Integral operator on $L^p$ is compact Let $(X,\Omega,\mu)$ be an arbitrary measure space, $1<p<\infty$ , and $\frac{1}{p}+ \frac{1}{q} = 1$. If $k:X. X\to \Bbb C$ is an $\Omega.\Omega-$ measurable function such that $$M = [\int (\int |k(x,y)|^p d\mu(x))^{\frac{q}{p}} d\mu(y)]^{\frac{1}{q}}< \infty$$ and if $(Kf)(x)=\int k(x,y)f(y)d\mu(y) $, then $K$ is compact operator on $L^p(\mu)$ and $||K||\leq M$. 
I know that on reflexive spaces completely continuous operators are compact. Thus I started to show $K$ is completely continuous, but after a while I got stuck. Please help me. Thanks in advance.
 A: In the following, I will assume that $\mu$ is $\sigma$-finite for
simplicity. One can probably drop that assumption with a bit of extra
work. Also, I do not claim that my proof is one of the simpler ones.
Let us define
$$
\left\Vert k\right\Vert _{L^{p,q}}:=\left(\int\left(\int\left|k\left(x,y\right)\right|^{p}d\mu\left(x\right)\right)^{q/p}\, d\mu\left(y\right)\right)^{1/q}
$$
for $k:X\times X\rightarrow\mathbb{C}$ measurable with respect to
the product $\sigma$-Algebra $\Omega\otimes\Omega$ and note that
the space
$$
L^{p,q}:=\left\{ k:X\times X\rightarrow\mathbb{C}\mid k\text{ measurable w.r.t. }\Omega\otimes\Omega\text{ and }\left\Vert k\right\Vert _{L^{p,q}}<\infty\right\} 
$$
is a normed vector space (with the $L^{p,q}$-norm defined above).
Observe that your assumption exactly means $\left\Vert k\right\Vert _{L^{p,q}}<\infty$.
Let us now define
$$
\Gamma:L^{p,q}\rightarrow\mathcal{B}\left(L^{p},L^{p}\right),k\mapsto K,
$$
where the operator $K$ for the function $k$ is defined as in your
post (see also below).
For $f\in L^{p}$, $g\in L^{q}$, we have
\begin{eqnarray*}
\int\left|g\left(x\right)\right|\int\left|k\left(x,y\right)f\left(y\right)\right|\, dy\, dx & = & \int\left|f\left(y\right)\right|\int\left|g\left(x\right)\right|\left|k\left(x,y\right)\right|\, dx\, dy\\
 & \leq & \int\left|f\left(y\right)\right|\cdot\left\Vert g\right\Vert _{q}\cdot\left(\int\left|k\left(x,y\right)\right|^{p}\, dx\right)^{1/p}\, dy\\
 & \leq & \left\Vert g\right\Vert _{q}\cdot\left\Vert f\right\Vert _{p}\cdot\left(\int\left(\int\left|k\left(x,y\right)\right|^{p}\, dx\right)^{q/p}\, dy\right)^{1/q}\\
 & = & \left\Vert g\right\Vert _{q}\cdot\left\Vert f\right\Vert _{p}\cdot\left\Vert k\right\Vert _{L^{p,q}}<\infty.
\end{eqnarray*}
As this holds for all $g\in L^{q}$, we conclude that $\left(\Gamma k\right)f=\left(x\mapsto\int k\left(x,y\right)f\left(y\right)\, dy\right)\in L^{p}$
for all $f\in L^{p}$ with $\left\Vert \left(\Gamma k\right)f\right\Vert _{p}\leq\left\Vert f\right\Vert _{p}\cdot\left\Vert k\right\Vert _{L^{p,q}}$,
which implies that $\Gamma$ is a bounded linear operator with norm
$\left\Vert \Gamma\right\Vert \leq1$.
Compactness is a more complicated buisness.


EDIT: I just discovered a simplified proof which goes as follows: Let 
$$
k_{n}:=k\cdot\chi_{X_{n}\times X_{n}}\cdot\chi_{\left|k\right|^{-1}\left(\left[0,n\right]\right)},
$$
where $X=\bigcup_{n\in\mathbb{N}}X_{n}$, each $X_{n}$ is of finite
measure and $X_{n}\subset X_{n+1}$ holds for all $n\in\mathbb{N}$. The existence of such a sequence $(X_n)_n$ is ensured by $\sigma$-finiteness of $\mu$, which I assume (as noted above).
Then $k_{n}\left(x,y\right)\to k\left(x,y\right)$ pointwise and $\left|k_{n}\left(x,y\right)\right|\leq\left|k\left(x,y\right)\right|$,
which yields
$$
\left\Vert k_{n}-k\right\Vert _{L^{p,q}}\xrightarrow[n\rightarrow\infty]{}0
$$
by dominated convergence. This implies $\Gamma k_{n}\to\Gamma k$
with respect to the operator norm, so that it suffices to establish
compactness of $\Gamma k_{n}$ (because the set of compact operators is closed w.r.t. the convergence in operator norm). This means that we can assume w.l.o.g.
that $k$ is bounded by $L>0$ and vanishes outside some set $Y\times Y$
where $Y$ is of finite measure.
