# Continuity of the functional calculus form of the Spectral Theorem

I'm trying to understand the proof of the functional calculus form of the Spectral Theorem. From what I understand the construction of a definition of $f(A)$ for $f$ measurable and $A$ $\in L(H)$ self-adjoint begins as follows.

Assume we know the proof of the continuous functional calculus so that $f(A) \in L(H)$ is defined for $f$ continuous on the spectrum of $A$.

We can define a function $T_\psi: C(\sigma(A)) \to \mathbb{C}$ by $$T_\psi(f) = (\psi,f(A)\psi)$$ for a set $\psi \in H.$

This gives a positive linear functional on $C(\sigma(A))$, and from the Riesz-Markov theorem there is a measure $\mu_\psi$ on the compact set $\sigma(A)$ s.t

$$T_\psi(f) = (\psi,f(A)\psi)=\int_{\sigma(A)}fdu_\psi.$$

We note that the right hand side is defined for any $f \in \mathbb{B}(\mathbb{R})$, a bounded measurable function, and so we define:

$(\psi,g(A)\psi):=\int_{\sigma(A)}gdu_\psi$ $\forall g \in \mathbb{B}(\mathbb{R})$.

*Now this is a definition, the left hand side is not an inner product because the second term is not yet defined.

We next extend the definition:

$\forall \psi, \phi \in H$, $(\phi, g(A)\psi) :=$

$\frac{1}{4}[(\phi + \psi, g(A)(\phi + \psi)) - (\phi - \psi, g(A)(\phi - \psi)) + (\phi + i\psi, g(A)(\phi + i\psi)) - (\phi - i\psi, g(A)(\phi - i\psi))]$

Lastly we can define a function $T_1: H \to \mathbb{C}$ by $T_1(\phi) = (\phi, g(A)\psi)$ for a set $\psi, g$.

$T_1$ is a continuous linear function, and so by the Riesz rep. theorem we have $\exists!$ $h_{\psi,g}\in H$ s.t $\forall \phi$ $(\phi, g(A)\psi) = (\phi, h_{\psi,g})$.

Now, the functional calculus theorem states that exists, and unique with the following properties, $\Phi: \mathbb{B}(\mathbb{R}) \to L(H)$ s.t

1. $\Phi$ is a $*$-homomorphism.

2. $\|\Phi(f)\| \leq \|f\|_{\infty}$

among other properties I won't list.

I need help showing 1. and 2.

For 2. I thought of trying to show that $\forall f \in \mathbb{B}(\mathbb{R})$ $f(A)$ is self adjoint but kind of got lost in the details of this. Is it true?

If it is, then we have

$\|\Phi(f)\| = sup_{\|\phi\|=1}[\|f(A)\phi\|] = sup_{\|\phi\|=1}(\phi,f(A)\phi) = \int_{\sigma(A)} f du_\phi \leq \|f\|_{\infty}\mu_\phi(\sigma(A)) = \|f\|_{\infty}$ where the last equality should hold by the Riesz Markov theorem.

For 1. I'm not sure where to begin.

• To future reads: I believe $f(A)$ doesn't need to be self-adjoint necessarily. My mistake was assuming $f$ was real. Can you provide a proof for the continuity of $\Phi$? – Mariah Oct 8 '17 at 20:29
• sorry,I can't see $\Phi$ has anything to do with the previous contents. – C.Ding Oct 9 '17 at 10:32
• Dose $B(\mathbb{R})$ mean all the bounded real functions? – C.Ding Oct 9 '17 at 10:56
• @C.Ding $\Phi(f) = f(A)$. Is that what you mean in your first question? To your second question; I don't think we necessarily need to limit to real bounded measurable functions. All he says $\mathbb{B}(\mathbb{R})$ means is the bounded Borel functions on $\mathbb{R}$ – Mariah Oct 9 '17 at 11:50
• @Mariah: It is worthwhile to abstract away a little to not get drowned in the details. In fact, you can prove that any $*$-homomorphism between $C^{*}$-algebras is norm-decreasing and so automatically continuous so this gives you $(2)$ and the continuity of $\Phi$ and leaves you only with proving that $\Phi$ is a $*$-homomorphism. The point is that those properties of $\Phi$ have nothing to do with the bounded functional calculus and don't require the specific form of $\Phi$ to prove them although you can also prove that directly for your specific $\Phi$. – levap Oct 9 '17 at 15:28

There is a lot of problems in your question, so I don't know where to begin my answer. But I believe your question will be answered if you answer my questions step by step.

First, the definition of $g(A)$ has some prblems. Why dose there exist $g(A)$ such that $(\psi,g(A)\psi):=\int_{\sigma(A)}gdu_\psi$?

Second, what's the definition of $f(A)$ for $f\in B(\mathbb{R})$?

Maybe the problem in your question is: Why dose there exist such $h_{\psi, g}$? It is because $T_1$ is bounded. So once $\Phi$ is defined, it is continuous.

For 1, (a) $\Phi$ is linear.

(b)To prove $\Phi$ preserves involution, we only need to verify $\Phi(f)$ is hermitian if $f$ is real because of (a). Since $(\phi, \Phi(f)\phi)=\int f d\mu_\phi$ is real for all $\phi$, $\Phi(f)$ is hermitian.

