Multiplication operators on $L^2$ Let $X$ be a $\sigma$-finite measure space, and let $g$ a measurable complex-valued function $X$, which lies in $L^\infty(X)$. I would like to determine sufficient and necessary properties for the operator $T:L^2(X)\rightarrow L^2(X)$ by $f\mapsto gf$ to be self-adjoint, an isometry (i.e. inner-product preserving), to be surjective, and to be injective. I feel that it is self-adjoint iff $g$ is real-valued, an isometry iff it takes values in the unit circle, surjective or injective iff it is constant. Is this true?
 A: This should be a comment, but it is too long.
Suppose $g$ is not essentially bounded from below (i.e., for every $c>0$, the set $\left\{x\in X : \left|g(x)\right|\leq c\right\}$ has positive measure). If $g=0$ is on a set of positive measure $E$, then since $X$ is $\sigma$-finite, we may assume that $\mu(E)<\infty$. Then $\chi_{E}\in L^{2}(X)$, but $Tf\neq\chi_{E}$ for any $f\in L^{2}(X)$, since $gf=0$ for a.e. $x\in E$. So assume that $\left|g\right|>0$ a.e.
Using our hypotheses together with the $\sigma$-additivity of measure, we may find a subsequence $n_{k}$ of positive integers such that the measurable set
$$E_{k}:=\left\{x\in X:2^{-(n_{k}+1)}\leq\left|g(x)\right|<2^{-n_{k}}\right\},\qquad k\in\mathbb{N}$$
has positive measure. Appealing to $\sigma$-additivity and replacing $E_{k}$ with a smaller subset, w.l.o.g. assume that $\mu(E_{k})<\infty$. Define a measurable function $h$ on $X$ by
$$h:=\sum_{k=1}^{\infty}\dfrac{2^{-n_{k}}}{(\mu(E_{k}))^{1/2}}\chi_{E_{k}}$$
$h\in L^{2}(X)$, as
$$\int_{X}\left|h\right|^{2}\mathrm{d}\mu=\sum_{k=1}^{\infty}\int_{E_{k}}\dfrac{2^{-2n_{k}}}{\mu(E_{k})}\mathrm{d}\mu=\sum_{k}2^{-2n_{k}}<\infty$$
I claim that $h\notin T(L^{2}(X))$. Otherwise, $f=g^{-1}h$ a.e., but
$$\int_{X}\left|g^{-1}h\right|^{2}\mathrm{d}\mu\geq\sum_{k}\dfrac{2^{-2n_{k}}}{\mu(E_{k})}\int_{E_{k}}\left|g^{-1}\right|^{2}\mathrm{d}\mu\geq\sum_{k}\dfrac{2^{-2n_{k}}}{\mu(E_{k})}\int_{E_{k}}2^{2(n_{k}+1)}=\sum_{k}4=\infty,$$
which is a contradiction.
