The Multiplication Operator $M_f: L^2(\mu) \to L^2(\mu)$ such that $M_f g = fg$ (Rudin)

A modification to the counter example in @OliverDiaz answer can be used to show that semi-finiteness of the measure $\mu$ is also a necessary condition for $\|M_f\|=\|f\|_\infty$.

Suppose $\mu$ is not semi-finite. This means that there exists a set $A$ with $\mu(A)=\infty$ such that for any measurable set $B$, if $B\subset A$ and $\mu(B)>0$, then $\mu(B)=\infty$. Define $f=\mathbb{1}_A$, where $\mathbb{1}_A$ is the indicator function of the set $A$. It is easy to check that $\|f\|_\infty=1$. For any $g\in L_2$ $$\Big|\int_A g(x) \mu(dx)\Big|^2=|M_fg|^2\leq\|g\|^2_2<\infty$$ We claim that $\mu(\{x\in A:|g(x)|>0\})=0$. Otherwise, since $\{x\in A: |g(x)|>0\}=\bigcup^\infty_{n=1}\big\{x\in A: |g(x)|>\frac1n\big\}$, there is $n\in\mathbb{N}$ such that $\mu(\{x\in A: |g(x)|>\tfrac1n\})>0$. This implies that $$\infty=\mu(\{x\in A: |g(x)|>\tfrac1n\})\leq n\int_A |g(x)|\,\mu(dx)=n M_f(|g|)\leq n\|g\|_2<\infty$$ which is not possible.

If follows that $M_fg=0$ for all $g\in L_2$; therefore $0=\|M_f\|<1=\|f\|_\infty$.

This discusses a counterexample to show that if $\mu$ is not $\sigma$--finite (or more generally, semi finite), then $\|M_f\|=\|f\|_\infty$ may fail.

(Conway, J., A Course in functional analysis, 2nd edition, Springer , p. 28) Consider the space $([0,1],\mathscr{B}([0,1])$ its the measure $\mu=\lambda +\infty\delta_0$, here $\lambda$ is Lebesgue's measure, and $\delta_0$ is the measure that gives mass $1$ to $\{0\}$. Then $\mu(A)=\lambda(A)$ if $A$ is Borel measurable and does not contain $0$, and $\infty$ other wise. Define $f=\mathbb{1}_{\{0\}}$. Clearly $\mu$ is not semi-finite $\mu(\{0\})=\infty$), and $\|f\|_\infty=1$. For any $g\in L_2$ $$|g(0)|^2\mu^2(\{0\})=|M_fg|^2\leq \|g\|^2_2= \int |g|^2d\mu<\infty$$ This means that $g(0)=0$ for all $g\in L_2$. Therefore $M_f=0$, and $\|M_f\|<\|f\|_\infty$.

The following addresses question 2. First we show the following result.

Lemma: If $\mu$ is semi finite, and $f$ is a measurable function such that $M_f:L_2\rightarrow L_2$, given by $g\mapsto gf$, is a bounded operator, then $f\in L_\infty$.

Proof: Suppose $f\notin L_\infty(\mu)$, and set $E_n=\{n\leq |f|<n+1\}$. Then, there are infinitely many $E_n$'s (say $E_{n_k}$) that have positive measure. Let $A_{n_k}\subset E_{n_k}$ with positive finite measure. Define $$g=\sum_k\frac{1}{k\sqrt{\mu(A_{n_k})}}\mathbb{1}_{A_{n_k}}$$ Then $g\in L_2(\mu)$ for $\int|g|^2\,d\mu=\sum_k\frac{1}{k^2}$. On the other hand, $$\int|f g|^2\,d\mu=\sum_k\frac{1}{k^2\mu(A_{n_k})}\int_{A_{n_k}}|f|^2\,d\mu\geq \sum_k1=\infty$$ in contradiction to $g\,f\in L_2(\mu)$. Therefore, $f\in L_\infty(\mu)$.

We now prove the following:

Under the assumptions of the Lemma above, $M_f:L_2\rightarrow L_2$ is onto iff there is $c>0$ such that $c<|f|<\|f\|_\infty$ $\mu$-a.s.

If $0<c<|f|\leq\|f\|_\infty$ then $1/f\in L_\infty$. Clearly $M_f$ is bijective, with $M^{-1}_f=M_{1/f}$.

Conversely, suppose $f\in L_\infty$ is such that $M_f$ is onto.
Claim: $\mu(f=0)=0$. Otherwise, there would be a set $E$ of positive finite measure on which $|g|>0$. Then, the function $g=\mathbb{1}_E\in L_2$, $\|g\|_2>0$, but for no $h\in L_2$ is $M_fh=g$. This proves the claim.
Now, since $0<|f|$ $\mu$-a.s., it follows that $M_f$ is one-to-one, for $M_fh=M_gh'$ iff $fh=fh'$ $\mu$-a.s and so, iff $h=h'$ $\mu$-a.s. An application of the open map theorem (Theorem 5.10 in Rudin's aforementioned book) implies that $M_f$ has an inverse $M^{-1}_f:L_2\rightarrow L_2$, which is also a bounded operator. Notice that the inverse $M^{-1}_f$ is given by $h\mapsto h/f$, that is $M^{-1}_f$ is the product operator $M_{1/f}$. As $M_{1/f}$ is a bounded operator from $L_2$ into itself, the Lemma we proved above implies that $1/f\in L_\infty$. This shows that with $c=1/\|1/f\|_\infty>0$, $$ c\leq |f|\leq \|f\|_\infty,\qquad\mu-\text{almost everywhere}$$

Comment: In the solution of @JoseAvilez, only semifiniteness of $\mu$ is used. That is, if $\|f\|_\infty\neq0$ and $0<\varepsilon<\|f\|_\infty$, then $\mu(|f|>\|f\|_\infty-\varepsilon)>0$ and so, there is a measurable set $E\subset \{|f|>\|f\|-\varepsilon\}$ with $0<\mu(E)<\infty$. The rest of the argument as in that solution.