Let $M_f: L^p(I) \to L^p(I)$ be the multiplication operator, i.e. that sends $g \to f \cdot g$. $I$ is assumed to be a connected subset of the real line (not necessarily with finite measure), $f$ to be an element of $L^{\infty}(I)$ and $p < \infty$.
Suppose $\inf\limits_{x \in I} |f(x)| = c > 0$. Then any $h \in L^p(I)$ has a preimage under this map, since it is given by $\frac{h(x)}{f(x)}$ and $$\int_I \frac{|h(x)|^p}{|f(x)|^p} dx \leq \frac{1}{c^p} \int_I |h(x)|^p dx < \infty$$
That is, if $f$ is bounded away from zero then $M_f$ is surjective. I have been trying pretty hard to prove the converse, which I believe is true but without any luck so far.
Could you help me prove surjectivity of $M_f$ implies that $f$ is bounded away from zero? Any hints are very welcome. Thank you.
Hint: Let $A_n=\{x: \frac 1{n+1} \leq |f(x)| <\frac 1 n\}$. Suppose $\mu (A_n) >0$ for all $n$. Let $g=\sum a_{n}\chi_{A_{n}}$ where $a_{n}=n^{-1-1/p}(\mu(A_{n}))^{-1/p}$. Check that $g \in L^{p}(I)$. If $g=fh$ for some $h$ then $|h| \geq n|a_n|$ in $A_n$ Use this to check that $h \notin L^{p}(I)$. So $M_f$ is not surjective.
We have proved that if $M_f$ is surjective then $\mu (A_n) >0$ for all $n$ is not possible. Modify this argument to show that $\mu (A_n) >0$ for infinitely many $n$ is also not possible. Now it follows that $|f| \geq \frac 1n$ almost everywhere for some $n$.