Now let $\left(f_{n}\right)_{n\in\mathbb{N}}$ be a sequence in $L^{p}$
with $f_{n}\rightharpoonup f$. For each $x$, we have $\left|k\left(x,y\right)\right|\leq L\cdot\chi_{Y}\left(y\right)\in L^{q}\left(\mu\right)$.
This implies
$$
\left(\left(\Gamma k\right)f_{n}\right)\left(x\right)=\int k\left(x,y\right)\cdot f_{n}\left(y\right)\, d\mu\left(y\right)\xrightarrow[n\rightarrow\infty]{}\int k\left(x,y\right)\cdot f\left(y\right)\, d\mu\left(y\right)=\left(\left(\Gamma k\right)f\right)\left(x\right).
$$
Furthermore,
$$
\left|\left(\left(\Gamma k\right)f_{n}\right)\left(x\right)\right|\leq\int\left|k\left(x,y\right)\right|\cdot\left|f_{n}\left(y\right)\right|\, d\mu\left(y\right)\leq L\cdot\left\Vert \chi_{Y}\right\Vert _{q}\cdot\left\Vert f_{n}\right\Vert _{p}\leq C<\infty,
$$
because the $L^{p}$-norm of a weakly convergent sequence is bounded.
But also $\left(\left(\Gamma k\right)f_{n}\right)\left(x\right)=0$
for $x\notin Y$, because of $k\left(x,y\right)=0$ for $x\notin Y$.
Hence,
$$
\left|\left(\left(\Gamma k\right)f_{n}\right)\left(x\right)\right|\leq C\cdot\chi_{Y}\left(x\right)\in L^{p}\left(\mu\right).
$$
By dominated convergence, this yields
$$
\left\Vert \left(\Gamma k\right)f_{n}-\left(\Gamma k\right)f\right\Vert _{p}\xrightarrow[n\rightarrow\infty]{}0,
$$
so that $\Gamma k$ is a compact operator, because $L^{p}\left(\mu\right)$
is reflexive and $\Gamma k$ is completely continuous.


My original (more complicated) proof was the following:
First assume that $k=\sum_{i=1}^{n}\alpha_{i}\cdot\chi_{A_{i}\times B_{i}}$
with $A_{i},B_{i}\in\Omega$ of finite measure. This implies
$$
\left(\left(\Gamma k\right)f\right)\left(x\right)=\sum_{i=1}^{n}\alpha_{i}\cdot\int\chi_{A_{i}\times B_{i}}\left(x,y\right)\cdot f\left(y\right)\, dy=\sum_{i=1}^{n}\alpha_{i}\int_{B_{i}}f\left(y\right)\, dy\cdot\chi_{A_{i}}\left(x\right)
$$
and hence $\left(\Gamma k\right)f\in\left\langle \chi_{A_{1}},\dots,\chi_{A_{n}}\right\rangle $
for all(!) $f\in L^{p}$, so that the image of $\Gamma k$ is finite
dimensional. It is then easy to see that $\Gamma k$ is a compact
operator.
Finally, it is well-known that the set of compact operators is closed
with respect to the operator norm. Hence, it suffices to show that
functions of the form $k=\sum_{i=1}^{n}\alpha_{i}\cdot\chi_{A_{i}\times B_{i}}$
with $A_{i},B_{i}\in\Omega$ of finite measure are dense in $L^{p,q}$.
To this end, we use the following
Lemma: For every $M\in\Omega\otimes\Omega$ with $\chi_{M}\in L^{p,q}$
and $\varepsilon>0$, we have $\left\Vert \chi_{M}-\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}\times B_{i}}\right\Vert _{L^{p,q}}<\varepsilon$
for suitable $n\in\mathbb{N}$, $\alpha_{1},\dots,\alpha_{n}\in\mathbb{R}$
and $A_{1},\dots,A_{n},B_{1},\dots,B_{n}\in\Omega$ of finite measure.
Proof: Assume for the moment that $\mu\left(X\right)<\infty$. Now,
let
$$
V:=\left\langle \chi_{A\times B}\mid A,B\in\Omega\text{ with }\mu\left(A\right),\mu\left(B\right)<\infty\right\rangle ,
$$
the vector space spanned by all characteristic functions of "rectangular"
sets and $W:=\overline{V}\leq L^{p,q}$, where the closure is taken
with respect to the $L^{p,q}$-norm defined above. Set
$$
\mathcal{G}:=\left\{ M\in\Omega\otimes\Omega\,\mid\,\chi_{A}\in W\right\} .
$$
We will show that $\mathcal{G}$ is a $\lambda$-system. As it trivially
contains the $\pi$-system $\left\{ A\times B\mid A,B\in\Omega\right\} $
of all "rectangular" sets $A\times B$ with $A,B\in\Omega$ (in
particular, $\emptyset$ and $X\times X$), this will establish $\mathcal{G}=\Omega\otimes\Omega$.
Let $M,N\in\mathcal{G}$ with $N\subset M$ and $\varepsilon>0$.
