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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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    $\mathcal M$ might be the set of chairs in Asia, for all you know...2012-03-07

3 Answers 3

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

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    Trying to do the semifinite part as well but I am not really sure what to do. Any suggestions, I read the hint above but not sure how to apply it2016-04-26
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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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    I see so by the last inequality you have and what we prove for the first part then we have equality or is there a step I am missing in between there?2016-04-27
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The last inequality from Robert Israel implies that for every $\epsilon > 0$, $\|T\| \ge \|g\|_\infty - \epsilon$, and consequently $\|T\| \ge \|g\|_\infty$.

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    Yes, but I had no rep so was unable to post a comment.2017-07-24