\) is in the complement of \(\mathbb{R}\) in \(\mathbb{C}\), and thus proves \(\sigma(f(A) \subseteq \{ f(\lambda) : \lambda \in \sigma(A) \}\) holds in this case.

Now let us assume that \(\lambda_0\) lies in \(\mathbb{R}\) in \(\mathbb{C}\). This implies that \(\lambda_0\) is real-valued.

For this case, consider then the function \(g\) on \(\sigma(A)\) defined by

As \(f\) is continuous and \(f(\mu) \neq \lambda_0\) for all \(\mu \in \sigma(A)\), it follows that \(g\) is continuous. Furthermore, as \(f\) is real-valued and \(\lambda_0\) lies in \(\mathbb{R}\) in \(\mathbb{C}\), it follows that \(g\) is real-valued. So \(g\) is an element of \(C^0(\sigma(A); \mathbb{R})\).

Now obviously for all \(\lambda \in \sigma(A)\) we have

As a result of Part 1 this implies

A similar argument using \(1 = g(\lambda) (f(\lambda) - \lambda_0)\) implies

In both cases \(g(A)\) is bounded as a result of Proposition. This implies that the bounded operator \(g(A)\) is the inverse of \((f(A) - \lambda_0 \mathbf{1})\). This in turn implies that \(\lambda_0\) is not in the spectrum \(\sigma(f(A))\) of \(f(A)\).

This proves the contrapositive of \(\sigma(f(A) \subseteq \{ f(\lambda) : \lambda \in \sigma(A) \}\) holds when \(\lambda_0\) is in \(\mathbb{R}\) in \(\mathbb{C}\), and thus proves \(\sigma(f(A) \subseteq \{ f(\lambda) : \lambda \in \sigma(A) \}\) holds in this case.

Next let us prove that \(\{ f(\lambda) : \lambda \in \sigma(A) \} \subseteq \sigma(f(A)\).

Assume that there exists some \(\lambda_0\) such that \(\lambda_0 = f(\mu)\) for some \(\mu\) in the spectrum \(\sigma(A)\) of \(A\). Our goal is then to prove that \(f(\mu)\) is in \(\sigma(f(A))\). We will do so using proof by contradiction, assuming \(f(\mu)\) is not in \(\sigma(f(A))\) and proving this leads to a contradiction.

To that end, assume that \(f(\mu)\) is in the resolvent set of \(A\). Hence, \(f(A) - f(\mu) \mathbf{1}\) is invertible in \(\mathcal{B}(\mathbf{H})\).

As mentioned in Part 1 there exists a sequence \(\{ s_i \}_{i \in \mathbb{N}}\) in the set of real-valued polynomials on \(\sigma(A)\) such that \(s_i \rightarrow f\) uniformly.

As \(s_i \rightarrow f\) uniformly Proposition implies that \(s_i(A) \rightarrow f(A)\) relative to the operator norm. Hence, for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i \ge N\) one has

As a result of \(s_i \rightarrow f\) uniformly and \(s_i(A) \rightarrow f(A)\) relative to the operator norm, one can conclude that for any \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(i \ge N\) one has

Now by assumption \(f(A) - f(\mu)\mathbf{1}\) is invertible. Thus as a result of Lemma we can select \(\epsilon\) so small that for the associated \(N\) and \(i \ge N\) one forces \(s_i(A) - s_i(\mu)\mathbf{1}\) to invertible. However, this contradicts the Spectral Mapping Theorem.

For the polynomial \(s_i\) the Spectral Mapping Theorem states

This implies that \(s_i(\mu)\) should be in the spectrum of \(s_i(A)\) which implies \(s_i(A) - s_i(\mu)\mathbf{1}\) is not invertible. However, we just proved that under the assumption \(f(\mu)\) is not in \(\sigma(f(A))\) it follows that \(s_i(A) - s_i(\mu)\mathbf{1}\) is invertible. Hence, our assumption that \(f(\mu)\) is not in \(\sigma(f(A))\) is false and \(f(\mu)\) is in \(\sigma(f(A))\).

Hence, we have proven the second desired result \(\{ f(\lambda) : \lambda \in \sigma(A) \} \subseteq \sigma(f(A)\).

Now we have proven

Together these imply

the final desired result.\(\blacksquare\)

Stage 2: An Operator-Valued Riesz Representation Theorem

We are finally in a position to kick off Stage 2. In this stage we will prove that for a real-valued, continuous function \(f\) on the spectrum \(\sigma(A)\) of \(A\), the operator \(f(A)\) derived using real-valued functional calculus can be represented as integration against a projection-valued measure. This will essentially amount to an operator-valued version of the Riesz Representation Theorem.

Theorem (Riesz Representation) Let \(X\) be a compact metric space and let \(C^0(X; \mathbb{R})\) be the space of continuous, real-valued functions on \(X\). Suppose \(\Lambda : C^0(X; \mathbb{R}) \rightarrow \mathbb{R}\) is a linear function with the property that \(\Lambda(f)\) is non-negative whenever all the values of \(f\) are non-negative. Then there exists a unique, real-valued, positive measure \(\mu\) on the Borel \(\sigma\)-algebra of \(X\) for which

\[\Lambda(f) = \int_X f \, d\mu\]

for all \(f \in C^0(X; \mathbb{R})\).

However, before considering an operator-valued version, let us consider how we can apply the non-operator-valued version to the situation at hand.

The hypothesis of the Spectral Theorem for Bounded, Self-Adjoint Operators presents us with an element \(A \in \mathcal{B}(\mathbf{H})\) that is self-adjoint. For any continuous, real-valued function \(f\) on the spectrum \(\sigma(A)\) of \(A\), i.e. any \(f \in C^0(\sigma(A) ; \mathbb{R})\), the real-valued functional calculus of Proposition defines a linear map \(f \mapsto f(A)\) from \(C^0(\sigma(A) ; \mathbb{R})\) into \(\mathcal{B}(\mathbf{H})\). Furthermore, if \(f\) is non-negative, then as a result of Part 3 of Proposition, it follows that \(f(A)\) is a non-negative bounded operator, i.e. \(0 \le \left< \psi, f(A)\psi \right>\) for all \(\psi \in \mathbf{H}\).

Before defining our candidate functional, let us verify that the quantity \(\left< \psi, f(A) \psi \right>\) is in fact a real number for any \(f \in C^0(\sigma(A) ; \mathbb{R})\) and any \(\psi \in \mathbf{H}\). As a result of Part 2 of Proposition, for any \(f \in C^0(\sigma(A) ; \mathbb{R})\) the operator \(f(A)\) is self-adjoint, i.e. \(f(A)^* = f(A)\). Hence, for any \(\psi \in \mathbf{H}\)

where the first line follows from the conjugate symmetry of the Hilbert space inner product, the second from the definition of the adjoint, and the third from the self-adjointness of \(f(A)\). As a complex number equal to its own complex conjugate is real, we conclude

for all \(f \in C^0(\sigma(A) ; \mathbb{R})\) and all \(\psi \in \mathbf{H}\).

Hence, for any \(\psi \in \mathbf{H}\) the function \(\Lambda_\psi : C^0(\sigma(A) ; \mathbb{R}) \rightarrow \mathbb{R}\) defined by

is well-defined, in the sense that it does indeed take values in \(\mathbb{R}\), and satisfies the hypotheses required by the Riesz Representation Theorem. It is linear, as \(f \mapsto f(A)\) is linear by Proposition and the inner product is linear in its second argument, and it is non-negative whenever all the values of \(f\) are non-negative, as established in the first paragraph above. Hence, we can apply the Riesz Representation Theorem and conclude that for any \(\psi \in \mathbf{H}\) there exists a measure \(\mu_\psi\) such that

for all \(f \in C^0(\sigma(A) ; \mathbb{R})\). Note how similar this is to the equality

that appears when Theorem (Operator-Valued Integration) is formulated on \(X \equiv \sigma(A)\). This will turn out to be more than a similarity. The remainder of Stage 2 will be dedicated to proving that these are indeed the same equation.

To that end let us make the following definition

Definition Let \(A\) in \(\mathcal{B}(\mathbf{H})\) be self-adjoint. For any bounded, measurable, complex-valued function \(f\) on the spectrum \(\sigma(A)\) of \(A\) let us define a map \(Q_f : \mathbf{H} \rightarrow \mathbb{C}\) by

\[Q_f(\psi) \equiv \int_{\sigma(A)} f(\lambda) \, d\mu_\psi(\lambda),\]

where \(\mu_\psi\) is the measure on \(\sigma(A)\) derived from our map \(\Lambda_\psi\) and the Riesz Representation Theorem.

Before proceeding, let us record a basic property of the measures \(\mu_\psi\) just defined. This property is used repeatedly in what follows—in particular it is the hypothesis required to apply the Bounded Convergence Theorem to \(\mu_\psi\)—and so we isolate it here rather than re-deriving it at each point of use.

Lemma (The Associated Measures are Finite) Let \(A \in \mathcal{B}(\mathbf{H})\) be self-adjoint and, for any \(\psi \in \mathbf{H}\), let \(\mu_\psi\) be the measure on \(\sigma(A)\) of Definition. Then

\[\mu_\psi(\sigma(A)) = \left\| \psi \right\|^2.\]

In particular \(\mu_\psi(\sigma(A)) < \infty\), i.e. \(\mu_\psi\) is a finite measure on \(\sigma(A)\).

Proof Recall from Definition that for any \(f \in C^0(\sigma(A); \mathbb{R})\) one has

The constant function \(f(\lambda) = 1\) is continuous and real-valued on \(\sigma(A)\), and so is an element of \(C^0(\sigma(A); \mathbb{R})\). Furthermore, viewing \(f(\lambda) = 1\) as the constant polynomial \(p(\lambda) = 1\), the real-valued functional calculus of Proposition gives \(f(A) = \mathbf{1}\), the identity operator on \(\mathbf{H}\). Hence, one has

where the final line follows from the definition of the norm on \(\mathbf{H}\). As \(\psi\) is an element of the Hilbert space \(\mathbf{H}\), its norm \(\left\| \psi \right\|\) is a finite real number, and thus

i.e. \(\mu_\psi\) is a finite measure on \(\sigma(A)\), the desired result.\(\blacksquare\)

It turns out that \(Q_f\) is a bounded quadratic form, as proven in the following Proposition

Proposition Let \(A\) in \(\mathcal{B}(\mathbf{H})\) be self-adjoint. For any bounded, measurable, complex-valued function \(f\) on the spectrum \(\sigma(A)\) of \(A\), let \(Q_f : \mathbf{H} \rightarrow \mathbb{C}\) be its associated map

\[Q_f(\psi) \equiv \int_{\sigma(A)} f(\lambda) \, d\mu_\psi(\lambda),\]

via Definition. This map \(Q_f\) is a bounded quadratic form.

Proof Let \(\mathcal{F}\) be the set of all bounded, Borel-measurable, complex-valued functions \(f\) on the spectrum \(\sigma(A)\) of \(A\) such that \(Q_f\) is a bounded quadratic form. It turns out that \(\mathcal{F}\) is a vector space.

Explicitly, consider \(f,g \in \mathcal{F}\) and \(\alpha, \beta \in \mathbb{C}\). To prove that \(\mathcal{F}\) is a vector space we must prove that \(\alpha f + \beta g\) is a member of \(\mathcal{F}\). It is to this we now turn.

The proof that \(\mathcal{F}\) is a vector space essentially relies on the fact that \(f \mapsto Q_f\) is linear. Explicitly,

Now to prove that \(\mathcal{F}\) is a vector space we must prove that

is bounded quadratic form and thus an element of \(\mathcal{F}\).

To prove that \(Q_{\alpha f + \beta g} = \alpha Q_f + \beta Q_g\) is a bounded quadratic form we must prove that

1. \(Q_{\alpha f + \beta g}(\lambda\psi) = \lvert\lambda\rvert^2 Q_{\alpha f + \beta g}(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\).

2. The map \(L_{\alpha f + \beta g} : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

   \[\begin{align} L_{\alpha f + \beta g}(\phi, \psi) &\equiv \frac{1}{2} \left[ Q_{\alpha f + \beta g}(\phi + \psi) - Q_{\alpha f + \beta g}(\phi) - Q_{\alpha f + \beta g}(\psi) \right] \\ &-\frac{i}{2} \left[ Q_{\alpha f + \beta g}(\phi + i\psi) - Q_{\alpha f + \beta g}(\phi) - Q_{\alpha f + \beta g}(i\psi) \right] \end{align}\]

   is a sesquilinear form on \(\mathbf{H}\).

3. That there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

   \[|Q_{\alpha f + \beta g}(\phi)| \le C \|\phi\|^2,\]

   where \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

Let us prove these one by one.

First let us prove \(Q_{\alpha f + \beta g}(\lambda\psi) = \lvert\lambda\rvert^2 Q_{\alpha f + \beta g}(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\).

This follows from the fact that \(f\) and \(g\) are in \(\mathcal{F}\) and thus \(Q_f\) and \(Q_g\) are bounded quadratic forms. Explicitly,

which implies

the first desired result.

Next let us prove the map \(L_{\alpha f + \beta g} : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

is a sesquilinear form on \(\mathbf{H}\).

Again this follows from the fact that \(f\) and \(g\) are in \(\mathcal{F}\) and thus \(Q_f\) and \(Q_g\) are bounded quadratic forms. Explicitly, linearity implies

with the obvious definitions of \(L_f\) in terms of \(Q_f\) and \(L_g\) in terms of \(Q_g\).

Now as \(Q_f\) and \(Q_g\) are bounded quadratic forms, \(L_f\) and \(L_g\) are sesquilinear forms. Hence, they are conjugate linear in the first factor and linear in the second factor. Thus \(L_{\alpha f + \beta g}\) is conjugate linear in the first factor and linear in the second factor. Hence, \(L_{\alpha f + \beta g}\) is a sesquilinear form, the desired result.

