Multiplication operator 
Let $M_{\phi}$ be a multiplication operator $M_{\phi}:L^{2}\left(\mu\right)\rightarrow L^{2}\left(\mu\right)$ defined by $M_{\phi}f=\phi f$. 
  
  
*
  
*Show that $\ker M_{\phi}=0$ if and only if $\mu\left(\left\{ x:\phi\left(x\right)=0\right\} \right)=0$. 
  
*Give necessary and sufficient conditions on $\phi$ that $\mbox{ran}M_{\phi}$ be closed.

 A: *

*Let $N=\{\phi=0\}$. We have $\phi f=0$ a.e. if and only if
$f=0$ a.e. on $N^c$. If the latter set has null complement, then $f=0$ in $L^2$, which means $M\phi$ is injective. Otherwise, there is a nonzero function $f$ which vanishes a.e. on $N^c$.


*Decompose $L^2(\mu)=L^2(\mu,N)\oplus L^2(\mu,N^c)$. Since $M_\phi$ kills the first component, its range is the same as the range of restriction to $L^2(\mu,N^c)$. This restriction is injective by part 1. Thus, it is closed iff it has a left inverse, or equivalently, if it has a positive lower bound $c$: namely $\|M_\phi f\|\ge c\|f\| $ for all $f\in L^2(\mu, N^c)$. The latter is equivalent to $|\phi|\ge c$ on $N^c$.
A: Just to elaborate on the answer given here
Suppose $(X,\mathscr{F},\mu)$ is $\sigma$-finite and that $X=\bigcup_nE_n$ where $0<\mu(E_n)<\infty$, and $E_n$ are disjoint. In addition, we assume that $\phi\in L_\infty(\mu)$. (see remarks at end of answer)

Q1. Suppose $\operatorname{ker}(M_\phi)=\{0\}$. If $\mu(\{\phi=0\})>0$, then there is $n$ such that $0<\mu(E_n\cap\{\phi=0\})<\infty$. Let $A=E_n\cap\{\phi=0\}$ and let $f=\mathbb{1}_A$. Then $f\in L_2$ and $\|f\|_2=\sqrt{\mu(A)}>0$; however, $M_\phi f=\phi f=0$ $\mu$-a.s., which gives  a contradiction.
Conversely, suppose $\mu(\{\phi=0\})=0$. If $f\in L_2$ and $\|f\|_2>0$, then there is $a>0$ such that $0<\mu(|f|>a)$ and so, for some $n$, $0<\mu(E_n\cap\{|f|>a\})<\infty$. Let $B=E_n\cap\{|f|>a\}$. It follows that
$$\|M_\phi f\|_2\geq \int_B|f\phi|^2\,d\mu\geq a\int_B|\phi|^2\,d\mu>0$$
Thus, $\operatorname{ker}(M_\phi)=\{0\}$.

Observation: $M_\phi$ is a normal operator on $L_1$, and $(M_\phi)^* =M_{\bar{\phi}}$:
$$(M_\phi f, g)=\int \phi f\, \overline{g}\,d\mu=\int f\overline{\bar{\phi} g}\,d\mu=(f, \bar{\phi}g)=(f,M_{\bar{\phi}}g)$$

Q2. If $\phi\equiv0$, then $\{0\}=\operatorname{ran}(M_\phi)$ and there is nothing to do: $\operatorname{ran}(M_\phi)=\{0\}$ which is closed.
Suppose $\|\phi\|_\infty>0$. Set $N=\{\phi=0\}$. Since
$$f=f\mathbb{1}_N + f\mathbb{1}_{N^c}$$
We can split $L_2$ as the sum of functions with support in $N$, and with support in $N^c$. This decomposition is unique, for if
$$f=g+h=f\mathbb{1}_N + f\mathbb{1}_{N^c}$$
and $\{g\neq0\}\subset N$ and $\{h\neq0\}\subset N^c$, then
$|g-f\mathbb{1}f_N|=|h-f\mathbb{1}_{N^c}|$ and so
$$\|g-f\mathbb{1}f_N\|^2_2=\int|h-f\mathbb{1}f_{N^c}||h-f\mathbb{1}_{N^c}|\,d\mu=0=\|h-f\mathbb{1}f_{N^c}\|^2_2$$
Let $H_1=\{f\in L_2:\{f\neq0\}\subset N\}$ and $H_2=\{f\in L_2:\{f\neq0\}\subset N^c\}$. Then
$$L_2=H_1\oplus H_2$$
Notice that both $H_1$ and $H_2$ are closed subsets of $L_2(\mu)$, and that  $M_\phi(H_1)=\{0\}$ (why?), which means that
$$\operatorname{ran}(M_\phi)=M_\phi(L_2(\mu))=M_\phi(H_2)$$
From (Q1) we have that the restriction $M'_\phi$ of $M_\phi$ to $H_2$ is injective. As $M'_\phi$ is normal, it follows that $M'_\phi(H_2)$ is dense in $H_2$ (for example problem).

