TITLE: Why is $A$ a compact operator?
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: 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 $T$.