Finally, let us prove that there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

where \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

Again this follows from the fact that \(f\) and \(g\) are in \(\mathcal{F}\) and thus \(Q_f\) and \(Q_g\) are bounded quadratic forms. Explicitly, the norm definition and linearity imply

where we have used the fact that \(Q_f\) and \(Q_g\) are bounded quadratic forms to infer that there exist constants \(C_f\) and \(C_g\) in \(\mathbb{R}\) such that

along with the definition

This implies

the final desired result. This completes the proof that \(Q_{\alpha f + \beta g}\) is a bounded quadratic form. This in turn implies that \(\alpha f + \beta g\) is an element of \(\mathcal{F}\) which in turn implies \(\mathcal{F}\) is a vector space.

Next let us prove that \(C^0(\sigma(A); \mathbb{R})\) is a subset of \(\mathcal{F}\).

By definition \(\mathcal{F}\) is a subset of the set of bounded, Borel-measurable, complex-valued functions on the spectrum \(\sigma(A)\) of \(A\), where \(A\) is self-adjoint. So, let us first prove that \(C^0(\sigma(A); \mathbb{R})\) is a subset of this set of bounded, Borel-measurable, complex-valued functions.

Recall that by Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a non-empty, compact metric measurable space, and is a subset of \(\mathbb{R} \subset \mathbb{C}\). In addition, the Boundedness Theorem implies that any element of \(C^0(\sigma(A); \mathbb{R})\) is bounded.

As \(C^0(\sigma(A); \mathbb{R})\) is continuous, for any \(f\) in \(C^0(\sigma(A); \mathbb{R})\) the pre-image of any open set in \(\mathbb{R}\) is open in \(\sigma(A)\). As \(\mathbb{R} \subset \mathbb{C}\) has the subset topology, for any \(f\) in \(C^0(\sigma(A); \mathbb{R})\) the pre-image of any open set in \(\mathbb{C}\) is open in \(\sigma(A)\). Hence, any element of \(C^0(\sigma(A); \mathbb{R})\) is Borel-measurable when considered as a complex-valued function on the spectrum \(\sigma(A)\) of \(A\).

So with that we have proven that \(C^0(\sigma(A); \mathbb{R})\) is a subset of the set of bounded, Borel-measurable, complex-valued functions on the spectrum \(\sigma(A)\) of \(A\).

The final step to prove \(C^0(\sigma(A); \mathbb{R}) \subset \mathcal{F}\) is to prove that any \(f\) in \(C^0(\sigma(A); \mathbb{R})\) results in a bounded quadratic form \(Q_f\).

Now tracing definitions we find that for any \(f\) in \(C^0(\sigma(A); \mathbb{R})\) one has

where bounded operator \(f(A)\) is the image of \(f\) under the real-valued functional calculus of Proposition. Thus as a result of Proposition

Proposition If \(A \in \mathcal{B}(\mathbf{H})\), one can construct a bounded quadratic form \(Q_A\) on \(\mathbf{H}\) by setting

\[Q_A(\psi) \equiv \left< \psi, A\psi \right>\]

for all \(\psi\) in \(\mathbf{H}\). The associated sesquilinear form \(L_A\) is then given by

\[L_A(\phi, \psi) = \left< \phi, A\psi \right>\]

for all \(\phi\) and \(\psi\) in \(\mathbf{H}\).

we can conclude that \(Q_f(\psi) = \left< \psi, f(A)\psi \right>\) is a bounded quadratic form, the final desired result required to prove that \(C^0(\sigma(A); \mathbb{R})\) is a subset of \(\mathcal{F}\).

Our next step is to prove that \(\mathcal{F}\) is closed under uniformly bounded pointwise limits. This essentially is a result of the fact that \(Q_f(\psi)\) is continuous with respect to such limits.

Explicitly, consider a sequence \(\{ f_i \}_{i \in \mathbb{N}}\) of elements in \(\mathcal{F}\) that are uniformly bounded, i.e. there exists a real-valued constant \(M\) such that

for all \(i \in \mathbb{N}\) and all \(\lambda \in \sigma(A)\), and that converge pointwise, i.e. there exists a map \(f : \sigma(A) \rightarrow \mathbb{C}\) that satisfies

pointwise for each \(\lambda \in \sigma(A)\) relative to the standard norm \(\lvert \cdot \rvert\) on \(\mathbb{C}\). Our goal is then to prove that \(f\) is in \(\mathcal{F}\).

As \(\mathcal{F}\) is a subset of the set of all bounded, Borel-measurable, complex-valued functions on \(\sigma(A)\), our first task is to prove that \(f\) is a bounded, Borel-measurable, complex-valued function.

Each \(f_i\) is an element of \(\mathcal{F}\), and thus a bounded, Borel-measurable, complex-valued function on \(\sigma(A)\). By Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a compact metric measurable space. Hence, identifying \(X\) with \(\sigma(A)\), the sequence \(\{ f_i \}_{i \in \mathbb{N}}\) and its pointwise limit \(f\) satisfy the hypotheses of Lemma (Pointwise Limits of Uniformly Bounded, Borel-Measurable Functions). That Lemma directly implies that \(f\) is bounded and Borel-measurable, the desired result.

Next we have to prove that \(f\) is in \(\mathcal{F}\). The definition of \(\mathcal{F}\) implies that this is equivalent to proving that \(Q_f\) is a bounded quadratic form.

To prove that \(Q_f\) is a bounded quadratic form we must prove that

1. \(Q_f(\lambda\psi) = \lvert\lambda\rvert^2 Q_f(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\).

2. The map \(L_f : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

   \[\begin{align} L_f(\phi, \psi) &\equiv \frac{1}{2} \left[ Q_f(\phi + \psi) - Q_f(\phi) - Q_f(\psi) \right] \\ &-\frac{i}{2} \left[ Q_f(\phi + i\psi) - Q_f(\phi) - Q_f(i\psi) \right] \end{align}\]

   is a sesquilinear form on \(\mathbf{H}\).

3. That there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

   \[|Q_f(\phi)| \le C \|\phi\|^2,\]

   where \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

The key “engine” in proving the \(Q_f\) is a bounded quadratic form is the Bounded Convergence Theorem

Theorem (Bounded Convergence Theorem) Let \(X\) be a set with finite measure \(\mu(X) < \infty\) and \(\{ f_i \}_{i \in \mathbb{N}}\) a sequence of uniformly bounded, complex valued functions on \(X\) that converge pointwise to \(f\), then

\[\int_X f \, d\mu = \lim\limits_{i \rightarrow \infty} \int_X f_i \, d\mu.\]

To apply this to the sequence \(\{ f_i \}_{i \in \mathbb{N}}\) at hand in order to prove

we must prove that

This is precisely the content of Lemma (The Associated Measures are Finite), which gives \(\mu_\psi(\sigma(A)) = \left\| \psi \right\|^2 < \infty\). In other words \(\sigma(A)\) has a finite measure \(\mu_\psi(\sigma(A))\).

That established we can apply the Bounded Convergence Theorem to conclude that

This implies

and it is this relation

that will allow us to prove that \(Q_f\) is a bounded quadratic form.

To wit, first let us prove \(Q_f(\lambda\psi) = \lvert\lambda\rvert^2 Q_f(\psi)\) for all \(\psi \in \mathbf{H}\) and \(\lambda \in \mathbb{C}\).

This follows from our previous result along with the fact that the \(Q_{f_i}(\psi)\) are bounded quadratic forms. We have

which is the desired result.

Now let us prove the map \(L_f : \mathbf{H} \times \mathbf{H} \rightarrow \mathbb{C}\) defined by

is a sesquilinear form on \(\mathbf{H}\).

Linearity implies

As the \(L_{f_i}(\phi, \psi)\) are conjugate linear in the first factor and linear in the second factor the same is true of \(L_f(\phi, \psi)\) and thus it is a sesquilinear form on \(\mathbf{H}\), the desired result.

Finally, let us prove that there exists a constant \(C\) in \(\mathbb{R}\) such that for all \(\phi\) in \(\mathbf{H}\)

where \(\mid \cdot \mid\) is the norm on \(\mathbb{C}\) and \(\|\cdot\|\) is the norm on \(\mathbf{H}\).

This follows from Lemma (The Associated Measures are Finite), which gives

for any \(\phi \in \mathbf{H}\), together with the fact that \(f\) is bounded. Together these imply

which gives the desired result, there exists a real constant

such that for all \(\phi\) in \(\mathbf{H}\) one has

This completes our proof that \(f\) is in \(\mathcal{F}\) and thus our proof that \(\mathcal{F}\) is closed under uniformly bounded pointwise limits.

Finally to complete the proof of Proposition we will prove that \(\mathcal{F}\) is the space of all bounded, Borel-measurable, complex-valued functions on \(\sigma(A)\).

Our proof of this result requires that we first prove some “utility” lemmas as stepping stones. The first of these “utility” lemmas is the following:

Lemma Let \(X\) be a compact metric measurable space with a measure \(\mu_X\). Let \(\mathcal{L}_0\) be the set of all measurable subsets \(E\) of \(X\) with an indicator function \(1_E\) that is a uniformly bounded limit of a sequence of continuous functions. Then \(\mathcal{L}_0\) is an algebra and contains all open sets in \(X\).

Proof This proof consists of two parts

Part 1: Prove that \(\mathcal{L}_0\) is an algebra of sets.

Part 2: Prove that \(\mathcal{L}_0\) contains all open sets in \(X\).

Let us first prove Part 1, that \(\mathcal{L}_0\) is an algebra of sets.

Recall that an algebra of sets is defined as follows:

Definition (Algebra of Sets) Given a set \(S\) and a set of subsets \(\mathcal{S}\) of \(S\), the set of subsets \(\mathcal{S}\) is an algebra of sets if it

1. Contains the empty set, i.e. \(\emptyset \in \mathcal{S}\).
2. Closed under complements, i.e. if \(s \in \mathcal{S}\), then \((S \backslash s) \in \mathcal{S}\).
3. Closed under finite union, i.e. if \(r,s \in \mathcal{S}\), then \(s \cup r \in \mathcal{S}\).

Let us one-by-one prove each of these properties holds for \(\mathcal{L}_0\).

Property 1: Let us start by proving that \(\emptyset \in \mathcal{L}_0\). To prove this we must prove Property 1.1: that \(\emptyset\) is measurable and Property 1.2: that \(1_\emptyset\) is the pointwise limit of a sequence of uniformly bounded continuous functions.

Property 1.1: Let us first prove that \(\emptyset\) is measurable.

The definition of a \(\sigma\)-algebra requires that the empty set \(\emptyset\) is an element of any \(\sigma\)-algebra, in particular the empty set \(\emptyset\) is an element of the \(\sigma\)-algebra of \(X\).

The definition of a measure implies that \(\mu_X\) maps the \(\sigma\)-algebra of \(X\) to the extended real numbers.

These facts together imply that \(\mu_X(\emptyset)\) is defined, which is none other than the desired result, \(\emptyset\) is measurable.

Property 1.2: Let us now prove that \(1_\emptyset\) is the pointwise limit of a sequence of uniformly bounded continuous functions.

To prove that \(1_\emptyset\) is the pointwise limit of a sequence of uniformly bounded continuous functions, we must prove that Property 1.2.1: there exists a sequence of function \(\{ f_n \}_{n \in \mathbb{N}}\) in \(C^0(X; \mathbb{R})\) and a real-valued constant \(C\) such that

for all \(x \in X\) and all \(n \in \mathbb{N}\); and Property 1.2.2: for any real number \(\epsilon > 0\) there exist a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\).

Below we will prove that the set of constant functions

on \(X\) has all these desired properties.

Property 1.2.1: Let us next prove that the sequence of functions \(\{ f_n \}_{n \in \mathbb{N}}\) above is in \(C^0(X; \mathbb{R})\) and a real-valued constant \(C\) exists such that

for all \(x \in X\) and all \(n \in \mathbb{N}\).

We begin with the obvious result that the constant functions

on \(X\) are elements of \(C^0(X; \mathbb{R})\).

Also a moment’s thought reveals that if we take \(C = 1\), then we have

for all \(x \in X\) and all \(n \in \mathbb{N}\). Hence, the sequence satisfies all the desired properties required by Property 1.2.1.

Property 1.2.2: Let us now prove that for any real number \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\).

Recall that the definition of an indicator function implies that

for all \(x \in X\). Hence, we must prove that for any real number \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\). As \(f_n(x) \equiv 1 / (n + 1)\), this is true as a result of the Archimedean Property

Theorem (Archimedean Property) For any real number \(\epsilon\) such that \(\epsilon > 0\) there exist a natural number \(N\) such that for all \(n \ge N\) one has \(1/n < \epsilon\).

Hence, the sequence \(f_n(x) \equiv 1 / (n + 1)\) satisfies all the desired properties required by Property 1.2.2. This completes the proof of Property 1.

Property 2: Let us next prove that \(\mathcal{L}_0\) is closed under complements. In other words if \(E \in \mathcal{L}_0\), then \(E^c \equiv (X \backslash E) \in \mathcal{L}_0\). To prove this we must prove Property 2.1: that \(E^c\) is measurable, and Property 2.2: that \(1_{E^c}\) is a pointwise limit of a sequence of uniformly bounded continuous functions.

Property 2.1: Let us prove that if \(E \in \mathcal{L}_0\), then \(E^c\) is measurable

By hypothesis \(E \in \mathcal{L}_0\). The definition of \(\mathcal{L}_0\) then implies that \(E\) is measurable and thus in the \(\sigma\)-algebra of \(X\). As \(E\) is in the \(\sigma\)-algebra of \(X\), the \(\sigma\)-algebra definition implies that \(E^c\) is also in the \(\sigma\)-algebra of \(X\). Hence, the definition of a measure and the fact that we have a measure \(\mu\) on \(X\) imply that \(E^c\) is measurable, the desired result.

Property 2.2: Now let us prove that \(1_{E^c}\) is a pointwise limit of a sequence of uniformly bounded continuous functions.

By hypothesis \(E \in \mathcal{L}_0\). Hence, (1) the definition of \(\mathcal{L}_0\) implies that there exists a sequence of functions \(\{ f_n \}_{n \in \mathbb{N}}\) in \(C^0(X; \mathbb{R})\) and a real-valued constant \(C\) such that

for all \(x \in X\) and \(n \in \mathbb{N}\) and (2) for every real number \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\).

