Why is $A$ a compact operator?

Question

Let $X$ be a compact space and let $\mu$ be a positive Borel measure on X. Let $T\in \mathscr{B}(L^p(\mu),C(X))$ where $1\lt p \lt \infty$.

Show that if $A:L^p(\mu)\rightarrow L^p(\mu)$ defined by $Af=Tf$, then $A$ is a compact operator.

Here is an example: let $X=[0,1]$,$\mu$ is the lebesgue measure on $[0,1]$, p=2.

$\forall f\in L^2(0,1)$, define $T(f)=\int_0^1{f(x)e^{-ixy}}dx,y\in[0,1]$. Then $$A:L^2(0,1)\rightarrow L^2(0,1)$$ $$f(x)\rightarrow \int_0^1{f(x)e^{-ixy}}dx, y\in[0,1]$$ is a compact operator.(this is a integral operator)

Any help would be appreciated!

Answer 1

You need to assume that $\mu$ is finite, otherwise take $\mu$ the Lebesgue measure on $\mathbb{R}$, let $X$ be the one-point compactification of $\mathbb{R}$ and (identifying $L^p(X)$ with $L^p(\mathbb{R})$) let $Af:=f*\rho$, where $\rho\in C^\infty_c(\mathbb{R})$ and $\int\rho=1$. Then $A$ maps $L^p(X)\to C(X)$ continuously (we use the fact that $f*\rho$ vanishes at infinity) but it is not compact from $L^p(X)$ to $L^p(X)$ (why?).

So let us assume $\mu(X)<\infty$. Let $(f_n)\subseteq L^p(\mu)$ be a bounded sequence. Since $L^p(\mu)$ is reflexive, we have $f_n\rightharpoonup f$ for some $f\in L^p(X)$ (up to subsequences), so we also have $Af_n\rightharpoonup Af$. Now for any $x\in X$ $$ (Af_n)(x)\to (Af)(x), $$ since evaluation at a point is a continuous functional on $C(X)$ and $Af_n$ is converging weakly to $Af$. Moreover $\|Af_n\|_\infty\le\|A\|\|f_n\|_p$, so we can apply the dominated convergence theorem to conclude that $$ \int_X |Af_n-Af|^p\,d\mu\to 0 $$ as $n\to\infty$ (here we are using the finiteness of $\mu$). So $Tf_n\to Tf$ in $L^p(X)$, proving the compactness of $T$.

Answer 2

I take it you are assuming that $\mu(X)<\infty$; I don't think it works otherwise.

In this case, compose $T$ with point evaluation at $x\in X$ to obtain the bounded linear functional on $L^p$ that sends $f\mapsto (Tf)(x)$. This shows that there exists $K(x,\cdot)\in L^q$, $1/p+1/q=1$, with $\|K(x,\cdot)\|_q\le \|T\|$, such that $$ (Tf)(x)= \int K(x,t)f(t)\, d\mu(t) . $$ Now for $p=2$ everything is clear: $K\in L^2(X\times X)$, so the operator is Hilbert-Schmidt and in particular compact. For general $1<p<\infty$, a similar criterion is available; see the remark at the bottom of pg. 7 of these notes.