Bound of norm of the operator $T(f)=fg$ on $L^p$ space

Question

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.

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-Álvarez, Mar 07 '12 at 07:10
$\mathcal M$ might be the set of chairs in Asia, for all you know... — Mariano Suárez-Álvarez, Mar 07 '12 at 07:19

score 3 · Answer 1 · edited Apr 08 '15 at 05:37

3

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.

edited Apr 08 '15 at 05:37

Community

1

answered Mar 07 '12 at 07:12

Mariano Suárez-Álvarez

135,076

first part is fine.....how about the semifinite part? – user24367 Mar 07 '12 at 07:18
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-Álvarez Mar 07 '12 at 07:20
Try approximation. – abatkai Mar 07 '12 at 07:32
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 it – Wolfy Apr 26 '16 at 23:49

score 3 · Answer 2 · answered Mar 07 '12 at 07:46

3

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

answered Mar 07 '12 at 07:46

Robert Israel

448,999

Where would we go from here by your hint? – Wolfy Apr 26 '16 at 23:38
Given $\epsilon > 0$, take such a set $A$ and let $f$ be the indicator function of $A$. Then $|Tf|p \ge (|g|\infty - \epsilon) |f|_p$. – Robert Israel Apr 27 '16 at 00:07
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? – Wolfy Apr 27 '16 at 00:17

score 0 · Answer 3 · answered Jul 23 '17 at 22:07

0

The last inequality from Robert Israel implies that for every $\epsilon > 0$, $\|T\| \ge \|g\|_\infty - \epsilon$, and consequently $\|T\| \ge \|g\|_\infty$.

answered Jul 23 '17 at 22:07

xidil

1

1

should this be in the comment instead? – Siong Thye Goh Jul 23 '17 at 22:37
Yes, but I had no rep so was unable to post a comment. – xidil Jul 24 '17 at 00:11

Bound of norm of the operator $T(f)=fg$ on $L^p$ space

3 Answers3