Here is a homework excercise.

Let $(X,\Omega,\mu)$ be a $\sigma$-finite measure space,$1\leq p <\infty.$ and suppose that $k:X\times X\rightarrow \mathbb{C}$ is an $\Omega \times \Omega$ measurable function such that for $f\in L^p(\mu)$ we have $k(x,\cdot)f(\cdot)\in L^1(\mu),a.e.x$ and $(Kf)(x)=\int k(x,y)f(y)d{\mu(y)}$ defines an element $Kf$ of $L^p(\mu)$.

Show that $K:L^p(\mu)\rightarrow L^p(\mu)$ is a bounded operator.

I think we can use the Closed Graph Theorem. suppose that $f_n\rightarrow 0,Kf_n\rightarrow g$ in $L^p$,we only need to prove $g=0$ in $L^p$. Since $Kf_n\rightarrow g$ in $L^p$, then without loss of generality, we can get $Kf_n(x)\rightarrow g(x)$ a.e. Then I want to show $Kf_n(x)\rightarrow 0$ a.e. using $f_n\rightarrow 0$ in $L^p$. But I need to prove that $k(x,\cdot)\in L^q(\mu)$ a.e. $x$ ($1/p+1/q=1$).

How to prove $k(x,\cdot)\in L^q(\mu)$ a.e. $x$?

  • $\begingroup$ I don't think that what you say suffices actually suffices. In any case, that's not what CGT says - to apply CGT you can assume $f_n\to f$ and $Kf_n\to g$ and show $Kf=g$. $\endgroup$ – David C. Ullrich May 18 '16 at 14:16
  • $\begingroup$ @DavidC.Ullrich It suffices to look at $f_n \to 0$ when $K$ is globally defined. $K(f_n - f) \to g - Kf$. $\endgroup$ – Daniel Fischer May 18 '16 at 14:55
  • $\begingroup$ @DanielFischer Well that's sort of obvious. I just assumed it was wrong without thinking about it, since I don't think I've ever seen this pointed out before. $\endgroup$ – David C. Ullrich May 18 '16 at 15:14
  • $\begingroup$ Is there any progression? $\endgroup$ – izimath Apr 23 at 4:00

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