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.
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.