TITLE: The Multiplication Operator $M_f: L^2(\mu) \to L^2(\mu)$ such that $M_f g = fg$ (Rudin)
QUESTION [11 upvotes]: This post is related but does not answer my question.
Question 2 is partly unaddressed. Perhaps we can split into two cases, (i) where $\mu(\{f = 0\}) = 0$ and (ii) $\mu(\{f = 0\}) > 0$? If $\mu(\{f = 0\}) = 0$, then $g = h/f$ a.e.

Problem $5.17$ (paraphrased), Rudin's Real and Complex Analysis.

If $\mu$ is a positive measure, each $f\in L^\infty(\mu)$ defines a multiplication operator $M_f$ on $L^2(\mu)$ into $L^2(\mu)$, such that $M_f(g) = fg$. Prove that $\|M_f\| \le \|f\|_\infty$.

I have shown that $\|M_f\| \le \|f\|_\infty$.
Question 1:

For which measures $\mu$ is it true that $\|M_f\| = \|f\|_\infty$ for all $f\in L^\infty(\mu)$?

Question 2:

For which $f\in L^\infty(\mu)$ does $M_f$ map $L^2(\mu)$ onto $L^2(\mu)$?


My work:

We need to find measures $\mu$ for which the reverse inequality, i.e. $\|M_f\| \ge \|f\|_\infty$ also holds, for all $f\in L^\infty(\mu)$. More explicitly, we need
$$\sup\{\|M_f g\|_2: g\in L^2(\mu), \|g\|_2 \le 1\} = \|f\|_\infty \quad (f\in L^\infty(\mu))$$
where $$\|f\|_\infty = \inf\{a\in\mathbb R: \mu(\{|f| > a\}) = 0\}$$
Consider $h\in L^2(\mu)$. We want $g\in L^2(\mu)$ such that $M_f g = fg = h$. If $1/f \in L^\infty(\mu)$, the map is surjective. Is the converse true? If not, could one help me get an iff condition without assuming anything on $\mu$ besides that it is a positive measure?

I'd appreciate any help, thank you!

Update:
The proof below by Jose Avilez, and Jean L., Theorem $1.5$ tells us that $\|M_f\| = \|f\|_\infty$ for all $f\in L^\infty(\mu)$ iff $\mu$ is semi-finite.

REPLY [3 votes]: This is to address a question by the OP regarding the case where $\mu$ is not semi finite and $M_f:L_2\rightarrow L_2$ is onto.
Define
$$\mu_0(E)=\sup\{\mu(F): F\subset E, \, \mu(F)<\infty\}$$
This defines a measure on $(X,\mathscr{F})$:

Clearly $\mu_0(\emptyset)=0$, $\mu_0(E)\geq0$ for all $E\in\mathscr{F}$,  $\mu_0(E_1)\leq \mu_0(E_2)$ whenever $E_1\subset E_2$.
$\mu_0$ is $\sigma$-aditive: Suppose $\{E_n:n\in\mathbb{N}\}$ are pairwise disjoint, and w.l.o.g. $\mu(E_n)>0$. If $A\subset\bigcup_nE_n$ and $\mu(A)<\infty$ then
$$\mu(A)=\sum^\infty_{n=1}\mu(A\cap E_n)\leq\sum^\infty_{n=1}\mu_0(E_n)$$
Hence $\mu_0(\bigcup_nE_n)\leq\sum^\infty_{n=1}\mu_0(E_n)$. If $\mu_0(E_{n'})=\infty$ for some $n'$, then $\infty=\mu_0(E_{n'})\leq \mu_0(\bigcup_nE_n)=\sum^\infty_{n=1}\mu_0(E_n)$. Suppose $\mu_0(E_n)<\infty$ for all $n$. Choose $A_n\subset E_n$ such that $\mu_0(E_n)<\mu(A_n)+2^{-n}\varepsilon<\infty$. Then
$$\sum^N_{n=1}\mu_0(E_n)\leq\sum^N_{n=1}\mu(A_n)+\varepsilon(1-2^{-N})=\mu(\bigcup^N_{n=1}A_n)+\varepsilon(1-2^{-N})\leq\mu_0(\bigcup_nE_n)+\varepsilon$$
It follows that $\mu_0(\bigcup_nE_n)=\sum_n\mu_0(E_n)$.

Notice that if $E$ is an atom of $\mu$ with infinite mass ($\mu(E)=\infty$), then $\mu_0(E)=0$. Also, it is clear that if $\mu(F)<\infty$, then $\mu_0(F)=\mu(F)$.
Claim:  If $\mu(F)=\infty$ and $F$ does not contain an atom of infinite mass, then $\mu_0(F)=\infty$. To see this,  let $E_n\subset F$ such that $\mu(E_n)<\infty$ and $\mu(E_n)\xrightarrow{n\rightarrow\infty}\mu_0(F)$. Substituting $E_n$ with $\bigcup^n_{k=1}E_k$ if necessary, we may assume that $E_n$ is monotone nondecreasing. Setting $E=\bigcup_nE_n$, we have that $\mu_0(F)=\lim_n\mu(E_n)=\mu(E)$.
If $\mu(F\setminus E)>0$, there is $A\subset F\setminus E$ such that $0<\mu(A)<\infty$. It follows that $\mu_0(E)<\mu(E)+\mu(A)=\mu(E\cup A)<\infty$ in contradiction to the definition of $\mu_0(E)$. Hence $\mu_0(F\setminus E)=0$ and so, $\mu(E)=\mu(F)=\infty$.
A consequence of the claim above is that $\mu_0$ is a semi-finite measure.
Let $\nu:\mathscr{F}\rightarrow\overline{\mathbb{R}}$ defined as
$\nu(A)=\infty$ if $A\in\mathscr{F}$ contains an atom of infinite $\mu$ mass, and $0$ otherwise. It is easy to check that  $\nu$ is a measure, and that
$$\mu=\mu_0+\nu$$


Notice that $L_2(\mu)=L_2(\mu_0)$.
If $A$ is an atom of inifinite $\mu$-mass, then as the answer of JeanL shows, $g\mathbb{1}_A=0$ for all $g\in L_2(\mu)$; hence, $\int fg\,d\nu=\int_A fg\,d\nu +\int_{A^c}fg\,d\mu=0$.

All this  shows that it is enough to concentrate on the measure $\mu_0$. An we can proceed as in the solution presented for the case where $\mu$ is semi-finite.