Let us define a sequence of functions \(\{ g_n \}_{n \in \mathbb{N}}\) on \(X\) by

Obviously, the \(g_n\) are in \(C^0(X; \mathbb{R})\). The definition of the \(g_n\), the properties of the \(f_n\), and the norm definition imply

where we have made the definition \(D \equiv 1 + C\). Thus we have proven that there exists a sequence of functions \(\{ g_n \}_{n \in \mathbb{N}}\) in \(C^0(X; \mathbb{R})\) and a real-valued constant \(D\) such that

for all \(x \in X\) and all \(n \in \mathbb{N}\). In other words our sequence \(\{ g_n \}_{n \in \mathbb{N}}\) is a sequence of uniformly bounded continuous functions.

Let \(\epsilon > 0\) be an arbitrary real number and consider if we can find a natural number \(N\) such that

for all \(x \in X\) and all \(n \ge N\).

To this end for an arbitrary \(g_n\) and arbitrary \(x \in X\) consider

where in the second line we used \(1_{E^c}(x) = 1 - 1_{E}(x)\) which follows easily from the definitions of the indicator function and the sets \(E\) and \(E^c\).

However, as \(E \in \mathcal{L}_0\) the definition of \(\mathcal{L}_0\) implies that for an arbitrary real-valued number \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\).

However, as we have proven that

for an arbitrary \(n \in \mathbb{N}\) and arbitrary \(x \in X\), then it follows that for an arbitrary real-valued number \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\).

Thus with this we have proven that \(1_{E^c}\) is the pointwise limit of a sequence of uniformly bounded functions, the desired Property 2.2 result.

In proving Property 2.1 and Property 2.2 we can thus conclude that \(\mathcal{L}_0\) is closed under complements, the desired Property 2 result.

Property 3: Next let us prove that \(\mathcal{L}_0\) is closed under finite union, i.e. if \(E_1,E_2 \in \mathcal{L}_0\), then \(E_1 \cup E_2 \in \mathcal{L}_0\). Proving Property 3 is tantamount to proving Property 3.1: that \(E_1 \cup E_2\) is measurable and Property 3.2: that \(1_{E_1 \cup E_2}\) is a pointwise limit of a sequence of uniformly bounded continuous functions.

Property 3.1: Let us next prove that \(E_1 \cup E_2\) is measurable.

By hypothesis \(E_1,E_2 \in \mathcal{L}_0\). The definition of \(\mathcal{L}_0\) implies that both \(E_1\) and \(E_2\) are measurable and thus in the \(\sigma\)-algebra of \(X\). As \(E_1\) and \(E_2\) are in the \(\sigma\)-algebra of \(X\), the \(\sigma\)-algebra definition implies that \(E_1 \cup E_2\) is also in the \(\sigma\)-algebra of \(X\). Hence, the definition of a measure and the fact that we have a measure \(\mu\) on \(X\) imply that \(E_1 \cup E_2\) is also measurable, the desired result.

Property 3.2: Finally let us prove that \(1_{E_1 \cup E_2}\) is a pointwise limit of a sequence of uniformly bounded continuous functions.

By hypothesis \(E_1,E_2 \in \mathcal{L}_0\). The definition of \(\mathcal{L}_0\) implies that (1) there for each \(E_i\) exists a sequence \(\{ f^i_n \}_{n \in \mathbb{N}}\) in \(C^0(X; \mathbb{R})\) and a real-valued constant \(C_i\) such that

for all \(n \in \mathbb{N}\) and all \(x \in X\) and (2) for each \(E_i\) and any real number \(\epsilon_i > 0\) there exists a natural number \(N_i\) such that for all \(n \ge N_i\) we have

for all \(x \in X\).

With the easily verifiable fact

in mind, let us define the sequence of functions \(\{ g_n \}_{n \in \mathbb{N}}\) on \(X\) by

As the \(f^i_n\) are in \(C^0(X; \mathbb{R})\), it obviously follows that the \(g_n\) are also in \(C^0(X; \mathbb{R})\).

The norm definition along with the fact that the \(f^i_n\) are bounded implies

where we have defined \(D \equiv C_1 + C_2 + C_1 C_2\).

Thus we have proven that there exists a sequence of functions \(\{ g_n \}_{n \in \mathbb{N}}\) in \(C^0(X; \mathbb{R})\) and a real-valued constant \(D\) such that

for all \(x \in X\) and all \(n \in \mathbb{N}\). In other words the \(g_n\) are a sequence of uniformly bounded continuous functions.

Let \(\epsilon > 0\) be an arbitrary real number and consider if we can find a natural number \(N\) such that

for all \(x \in X\) and all \(n \ge N\).

This is always possible as for all \(x \in X\) the limit \(f^1_n(x) \rightarrow 1_{E_1}(x)\) and the limit \(f^2_n \rightarrow 1_{E_2}\) exist as real finite numbers. Hence,

for all \(x \in X\). Thus for an arbitrary real number \(\epsilon > 0\) there exists a natural number \(N\) such that for all \(n \ge N\) one has

for all \(x \in X\). This completes the proof of Property 3.2 and the fact that \(\mathcal{L}_0\) is an algebra of sets.

Let us next prove Part 2, that \(\mathcal{L}_0\) contains all open sets in \(X\).

We will prove that \(\mathcal{L}_0\) contains all open sets in \(X\) by Step 1: proving that \(\mathcal{L}_0\) contains all closed sets in \(X\), then Step 2: relying on the fact that any open set is the complement of a closed set along with the fact that we proved that \(\mathcal{L}_0\) is closed under complements.

Step 1: Let us first prove that \(\mathcal{L}_0\) contains all closed sets in \(X\). To prove that \(\mathcal{L}_0\) contains all closed sets in \(X\) we must Step 1.1: prove that any closed set \(\overline{E}\) in \(X\) is measurable and Step 1.2: prove that \(1_{\overline{E}}\) is the pointwise limit of a sequence of uniformly bounded continuous functions.

Step 1.1: Let us first prove that any closed set \(\overline{E}\) in \(X\) is measurable.

Note that by hypothesis \(X\) is a metric measurable space. By definition this implies that \(X\) is a measure space with a Borel regular measure. By definition a Borel regular measure is a measure in which any Borel set is measurable. By definition a Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and complement.

As any closed set \(\overline{E}\) in \(X\) is, by definition, the complement of an open set, and all open sets are Borel sets, it follows that any closed set \(\overline{E}\) is a Borel set and thus measurable, the desired Step 1.1 result.

Step 1.2: Let us next prove that \(1_{\overline{E}}\) is the pointwise limit of a sequence of uniformly bounded continuous functions.

To this end let us review a definition we will require, that of a “bump function”

Definition (Bump Function) Given a topological manifold \(X\), a closed subset \(\overline{E}\) of \(X\), and an open subset \(U\) of \(X\) that contains \(\overline{E}\), i.e. \(\overline{E} \subset U\), then a function \(f\) in \(C^0(X; \mathbb{R})\) is called a bump function for \(\overline{E}\) supported in \(U\) if

1. For every \(x \in X\) one has \(0 \le f(x) \le 1\).
2. For every \(x \in \overline{E}\) one has \(f(x) = 1\).
3. \(f\) is supported on \(U\).

The existence of a bump function is guaranteed by the following theorem

Theorem (Existence of Bump Functions) Let \(X\) be a normal topological space. For any closed set \(\overline{E}\) in \(X\) and any open set \(U\) containing \(\overline{E}\), there exists a bump function for \(\overline{E}\) supported in \(U\).

With this in mind consider our \(\overline{E}\). One can always arrange for the existence of a sequence of open sets \(\{ U_i \}_{i \in \mathbb{N}}\) in \(X\) such that \(U_{i + 1} \subset U_{i}\) and \(\overline{E} \subset U_i\) for all \(i \in \mathbb{N}\). By way of Existence of Bump Functions we are guaranteed the existence of a sequence of bump functions \(\{ f_i \}_{i \in \mathbb{N}}\) with \(f_i\) being a bump function for \(\overline{E}\) supported in \(U_i\).

The definition of a bump function implies that for \(C \equiv 1\) we have

for all \(x \in X\) and all \(i \in \mathbb{N}\). Thus the sequence \(\{ f_i \}_{i \in \mathbb{N}}\) is uniformly bounded.

Furthermore, by construction for an arbitrary real number \(\epsilon > 0\) we can find a natural number \(N\) such that for all \(i \ge N\) we have

for all \(x \in X\), this completes the proof of Step 1.2 and also the proof of Step 1. Thus we have established that \(\mathcal{L}_0\) contains all closed sets.

Step 2: Let us now prove that \(\mathcal{L}_0\) contains all open sets.

We just established that \(\mathcal{L}_0\) contains all closed sets. However, we previously established that \(\mathcal{L}_0\) is closed under complement, i.e. if a set is in \(\mathcal{L}_0\) then its complement is also in \(\mathcal{L}_0\). By definition any open set is the complement of a closed set. Hence, as any closed set is in \(\mathcal{L}_0\) it thus follows that any open set, as it’s the complement of a closed set, must also be in \(\mathcal{L}_0\), the desired Step 2 result and the conclusion of the Lemma proof.\(\blacksquare\)

The next of these “utility” lemmas is the following:

Lemma Let \(X\) be a compact metric measurable space and \(C^0(X; \mathbb{R})\) the set of continuous real-valued functions on \(X\). Let \(\mathcal{F}\) be the set of bounded, measurable, complex-valued functions on \(X\) such that (1) \(\mathcal{F}\) is a complex vector space, (2) \(\mathcal{F}\) contains \(C^0(X; \mathbb{R})\), and (3) \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences. Finally let \(\mathcal{L}_1\) be the set of all measurable sets \(E\) in \(X\) such that the indicator function \(1_E\) belongs to \(\mathcal{F}\). Then \(\mathcal{L}_1\) contains all Borel sets in \(X\).

Proof By definition \(\mathcal{L}_0\) of Lemma is the set of all measurable subsets \(E\) of \(X\) with an indicator function \(1_E\) that is a uniformly bounded limit of a sequence of continuous, real-valued functions.

Let us first prove that \(\mathcal{L}_0 \subseteq \mathcal{L}_1\). Consider any \(E\) in \(\mathcal{L}_0\). Then there exists a sequence \(\{ f_n \}_{n \in \mathbb{N}}\) in \(C^0(X; \mathbb{R})\), uniformly bounded, that converges pointwise to \(1_E\). By hypothesis \(C^0(X; \mathbb{R})\) is a subset of \(\mathcal{F}\), so each \(f_n\) is in \(\mathcal{F}\). By hypothesis \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences, so \(1_E\), being such a limit, is in \(\mathcal{F}\). Hence \(E\) is in \(\mathcal{L}_1\), the desired result \(\mathcal{L}_0 \subseteq \mathcal{L}_1\).

Next let us prove that \(\mathcal{L}_1\) is a monotone class, i.e. that \(\mathcal{L}_1\) is closed under countable increasing unions and countable decreasing intersections.

Let \(\{ E_n \}_{n \in \mathbb{N}}\) be an increasing sequence of sets in \(\mathcal{L}_1\), i.e. \(E_1 \subseteq E_2 \subseteq \cdots\) with each \(1_{E_n} \in \mathcal{F}\), and let \(E \equiv \bigcup_{n \in \mathbb{N}} E_n\). For any \(x \in X\), if \(x \in E\) then \(x \in E_{n_0}\) for some \(n_0\), and since the \(E_n\) are increasing, \(1_{E_n}(x) = 1\) for all \(n \ge n_0\); if \(x \notin E\) then \(1_{E_n}(x) = 0\) for all \(n\). Hence \(1_{E_n}(x) \rightarrow 1_E(x)\) for every \(x \in X\), i.e. \(1_{E_n} \rightarrow 1_E\) pointwise. Furthermore this convergence is uniformly bounded, as \(\lvert 1_{E_n}(x) \rvert \le 1\) for all \(n\) and \(x\). As \(\mathcal{F}\) is closed under uniformly bounded pointwise limits (property (3) of \(\mathcal{F}\)), we conclude \(1_E \in \mathcal{F}\), i.e. \(E \in \mathcal{L}_1\).

An entirely analogous argument applies to a decreasing sequence \(\{ E_n \}_{n \in \mathbb{N}}\) in \(\mathcal{L}_1\), i.e. \(E_1 \supseteq E_2 \supseteq \cdots\): with \(E \equiv \bigcap_{n \in \mathbb{N}} E_n\), one again has \(1_{E_n} \rightarrow 1_E\) pointwise and uniformly bounded by \(1\), so property (3) of \(\mathcal{F}\) gives \(1_E \in \mathcal{F}\), i.e. \(E \in \mathcal{L}_1\).

Together these prove that \(\mathcal{L}_1\) is a monotone class.

With this established, we can apply the Monotone Class Theorem

Theorem (Monotone Class Theorem) Let \(\mathcal{A}\) be an algebra of subsets of a set \(X\), and let \(\mathcal{M}\) be a monotone class, i.e. a collection of subsets of \(X\) closed under countable increasing unions and countable decreasing intersections, such that \(\mathcal{A} \subseteq \mathcal{M}\). Then \(\sigma(\mathcal{A}) \subseteq \mathcal{M}\), where \(\sigma(\mathcal{A})\) is the smallest \(\sigma\)-algebra containing \(\mathcal{A}\).

Identifying \(\mathcal{A}\) with \(\mathcal{L}_0\)—an algebra, as established in Lemma—and \(\mathcal{M}\) with \(\mathcal{L}_1\)—a monotone class containing \(\mathcal{L}_0\), as we just proved—the Monotone Class Theorem implies

where \(\sigma(\mathcal{L}_0)\) is the smallest \(\sigma\)-algebra containing \(\mathcal{L}_0\).

