# Showing that an integral operator on $L^p$ spaces has a certain norm

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 $$1 and let $$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.

• @ChristianRemling Can you post an answer giving an example where $K$ is unbounded even where it is defined? – Keshav Srinivasan Feb 3 at 17:48
• @Ian So do you think the statement I’m trying to prove is actually true? – Keshav Srinivasan Feb 3 at 20:38

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