*

*If $M_\phi(L_2(\mu))=M'_\phi(H_2)$ is is closed, then $M'_\phi(H_2)=H_2$. It follows (the open map theorem for example) that $M'_\phi:H_2\rightarrow H_2$ is invertible and that $(M'_\phi)^{-1}$ is a bounded operator. Let $\kappa:=\|M'_\phi\|$. Then, $\|(M'_\phi)^{-1}f\|_2\leq \kappa\|f\|_2$ for all $f\in H_2$, or equivalently
$$\|f\|_2\leq \kappa\|M_\phi f\|_2,\qquad f\in H_2$$
We now show that this implies that $\mu(N^c\cap|\phi|< \kappa^{-1})=0$. Indeed, if $\mu(N^c\cap\{|\phi|< \kappa^{-1}\})>0$, then for some $n\in\mathbb{N}$ we have that
$$0<\mu(E_n\cap N^c\cap\{|\phi|< \kappa^{-1}\})<\infty$$
Let $B=E_n\cap N^c\cap\{|\phi|< \kappa^{-1}\}$. Then $\mathbb{1}_B\in H_2$ and $\|\mathbb{1}_B\|_2>0$. Consequently
$$\mu(B)=\|\mathbb{1}_B\|^2_2\leq \kappa^2\|M'_\phi\mathbb{1}_B\|^2_2=\kappa^2\int_B|\phi|^2\,d\mu<\kappa^2\kappa^{-2}\mu(B)=\mu(B)$$
which is a contradiction. This shows that there is $\kappa>0$ such that $\phi\geq \kappa^{-1}$ $\mu$-a.s. on $N^c$.


*Conversely, suppose there is $\kappa>0$ such that $|\phi|\geq \kappa^{-1}$ $\mu$-a.s. on $N^c$. Denote by $\mathbb{1}_{N^c}\cdot\mu$ the restriction of $\mu$ to measurable subsets of $N^c$.  Then, $1/\phi \in L_\infty(\mathbb{1}_{N^c}\cdot\mu)$ (why?), and the operator
$M'_\phi:H_2\rightarrow H_2$ is invertible with inverse $(M'_\phi)^{-1}=M'_{1/\phi}:f\mapsto \frac{1}{\phi}f\mathbb{1}_{N^c}$. Consequently,
$$\operatorname{ran}(M_\phi)=M_\phi(L_2(\mu))=M'_\phi(H_2)=H_2$$ which is a closed subspace of $L_2(\mu)$.
Putting things together, we have that  $\operatorname{ran}(M_\phi)$ is closed if and only if there is a constant $c>0$ such that $\mu(\{\phi\neq0\}\cap\{|\phi|< c\})=0$, i.e., if and only if $\phi$ is bounded away from $0$ on $\{\phi\neq0\}$.

Final comments:

*

*The assumption of $\sigma$-finiteness can be replaced by the assumption that every set of positive $\mu$ measure contains a set of positive finite measure.


*The condition $\phi\in L_\infty$ seems to be necessary and sufficient. Indeed, if $\phi f\in L_2$ for all $f\in L_2$, then $\phi\in L_\infty$. Suppose not, and set $E_n=\{n\leq |\phi|<n+1\}$. 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 $$f=\sum_k\frac{1}{k\sqrt{\mu(A_{n_k})}}\mathbb{1}_{A_k}$$
Then $f\in L_2(\mu)$ for $\int|f|^2\,d\mu=\sum_k\frac{1}{k^2}$. On the othwr hand
$$\int|f\phi|^2\,d\mu=\sum_k\frac{1}{k^2\mu(A_{n_k})}\int_{A_k}|\phi|\,d\mu\geq \sum_k1=\infty$$
in contradiction to $\phi\,f\in L_2(\mu)$. This shows that if $M_\phi$ is defined on $L_2(\mu)$, then  it is the case that $\phi\in L_\infty$.
There are other versions of multiplication operators $M_\phi$ where $\phi\notin L_\infty$, and are defined an a smaller domain $D_\phi=\{f\in L_2(\mu): \phi f\in L_2$}.


*In general, if $\phi\in L_\infty$, $\|M_\phi f\|_2\leq \|\phi\|_\infty\|f\|_2$. $\sigma$-finiteness guarantees that $\|M_\phi\|=\|\phi\|_\infty$ (see the aforementioned example 1.5 pp. 28 in Conway's, a course on functional analysis)
