# Extension of functional calculus of continuous functions

On Reed & Simon's book, we can find the following theorem, which is called the continuous functional calculus.

Notation: $$\sigma(A)$$ is the spectrum of the operator $$A$$, $$\mathscr{L}(\mathscr{H})$$ is the space of all bounded linear maps on a Hilbert space $$\mathscr{H}$$ and $$C(\sigma(A))$$ is the space of all continuous functions defined on $$\sigma(A)$$.

Theorem: Let $$A$$ be a bounded self-adjoint operator on a Hilbert space $$\mathscr{H}$$. Then there exists a unique map $$\phi: C(\sigma(A)) \to \mathscr{L}(\mathscr{H})$$ with the following properties:

(a) $$\phi$$ is an algebraic $$*$$-homomorphism, that is: $$\phi(fg) = \phi(f)\phi(g) \quad \phi(\lambda f) = \lambda \phi(f) \quad \phi(1) = I \quad \mbox{and} \quad \phi(\bar{f}) = \phi(f)^{*}$$

(b) $$\phi$$ is continuous, that is, $$||\phi(f)||_{\mathscr{L}(\mathscr{H})} \le C||f||_{\infty}$$ for some $$C>0$$.

(c) If $$f(x) = x$$ then $$\phi(A) = A$$.

(d) If $$A\psi = \lambda \psi$$, then $$\phi(f)\psi = f(\lambda)\psi$$, $$\psi \in \mathscr{H}$$,

(e) $$\sigma[\phi(f)] = \{f(\lambda): \lambda \in \sigma(A)\}$$

(f) If $$f \ge 0$$ then $$\phi(f) \ge 0$$

(g) $$||\phi(f)|| = ||f||_{\infty}$$.

The proof of this theorem is given by the authors and I have no problem with it.

Later in the book, the authors discuss the extension of this theorem for bounded Borel measurable functions. In this latter case, $$\phi$$ is now a map $$\hat{\phi}: B(\mathbb{R}) \to \mathscr{L}(\mathscr{H})$$, where $$B(\mathbb{R})$$ denotes the space of all bounded Borel measurable functions on $$\mathbb{R}$$. All properties (a) - (g) remain unchanged for $$\hat{\phi}$$. The discussion uses spectral measures to prove this Borel functional calculus. However, the authors state that the Borel functional calculus can be proven directly by extending the above continuous functional calculus.

Question: How do I extend the above stated continuous functional calculus to prove the Borel functional calculus? The authors give no hint on how to do it and I have no clue on how to approach such extension. By their comment, I'd expect that some natural extension exists but I couldn't even sketch one by myself.

• See if you can use positivity and extend to intervals, and, finally, to Borel sets. Sep 15 at 1:45
• The extension $\hat \phi$ does not satisfy all properties (a)-(g). Both (e) and (g) fail in general. Sep 15 at 7:40

First of all, let me recall the main theorems about the functional calculus in order to show which properties are needed for existence and uniqueness and which properties are consequences of the definition. The continous functional calculus states the following:

Let $$A\in\mathcal{L}(\mathcal{H})$$ be a self-adjoint operator. Then there is a unique map $$\mathcal{P}_{A}:C(\sigma(A),\mathbb{C})\to\mathcal{L}(\mathcal{H})$$ $$\hspace{4.7cm}f \mapsto \mathcal{P}_{A}(f)=:f(A)$$ called "continuous functional calculus of $$A$$", such that the following properties are fulfilled for all $$f,g\in C(\sigma(A),\mathbb{C})$$ and $$\lambda\in\mathbb{C}$$:

1. $$\operatorname{id}_{\sigma(A)}(A)=A$$ and $$1(A)=\operatorname{id}_{\mathcal{H}}$$, where $$1$$ denotes the constant $$1$$-function.
2. $$\mathcal{P}_{A}$$ is continuous.
3. $$\mathcal{P}_{A}$$ is a "$$\star$$-algebra homomorphism", which means that it is an alegbra homomorphism, i.e. $$(\lambda f+g)(A)=\lambda f(A) + g(A)$$ multiplicative, i.e. $$(f\cdot g)(A)=f(A)g(A)$$ and the adjoint-operation commutes with complex-conjugation, i.e. $$\overline{f}(A)=f(A)^{\ast}.$$

To proof this, one first of all defines the map for polynomials in the obvious way and then extends the construction to all continous functions via the Weierstrass approximation theorem. Now, using the above properties as well as the definition, one can show that the continous functional calculus has some further properties:

Let $$A\in\mathcal{L}(\mathcal{H})$$ be a self-adjoint operator. Then the continous functional calculus $$\mathcal{P}_{A}$$ has the following additional properties:

1. $$\Vert f(A)\Vert = \Vert f\Vert_{\infty}$$.
2. If $$f\geq 0$$ then $$f(A)$$ is a positive operator.
3. If $$A\psi=\lambda\psi$$ then $$f(A)\psi=f(\lambda)\psi$$.
4. $$\sigma(f(A))=f(\sigma(A))$$.

