A question about spectral theorem The following is a discussion about spectral theorem of Folland's Harmonic analysis page 18.
Suppose $A$ is a unital commutative C*- subalgebra of $B(H)$ and $u,v\in H$. Put $\Sigma = \sigma(A)$ . Define $\phi_1: C(\Sigma)\to {\Bbb C} $ such that $\phi_1(f)=(T_f u,v)$ where $T_f$ is the inverse Gelfand transform of $f$. Clearly $\phi\in C(\Sigma)^*$. There is a unique measure $\mu_{u,v}$ such that $(T_fu,v)=\int f d\mu_{u,v}$.
Now he uses above information to extend the representation $\phi: C(\Sigma)\to B(H)$ to the larger algebra $B(\Sigma)$(the space of bounded Borel measurable functions on $\Sigma$) namely $\bar\phi$. 
If $f\in B(\Sigma)$ , then $|\int fd\mu_{u,v}|\leq ||f||||u||||v||$.
Now he concludes that there is a unique $T_f\in B(H)$ such that $(T_f u,v) = \int f d\mu_{u,v}$, while I can not understand why there is such $T_f$. Please help me. Thanks in advance.
 A: What you know is that, for every $u,v \in H$, there exists a unique Borel measure $\mu_{u,v}$ such that
$$
          (T_{f}u,v) = \int fd\mu_{u,v}, \;\;\; f \in C(\Sigma).
$$
Because of this uniqueness of $\mu_{u,v}$, it follows that
$$
       \mu_{\alpha u+\beta u',v}=\alpha\mu_{u,v}+\beta\mu_{u',v},\\
       \mu_{u,\alpha v+\beta v'}=\overline{\alpha}\mu_{u,v}+\overline{\beta}\mu_{u,v'}
$$ for all scalars $\alpha$, $\beta$ and $u,u',v,v'\in H$.
Therefore, if $f \in \mathcal{B}(\Sigma)$ is fixed, then the following is a sesquilinear form on $H$:
$$
           B(u,v) = \int f d\mu_{u,v},\;\;\; u,v \in H.
$$
And, the bound you've established shows that $B$ is a bounded form with
$$
              |B(u,v)| \le \|f\|\|u\|\|v\|, \;\;\; u,v\in H.
$$
By the Lax-Milgram Theorem there exists a unique operator $T_{f}\in\mathcal{B}(H)$ such that $B(u,v) = (T_{f}u,v)$ for all $u,v\in H$. Furthermore, $\|T_{f}\|\le \|f\|$. The existence of $T_{f}\in\mathcal{B}(H)$ gives
$$
                 (T_{f}u,v) = \int f\,d\mu_{u,v},\;\;\; f \in \mathcal{B}(\Sigma),\; u,v\in H.
$$
I think this is what you want.
