Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that: if $g \in L^{\infty}$, the operator $T$ defined by $Tf = fg$ is bounded on $L^{p}$ for $1\leq p\leq \infty$. Its operator norm is at most $||g||_{\infty}$, with equality if $\mu$ is semifinite, where $\mu$ is the measure on $\mathcal{M}$, the measure space.

My approach: I consider $\hat{g(x)}(f) = f(g(x))$, which is a linear operator on $L^{p}$. Clearly $||\hat{g(x)}|| = ||g(x) || \leq ||g||_{\infty}$ , which gives me the first part of the proof. I am clueless about the second part, involving semifinite measure.

share|improve this question
How are you computing $f(g(x))$? $g$ takes values in $\mathbb R$ (or $\mathbb C$, I guess) while $f$ is defined on $\mathcal M$... –  Mariano Suárez-Alvarez Mar 7 '12 at 7:10
$g$ is in $L^{\infty}$ –  user24367 Mar 7 '12 at 7:13
$\mathcal M$ might be the set of chairs in Asia, for all you know... –  Mariano Suárez-Alvarez Mar 7 '12 at 7:19

2 Answers 2

If $f\in L^p$, then $$\lVert{Tf}\rVert_p^p=\int|fg|^p=\int|f|^p|g|^p\leq\lVert g\rVert_\infty^p\int|f|^p=\lVert g\rVert_\infty^p \lVert g\rVert_p^p,$$ because $|g|\leq\lVert g\rVert_\infty$, so $\lVert Tf\rVert_p\leq\lVert g\rVert_\infty\lVert f\rVert_p$ and therefore $\lVert T\rVert\leq\lVert g\rVert_\infty$.

share|improve this answer
first part is fine.....how about the semifinite part? –  user24367 Mar 7 '12 at 7:18
Well, now that you see how to do the first part (your approach did not make any sense), you might try to do something with the second part yourself :) –  Mariano Suárez-Alvarez Mar 7 '12 at 7:20
Try approximation. –  abatkai Mar 7 '12 at 7:32

Hint for the second part: you need $\mu$ to be semifinite so that for any $\epsilon > 0$, there is a set $A$ with $0 < \mu(A) < \infty$ on which $|g(x)| > \|g\|_\infty - \epsilon$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.