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Let $(X, \mathcal{S}, \mu)$ be a finite measure space, $1<p<\infty$, and $1/p+1/q=1$. If $k: X\times X\to\mathbb{F}$ is an $\mathcal{S}\times\mathcal{S}$ measurable function such that \begin{align*} \sup\bigg\{\int_X|k(x, y)|^q\,d\mu(y):\,x\in X\bigg\}<\infty, \end{align*} then \begin{align*} (Kf)(x)=\int_X k(x, y)f(y)\,d\mu(y) \end{align*} defines a compact operator on $L^p(\mu)$.

I have seen the proof of this result but my hypothesis is a little different and my background in topology is pretty weak. I would like to show that $Kf$ is completely continuous so that I can use the result that any completely continuous operator on a reflexive space is compact. I am wondering if someone wouldn't mind taking a "heres why $Kf$ is completely continuous for dummies" approach. Thanks in advance.

The Mad Scientist
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1 Answers1

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Let $$c= \sup\bigg\{\int_X|k(x, y)|^q\,d\mu(y):\,x\in X\bigg\}.$$ Suppose that $f_n\to0$ weakly. This means that for every $g\in L^q$, $$ \int_X f_n\overline g\,d\mu\to0. $$ In particular, for each $x$ $$ \int_Xk(x,y)\,f_n(y)\,d\mu(y)\to0 $$ Since a weakly convergent sequence is bounded, there exists $b>0$ with $\|f_n\|_p<b$ for all $n$. Then, by Hölder, $$ \bigg|\int_Xk(x,y)\,f_n(y)\,d\mu(y)\bigg|\leq\Bigg(\int_X|k(x,y)|^q\,d\mu(y)\bigg)^{1/q}\|f_n\|_p<bc^{1/q}. $$ Because we are in a finite-measure space, a bounded function is integrable. Then, using Dominated Covergence, \begin{align} \|Kf_n\|_p^p&=\int_X\bigg|\int_X k(x,y)\,f_n(y)\,d\mu(y)\bigg|^p\,d\mu(x)\to0. \end{align} so $K$ is completely continuous, and as $L^p$ is reflexive, $K$ is compact.

Martin Argerami
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