Multiplication operator is injective if and only if its range is dense I was working on couple of questions regarding multiplication operators and I couldn't figure out how to proceed for this one.
Let $f \in L^{\infty} ([a,b])$ be continuous and consider the multiplication operator $M_f$ defined as $M_f : L^{2} ([a,b]) \to L^{2} ([a,b])$.
$M_f(h) = fh$ for $h \in L^2 ([a,b])$.   I want to show that $M_f$ is injective if and only if $R(M_f)$ is dense in $L^2([a,b])$.
We know that $M_f$ is injective if and only if $f \neq 0$ a.e. in $[a,b]$. I have managed to show this.
I believe showing that  $f \neq 0$ a.e. in $[a,b]$ if and only if $R(M_f)$ is dense in $L^2([a,b])$ is a good way to approach this problem but I couldn't figure out how to proceed from here. I would appreciate any help!
 A: 
Lemma: Let $T\in B(X)$ be a bounded operator on a Banach space $X$. Then $T$ is invertible if and only if it has dense range and it is bounded below.

Let $f,g\in C[a,b]$ be continuous functions. Then it is easily verified that $M_f\cdot M_g=M_{fg}$.
Now if $f\neq0$ a.e. (which means $f\neq0$ everywhere, since $f$ is continuous), we can define $g=\frac{1}{f}$ and this is a continuous function on $[a,b]$. Note that $fg=1$, so $M_fM_g=I=M_gM_f$, i.e. $M_f$ is invertible and by the above lemma $M_f$ has dense range.
Conversely assume that $M_f$ has dense range and set $E=\{t\in[a,b]:f(t)=0\}$. Then for the characteristic function we have that $\chi_E\in L^2[a,b]$, so there exists a sequence $(g_n)\subset L^2[a,b]$ so that $f\cdot g_n\xrightarrow{\|\cdot\|_2}\chi_E$, so
$$\lim_{n\to\infty}\int_a^b|\chi_E(t)-f(t)g_n(t)|^2dt=0$$
But we have that
$\int_a^b|\chi_E(t)-f(t)g_n(t)|^2dt=\int_E|1-f(t)g_n(t)|^2dt+\int_{E^c}|f(t)g_n(t)|^2=\mu(E)+\int_{E^c}|f(t)g_n(t)|^2dt\geq\mu(E)$
where we used the fact that $f(t)=0$ for all $t\in E$ in the middle equality. This shows that $\mu(E)=0$, so $f\neq0$ almost everywhere and since $f$ is continuous this shows that $f\neq0$ everywhere.
Note to OP: Please do not delete your questions after someone gives an answer. It is unfair to the effort put in by other users.
A: Three observations:
1.$M^*_f=M_{\overline{f}}$. Here $*$ means conjugation of operator, $\overline{f}$ means conjugation of function $f$.
2.$\textrm{Ran}^{\perp}M=\textrm{Ker}M^*$
3.A subspace of a Hilbert space is dense if and only if it’s orthogonal complement is zero.
Together with above, we see $$\textrm{Ran}^{\perp}M_f=\text{Ker}M_{\overline{f}}.$$
So $\textrm{Ran}M_f$ is dense if and only if $\textrm{Ker}M_{\overline{f}}=0$, this is equivalent to $\overline{f}\neq 0$ a.e. by your attempt.
