In a course I've seen the proof of the following theorem:
Let $A\in\mathcal{L}(\mathcal{H})$ be a self-adjoint operator. Then there is a unique map
$\phi_{A}:\mathcal{B}(\sigma(T))\to\mathcal{L}(\mathcal{H})$
such that:
$\phi_{A}$ is a $\star$-algebra homomorphism
$\phi_A(\operatorname{id}_{\sigma(A)})=A$
$\phi_{A}$ is continuous.
Let $(f_{n})_{n\in\mathbb{N}}\subseteq\mathcal{B}(\sigma(A))$ be a bounded sequence s.t. $f_{n}(t)\to f(t)$ for all $t\in\sigma(A)$ for some $f\in\mathcal{B}(\sigma(A))$. Then $\langle f_{n}(A)v,w\rangle\to \langle f(A)v,w\rangle$
I give a sketch of the proof of the existence that we have seen:
Let $f\in C(\sigma(A))$. Consider the continuous linear functional $L_{v,w}(f):=\langle f(A)v,w\rangle$.
By the Riesz theorem(for $C_c$), there are complex measures $\mu_{v,w}$ such that
$$\langle f(A)v,w\rangle=\int_{\sigma(A)}\,f\,\mathrm{d}\mu_{v,w}.$$
Now we want to define $f(A)$ for $f\in\mathcal{B}(\sigma(A))$ as the operator that satisfies:
$$\langle f(A)v,w\rangle=\int_{\sigma(A)}\,f\,\mathrm{d}\mu_{v,w}.$$
Using Riesz theorem(for Hilbert spaces) it is wasy to show that this operator exists.
Now define $$\phi_A(f)=f(A)$$ and show that it satisfies all the properties.
The details were left as an exercise. I've been able to prove that $\phi_A$ fullfuills all the properties except that $\phi_A(fg)=\phi_A(f)\phi_A(g)$. I think that should be easy but I can't figure it out. Can you explain me how to prove it or give me same hint? Thank you
Best Answer
As I said in the comments, you can show that $$\phi(fg) = \phi(f) \phi(g)$$ for $f,g$ simple functions. This then implies that $$||\phi(s)|| \le ||s||_\infty \tag {1} $$ for any simple function $s$. Indeed, for $x \in H$, \begin{align*} ||\phi(s) x||^2 &= (\phi(s) \phi(s)^* x,x) = (\phi(s) \phi(s^*)x,x) = (\phi(|s|^2)x,x) \\ &= \int_{\sigma(T)} |s|^2 \ \mu_{x,x} \le \|s\|_{\infty}^2 \mu_{x,x} (\sigma(T)) \\ & = \|s\|_\infty^* \|x\|^2. \end{align*} [In fact, one can show equality in (1) but we don't need that.]
Let $f,g$ be bounded borel functions on $\sigma(T)$. We can approximate $f$ and $g$ in the supremum norm by a sequence of simple functions $(f_n) $ and $(g_n)$ respectively. Then by (1) the sequences $(\phi(f_n))$ and $(\phi(g_n))$ are Cauchy in $B(H)$ and thus converge. Since $\phi(f_n) \xrightarrow{WOT} \phi(f) $ and $\phi(g_n) \xrightarrow{WOT} \phi (g) $, we infer that $\phi(f_n) \to \phi (f) $ and $\phi(g_n) \to \phi(g)$ in the operator norm. Hence, $$ \phi(fg) = \lim \phi(f_n g_n) = \lim \phi (f_n) \phi (g_n) = \phi(f) \phi(g) .$$