(c)To prove $\Phi$ preservers product, we have to prove:for all $f, g\in B(\mathbb{R})$, $$(\phi, \Phi(fg)\phi)=(\Phi(\bar f)\phi, \Phi(g)\phi)=\sum_{k=0}^3 i^k(\phi_k, \Phi(g)\phi_k)~~(polarization),$$ i.e., \begin{align}\label{eq} \int fg d\mu_\phi=\sum_{k=0}^3 i^k\int gd\mu_{\phi_k}.\tag{1}\end{align}

(c.1)Fixed $f\in C(\mathbb{R})$. Since the above equation holds for all continous function $g$, it also holds for all bounded measurable function $g$.

(c.2)Because of (c.1), for all continuous function $f$ and bounded measurable function $g$ $$\overline{(\phi, \Phi(\bar f\bar g)\phi)}=\overline{(\Phi( f)\phi, \Phi(\bar g)\phi)},$$ i.e. $$(\phi, \Phi(gf)\phi)=(\Phi(\bar g)\phi, \Phi(f)\phi),$$ that is, \begin{align} \int gf d\mu_\phi=\sum_{k=0}^3 i^k\int fd\mu_{\phi'_k}.\end{align} So it also holds for all bounded measurable function $f$. Done!

Details for (c.1): By Riesz Representation theorm,

Theorem 1. $$C(X)^*\cong M(X),$$ where $X$ is a compact space and $M(X)$ denotes all the complex regular Borel measure on $X$.

Another useful theory(V.4.1 in Conway's text):

Theorem 2. Suppose $X$ is a normed space, then the closed unit ball of $X$ is $\sigma(X^{**},X^*)$ dense in the closed unit ball of $X^{**}$.

For every bounded measurable function $g\in B(\sigma(A))\subset C(\sigma(A))^{**}$, there is a net $g_i\in C(\sigma(A))$ such that $$g_i\xrightarrow{\sigma(C(\sigma(A))^{**},C(\sigma(A))^*)}g$$ by theorem 2. That is precisely $$\int g_i d \mu\to\int g d \mu$$ for every $\mu \in M(\sigma(A))$ by theorem 1. Note that $fd\mu\in M(\sigma(A))$ for $f\in B(\sigma(A))$ and $\mu\in M(\sigma(A))$.

• the left hand side isn't an inner product, to my understanding. It is notation. The definition of $(\phi, g(A)\psi)$ later as a "polarization" form gives me an extension of the inner product. This is how I understood this construction. Does that make sense? – Mariah Oct 9 '17 at 11:40
• You mean $\langle \phi,\psi\rangle=(\phi, g(A)\psi)$ is an inner product? – C.Ding Oct 9 '17 at 11:47
• see here, page 4: math.mcgill.ca/jakobson/courses/ma667/… – Mariah Oct 9 '17 at 11:49
• He means $(\phi, g(A)\psi)$ is merely notation, the extension is also a notion. Do you agree? – C.Ding Oct 9 '17 at 11:59
• This depend on the definition of the inner product "(\cdot, \cdot)".The inner product in the question is linear for the second term and conjugate linear for the first item. – C.Ding Oct 11 '17 at 10:59

Throughout the post you mean $B(\mathbb{R})$, right? The proof for 2. seems O.K.

For 1., observe that it is naturally true for polynomials, and so it is true for all continuous functions since every continuous function on a compact set is a limit of polynomials. Now it is true for all of $B(\mathbb{R}$) since $C(\mathbb{R})$ is dense in $B(\mathbb{R})$.

Let $f,g\in C(\mathbb{R})$. Since $\sigma (A)$ is compact, there exists polynomials $p_n$ s.t. $p_n\longrightarrow f$, and $q_n$ s.t. $q_n\longrightarrow g$ uniformly. Therefore, $<\phi,fg(A)\phi>=<\phi,\underset{n,m\longrightarrow \infty}{lim}p_nq_n(A)\phi>=\underset{n,m\longrightarrow \infty}{lim}<\phi,p_nq_n(A)\phi>=$ $\underset{n,m\longrightarrow \infty}{lim}<\phi,p_n(A)q_n(A)\phi>=...=<\phi,f(A)g(A)\phi>$ and from here we conclude that $fg(A)=f(A)g(A)$ for every $f,g\in C(\mathbb{R})$.

Let $f,g\in B(\mathbb{R})$. Since $C(\mathbb{R})$ is dense in $B(\mathbb{R})$ with respect to $L^1$ convergence, there exist $f_n,g_n\in C(\mathbb{R})$ bounded s.t. $f_n,g_n$ converge to $f,g$ respectively in $L^1$. Now similarly, the same equalities as the previous case hold. Note that this time the equality $<\phi,\underset{n,m\longrightarrow \infty}{lim}f_ng_n(A)\phi>=\underset{n,m\longrightarrow \infty}{lim}<\phi,f_ng_n(A)\phi>$ holds because since $f_n,g_n$ are bounded and $\mu_\phi$ is finite, we can use the bounded convergence theorem.

• more easily said than done! If you feel like elaborating and writing it out I'll be happy to accept your answer. BTW, so you say $f(A)$ is indeed self adjoint for any measurable bounded $f$? – Mariah Oct 2 '17 at 14:27
• I added some details. Where do you use $f(A)$'s self-adjointness? – joeyg Oct 3 '17 at 14:20
• I used $f(A) = f(A)^*$ in my proof of property 2. – Mariah Oct 3 '17 at 14:49
• I don't see where. – joeyg Oct 6 '17 at 11:51
• $||\Phi(f)|| = sup_{||\phi||=1}[||f(A)\phi||] = sup_{||\phi||=1}(\phi,f(A)\phi)$ On the second equality; it doesn't hold generally I believe. – Mariah Oct 6 '17 at 13:37