However, as \(\mathcal{L}_0\) contains all open sets in \(X\) (Lemma), and the Borel \(\sigma\)-algebra of \(X\) is by definition the smallest \(\sigma\)-algebra containing all open sets in \(X\), it follows that the Borel \(\sigma\)-algebra of \(X\) is contained in \(\sigma(\mathcal{L}_0)\). Combined with our previous result this gives

Hence, \(\mathcal{L}_1\) contains all Borel sets, the desired result.\(\blacksquare\)

Before proving the final of these “utility” lemmas, we isolate the following elementary fact about pointwise limits of Borel-measurable functions, which we will have need of shortly.

Lemma (Pointwise Limits of Uniformly Bounded, Borel-Measurable Functions) Let \(X\) be a compact metric measurable space. Let \(\{ f_i \}_{i \in \mathbb{N}}\) be a sequence of bounded, Borel-measurable, complex-valued functions on \(X\) that are uniformly bounded, i.e. there exists a real-valued constant \(M\) such that

\[\lvert f_i(x) \rvert \le M\]

for all \(i \in \mathbb{N}\) and all \(x \in X\), and that converge pointwise to a function \(f : X \rightarrow \mathbb{C}\). Then \(f\) is bounded and Borel-measurable.

Proof Let us first prove that \(f\) is bounded. For any \(x \in X\) we have

where we have used that \(f_i(x) \rightarrow f(x)\) and the definition of a norm. This proves \(\lvert f(x) \rvert \le M\) for all \(x \in X\), i.e. \(f\) is bounded.

Next let us prove that \(f\) is Borel-measurable.

As \(\{ f_i \}_{i \in \mathbb{N}}\) converges to \(f\) pointwise, and as taking real and imaginary parts is continuous, the real-valued sequences \(\{ \text{Re}(f_i) \}_{i \in \mathbb{N}}\) and \(\{ \text{Im}(f_i) \}_{i \in \mathbb{N}}\) converge pointwise to \(\text{Re}(f)\) and \(\text{Im}(f)\) respectively. Explicitly, for any \(x \in X\),

as \(i \rightarrow \infty\), and similarly

As \(\{ f_i \}_{i \in \mathbb{N}}\) is uniformly bounded by \(M\), so too are \(\{ \text{Re}(f_i) \}_{i \in \mathbb{N}}\) and \(\{ \text{Im}(f_i) \}_{i \in \mathbb{N}}\), since \(\lvert \text{Re}(f_i(x)) \rvert \le \lvert f_i(x) \rvert \le M\) and likewise for the imaginary part. This allows us to define, for each \(i \in \mathbb{N}\), the real numbers

both of which are well-defined, as suprema of bounded sets of real numbers.

By an identical argument to before—now applied to the real-valued sequence \(\{ \text{Re}(f_n(x)) \}_{n \in \mathbb{N}}\) converging to \(\text{Re}(f(x))\)—for any \(\epsilon > 0\) there exists a natural number \(N\) such that for \(i \ge N\) one has

which implies

An identical argument applied to the real-valued sequence \(\{ \text{Im}(f_n(x)) \}_{n \in \mathbb{N}}\) converging to \(\text{Im}(f(x))\) gives

Now looking at the sequences

we see that both are non-increasing, since each is a supremum taken over a shrinking set of indices \(\{ n : n \ge i \}\) as \(i\) increases. Hence, we can apply the Monotone Convergence Theorem (Non-Increasing Case)

Theorem (Monotone Convergence Theorem, Non-Increasing Case) If \(\{ a_i \}_{i \in \mathbb{N}}\) is a non-increasing sequence, i.e. \(a_i \ge a_{i + 1}\) for all \(i \in \mathbb{N}\), then

\[\lim\limits_{i \rightarrow \infty} a_i = \inf_{i \in \mathbb{N}} a_i.\]

to each of these sequences, giving

We will next prove that \(g_i\) and \(h_i\) are Borel-measurable, and use this and the two equations above to prove that \(\text{Re}(f)\) and \(\text{Im}(f)\)—and hence \(f\) itself—are Borel-measurable.

Note that a real-valued function is Borel-measurable if the preimage of any set of the form \((a, \infty)\) is a Borel set. With that in mind consider the preimage of \((a, \infty)\) under \(g_i\)

The supremum of a set of numbers is strictly greater than \(a\) if and only if at least one of the numbers in the sequence is strictly greater than \(a\). Hence,

As each \(f_n\) is by hypothesis Borel measurable, and the Borel \(\sigma\)-algebra on \(\mathbb{C}\) is generated by the Cartesian product of Borel sets on \(\mathbb{R}\), each \(\text{Re}( f_n(x) )\) is Borel measurable. Hence, sets of the form

are Borel measurable sets. As the \(\sigma\)-algebra of Borel sets is closed under countable unions, it then follows that

is a Borel measurable set. Hence, the preimage of any set of the form \((a, \infty)\) under \(g_i\) is a Borel measurable set, and thus \(g_i\) is Borel measurable. An identical argument, with \(\text{Im}(f_n)\) in place of \(\text{Re}(f_n)\), proves that \(h_i\) is Borel measurable.

Now let us prove that \(\text{Re}(f)\) and \(\text{Im}(f)\) are Borel measurable.

Recall that a real-valued function is Borel measurable if the preimage of any set of the form \((-\infty, a)\) is a Borel set. With this in mind consider the preimage of \((-\infty, a)\) under \(\text{Re}(f)\)

Now the infimum of a sequence is strictly less than \(a\) if and only if at least one term in the sequence is strictly less than \(a\). Hence,

However, we proved that \(g_i\) is Borel measurable. Thus

are Borel measurable sets. As the \(\sigma\)-algebra of Borel sets is closed under countable unions, it then follows that

is a Borel measurable set. Hence, the preimage of any set of the form \((-\infty, a)\) under \(\text{Re}(f)\) is a Borel measurable set, and thus \(\text{Re}(f(x))\) is Borel measurable. An identical argument, with \(h_i\) in place of \(g_i\), proves that \(\text{Im}(f(x))\) is Borel measurable.

As the Borel \(\sigma\)-algebra on \(\mathbb{C}\) is generated by the Cartesian product of Borel sets on \(\mathbb{R}\), a complex valued function is Borel-measurable if and only if both its real part and its imaginary part are real-valued Borel measurable functions. As we have just shown \(\text{Re}(f)\) and \(\text{Im}(f)\) are both real-valued Borel measurable functions, it follows that \(f\) is Borel-measurable.

Together, boundedness and Borel-measurability of \(f\) give the desired result.\(\blacksquare\)

The final of these “utility” lemmas is the following:

Lemma Let \(X\) be a compact metric measurable space and \(C^0(X; \mathbb{R})\) the set of continuous real-valued functions on \(X\). Let \(\mathcal{F}\) be the set of bounded, measurable, complex-valued functions on \(X\) such that (1) \(\mathcal{F}\) is a complex vector space, (2) \(\mathcal{F}\) contains \(C^0(X; \mathbb{R})\), and (3) \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences. Then \(\mathcal{F}\) consists of all bounded, Borel-measurable functions on \(X\).

Proof By hypothesis \(\mathcal{F}\) is the set of bounded, measurable, complex-valued functions on \(X\) such that (1) \(\mathcal{F}\) is a complex vector space, (2) \(\mathcal{F}\) contains \(C^0(X; \mathbb{R})\), and (3) \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences.

Let us first prove that any function \(f\) in \(\mathcal{F}\) is bounded and Borel-measurable.

For any \(f \in \mathcal{F}\), the definition of \(\mathcal{F}\) implies that \(f\) is bounded and measurable. As \(X\) is a metric measurable space, its measure is a Borel regular measure. A Borel regular measure is a measure in which any Borel set is measurable. Hence, in the definition of \(\mathcal{F}\) when we state that \(f\) is measurable we mean that \(f\) is Borel-measurable. Thus, any \(f \in \mathcal{F}\) is bounded and Borel-measurable.

Let us now prove that any bounded, Borel-measurable, complex-valued function \(f\) on \(X\) is in \(\mathcal{F}\).

By Lemma, the set \(\mathcal{L}_1 \equiv \{ E \subseteq X \text{ measurable} : 1_E \in \mathcal{F} \}\) contains all Borel sets of \(X\). As noted above, on \(X\) “measurable” and “Borel-measurable” coincide; hence every measurable set of \(X\) is a Borel set, and so for any measurable set \(E\) of \(X\), \(1_E \in \mathcal{F}\).

Consider now an arbitrary simple function \(s\) on \(X\), i.e.

where \(\alpha_i \in \mathbb{C}\) and the \(E_i\) are pairwise disjoint measurable sets of \(X\). By the previous paragraph, each \(1_{E_i}\) is in \(\mathcal{F}\). As \(\mathcal{F}\) is by hypothesis a complex vector space, it is closed under finite linear combinations, so \(s \in \mathcal{F}\). Hence every simple function on \(X\) is in \(\mathcal{F}\).

Now consider our arbitrary bounded, Borel-measurable, complex-valued function \(f\) on \(X\). By the Complex-Valued Simple Approximation Theorem, there exists a sequence \(\{ s_i \}_{i \in \mathbb{N}}\) of complex-valued simple functions on \(X\) that converges uniformly to \(f\) on \(X\). By the previous paragraph, each \(s_i\) is in \(\mathcal{F}\).

We claim the \(s_i\) are uniformly bounded. As \(f\) is bounded, there exists \(M_f \in \mathbb{R}\) such that \(\lvert f(x) \rvert \le M_f\) for all \(x \in X\). As \(s_i \rightarrow f\) uniformly, there exists \(N \in \mathbb{N}\) such that for all \(i \ge N\) and all \(x \in X\)

so that \(\lvert s_i(x) \rvert < M_f + 1\) for all \(i \ge N\) and all \(x \in X\). Each of the finitely many remaining \(s_1, \ldots, s_{N-1}\) is individually a simple function, hence bounded. Setting

we have \(\lvert s_i(x) \rvert \le M\) for all \(i \in \mathbb{N}\) and all \(x \in X\), i.e. the \(s_i\) are uniformly bounded.

As \(s_i \rightarrow f\) uniformly, in particular \(s_i \rightarrow f\) pointwise. Thus \(\{ s_i \}_{i \in \mathbb{N}}\) is a uniformly bounded sequence in \(\mathcal{F}\) that converges pointwise to \(f\). As \(\mathcal{F}\) is by hypothesis closed under pointwise limits of uniformly bounded sequences, \(f \in \mathcal{F}\).

Thus, any bounded, Borel-measurable, complex-valued function \(f\) on \(X\) is in \(\mathcal{F}\).

So we’ve proven that any function \(f\) in \(\mathcal{F}\) is bounded and Borel-measurable and we’ve proven that any bounded, Borel-measurable, complex-valued function \(f\) on \(X\) is in \(\mathcal{F}\). Thus, \(\mathcal{F}\) is the set of bounded, Borel-measurable, complex-valued functions on \(X\), the desired result.\(\blacksquare\)

With these “utility” lemmas established, we can once again consider the Proposition we were in the process of proving.

As one will recall there \(\mathcal{F}\) was the set of all bounded, Borel-measurable, complex-valued functions \(f\) on the spectrum \(\sigma(A)\) of \(A\) such that \(Q_f\) is a bounded quadratic form.

Previously we established that (1) \(\mathcal{F}\) is a complex vector space, (2) \(\mathcal{F}\) contains \(C^0(\sigma(A); \mathbb{R})\), and (3) \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences. By Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a non-empty, compact metric measurable space.

Now, with the metric and measure on \(\sigma(A)\) arising from \(\mathbb{C}\) as \(\sigma(A) \subset \mathbb{C}\), we find that \(\sigma(A)\) becomes a metric measurable space.

All of this together allows us apply Lemma, \(\sigma(A)\) taking the place of \(X\) and this \(\mathcal{F}\) the place of the identically named \(\mathcal{F}\) of the lemma. Doing so we can conclude that \(\mathcal{F}\) consists of all bounded, Borel-measurable functions on \(\sigma(A)\).

As \(\mathcal{F}\) consists of all bounded, Borel-measurable functions on \(\sigma(A)\), it follows that \(Q_f\) is quadratic form not only on some subset of functions, but all bounded, Borel-measurable functions on \(\sigma(A)\), proving that \(Q_f\) satisfies all the properties of Definition, the desired result of Proposition.\(\blacksquare\)

With this proposition resolved, let us introduce another definition that will be of use later. It essentially amounts to a means of defining an operator \(f(A)\) from a bounded measurable function \(f\) on \(\sigma(A)\); this is in contrast to the identically notated operator \(f(A)\) defined in Proposition which requires \(f\) be an element of \(C^0(\sigma(A); \mathbb{R})\).

Definition Let \(A\) in \(\mathcal{B}(\mathbf{H})\) be self-adjoint. For a bounded, measurable, complex-valued function \(f\) on the spectrum \(\sigma(A)\) of \(A\), let \(f(A)\) be the operator associated to the quadratic form \(Q_f\) of Definition by Proposition. This means that \(f(A)\) is the unique operator such that

\[\left< \psi, f(A)\psi \right> = Q_f(\psi) = \int_{\sigma(A)} f(\lambda) \, d\mu_\psi(\lambda)\]

for all \(\psi \in \mathbf{H}\).

As a first use of this definition we can prove the following lemma

Lemma Let \(A\) in \(\mathcal{B}(\mathbf{H})\) be self-adjoint, \(f\) a bounded, measurable, complex-valued function on the spectrum \(\sigma(A)\) of \(A\), and \(f(A)\) the operator associated to \(f\) by way of the previous Definition. If \(f\) is real-valued, then \(f(A)\) is self-adjoint.

Proof By hypothesis \(f\) is a bounded, measurable, real-valued function on the spectrum \(\sigma(A)\) of \(A\). By definition \(Q_f\) acting on an arbitrary \(\psi \in \mathbf{H}\) is given by

where, as stated in Definition, the measure \(\mu_\psi\) is the measure on \(\sigma(A)\) derived from the Riesz Representation Theorem. As a result of the Proposition we just proved, \(Q_f\) is a bounded quadratic form.

Now, as \(\mu_\psi\) is the measure derived from the Riesz Representation Theorem, it is a real-valued, positive measure on the Borel \(\sigma\)-algebra of \(\sigma(A)\).

