# Why is the $*$-homomorphism of Borel functional calculus bounded?

This is probably a very easy question, but I'm stuck at it for hours now. In what follows, $$\mathcal{B}(\mathbb{R})$$ is the space of all bounded Borel measurable functions on $$\mathbb{R}$$. Let $$\mathcal{H}$$ be a Hilbert space, $$A$$ a bounded self-adjoint operator on $$\mathcal{H}$$ and $$\psi \in \mathcal{H}$$ be fixed. By Riesz-Markov Theorem, there exists a measure $$\mu_{\psi}$$ on $$\sigma(A)$$ (the spectrum of $$A$$) such that: $$\langle \psi, f(A)\psi\rangle = \int_{\sigma(A)}f(\lambda)d\mu_{\psi}(\lambda)$$ for every continuous function $$f$$ defined on $$\sigma(A)$$. Thus, if $$g\in \mathcal{B}(\mathbb{R})$$ we can define $$g(A)$$ by the rule: $$\langle \psi, g(A),\psi\rangle = \int_{\sigma(A)}g(\lambda)d\mu_{\psi}(\lambda)$$ for every $$\psi \in \mathcal{H}$$. The polarization identity allows us to obtain $$\langle g(A)\psi,\phi\rangle$$ from the above identity. If $$T_{\psi}$$ is a bounded linear operator $$T_{\psi}:\mathcal{H}\to \mathcal{H}$$ given by $$\phi \mapsto T_{\psi}(\phi) := \langle g(A)\psi, \phi\rangle$$, then, by Riesz Representation Theorem there exists $$\varphi \in \mathcal{H}$$ such that: $$T_{\psi}(\phi) = \langle \varphi, \phi \rangle$$ for every $$\phi \in \mathcal{H}$$. Thus, we may define $$g(A):\mathcal{H}\to \mathcal{H}$$ as the linear map satisfying: $$g(A)\psi := \varphi$$

Question: $$g(A)$$ is linear by construction but how can I prove it is bounded? Also, using only the above construction, if $$g$$ is real valued, does it follow that $$g(A)$$ is self-adjoint?

• Aren't all $*$-homomorphisms between $C^*$-algebras bounded? Aug 24, 2020 at 19:24
• Yes, but I haven't proved it is a $*$-homomorphism yet. I just proved it for $C(\sigma(A))$, not for $\mathcal{B}(\mathbb{R})$. Aug 24, 2020 at 19:26
• Just to clarify, the $*$-homomorphism will be the map $\phi: \mathcal{B}(\mathbb{R}) \to B(\mathcal{H})$, where $B(\mathcal{H})$ is the set of all bounded linear operators from $\mathcal{H}$ to itself, right? Thus, I need to prove that $g(A) = \phi(g)$ is bounded, so that the above map makes sense. Aug 24, 2020 at 19:29
• Oh, I misunderstood. I thought you want to prove that $\phi$ itself is bounded. Aug 24, 2020 at 19:33

You can define $$\mu_{\phi,\psi}$$ by the Riesz-Markov theorem analogically (i.e. as the measure corresponding to $$f \mapsto \langle f(A)\phi, \psi\rangle$$). We have $$|\langle f(A) \phi, \psi\rangle| \leq ||f||_\infty ||\phi|| \;||\psi||.$$ Hence, $$||\mu_{\phi,\psi}|| \leq ||\phi|| \; ||\psi||$$. Note that this measure is actually the polarization of measures $$\mu_\phi$$ and $$\mu_\psi$$.
Now fix $$g \in \mathcal{B}(\mathbb{R})$$. We have $$|T_\psi(\phi)| = \left| \int_{\sigma(A)} g(\lambda) d \mu_{\phi,\psi}(\lambda) \right| \leq ||g||_\infty ||\mu_{\phi,\psi}|| \leq ||g||_\infty ||\phi|| \; ||\psi||$$ and so $$||T_\psi|| \leq ||g||_\infty ||\psi||$$. Finally $$||g(A) \psi|| = ||\varphi|| = ||T_\psi|| \leq ||g||_\infty ||\psi||$$ which proves that $$||g(A)|| \leq ||g||_\infty$$.
For your second question, we first show that $$\mu_\phi$$ is a nonegative measure. This holds as for each nonnegative $$f \in C(\sigma(A))$$ we have that $$f(A)$$ is positive. Hence, $$\int_{\sigma(A)} f d\mu_\phi = \langle f(A)\phi,\phi \rangle \geq 0.$$
Now for real valued $$g \in \mathcal{B}(\mathbb{R})$$ we have $$\langle g(A)\phi,\phi \rangle = \int_{\sigma(A)}g d \mu_\phi \in \mathbb{R}.$$ Hence $$g(A)$$ is self-adjoint.
• Thanks for the answer! Just to clarify, what's your definition of $||\mu_{\phi,\psi}||$? Aug 24, 2020 at 22:05
• It is analogous to the definition of $\mu_\phi$: The mapping $C(\sigma(A)) \rightarrow B(\mathcal{H})$, $f \mapsto \langle f(A) \phi, \psi \rangle$ is a continuous linear functional, which is represented (by the Riesz-Markov theorem) by the measure $\mu_{\phi,\psi}$. @MathMath Aug 25, 2020 at 6:36
• Great. I got it. Thanks. Do you know if I can prove the result not using $\mu_{\psi,\phi}$? Because I defined $\langle f(A)\phi,\psi\rangle$ by means of the polarization identity and I couldn't be able to get the same estimate using it. However, if I approach as you did, I'd have to show that the $\mu_{\psi,\phi}$ you got by Riesz-Markov is equivalent to my definition using the polarization identity, and this would be very boring to do! Aug 25, 2020 at 18:11
• Showing that the definitions are equivalent is not that hard or long. It follows as $\langle f(A) \phi, \psi \rangle$ is in fact the polarization of $\langle f(A) \phi,\phi \rangle$ and as the Riesz reprezentation is linear. Defining $\mu_{\phi,\psi}$ will also come very handy later for proving some of the properties of borel calculus. @MathMath Aug 25, 2020 at 22:31