# 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 $\Phi$? Commented Oct 8, 2017 at 20:29
• sorry,I can't see $\Phi$ has anything to do with the previous contents. Commented Oct 9, 2017 at 10:32
• Dose $B(\mathbb{R})$ mean all the bounded real functions? Commented Oct 9, 2017 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 $\mathbb{R}$ Commented Oct 9, 2017 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 $\Phi$. Commented Oct 9, 2017 at 15:28

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 $f$? Commented Oct 2, 2017 at 14:27
• I added some details. Where do you use $f(A)$'s self-adjointness? Commented Oct 3, 2017 at 14:20
• I used $f(A) = f(A)^*$ in my proof of property 2. Commented Oct 3, 2017 at 14:49
• I don't see where. Commented Oct 6, 2017 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. Commented Oct 6, 2017 at 13:37

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? Commented Oct 9, 2017 at 11:40
• You mean $\langle \phi,\psi\rangle=(\phi, g(A)\psi)$ is an inner product? Commented Oct 9, 2017 at 11:47
• see here, page 4: math.mcgill.ca/jakobson/courses/ma667/… Commented Oct 9, 2017 at 11:49
• He means $(\phi, g(A)\psi)$ is merely notation, the extension is also a notion. Do you agree? Commented Oct 9, 2017 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. Commented Oct 11, 2017 at 10:59