Furthermore, by Lemma (The Spectrum is a Compact Metric Measurable Space) the spectrum \(\sigma(A)\) of \(A\) is in \(\mathbb{R}\).

Thus, \(f\) being real-valued, \(\mu_\psi\) being real-valued, and the spectrum \(\sigma(A)\) of \(A\) being a subset of \(\mathbb{R}\) together imply that the integral defining \(Q_f(\psi)\)

and thus \(Q_f(\psi)\) is real-valued for any \(\psi \in \mathbf{H}\).

As \(Q_f(\psi)\) is real-valued for all \(\psi \in \mathbf{H}\), Proposition implies that \(f(A)\) is self-adjoint, the desired result.\(\blacksquare\)

The next proposition proves the analog of multiplicativity from Proposition for operators \((fg)(A)\), \(f(A)\), and \(g(A)\) that arise from bounded measurable functions \(f\) and \(g\) by way of Definition.

Proposition Let \(A \in \mathcal{B}(\mathbf{H})\) be self-adjoint and \(f\) and \(g\) be bounded, measurable, complex-valued functions on the spectrum \(\sigma(A)\) of \(A\), then

\[(fg)(A) = f(A) g(A),\]

where \((fg)(A)\), \(f(A)\), and \(g(A)\) are operators that arise respectively from \((fg)\), \(f\), and \(g\) by way of Definition.

Proof Let \(\mathcal{F}_1\) be the set of bounded, measurable, complex-valued functions \(f\) such that

for all \(g\) in \(C^0(\sigma(A); \mathbb{R})\).

We claim that \(\mathcal{F}_1\) is a complex vector space. To prove this consider any \(f_1\) and \(f_2\) in \(\mathcal{F}_1\) and any \(\alpha_1\) and \(\alpha_2\) in \(\mathbb{C}\). Then one has

where the first line is the distributive property over functions, the second from linearity of the integral in Definition, the third from properties of \(\mathcal{F}_1\) along with the fact \(f_1, f_2 \in \mathcal{F}_1\) and \(g \in C^0(\sigma(A); \mathbb{R})\), and the final is the distributive property over operators. These imply

where the second equality follows from linearity. This implies

which is nothing more than the statement that \(\mathcal{F}_1\) is a complex vector space.

We further claim that \(C^0(\sigma(A); \mathbb{R})\) is a subset of \(\mathcal{F}_1\). To prove this consider arbitrary \(f\) and \(g\) in \(C^0(\sigma(A); \mathbb{R})\). If \(f(A)\) and \(g(A)\) were the operators defined by Proposition then the desired property

would follow from the multiplicativity property of Proposition. So if we can prove that in the case \(f,g \in C^0(\sigma(A); \mathbb{R})\), the operators \(f(A)\) and \(g(A)\) defined by Definition are the same as those defined by Proposition, then we will have proven the claim.

If \(f,g \in C^0(\sigma(A); \mathbb{R})\), the operators \(f(A)\) and \(g(A)\) defined by Definition are the unique operators \(f(A)\) and \(g(A)\) such that

for all \(\psi \in \mathbf{H}\). If \(f,g \in C^0(\sigma(A); \mathbb{R})\), the operators \(f(A)\) and \(g(A)\) defined by Proposition are the unique operators \(f(A)\) and \(g(A)\) such that

for all \(\psi \in \mathbf{H}\). Note here we have used the equation to write these operators in a form similar to that above. Hence, in both cases the operators \(f(A)\) and \(g(A)\) are the same. This then proves the claim that \(C^0(\sigma(A); \mathbb{R})\) is a subset of \(\mathcal{F}_1\).

Our next claim is that the map \(f \mapsto Q_f(\psi)\) is continuous under uniformly bounded pointwise convergence for any \(\psi \in \mathbf{H}\). It turns out we actually proved this as part of our Proposition proof.

In Proposition we proved that \(\mathcal{F}\) the space of all bounded, Borel-measurable, complex-valued functions \(f\) such that \(Q_f\) is a quadratic form is closed under uniformly bounded pointwise limits and that \(\mathcal{F}\) is the space of all bounded, Borel-measurable, complex-valued functions. Hence, \(f \mapsto Q_f(\psi)\) is continuous under uniformly bounded pointwise convergence for any \(\psi \in \mathbf{H}\) when \(f\) is a bounded, Borel-measurable, complex-valued function, the desired result.

Now the Polarization Identity states

Proposition (Polarization Identity) If \(L\) is a sesquilinear form on the Hilbert space \(\mathbf{H}\), then for any \(\phi, \psi \in \mathbf{H}\) the value of \(L(\phi, \psi)\) can be determined from the values of \(L\) “on the diagonal” (i.e. the values of \(L(\xi, \xi)\) for various \(\xi \in \mathbf{H}\)) as follows:

\[\begin{align} L(\phi, \psi) &= \frac{1}{2} \left[ L(\phi + \psi, \phi + \psi) - L(\phi, \phi) - L(\psi, \psi) \right] \\ &-\frac{i}{2} \left[ L(\phi + i\psi, \phi + i\psi) - L(\phi, \phi) - L(i\psi, i\psi) \right]. \end{align}\]

This result—along with the Proposition which for quadratic form \(Q\) expresses its sesquilinear form \(L\) on the diagonal in terms of the quadratic form itself as follows

for all \(\psi \in \mathbf{H}\)—allows us to write a quadratic form’s sesquilinear form “off the diagonal” purely in terms of the quadratic form itself.

So, as Proposition proves \(Q_f\) is a bounded, quadratic form for any bounded, Borel-measurable, complex-valued function \(f\), we can write the sesquilinear form \(L_f\) associated to \(Q_f\) purely in terms of \(Q_f\).

As we can express \(L_f\) purely in terms of \(Q_f\) it follows from our previous result that the map \(f \mapsto L_f(\phi, \psi)\) is continuous under uniformly bounded pointwise convergence for any \(\phi,\psi \in \mathbf{H}\) when \(f\) is a bounded, Borel-measurable, complex-valued function.

Now by definition for any \(f \in \mathcal{F}_1\) and any \(g \in C^0(\sigma(A); \mathbb{R})\) we have

This implies that for any \(\psi \in \mathbf{H}\) we have

Now the Definition implies that the operator \((fg)(A)\) satisfies

for all \(\psi \in \mathbf{H}\).

In addition the Definition and the previous result allowing us to express \(L_f\) in terms of \(Q_f\) gives

Hence, we have

Thus \(\left< \psi, (fg)(A)\psi \right> = \left< \psi, f(A)g(A)\psi \right>\) implies

for all \(\psi \in \mathbf{H}\), all \(f \in \mathcal{F}_1\), and all \(g \in C^0(\sigma(A); \mathbb{R})\).

With this identity in hand let us now prove that \(\mathcal{F}_1\) is closed under pointwise limits of uniformly bounded sequences.

To this end let \(\{ f_i \}_{i \in \mathbb{N}}\) be a sequence in \(\mathcal{F}_1\), uniformly bounded by some \(M \in \mathbb{R}\), that converges pointwise to a function \(f\) on \(\sigma(A)\). By Lemma (Pointwise Limits of Uniformly Bounded, Borel-Measurable Functions), \(f\) is itself a bounded, Borel-measurable, complex-valued function on \(\sigma(A)\), and so is a candidate for membership in \(\mathcal{F}_1\). Let \(g \in C^0(\sigma(A); \mathbb{R})\) and \(\psi \in \mathbf{H}\) be arbitrary.

As each \(f_i\) belongs to \(\mathcal{F}_1\), the identity established above gives

for every \(i \in \mathbb{N}\). We now take the limit \(i \rightarrow \infty\) of each side separately.

For the lefthand side, note first that as \(\sigma(A)\) is compact by Lemma (The Spectrum is a Compact Metric Measurable Space) and \(g\) is continuous, the Boundedness Theorem implies \(g\) is bounded, say by \(M_g \in \mathbb{R}\). As \(f_i \rightarrow f\) pointwise we have \(f_i g \rightarrow fg\) pointwise, and

for all \(i \in \mathbb{N}\) and all \(\lambda \in \sigma(A)\), so the sequence \(\{ f_i g \}_{i \in \mathbb{N}}\) is uniformly bounded. Hence, the Bounded Convergence Theorem, applied to the measure \(\mu_\psi\), which is finite by Lemma (The Associated Measures are Finite), gives

For the righthand side, recall we proved above that the map \(h \mapsto L_h(\phi, \psi)\) is continuous under uniformly bounded pointwise convergence for any \(\phi, \psi \in \mathbf{H}\). Applying this to the uniformly bounded sequence \(\{ f_i \}_{i \in \mathbb{N}}\) converging pointwise to \(f\), with the fixed vectors \(\psi\) and \(g(A)\psi\), gives

As a sequence in \(\mathbb{C}\) has at most one limit, these two computations give

Now, by Definition the lefthand side satisfies \(Q_{fg}(\psi) = \left< \psi, (fg)(A)\psi \right>\), while the righthand side satisfies \(L_f(\psi, g(A)\psi) = \left< \psi, f(A)g(A)\psi \right>\) by the expression for \(L_f\) obtained above. Hence,

for all \(\psi \in \mathbf{H}\). As both \((fg)(A)\) and \(f(A)g(A)\) are elements of \(\mathcal{B}(\mathbf{H})\), the uniqueness clause of Proposition, applied to the bounded quadratic form \(Q_{fg}\), implies

As \(g\) was an arbitrary element of \(C^0(\sigma(A); \mathbb{R})\), this is precisely the statement that \(f \in \mathcal{F}_1\). Thus \(\mathcal{F}_1\) is closed under pointwise limits of uniformly bounded sequences.

As \(\sigma(A)\) is a compact metric measurable space by Lemma and \(\mathcal{F}_1\) (1) is a vector space, (2) contains \(C^0(\sigma(A); \mathbb{R})\), and (3) is closed under uniformly bounded pointwise limits, we can apply Lemma to conclude that \(\mathcal{F}_1\) consists of all bounded, Borel-measurable functions on \(\sigma(A)\).

The result just obtained for \(\mathcal{F}_1\) is multiplicativity for an arbitrary bounded, Borel-measurable \(f\) paired with a continuous \(g\). To obtain the full result we must also allow \(g\) to be an arbitrary bounded, Borel-measurable function. To this end let \(\mathcal{F}_2\) be the set of all bounded, Borel-measurable, complex-valued functions \(g\)—note that here it is the second argument that varies—such that

for all bounded, Borel-measurable, complex-valued functions \(f\) on \(\sigma(A)\). We will prove that \(\mathcal{F}_2\) (1) is a complex vector space, (2) contains \(C^0(\sigma(A); \mathbb{R})\), and (3) is closed under pointwise limits of uniformly bounded sequences, so that Lemma applies.

(1) Consider any \(g_1\) and \(g_2\) in \(\mathcal{F}_2\), any \(\alpha_1\) and \(\alpha_2\) in \(\mathbb{C}\), and any bounded, Borel-measurable \(f\). Then

where the first line is the distributive property over functions, the second and fifth follow from linearity of the map of Definition, the third from \(g_1, g_2 \in \mathcal{F}_2\), and the fourth is the distributive property over operators. Hence \(\alpha_1g_1 + \alpha_2g_2 \in \mathcal{F}_2\), i.e. \(\mathcal{F}_2\) is a complex vector space.

(2) Let \(g \in C^0(\sigma(A); \mathbb{R})\). Membership of \(g\) in \(\mathcal{F}_2\) requires that \((fg)(A) = f(A)g(A)\) for every bounded, Borel-measurable \(f\). This is precisely the statement that every such \(f\) belongs to \(\mathcal{F}_1\), which we established above. Hence \(C^0(\sigma(A); \mathbb{R}) \subseteq \mathcal{F}_2\).

(3) Let \(\{ g_i \}_{i \in \mathbb{N}}\) be a sequence in \(\mathcal{F}_2\), uniformly bounded by some \(M \in \mathbb{R}\), converging pointwise to \(g\). Let \(f\) be an arbitrary bounded, Borel-measurable, complex-valued function on \(\sigma(A)\), bounded by \(M_f \in \mathbb{R}\), and let \(\psi \in \mathbf{H}\) be arbitrary.

As \(g_i \rightarrow g\) pointwise, we have \(fg_i \rightarrow fg\) pointwise, and

for all \(i \in \mathbb{N}\) and all \(\lambda \in \sigma(A)\), so the sequence \(\{ fg_i \}_{i \in \mathbb{N}}\) is uniformly bounded. Hence, the Bounded Convergence Theorem, applied to the measure \(\mu_\psi\), which is finite by Lemma (The Associated Measures are Finite), gives

On the other hand, as each \(g_i\) is in \(\mathcal{F}_2\) we have \((fg_i)(A) = f(A)g_i(A)\), and thus

where \(f(A)^*\) is the adjoint of the bounded operator \(f(A)\). Recall we established above that for any bounded, Borel-measurable, complex-valued function \(h\) the sesquilinear form \(L_h\) associated to \(Q_h\) satisfies \(L_h(\phi, \psi) = \left< \phi, h(A)\psi \right>\), and that the map \(h \mapsto L_h(\phi, \psi)\) is continuous under uniformly bounded pointwise convergence for any \(\phi, \psi \in \mathbf{H}\). Applying this with \(\phi = f(A)^*\psi\) to the uniformly bounded sequence \(\{ g_i \}_{i \in \mathbb{N}}\) converging pointwise to \(g\) gives

As a sequence in \(\mathbb{C}\) has at most one limit, the two computations give

for all \(\psi \in \mathbf{H}\). Both \((fg)(A)\) and \(f(A)g(A)\) are elements of \(\mathcal{B}(\mathbf{H})\), so by the uniqueness clause of Proposition applied to the bounded quadratic form \(Q_{fg}\) we conclude

As \(f\) was arbitrary, \(g \in \mathcal{F}_2\), and thus \(\mathcal{F}_2\) is closed under pointwise limits of uniformly bounded sequences.

