Suppose that $X$ is a compact, Hausdroff space. Suppose that $\nu$ is a probability measure on $X$. Let $T: C(X) \to \mathbb{B}(L^2(X,\nu))$ be defined by $T(f)=M_f$, where $M_f(g)=fg$.
$M_f$ is well-defined
$\|M_f(g)\|_2^2=\int_X|fg|^2d\nu(x) \le \|f\|_{\infty}^2\int_X|g|^2d\nu(x)=\|f\|_{\infty}^2\|g\|_2^2 \implies \|M_f(g)\|_2 \le \|f\|_{\infty}\|g\|_2$. This also shows that $M_f$ is bounded above by $\|f\|_{\infty}$
$M_f$ is a star homomorphism.
This is because $M_{f_1f_2}(g)=f_1f_2(g)=M_{f_1}(f_2g)=M_{f_1}\circ M_{f_2}(g)$. Also $M_{f}^*=M_{\bar{f}}$. Hence $T(f^*)=T(f)^*$.
I want to see if $T$ is an isometry. It is equivalent to showing that $T$ is injective. Suppose that $T(f)=0$ for some $f \in C(X)$. Then $M_f(g)=0$ for all $g \in L^2(X,\nu)$. Hence, $fg=0$ for all $g \in L^2(X,\nu)$. Suppose that $f$ is not $0$. Let $E=\{x \in X:f(x) \ne 0$}. Then $E$ is measurable and $0<\nu(E) \le \nu(X)=1$. Since $f\chi_E=0$, we have $f(x)\chi_E(x)=0$ for $x \in E$. Therefore $f(x)=0$ for all $x \in E$. This is a contradiction. Therefore $\nu(E)=0$. Since $f$ is a continuous function, we must have that $f=0$. Thus $T$ is injective. Therefore, $T$ is an isometry.
The proof doesn't depend on the measure $\nu$ except for the fact that $\nu$ is a finite measure. Is this proof alright?
Thanks for the help!!
From $\nu(E)=0$, it seems that we cannot conclude $f=0$ by the continuity of $f$. This goes through if $\nu$ is the standard Euclidean Lebesgue measure.
The following reasoning assumes that if $f\in C(X)$ is such that $f=0$ $\nu$-a.e. then $f=0$ everywhere.
Actually $f$ is $L^{2}(X,\nu)$: We let $M=\max_{x\in X}|f(x)|$, then $\displaystyle\int_{X}|f(x)|^{2}d\nu(x)\leq M^{2}\int_{X}d\nu(x)=M^{2}\nu(X)=M^{2}<\infty$.
So the complex-conjugate $\overline{f}$ is also in $L^{2}$, hence $M_{f}(\overline{f})=0$, in particular, $|f|^{2}=f\cdot\overline{f}=0$, hence $f=0$.
If $\nu$ is such that $\nu(G)>0$ for all nonempty open sets $G$, then our proofs both go through.