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Let $X$ be a sigma-finite measure space, and let $k$ be a measurable function on $X\times X$. Suppose that $F(x)=\int |k(x,y)| dy$ and $G(y)=\int |k(x,y)| dx$ are in $L^\infty$. Let $1<p<\infty$ and let $K:L^p\rightarrow L^\infty$ be defined by $K(f)(x)=\int k(x,y) f(y) dy$. Show that $K$ is bounded with operator norm less than or equal to $||F||_\infty^{1/p}||G||_\infty^{1/q}$, where $\frac{1}{p}+\frac{1}{q}=1$.

I’m not sure how to approach this. I need to show that for any $f\in L^p$, we have $||K(f)||_\infty\leq||F||_\infty^{1/p}||G||_\infty^{1/q}||f||_p$. I was thinking of using Holder’s inequality, since that involves $p$ and $q$, but we have an $L_\infty$ norm here rather than an $L_1$ norm.

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  • $\begingroup$ @ChristianRemling Can you post an answer giving an example where $K$ is unbounded even where it is defined? $\endgroup$ – Keshav Srinivasan Feb 3 at 17:48
  • $\begingroup$ @Ian So do you think the statement I’m trying to prove is actually true? $\endgroup$ – Keshav Srinivasan Feb 3 at 20:38
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This is a corrected version of my comment. The claim isn't true (what is true is that the operator is bounded $L^p\to L^p$ by Schur's test). We can take $k(x,y)=|x-y|^{-3/4}$, $f(y)=y^{-3/4}$ on $X=(0,1)$ with Lebesgue measure. Then $$ (Kf)(x)\ge x^{-3/4} \int_0^x y^{-3/4}\, dy = 4x^{-1/2} . $$

This shows that the operator need not map to $L^{\infty}$ for $p<4/3$, but of course we can handle any $p<\infty$ in this way by slightly adapting the details.

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  • $\begingroup$ But if you restrict things to functions where $K$ does map to $L^\infty$, will it be a bounded operator? $\endgroup$ – Keshav Srinivasan Feb 3 at 23:33
  • $\begingroup$ @KeshavSrinivasan: No. Just take $f_n(y)=\chi_{(1/n,1)}(y)y^{-3/4}$ to see this. $\endgroup$ – user138530 Feb 4 at 16:52

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