Cyclic representation on $L^2(\mu)$ Show that if $(X,\Omega,\mu)$ is a $\sigma-$ finite measure space and $H=L^2(\mu)$, then $\pi:L^\infty(\mu)\to B(H)$ defined by $\pi(\phi)=M_\phi$ is a cyclic representation and find all the cyclic vectors.
I do not have any idea about it. I'm studying Functional analysis by myself and it is an Example of Conway's Functional Analysis. Please help me to understand it. Thanks in advance.
 A: As I already stated in the comment, the cyclic vectors are exactly those $f \in L^2$ for which $f(x) \neq 0$ holds for almost every $x \in X$.
It is easy to see that if $M := \{x \mid f(x) = 0\}$ has positive measure, then by $\sigma$-finiteness, there is some set $M' \subset M$ of finite positive measure. But then
$$
\overline{\{\phi \cdot f \mid \phi \in L^\infty\}} \subset \{g \in L^2 \mid g(x) = 0 \text{ for a.e. } x \in M\},
$$
so that $g = \chi_{M'} \in L^2 \setminus \overline{\{\phi \cdot f \mid \phi \in L^\infty\}}$, which shows that $f$ is not cyclic.

Now assume that $f(x) \neq 0$ almost everywhere. By considering
$$
\tilde{f} = |f| = \frac{\overline{f}}{|f|} \cdot f,
$$
we can assume w.l.o.g. that $f \geq 0$ (note that $\overline{f}/|f| \in L^\infty$).
Also, the linear span of all indicator functions $\chi_M$ with $M$ measurable, of finite measure is dense in $L^2$ (why?), so that it suffices to show that we can approximate each $\chi_M$ in the $L^2$ norm using functions of the form $\phi \cdot f$, $\phi \in L^\infty$.
Let $K_n := \{x \mid f(x) \geq 1/n\}$. Note that $M = \bigcup_n (K_n \cap M)$ (up to a set of measure zero), where the union is increasing. By (e.g.) dominated convergence, this implies
$$
\Vert \chi_M - \chi_{M \cap K_n} \Vert_2 = \sqrt{\mu(M \setminus K_n)} \to 0,
$$
so that it suffices to approximate each $\chi_{M \cap K_n}$.
But $\phi := \chi_{M \cap K_n} \cdot \frac{1}{f} \in L^\infty$ (why?) with
$$
\phi \cdot f = \chi_{M \cap K_n},
$$
which completes the proof.
