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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.

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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
2  
$\mathcal M$ 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 $\lVert T\rVert\leq\lVert g\rVert_\infty$.

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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 $|g(x)| > \|g\|_\infty - \epsilon$.

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