Now, let us extend this construction to the more general case of measurable functions:

Let $$A\in\mathcal{L}(\mathcal{H})$$ be a self-adjoint operator. Then there is a unique map $$\mathcal{P}_{A}:\mathcal{B}(\sigma(A),\mathbb{C})\to\mathcal{L}(\mathcal{H})$$ $$\hspace{4.7cm}f \mapsto \mathcal{P}_{A}(f)=:f(A)$$ called "measurable functional calculus of $$A$$", such that the following properties are fulfilled for all $$f,g\in C(\sigma(A),\mathbb{C})$$ and $$\lambda\in\mathbb{C}$$:

1. $$\operatorname{id}_{\sigma(A)}(A)=A$$ and $$1(A)=\operatorname{id}_{\mathcal{H}}$$, where $$1$$ denotes the constant $$1$$-function.
2. $$\mathcal{P}_{A}$$ is continuous.
3. $$\mathcal{P}_{A}$$ is a "$$\star$$-algebra homomorphism" as above.
4. Let $$(f_{n})_{n\in\mathbb{N}}\in\mathcal{B}(\sigma(A),\mathbb{C})^{\mathbb{N}}$$ be a bounded sequence s.t. $$f_{n}(t)\to f(t)$$ for all $$t\in\sigma(A)$$ for some $$f\in\mathcal{B}(\sigma(A),\mathbb{C})$$. Then $$\langle f_{n}(A)\varphi,\psi\rangle\to \langle f(A)\varphi,\psi\rangle$$ for all $$\varphi,\psi\in\mathcal{H}$$.

In the special case of $$f\in C(\sigma(A),\mathbb{C})\subset\mathcal{B}(\sigma(A),\mathbb{C})$$, the uniqueness of the functional calculus is garantueed by the properties 1.-3. Now, it is important to note that for uniqueness in the general case, we also need the 4. property.

To show existence, there are several possibilities. One option is to first of all use the continuous functional calculus above to construct a set of measures. Let $$f\in C(\sigma(A),\mathbb{C})$$. Then we define maps $$l_{\varphi,\psi}(f):=\langle f(A)\varphi,\psi\rangle$$. Using this, we define maps $$l_{\varphi,\psi}:C(\sigma(A),\mathbb{C})\to\mathbb{C}$$, which are linear and also continuous, since $$\vert l_{\varphi,\psi}(f)\vert\leq \Vert f\Vert_{\infty}\cdot \Vert\varphi\Vert\cdot\Vert\psi\Vert.$$ Hence, by the theorem of Frechet-Riesz, there are complex measures $$\mu_{\varphi,\psi}$$ such that $$\langle f(A)\varphi,\psi\rangle=\int_{\sigma(A)}\,f\,\mathrm{d}\mu_{\varphi,\psi}.$$ Now, the main point is that the right-hand side of this equation is also meaningful for $$f\in\mathcal{B}(\sigma(A),\mathbb{C})$$ and hence, we can use this measures to define the functional calculus for Borel-measurable functions. More precisely, we define for all $$f\in\mathcal{B}(\sigma(A),\mathbb{C})$$ the map $$(\varphi,\psi)\mapsto \int_{\sigma(A)}\,f\,\mathrm{d}\mu_{\varphi,\psi},$$ which clearly is a continuous sesquilinearform. Hence, we can use the theorem of Lax-Milgram (basically a similar statement as the theorem of Frechet-Riesz but for sesquilinear forms), which shows that there exists an operator, which we denote by $$f(A)$$, such that $$\langle f(A)\varphi,\psi\rangle=\int_{\sigma(A)}\,f\,\mathrm{d}\mu_{\varphi,\psi}.$$ Using this definition, it is straight-forward to verify properties 1. - 4. above. Furthermore, it is also clear the the measurable functional calculus reduces to the continuous functional calculus in the case we take a function $$f\in C(\sigma(A),\mathbb{C})\subset\mathcal{B}(\sigma(A),\mathbb{C})$$.

As before, there are some immediate consequences of this definition:

Let $$A\in\mathcal{L}(\mathcal{H})$$ be a self-adjoint operator. Then the continous functional calculus $$\mathcal{P}_{A}$$ has the following additional properties:

1. The defining property 4. can be improved: Let $$(f_{n})_{n\in\mathbb{N}}\in\mathcal{B}(\sigma(A),\mathbb{C})^{\mathbb{N}}$$ be a bounded sequence s.t. $$f_{n}(t)\to f(t)$$ for all $$t\in\sigma(A)$$ for some $$f\in\mathcal{B}(\sigma(A),\mathbb{C})$$. Then $$\Vert f_{n}(A)\psi\Vert\to\Vert f(A)\psi\Vert$$ for all $$\psi\in\mathcal{H}$$.
2. $$\Vert f(A)\Vert\leq \Vert f\Vert_{\infty}$$.

Hope this helps. Feel free to ask question, if something is unclear! :-)