Then there are $f,g\in V$ with $\left\Vert \chi_{M}-f\right\Vert _{L^{p,q}}<\frac{\varepsilon}{2}$
and $\left\Vert \chi_{N}-g\right\Vert _{L^{p,q}}<\frac{\varepsilon}{2}$.
As $V$ is a vector space, $f-g\in V$. But $N\subset M$ implies
$\chi_{M\setminus N}=\chi_{M}-\chi_{N}$ and hence 
$$
\left|\chi_{M\setminus N}-\left(f-g\right)\right|=\left|\left(\chi_{M}-f\right)+\left(g-\chi_{N}\right)\right|\leq\left|\chi_{M}-f\right|+\left|\chi_{N}-g\right|,
$$
which entails $\left\Vert \chi_{M\setminus N}-\left(f-g\right)\right\Vert _{L^{p,q}}\leq\left\Vert \chi_{M}-f\right\Vert _{L^{p,q}}+\left\Vert \chi_{N}-g\right\Vert _{L^{p,q}}<\varepsilon$.
As $\varepsilon>0$ was arbitrary, $\chi_{M\setminus N}\in\overline{V}=W$
and hence $M\setminus N\in\mathcal{G}$.
Now, let $\left(M_{n}\right)_{n\in\mathbb{N}}$ be a sequence in $\mathcal{G}$
with $M_{n}\subset M_{n+1}$ for all $n\in\mathbb{N}$ and $M:=\bigcup_{n\in\mathbb{N}}M_{n}$.
This implies $M-\chi_{M_{i}}\xrightarrow[n\rightarrow\infty]{}0$
pointwise. By dominated convergence (note that we assume $\mu\left(X\right)<\infty$
and that $p,q<\infty$), we conclude
$$
\left\Vert \chi_{M}-\chi_{M_{i}}\right\Vert _{L^{p,q}}\xrightarrow[n\rightarrow\infty]{}0.
$$
But $\chi_{M_{i}}\in W=\overline{V}$, which implies $M\in\overline{\overline{V}}=\overline{V}=W$
and hence $M\in\mathcal{G}$.
By the assumption above, $\mu$ is $\sigma$-finite. Hence, $X=\bigcup_{n\in\mathbb{N}}X_{n}$
with each $X_{n}$ of finite measure and $X_{n}\subset X_{n+1}$ for
all $n\in\mathbb{N}$. By applying the above to $M\cap\left(X_{n}\times X_{n}\right)$
for the restricted measure $\mu|_{X_{n}}$, we see $M\cap\left(X_{n}\times X_{n}\right)\in V$
for all $n\in\mathbb{N}$ and $M\in\Omega\otimes\Omega$ with $\chi_{M}\in L^{p,q}$.
But $\chi_{M\cap\left(X_{n}\times X_{n}\right)}\xrightarrow[n\rightarrow\infty]{}\chi_{M}$
pointwise and 
$$
\left|\chi_{M\cap\left(X_{n}\times X_{n}\right)}-\chi_{M}\right|\leq\chi_{M},
$$
so that dominated convergence yields $\left\Vert \chi_{M\cap\left(X_{n}\times X_{n}\right)}-\chi_{M}\right\Vert _{L^{p,q}}\xrightarrow[n\rightarrow\infty]{}0$.
As above, this implies the claim. $\square$
For the proof of the denseness claimed above, take any $k\in L^{p,q}$.
As $k$ is $\Omega\otimes\Omega$ measurable, there is a sequence
$\left(k_{n}\right)_{n\in\mathbb{N}}$ of the form $k_{n}=\sum_{i=1}^{m_{n}}\alpha_{i}^{\left(n\right)}\chi_{M_{i}^{\left(n\right)}}$
with $M_{i}^{\left(n\right)}\in\Omega\otimes\Omega$ and $\alpha_{i}^{\left(n\right)}\in\mathbb{C}$.
We can take the $\left(M_{i}^{\left(n\right)}\right)_{i=1,\dots,m_{n}}$
to be pairwise disjoint for all $n\in\mathbb{N}$ and $\alpha_{i}^{\left(n\right)}\neq0$.
Furthermore, we can replace $M_{i}^{\left(n\right)}$ by 
$$
M_{i}^{\left(n\right)}\cap\left\{ \left(x,y\right)\mid\left|k_{n}\left(x,y\right)\right|\leq2\left|k\left(x,y\right)\right|\right\} ,
$$
so that $\left|k_{n}\left(x,y\right)\right|\leq2\left|k\left(x,y\right)\right|$
holds for all $\left(x,y\right)$ (check that still $k_{n}\left(x,y\right)\to k\left(x,y\right)$).
By dominated convergence again, this yields $\left\Vert k_{n}-k\right\Vert _{L^{p,q}}\xrightarrow[n\rightarrow\infty]{}0$
and the disjointness of $\left(M_{i}^{\left(n\right)}\right)_{i=1,\dots,m_{n}}$
also implies $\chi_{M_{i}^{\left(n\right)}}\in L^{p,q}$ for all $i,n$,
so that we can approximate the $\chi_{M_{i}^{\left(n\right)}}$ by
finite linear combinations of "rectangular" indicator functions
as above. This completes the proof.
