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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.

Barreto
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    Maybe a proof by contraposition should work? Assume for simplicity $f \geq 0$ and that $f$ is zero on a non-null set with finite measure denoted by $M$. Then $\chi_M$ is in $L^p(I)$ but it can never be reached by $M_f$ since $M_f g \equiv 0$ on $M$. Does that work? – hal4math Mar 12 '22 at 22:04
  • Well, almost :) Note that $f$ not being bounded away from zero does not automatically imply that $f$ is 0 in a set with positive measure. @hal4math – Barreto Mar 12 '22 at 22:11
  • $\inf\limits_{x \in I} |f(x)| = c > 0$ does not even make sense. Elements of $L^{p}(I)$ are not ordinary functions. They are equivalence classes of functions. – Kavi Rama Murthy Mar 12 '22 at 23:30
  • I mean that for any set $A$ of full measure $\inf\limits_{x \in A} |f(x)| >c$. Does it make sense now? @KaviRamaMurthy – Barreto Mar 12 '22 at 23:34
  • No, it still doesn't make sense. – Kavi Rama Murthy Mar 12 '22 at 23:35
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    I think the essential infimum is meant. So the largest of all $c > 0$ such that $|f(x)| > c$ for almost every $x \in I$. @KaviRamaMurthy – hal4math Mar 12 '22 at 23:38
  • Why is there a 'complex analysis' tag? – Kavi Rama Murthy Mar 12 '22 at 23:39
  • Could you help me with the problem? @KaviRamaMurthy – Barreto Mar 12 '22 at 23:42
  • @hal4math Exactly, isn't what I wrote in the comment the essential infimum? – Barreto Mar 12 '22 at 23:42

1 Answers1

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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 $n$.

Kavi Rama Murthy
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