As \(\sigma(A)\) is a compact metric measurable space by Lemma and \(\mathcal{F}_2\) (1) is a vector space, (2) contains \(C^0(\sigma(A); \mathbb{R})\), and (3) is closed under uniformly bounded pointwise limits, we can apply Lemma to conclude that \(\mathcal{F}_2\) consists of all bounded, Borel-measurable, complex-valued functions on \(\sigma(A)\). This is none other than the desired result

for all bounded, Borel-measurable, complex-valued functions \(f\) and \(g\) on \(\sigma(A)\).\(\blacksquare\)

In what is the penultimate result required to prove the Spectral Theorem for Bounded, Self-Adjoint Operators we present the following theorem that covers all of the Spectral Theorem for Bounded, Self-Adjoint Operators except uniqueness of the projection-valued measure \(\mu^A\).

Theorem Suppose \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint. For any measurable subset \(E\) of the spectrum \(\sigma(A)\) of \(A\), define the operator \(\mu^A(E)\) by

\[\mu^A(E) \equiv 1_E(A),\]

where \(1_E(A)\) is the operator associated to the indicator function \(1_E\) on \(\sigma(A)\) by way of Definition. Then \(\mu^A\) is a projection-valued measure on \(\sigma(A)\) and satisfies

\[\int_{\sigma(A)} \lambda \, d\mu^A(\lambda) = A.\]

Proof This proof broadly consists of two parts Part 1: prove that \(\mu^A\) is a projection valued measure on \(\sigma(A)\) and Part 2: prove that \(\mu^A\) satisfies

Part 1: Let us first prove that \(\mu^A\) is a projection valued measure on \(\sigma(A)\). This involves Part 1.1: proving for each measurable set \(E\) on \(\sigma(A)\) it follows that \(\mu^A(E)\) is a bounded orthogonal projection, Part 1.2: proving for any measurable sets \(E_1\) and \(E_2\) on \(\sigma(A)\) that \(\mu^A(E_1 \cap E_2) = \mu^A(E_1) \mu^A(E_2)\), Part 1.3: proving that \(\mu^A(\emptyset) = 0\), where \(\emptyset\) is the empty set, and \(\mu^A(\sigma(A)) = \mathbf{1}\), where \(\mathbf{1}\) is the multiplicative identity element, and finally Part 1.4: proving that for pairwise disjoint measurable sets \(\{ E_i \}_{i \in \mathbb{N}}\) on \(\sigma(A)\) and any \(\psi \in \mathbf{H}\) we have

where the convergence of the sum is in the norm topology on \(\mathbf{H}\).

Part 1.1: Let us now prove that for each measurable set \(E\) on \(\sigma(A)\) it follows that \(\mu^A(E)\) is a bounded orthogonal projection.

To this end, first let us note that as a result of Lemma for any measurable subset \(E\) of the spectrum \(\sigma(A)\) of \(A\) the fact that \(1_E\) is a bounded, Borel-measurable, real-valued function on \(\sigma(A)\) allows us to conclude that \(1_E(A)\) is self-adjoint.

Next note that the fact that \(1_E 1_E = 1_E\) allows us to conclude that

At the same time the Proposition implies

Together these imply

By way of the orthogonal projection Definition the fact that \(1_E(A)\) is a bounded, self-adjoint operator that satisfies \(1_E(A)1_E(A) = 1_E(A)\) implies that \(1_E(A)\) is an orthogonal projection. Hence,

is also an orthogonal projection for any measurable subset \(E\) of the spectrum \(\sigma(A)\), the desired Part 1.1 result.

Part 1.2: Next let us prove that for any measurable sets \(E_1\) and \(E_2\) on \(\sigma(A)\) that \(\mu^A(E_1 \cap E_2) = \mu^A(E_1) \mu^A(E_2)\).

So to this end, consider any two measurable subsets \(E_1\) and \(E_2\) of the spectrum \(\sigma(A)\). Tracing definitions one has

Using the definition of \(\mu^A\) this implies

where the second line follows from the previous equation, the third from Proposition, and the final from the definition of \(\mu^A\). Hence,

the desired Part 1.2 result.

Part 1.3: Now let us prove that \(\mu^A(\emptyset) = 0\), where \(\emptyset\) is the empty set, and \(\mu^A(\sigma(A)) = \mathbf{1}\), where \(\mathbf{1}\) is the multiplicative identity element.

To this end, consider the empty set \(\emptyset\) which is a measurable subset of the spectrum \(\sigma(A)\). Tracing definitions one has

Definition then implies

for all \(\psi \in \mathbf{H}\). So as

is true for any \(\psi\) in \(\mathbf{H}\) we conclude that \(1_\emptyset(A)\) is the zero operator.

However, by definition

Thus, \(\mu^A(\emptyset)\) is the zero operator, the first of the desired Part 1.3 results.

Now consider the measurable set \(\sigma(A)\), the entire spectrum. Tracing definitions

Definition then implies for all \(\psi \in \mathbf{H}\) that

where the fourth line follows from the definition of \(\mu_\psi\) in terms of our map \(\Lambda_\psi\) by way of the Riesz Representation Theorem and the final line follows from the definition of \(\Lambda_\psi\). So for any \(\psi \in \mathbf{H}\) we have

which implies that

By definition

Hence, we have proven

the second and final Part 1.3 result.

Part 1.4: Now let us prove that for pairwise disjoint measurable sets \(\{ E_i \}_{i \in \mathbb{N}}\) on \(\sigma(A)\) and any \(\psi \in \mathbf{H}\) we have

where the convergence of the sum is in the norm topology on \(\mathbf{H}\).

In order to prove this, consider now any two disjoint measurable subsets \(E_1\) and \(E_2\) of the spectrum \(\sigma(A)\). The definition of \(\mu^A\) along with our previous result implies

where the first line follows from the fact that \(E_1\) and \(E_2\) are disjoint and the final line follows from our previous result, \(1_\emptyset(A)\) is the zero operator.

However, previously we found that for arbitrary measurable subsets \(E_1\) and \(E_2\) of the spectrum \(\sigma(A)\) of \(A\)

So if \(E_1\) and \(E_2\) are disjoint, this result along with the last result imply

So if \(E_1\) and \(E_2\) are disjoint, then

As \(\mu^A(E_1)\) is self-adjoint, this implies that for arbitrary \(\phi\) and \(\psi\) in \(\mathbf{H}\) one has

which is nothing more than the statement that for disjoint \(E_1\) and \(E_2\) and arbitrary \(\phi\) and \(\psi\) in \(\mathbf{H}\) we have

i.e. the ranges of \(\mu^A(E_1)\) and \(\mu^A(E_2)\) are orthogonal.

With this result as motivation, let us prove the following “utility” lemma

Lemma Let \(\{ P_i \}_{i \in \mathbb{N}}\) be a set of bounded orthogonal projections on a separable, complex Hilbert space \(\mathbf{H}\) that satisfy \(P_iP_j = 0\) for \(i \neq j\). Then for all \(\psi \in \mathbf{H}\) the sequence of partial sums

\[S_n\psi \equiv \sum_{i = 0}^n P_i\psi\]

converges to \(P\psi\) where \(P\) is a bounded orthogonal projection onto the smallest closed subspace containing the range of the \(P_i\).

Proof This proof broadly consists of three parts Part 1: proving that the sequence of partial sums \(S_n\psi\) converges and Part 2: proving that the sequence limit \(P\psi\) defines a bounded orthogonal projection operator \(P\), and Part 3: proving that the range of \(P\) is the smallest closed subspace containing the range of the \(P_i\).

Part 1: Let us first prove that the sequence of partial sums \(S_n\psi\) converges.

To this end for an arbitrary \(\psi \in \mathbf{H}\) let us first examine the norm of the partial sum \(S_n\psi\)

where on the third line we made use of the fact that \(P_iP_j = 0\) if \(i \neq j\). So we have proven

Now as the projections \(P_i\) satisfy \(P_iP_j = 0\) if \(i \neq j\), the partial sum

must have a magnitude less than or equal to that of \(\psi\), in other words

These last two results then imply

Which implies

As \(\|\psi\|^2 < \infty\) and \(0 \le \|P_i\psi\|^2\) for all \(i \in \mathbb{N}\), this implies that for any \(\psi \in \mathbf{H}\) the sequence \(\{ a_n(\psi) \}_{n \in \mathbb{N}}\) where \(a_n(\psi)\) is defined by

in a bounded, monotonically increasing sequence of real numbers. Thus the Monotone Convergence Theorem

Theorem (Monotone Convergence Theorem) Let \(\{ a_n \}_{n \in \mathbb{N}}\) be a monotone sequence of real numbers (either \(a_n \le a_{n+1}\) or \(a_n \ge a_{n+1}\) for all \(n\)). Then the sequence \(\{ a_n \}_{n \in \mathbb{N}}\) has a finite limit in \(\mathbb{R}\) if and only if \(\{ a_n \}_{n \in \mathbb{N}}\) is bounded.

implies that the bounded, monotonically increasing sequence of real numbers

has a finite limit in \(\mathbb{R}\) for any \(\psi \in \mathbf{H}\).

For an arbitrary \(\psi \in \mathbf{H}\) consider again the sequence

Using the result we just established, we will now prove that this sequence converges, the desired conclusion of Part 1.

We will do so by employing the fact that \(\mathbf{H}\) being a Hilbert space implies that \(\mathbf{H}\) is also a Banach space. Thus, a Cauchy sequence in \(\mathbf{H}\) converges in \(\mathbf{H}\). So, if we can prove the sequence \(S_n\psi\) is a Cauchy sequence, then we can conclude it converges.

To that end, consider \(n, m \in \mathbb{N}\) and without loss of generality assume that \(n > m\). One has

where the final equality follows from \(P_iP_j = 0\) when \(i \neq j\). Now as we proved the sequence

converges, it follows that as \(n,m \rightarrow \infty\) the tail sum goes to zero

This then implies that as \(n,m \rightarrow \infty\) one has

In other words \(S_n\psi\) is a Cauchy sequence.

As \(S_n\psi\) is a Cauchy sequence and \(\mathbf{H}\) is a Hilbert, and thus a Banach space, this implies that \(S_n\psi\) converges to some element in \(\mathbf{H}\). This and linearity allows us to define a linear operator \(P\) by

Part 2: Now let us prove that the sequence limit \(P\psi\) defines a bounded orthogonal projection operator \(P\). To prove that \(P\) is a bounded orthogonal projection operator we must Part 2.1: prove that \(P\) is an element of \(\mathcal{B}(\mathbf{H})\), Part 2.2: prove that \(P\) is self-adjoint, and Part 2.3: prove that \(PP=P\).

Part 2.1: Next let us prove that \(P\) is an element of \(\mathcal{B}(\mathbf{H})\).

By definition

As the \(P_i\) are elements in \(\mathcal{B}(\mathbf{H})\) they are linear. This then implies that \(P\) is linear.

Recall we previously proved that for any \(n\) in \(\mathbb{N}\) one has

Thus for any unit norm \(\psi\), i.e. \(\psi\) such that \(\|\psi\| = 1\), we have

and thus

for arbitrary \(n \in \mathbb{N}\).

Now the operator norm \(\|P\|\) of \(P\) is defined by

Hence, the definition of \(P\) along with our previous result and Part 1 of Proposition (Continuity of the Norm and Inner Product), which permits the exchange of the norm with the limit, implies

Thus \(\|P\| \le 1\), proving that \(P\) is bounded and thus an element of \(\mathcal{B}(\mathbf{H})\), the desired Part 2.1 result.

Part 2.2: Next let us prove that \(P\) is self-adjoint. It turns out this follows directly from the fact that each of the \(P_i\) is self-adjoint.

Tracing definitions, and using Part 2 of Proposition (Continuity of the Norm and Inner Product) to exchange the inner product with the limit, one has for arbitrary \(\phi, \psi \in \mathbf{H}\)

proving

for arbitrary \(\phi, \psi \in \mathbf{H}\). In other words \(P\) is self-adjoint, the desired result of Part 2.2.

Part 2.3: Now let us prove that \(P\) satisfies \(PP=P\). Again it turns out this follows directly from the fact that for each of the linear, bounded operators \(P_i\) we have \(P_iP_i = P_i\) and \(P_iP_j = 0\) if \(i \neq j\).

Tracing definitions one has for arbitrary \(\psi \in \mathbf{H}\)

Proving

for arbitrary \(\psi \in \mathbf{H}\), the desired Part 2.3 result which concludes the proof that \(P\) is a bounded orthogonal projection.

Part 3: Next we will prove the range of \(P\) is the smallest closed subspace containing the range of the \(P_i\). Proving this will require two parts Part 3.1: prove that the closed subspace containing the range of the \(P_i\) is a subset of the range of \(P\) and Part 3.2: prove that the range of \(P\) is a subset of the closed subspace containing the range of the \(P_i\).

Part 3.1: Let us begin by proving that the closed subspace containing the range of the \(P_i\) is a subset of the range of \(P\).

To facilitate this, let us define the notation

for the range of \(P\). Similarly, let us define the notation

for the closure of the subspace containing the range of the \(P_i\).

With this notation in hand, let us first prove that \(M\) is closed.

Let \(\{ \psi_i \}_{i \in \mathbb{N}}\) be a sequence in \(M\) that converges to \(\psi\) in \(\mathbf{H}\). As the \(\psi_i\) are in \(M \equiv \text{Range}(P)\), the fact that \(P\) is a bounded orthogonal projection operator implies that \(P\psi_i = \psi_i\) for all \(i \in \mathbb{N}\).

Now, as we proved that \(P\) is bounded the standard proposition Bounded Operators are Continuous proves that \(P\) is continuous. As \(P\) is continuous, we can pull \(P\) through limits; this implies

This implies

which implies that \(\psi\) is in the range of \(P\) and thus in \(M \equiv \text{Range}(P)\), proving \(M\) is closed.

To complete the proof of Part 3.1 we must prove that \(V \subseteq M\).

