Showing that the range is dense Take $E \subseteq \mathbb{R}^n$. Let $M_\phi$ be the multiplication operator, then fix $\phi \in L^\infty(E)$ and let $M_\phi:L^p(E) \rightarrow L^p(E)$ such that $M_\phi f(x)=\phi(x) f(x)$. I want to show that the range of $M_\phi$ is dense in $L^p(E)$ whenever $M_\phi$ is injective. Let $g \in L^p(E)$, the goal is construct $g_n \phi$ such that $\phi g_n \rightarrow g$ in the $L^p(E)$ norm. The thing is I have no idea how to proced, I know that since $M_\phi$ is injective this implies that $\phi \neq 0$ a.s., but how can I use this to show that the range is dense?
Please see my partial answer below.
 A: Here is what I have so far based on @CameronWilliams great suggestion. Let $g \in L^p(E)$ and construct the functions $g_n = \chi_{\{x:|\phi(x)| \geq n\}} \frac{g}{\phi}$. First, we ensure that this do indeed belong to $L^p(E)$. We have $$\int_{E} |g_n(x)|^p \, dx = \int_{\{x:|\phi(x)| \geq n\}} \left|\frac{g(x)}{\phi(x)}\right|^p \leq \frac{1}{n^p} \|g\|_p^p < \infty,$$ for every $n \geq 1$. Now that we are sure we are in $L^p(E)$ we consider the image of these functions under $M_\phi$. We must show that $M_\phi(g_n) \rightarrow g$ in the $L^p(E)$-norm. To do this, we have
$$\int_E |M_\phi(g_n(x))-g(x)|^p \, dx = \int_{E} |\chi_{\{x:|\phi(x)| \geq n\}}g(x)-g(x)|^p \, dx= \int_{\{x:|\phi(x)|<n\}} |g(x)|^p\,dx.$$ Now from here how can I show that this integral can be made as small as I want? I know that since $|g|^p$ is integrable, we can make $\int_X |g|^p < \epsilon$ when $m(X) < \delta$, where $m$ denotes the Lebesgue measure. But how can I ensure that my set $\{x:|\phi(x)| < n\}$ can have measure measure less than $\delta$?
A: For $1 \leq p < \infty$, you can proceed as follows :
Let us assume that $image(M_{\phi})$ is not dense in $L^p(E)$. Hence there exists $0 \neq v \notin \overline{image(M_{\phi})} \subset L^p(E)$. Consider a linear functional $\lambda$ that is defined on $\overline{image(M_{\phi})} + \mathbb{R}v$, such that $\lambda(\overline{image(M_{\phi})}) = 0$ and $\lambda(v) = 1$. Now, since $L^p(E)$ is a normed space and hence a locally convex topological vector space one can extend $\lambda$ to continuous linear functional on $L^p(E)$ and will again be denoted by $\lambda$. See : https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem#In_locally_convex_spaces. Now, we know that any such linear functional must be of the form $L^p(E) \ni f \mapsto \int_E f(x)g(x)dx$ for some $g \in L^q(E)$, where $q = p/(p-1)$. Thus we get following equality
$$
\int \phi(x)f(x)g(x) = 0       \quad \qquad \forall f \in L^p(E)
$$
Since $\phi \in L^{\infty}(E)$, we get $\phi.g \in L^q(E)$. Using again the fact that $L^q(E)^* = L^p(E)$, we get that $\phi.g = 0$. But(as you have also observed) $\phi$ is nonzero a.e., hence $g = 0$ a.e.. This implies that $\lambda = 0$, a contradiction.
Hence the assumption that $image(M_{\phi})$ is not dense in $L^p(E)$ is false.
I am not sure if this argument goes through for $p = \infty$.
Update : added on August 28.
It is enough to prove that there exists $\phi_n$ such that $M_{\phi}\phi_n \to 1$ in $L^{\infty}(E)$. For any other $g \in L^{\infty}(E)$, then the sequence $\phi_ng$ satisfies : $M_{\phi}(\phi_ng) \to g$.
Consider the set
$$
E_n = \lbrace x \in E | \phi(x) > 1/n \rbrace
$$
We break further analysis in two parts

*

*$E_N \underset{\mu}{=} \emptyset$ for some $N$. Then it is clear that the function $\phi^{-1}$ makes sense and $M_{\phi}(\phi^{-1}) = 1$.


*$\mu(E_n) \neq = 0$ for any $n$. Observe that $\mu(E_n) \to 0$. Define the function $\phi_n$ as follows : $\phi_n|_{E_n} = \phi|_{E_n}$ and $\phi_n|_{E\setminus E_n} \equiv 1/n$. Clearly $\phi^{-1}_n$ can be defined. It follows from construction that $\|M_{\phi}(\phi_n)-1\|_{L^{\infty}(E)} \leq 1/n$. Hence $M_{\phi}(\phi_n) \to 1$.