As \(P_iP_i = P_i\) and \(P_iP_j = 0\) for \(i \neq j\), for any \(\psi_i\) in \(\text{Range}(P_i)\) we have \(P_i\psi_i = \psi_i\) and \(P_j\psi_i = 0\) for \(i \neq j\). As a result we have

This implies \(P\psi_i = \psi_i\) which in turn implies that \(\psi_i\) is in \(M \equiv \text{Range}(P)\). However, as \(\psi_i\) is an arbitrary element in \(\text{Range}(P_i)\), this implies that \(\text{Range}(P_i) \subseteq M\). But, as the index \(i\) we used was arbitrary this implies that

However, as we proved, \(M\) is closed. Thus it must contain the closed linear span of all these ranges

which is none other than the desired result \(V \subseteq M\) of Part 3.1.

Part 3.2: Next let us prove that the range of \(P\) is a subset of the closed subspace containing the range of the \(P_i\).

Consider an arbitrary \(\psi \in \mathbf{H}\). The partial sum

is a linear combination of a finite number of vectors \(P_i\psi\) in

Obviously, \(S_n\psi \in V\). As \(V\) is by construction closed and \(S_n\psi \rightarrow P\psi\) relative to the norm on \(\mathbf{H}\), the limit \(P\psi\) must also live in \(V\). Thus, \(M \subseteq V\), the desired Part 3.2 result.

This completes the proof of Lemma.\(\blacksquare\)

Now back on the main thread, proving Part 1.4 of Theorem, consider a series \(\{ E_i \}_{i \in \mathbb{N}}\) of disjoint measurable subsets of the spectrum \(\sigma(A)\) of \(A\). We proved that the \(\mu^A(E_i)\) are bounded orthogonal projections on the separable, complex Hilbert space \(\mathbf{H}\) and that \(\mu^A(E_i)\mu^A(E_j) = 0\) if \(i \neq j\).

Hence, we can apply the Lemma we just proved to conclude that for any \(\psi \in \mathbf{H}\) the sequence of partial sums

converges to \(P\psi\) where \(P\) is a bounded orthogonal projection onto the smallest closed subspace containing the range of the \(\mu^A(E_i)\).

Now with this same series \(\{ E_i \}_{i \in \mathbb{N}}\) let us define \(E\) by

As the \(E_i\) are disjoint, the sequence

is uniformly bounded by \(1\), and converges pointwise to \(1_E\).

As both the \(1_{E_i}\) and \(1_E\) are bounded measurable functions on the spectrum \(\sigma(A)\) of \(A\), we can associate by way of the Definition the operators \(1_{E_i}(A)\) and \(1_E(A)\) to them.

As the sequence \(f_n\) is uniformly bounded and converges pointwise to \(1_E\), we can use the fact, determined within the proof of Proposition, that the map \(f \mapsto Q_f(\psi)\) is continuous under uniformly bounded pointwise convergence for any \(\psi \in \mathbf{H}\) to conclude that the convergence of \(f_n\) along with Definition imply

However, by definition \(\mu^A(E) \equiv 1_E(A)\) and \(\mu^A(E_i) \equiv 1_{E_i}(A)\), thus the previous equation implies

for arbitrary \(\psi \in \mathbf{H}\). This is nothing more than the statement that \(\mu^A\) is countable additive, the desired result of Part 1.4. This concludes the proof that \(\mu^A\) is a projection-valued measure.

Part 2: Now let us prove that \(\mu^A\) satisfies

To this end, consider the measurable set \(E\) on \(\sigma(A)\) and its bounded, measurable indicator function \(1_E\). As a result of Theorem (Operator-Valued Integration) we have the unique linear map

from the space of bounded, measurable, complex-valued functions on \(\sigma(A)\) into \(\mathcal{B}(\mathbf{H})\). By definition

and by definition

Hence, we have the unique linear map

Consider now any simple function \(s\) on \(\sigma(A)\), i.e. any finite linear combination of indicator functions \(1_{E_i}\) on \(\sigma(A)\)

where the \(E_i\) are pairwise disjoint measurable sets on \(\sigma(A)\) and the \(\alpha_i\) are in \(\mathbb{C}\). As the map

is linear we have for the arbitrary simple function \(s\)

Now, by taking limits of such simple functions this map extends to all bounded, Borel-measurable functions \(f\) as follows

In particular, \(\sigma(A)\) is compact as a result of Lemma (The Spectrum is a Compact Metric Measurable Space), thus the function \(f(\lambda) = \lambda\) on \(\sigma(A)\) is a bounded and obviously Borel-measurable function on \(\sigma(A)\). Hence, we can apply the map above to \(f(\lambda) = \lambda\).

We claim that the integral of \(f(\lambda) = \lambda\) against \(\mu^A\) is \(A\).

Proving this claim explicitly follows from (1) proving that the measures \(\mu^A_\psi\) and \(\mu_\psi\) agree, (2) proving that the operator \(f(A)\) of Definition agrees with the projection-valued integral of \(f\) with respect to \(\mu^A\), (3) proving that for continuous \(g\) the operator \(g(A)\) of Definition agrees with the operator \(g(A)\) of Proposition, and (4) finally specializing to the continuous function \(f(\lambda) = \lambda\) for the final result.

To (1) prove that the measures \(\mu^A_\psi\) and \(\mu_\psi\) agree we can unroll definitions. For any measurable \(E\) on \(\sigma(A)\) we have

where the first equality follows from the definition of \(\mu^A_\psi\) in Theorem (Projection-Valued Measure’s Associated Measure), the second from the definition of the operator \(\mu^A(E)\), the third from the characterization of \(1_E(A)\) in Definition, and the fourth from the Definition of \(Q_{1_E}(\psi)\). This implies

which, as \(E\) is arbitrary, is none other than the statement that the measures \(\mu^A_\psi\) and \(\mu_\psi\) agree.

Next let us (2) prove that the operator \(f(A)\) of Definition agrees with the projection-valued integral of \(f\) with respect to \(\mu^A\).

For any bounded, Borel-measurable, complex-valued function \(f\) on \(\sigma(A)\) as a result of Definition and Definition we have

for any \(\psi\) in \(\mathbf{H}\). Also, as a result of the defining property of a projection-valued measure \(\mu^A\), we have

from Theorem (Operator-Valued Integration). However, we just proved the measures \(\mu^A_\psi\) and \(\mu_\psi\) agree. Thus

From our previous equations, this implies that the operators

both represent the same bounded quadratic form \(Q_f\).

However, by the uniqueness clause of Proposition the operator representing a given bounded quadratic form \(Q_f\) is unique. Thus

Next we want to (3) prove that for continuous \(g\) in \(C^0(\sigma(A); \mathbb{R})\) the operator \(g(A)\) of Definition agrees with the operator \(g(A)\) of Proposition.

Directly before the statement of Definition we established that for any \(\psi\) in \(\mathbf{H}\) we can construct a map \(\Lambda_\psi : C^0(\sigma(A); \mathbb{R}) \rightarrow \mathbb{R}\) defined by

where \(g(A)\) arises from the real-valued functional calculus of Proposition, such that the application of Theorem (Riesz Representation) implies

where the measure \(\mu_\psi\) arises from the Theorem (Riesz Representation) and \(Q_g(\psi)\) arises from Definition.

However, the \(g(A)\) from Definition satisfies

So both the \(g(A)\) from Proposition and the \(g(A)\) from Definition both give rise to the same bounded quadratic form \(Q_g\). However, by the uniqueness clause of Proposition, the operator representing a given bounded quadratic form \(Q_g\) is unique. Thus, for \(g\) in \(C^0(\sigma(A); \mathbb{R})\) the \(g(A)\) from Proposition agrees with the \(g(A)\) from Definition, the desired result.

Finally, let us (4) prove that specializing to the continuous function \(f(\lambda) = \lambda\) we obtain the desired result, that the integral of \(f(\lambda) = \lambda\) against \(\mu^A\) is \(A\).

Recall that Lemma (The Spectrum is a Compact Metric Measurable Space) implies that \(\sigma(A)\) is a subset of \(\mathbb{R}\). Thus, \(f(\lambda) = \lambda\) is real-valued. Furthermore, as \(f(\lambda) = \lambda\) is a polynomial it is continuous. Also note that the same Lemma implies \(\sigma(A)\) is compact. Hence, as a result of Theorem (Boundedness Theorem), \(f(\lambda) = \lambda\) is bounded.

So the result of the last section on continuous functions applies to \(f(\lambda) = \lambda\) and the \(f(A)\) from Proposition agrees with the \(f(A)\) from Definition.

Now by way of the map defined in the Lemma (Spectral Mapping Theorem) the polynomial \(f(\lambda) = \lambda\) is mapped to the operator \(f(A) = A\). Recall also that the \(f(A)\) of Proposition is constructed such that when \(f\) is a polynomial it agrees with the \(f(A)\) of Lemma (Spectral Mapping Theorem). Hence, as \(f(\lambda) = \lambda\) is a polynomial the \(f(A)\) of Proposition is given by \(f(A) = A\).

So putting this all together we have from our second step that

where \(f(A)\) is the \(f(A)\) of Definition. Now using the results of this step we know that for \(f(\lambda) = \lambda\) the \(f(A)\) of Definition is equal to the \(f(A)\) of Proposition and the \(f(A)\) of Proposition is equal to \(A\), giving in full

the desired result.

So in summary we have proven that

the desired Part 2 result which completes the proof of Theorem.\(\blacksquare\)

The final result we need to prove to complete our proof of the Spectral Theorem for Bounded, Self-Adjoint Operators is to prove that the projection-valued measure \(\mu^A\) of Theorem is unique. It is to this we turn.

Theorem Let \(A \in \mathcal{B}(\mathbf{H})\) be self-adjoint and let \(\mu^A\) and \(\nu^A\) be two projection-valued measures on the spectrum \(\sigma(A)\) of \(A\) such that

\[\begin{align} \int_{\sigma(A)} \lambda \, d\mu^A(\lambda) &= A \\ \int_{\sigma(A)} \lambda \, d\nu^A(\lambda) &= A. \end{align}\]

Then \(\mu^A(E) = \nu^A(E)\) for all measurable subsets \(E\) of the spectrum \(\sigma(A)\) of \(A\), i.e. \(\mu^A\) and \(\nu^A\) are equivalent projection-valued measures.

Proof This proof broadly consists of four parts Part 1: prove that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on complex-valued polynomials, Part 2: prove that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on continuous complex-valued functions, Part 3: prove that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on bounded, measurable complex-valued functions, and Part 4 prove that \(\mu^A(E) = \nu^A(E)\) for all measurable subsets \(E\) of the spectrum \(\sigma(A)\) of \(A\).

Part 1: Let us first prove that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on complex-valued polynomials.

Consider an arbitrary complex valued polynomial \(p\) on \(\sigma(A)\) with \(\mathbb{C}\) valued coefficients. Generically \(p\) has the form

where the \(a_i\) take values in \(\mathbb{C}\). Operator-valued integration with respect to \(\mu^A\) gives

where the first equality made use of the definition of \(p\), the second equality made use of linearity of operator-valued integration following from Theorem (Operator-Valued Integration), the third equality made use of multiplicativity of operator-valued integration also from Theorem (Operator-Valued Integration), and the final equality made use of the hypothesis of this theorem.

Using the same logic, operator-valued integration with respect to \(\nu^A\) gives

Hence, we have proven the desired result

operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on complex-valued polynomials.

Part 2: Let us next prove that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on the set of continuous, complex-valued functions on \(\sigma(A)\).

Later in Part 2 we will have need of the fact that the set of complex-valued polynomials on \(\sigma(A)\) is dense in the set of continuous, complex-valued functions on \(\sigma(A)\) with respect to the supremum norm. So we begin by proving this fact.

Lemma Let \(A \in \mathcal{B}(\mathbf{H})\) be self-adjoint, then the set of complex-valued polynomials on \(\sigma(A)\) is dense in the set of continuous, complex-valued functions on \(\sigma(A)\) with respect to the supremum norm.

Proof By Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a non-empty, compact metric measurable space.

As \(\sigma(A)\) is a subset of \(\mathbb{C}\), we can restrict the standard metric on \(\mathbb{C}\)

where \(\lvert \cdot \rvert\) is the norm on \(\mathbb{C}\), to \(\sigma(A)\) making \(\sigma(A)\) a metric space.

Consider now \(\mathcal{P}(\sigma(A); \mathbb{C})\) the set of complex-valued polynomials on \(\sigma(A)\) and \(C^0(\sigma(A); \mathbb{C})\) the set of continuous, complex-valued functions on \(\sigma(A)\). Obviously, \(\mathcal{P}(\sigma(A); \mathbb{C})\) is a subset of \(C^0(\sigma(A); \mathbb{C})\) as all complex-valued polynomials on \(\sigma(A)\) are continuous.

Furthermore, \(\mathcal{P}(\sigma(A); \mathbb{C})\) is obviously an algebra as it has the required vector space properties, e.g. closure under addition, as well as the required bilinear product properties, e.g. left distributivity.

In addition it is also obvious that \(\mathcal{P}(\sigma(A); \mathbb{C})\) separates points. Explicitly, consider any two distinct points \(\lambda_1, \lambda_2 \in \sigma(A)\). There exists a polynomial \(p\) in \(\mathcal{P}(\sigma(A); \mathbb{C})\) such that \(p(\lambda_1) \neq p(\lambda_2)\). For example the polynomial

would suffice as \(p(\lambda_1) \neq p(\lambda_2)\) follows from the fact that \(\lambda_1\) and \(\lambda_2\) are distinct.

Another obvious point we will have need of is the fact that \(\mathcal{P}(\sigma(A); \mathbb{C})\) contains constant functions. Explicitly the constant function

where \(a_0 \in \mathbb{C}\), is an element of \(\mathcal{P}(\sigma(A); \mathbb{C})\).

The final point we will have need of is the fact that \(\mathcal{P}(\sigma(A); \mathbb{C})\) is closed under complex conjugation, i.e. if \(p(\lambda)\) is in \(\mathcal{P}(\sigma(A); \mathbb{C})\), then \(\overline{p(\lambda)}\) is also in \(\mathcal{P}(\sigma(A); \mathbb{C})\). This follows from noting that Proposition implies that the spectrum \(\sigma(A)\) of \(A\) is in \(\mathbb{R}\). Hence, a generic \(p\) in \(\mathcal{P}(\sigma(A); \mathbb{C})\) is of the form

where the \(a_i\) are complex-valued and \(\lambda\) is real, and this implies

Now as the \(\overline{a_i}\) are also in \(\mathbb{C}\), this implies that

is an element in \(\mathcal{P}(\sigma(A); \mathbb{C})\), proving that \(\mathcal{P}(\sigma(A); \mathbb{C})\) is closed under complex conjugation.

With all of this in place we are in a position to apply the Theorem (Stone–Weierstrass for Complex Numbers)

Theorem (Stone–Weierstrass for Complex Numbers) Let \(X\) be a compact metric space and let \(\mathcal{A}\) be an algebra in \(C^0(X; \mathbb{C})\), the space of continuous, complex-valued functions on \(X\). If \(\mathcal{A}\) contains the constant functions, separates points, and is closed under complex conjugation, then \(\mathcal{A}\) is dense in \(C^0(X; \mathbb{C})\) with respect to the supremum norm.

and conclude that \(\mathcal{P}(\sigma(A); \mathbb{C})\) is dense in \(C^0(\sigma(A); \mathbb{C})\) with respect to the supremum norm, the desired result.\(\blacksquare\)

Our next long term goal is to apply the Bounded Linear Transformation Theorem to prove that operator-valued integration with respect to \(\mu^A\) agrees with the same with respect to \(\nu^A\) when the integrand is an element of \(C^0(\sigma(A); \mathbb{C})\). In other words to prove

for an arbitrary \(f\) in \(C^0(\sigma(A); \mathbb{C})\). Let’s begin.

First, as mentioned after our application of the Theorem (Stone–Weierstrass for Complex Numbers), \(C^0(\sigma(A); \mathbb{C})\) is equipped with a norm, the supremum norm, and is thus a normed space.

Second, recall that as a result of Lemma (Bounded Operators form a Banach Space) we know that \(\mathcal{B}(\mathbf{H})\) is a Banach space with respect to the operator norm.

Now recall that Theorem (Operator-Valued Integration) implies that the maps

from the set of bounded, measurable complex-valued functions on \(\sigma(A)\) to \(\mathcal{B}(\mathbf{H})\) are linear. We want to prove \(\mathcal{P}(\sigma(A); \mathbb{C})\) is a subset of the domain of \(I_{\mu^A}\) and \(I_{\nu^A}\).

Recall that by Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a non-empty, compact metric measurable space.

Also recall that any complex-valued polynomial \(p\) on \(\sigma(A)\) is continuous, i.e. \(\mathcal{P}(\sigma(A); \mathbb{C}) \subset C^0(\sigma(A); \mathbb{C})\). Thus the norm \(\lvert p \rvert\) of any element of \(\mathcal{P}(\sigma(A); \mathbb{C})\) is a continuous real-valued function on \(\sigma(A)\). Hence, as a result of Theorem (Boundedness Theorem), \(\lvert p \rvert\) is bounded on \(\sigma(A)\). This is none other than the statement that \(\mathcal{P}(\sigma(A); \mathbb{C})\) is bounded on \(\sigma(A)\).

Finally, note that as \(\mathcal{P}(\sigma(A); \mathbb{C})\) is continuous and the \(\sigma\)-algebra on \(\sigma(A)\) is the Borel \(\sigma\)-algebra, it follows that any element of \(\mathcal{P}(\sigma(A); \mathbb{C})\) is measurable.

Hence, elements of \(\mathcal{P}(\sigma(A); \mathbb{C})\) are bounded, measurable, complex-valued functions on \(\sigma(A)\). Thus, \(\mathcal{P}(\sigma(A); \mathbb{C})\) is in the domain of the \(I_{\mu^A}\) and \(I_{\nu^A}\).

Furthermore, Property 2 of the Theorem (Operator-Valued Integration) implies that for any bounded, measurable, complex-valued function \(f\) on \(\sigma(A)\)

and similarly with \(\mu^A\) replaced by \(\nu^A\). Applying this to an arbitrary \(p\) in \(\mathcal{P}(\sigma(A); \mathbb{C})\), which we have just shown is bounded and measurable, gives

and similarly \(\| I_{\nu^A}(p) \| \le \|p\|\), where \(\|p\|\) denotes the supremum norm of \(p\). This is none other than the statement that the linear maps \(I_{\mu^A}\) and \(I_{\nu^A}\), restricted to \(\mathcal{P}(\sigma(A); \mathbb{C})\), are bounded using constant \(1\).

With that established, let us restrict the domain of \(I_{\mu^A}\) and \(I_{\nu^A}\) to be \(\mathcal{P}(\sigma(A); \mathbb{C})\) and only \(\mathcal{P}(\sigma(A); \mathbb{C})\) while retaining the same notation \(I_{\mu^A}\) and \(I_{\nu^A}\). Hopefully this is not too confusing.

Finally, we may make use of the Theorem (Bounded Linear Transformation Theorem) to uniquely extend the domain of \(I_{\mu^A}\) and \(I_{\nu^A}\) from \(\mathcal{P}(\sigma(A); \mathbb{C})\) to all of \(C^0(\sigma(A); \mathbb{C})\). (Note we will still use the same notation \(I_{\mu^A}\) and \(I_{\nu^A}\) for these extended maps.) As \(I_{\mu^A}\) and \(I_{\nu^A}\) are extensions, \(I_{\mu^A}\) agrees on \(\mathcal{P}(\sigma(A); \mathbb{C})\) with its non-extended version, and a similar statement is true of \(I_{\nu^A}\).

However, as we just established in Part 1 the maps \(I_{\mu^A}\) and \(I_{\nu^A}\) agree when restricted to \(\mathcal{P}(\sigma(A); \mathbb{C})\). Thus, as the extensions \(I_{\mu^A}\) and \(I_{\nu^A}\) obtained from Theorem (Bounded Linear Transformation Theorem) are unique, this implies that the extensions \(I_{\mu^A}\) and \(I_{\nu^A}\) agree everywhere on the domain \(C^0(\sigma(A); \mathbb{C})\).

Hence, we have proven the desired Part 2 result that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on the set of continuous, complex-valued functions on \(\sigma(A)\).

Part 3: Now let us prove that operator-valued integration with respect to \(\mu^A\) agrees with that with respect to \(\nu^A\) on the set of bounded, measurable complex-valued functions.

Let \(\mathcal{F}\) be the set of bounded, measurable, complex-valued functions on \(\sigma(A)\) such that operator-valued integration with respect to \(\mu^A\) agrees with the same with respect to \(\nu^A\).

Recall that in Part 2 we proved that \(C^0(\sigma(A); \mathbb{C})\) is a subset of \(\mathcal{F}\). Now, as \(C^0(\sigma(A); \mathbb{R})\) is a subset of \(C^0(\sigma(A); \mathbb{C})\), it obviously follows that \(C^0(\sigma(A); \mathbb{R})\) is a subset of \(\mathcal{F}\) too.

Also note that as a result of Theorem (Operator-Valued Integration) the maps \(I_{\mu^A}\) and \(I_{\nu^A}\)—now viewed as having domain consisting of all bounded, measurable, complex-valued functions on \(\sigma(A)\)—are linear. This implies that \(\mathcal{F}\) is a vector space over \(\mathbb{C}\). Explicitly, consider arbitrary \(f_1,f_2 \in \mathcal{F}\) and arbitrary \(\alpha_1, \alpha_2 \in \mathbb{C}\). Linearity of \(I_{\mu^A}\) implies

Similarly linearity of \(I_{\nu^A}\) implies

However, as \(f_1,f_2 \in \mathcal{F}\) the definition of \(\mathcal{F}\) implies

Hence, the last four equations imply

This implies that \(\alpha_1 f_1(\lambda) + \alpha_2 f_2(\lambda)\) is in element \(\mathcal{F}\) and thus \(\mathcal{F}\) is a vector space over \(\mathbb{C}\).

Next we want to prove that \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences.

To this end let \(\{ f_i \}_{i \in \mathbb{N}}\) be a sequence in \(\mathcal{F}\), uniformly bounded by some \(M \in \mathbb{R}\), that converges pointwise to a function \(f\) on \(\sigma(A)\). By Lemma (Pointwise Limits of Uniformly Bounded, Borel-Measurable Functions), \(f\) is itself a bounded, Borel-measurable, complex-valued function on \(\sigma(A)\), and so is a candidate for membership in \(\mathcal{F}\). Let \(\psi \in \mathbf{H}\) be arbitrary.

By the defining property of the map of Theorem (Operator-Valued Integration), applied to the projection-valued measure \(\mu^A\), we have for every \(i \in \mathbb{N}\)

where \((\mu^A)_\psi\) is the positive real-valued measure associated to \(\mu^A\) and \(\psi\) by way of Theorem (Projection-Valued Measure’s Associated Measure). Note that \((\mu^A)_\psi\) is a finite measure, as

As \(f_i \rightarrow f\) pointwise with \(\lvert f_i(\lambda) \rvert \le M\) for all \(i \in \mathbb{N}\) and all \(\lambda \in \sigma(A)\), the Bounded Convergence Theorem, applied to the finite measure \((\mu^A)_\psi\), gives

Combining the last two results we obtain

An entirely identical argument, with \(\mu^A\) replaced by \(\nu^A\) throughout, gives

However, as each \(f_i\) is an element of \(\mathcal{F}\), the definition of \(\mathcal{F}\) gives

for every \(i \in \mathbb{N}\), and hence

for every \(i \in \mathbb{N}\). So the two convergent sequences displayed above are in fact the same sequence of complex numbers. As a sequence in \(\mathbb{C}\) has at most one limit, their limits coincide, giving

for all \(\psi \in \mathbf{H}\).

Now, as \(f\) is a bounded, Borel-measurable, complex-valued function on \(\sigma(A)\), Theorem (Operator-Valued Integration) implies that both \(\int_{\sigma(A)} f \, d\mu^A\) and \(\int_{\sigma(A)} f \, d\nu^A\) are elements of \(\mathcal{B}(\mathbf{H})\). Hence the map

is a bounded quadratic form on \(\mathbf{H}\), and the equation above states that both \(\int_{\sigma(A)} f \, d\mu^A\) and \(\int_{\sigma(A)} f \, d\nu^A\) represent \(Q\). Thus, the uniqueness clause of Proposition implies

which is precisely the statement that \(f \in \mathcal{F}\). Thus \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences, the desired result.

Recall that by Lemma (The Spectrum is a Compact Metric Measurable Space), \(\sigma(A)\) is a non-empty, compact metric measurable space.

Furthermore, as \(\sigma(A)\) is a subset of \(\mathbb{C}\) it inherits the metric from \(\mathbb{C}\). Thus \(\sigma(A)\) is also a metric space.

Finally recall, as we have noted many times, \(\sigma(A)\) is also a measurable space as it is equipped with the Borel \(\sigma\)-algebra.

So we have \(\sigma(A)\) a compact metric measurable space by Lemma, \(\mathcal{F}\) a set of bounded, measurable, complex-valued functions on \(\sigma(A)\) such that (1) \(\mathcal{F}\) is a complex vector space, (2) \(\mathcal{F}\) contains \(C^0(\sigma(A); \mathbb{R})\), and (3) \(\mathcal{F}\) is closed under pointwise limits of uniformly bounded sequences. These are the exact conditions we require to apply the Lemma we previously proved.

Applying this Lemma to the case at hand, we can conclude that \(\mathcal{F}\) consists of all bounded, Borel-measurable functions on \(\sigma(A)\). However, by definition \(\mathcal{F}\) is the set of bounded, measurable, complex-valued functions on \(\sigma(A)\) such that operator-valued integration with respect to \(\mu^A\) agrees with the same with respect to \(\nu^A\). Thus, these last two facts imply that operator-valued integration with respect to \(\mu^A\) agrees with the same with respect to \(\nu^A\) on all bounded, Borel-measurable functions on \(\sigma(A)\), the desired result of Part 3.

Part 4: Finally let us prove that \(\mu^A(E) = \nu^A(E)\) for all measurable subsets \(E\) of the spectrum \(\sigma(A)\) of \(A\).

Consider any measurable subset \(E\) of the spectrum \(\sigma(A)\) of \(A\) along with its indicator function \(1_E\). As a result of Theorem (Operator-Valued Integration) we have

Now the indicator function \(1_E\) is obviously bounded, as

for all \(\lambda\) in \(\sigma(A)\); measurable, as the inverse image of any measurable set in \(\mathbb{C}\) is either the empty set or \(E\); and also complex-valued, as \(\mathbb{R}\) is a subset of \(\mathbb{C}\). Hence, the indicator function \(1_E\) is a bounded, Borel-measurable function on \(\sigma(A)\).

Now, as we proved in Part 3, operator-valued integration with respect to \(\mu^A\) agrees with the same with respect to \(\nu^A\) on all bounded, Borel-measurable functions on \(\sigma(A)\). So in the case of \(1_E\) this implies that

This along with our previous expressions for \(\mu^A(E)\) and \(\nu^A(E)\) then imply

the desired result of this final part.

So with this we have proven that under the hypotheses of the Theorem \(\mu^A(E) = \nu^A(E)\) for all measurable subsets \(E\) of the spectrum \(\sigma(A)\) of \(A\), i.e. \(\mu^A\) and \(\nu^A\) are equivalent projection-valued measures.\(\blacksquare\)

This also completes the Spectral Theorem for Bounded, Self-Adjoint Operators proof.\(\blacksquare\)

With the Spectral Theorem for Bounded, Self-Adjoint Operators proof complete, we can conclude by finally introducing the normative definition of the “functional calculus”

Definition (Functional Calculus) If \(A \in \mathcal{B}(\mathbf{H})\) is self-adjoint and \(f : \sigma(A) \rightarrow \mathbb{C}\) is a bounded measurable function on the spectrum \(\sigma(A)\) of \(A\), functional calculus defines an operator \(f(A)\) by

\[f(A) \equiv \int_{\sigma(A)} f(\lambda) \, d\mu^A(\lambda),\]

where \(\mu^A\) is the unique projection-valued measure of Theorem (Spectral Theorem for Bounded, Self-Adjoint Operators) associated to \(A\).

a tool which will often be used.