Let $\displaystyle U_1 \subset \mathbb R^{n_1}$ and $\displaystyle U_2 \subset \mathbb R^{n_2} $ measurable sets, $\displaystyle 1 < p,q < \infty $ and consider the measurable function $\displaystyle K:U_1 \times U_2 \to \mathbb R $ with $$\displaystyle \|K\|= \left( \int_{U_1} \Big( \int_{U_2} |\,K(x,y) |^{p^{'}} dy \Big)^{q/p^{'}} dx \right)^{1/q} < \infty ,$$ where $\displaystyle \frac{1}{p} + \frac{1}{p^{'}} =1$.
Prove that the operator $T:\displaystyle L^p (U_2) \to L^q (U_1) $, with $\displaystyle (Tf)(x)= \int_{U_2} K(x,y) f(y) dy $, is compact.
I tried to prove it by the definition of the compact operator but I didn't made it. Is there some other way to do it?
Hints.
$K(x,y)$ can be approximated in its operator norm by sums of the form $\sum_{i=1}^N a_i(x)b_i(y)$, where $a_i\in L^q(U_1)$ and $b_i\in L^{p'}(U_2)$.
If $K(x,y)$ is replaced by $a_i(x)b_i(x)$, where $a_i\in L^q(U_1)$ and $b_i\in L^{p'}(U_2)$, then the resulting operator is of rank $1$, and hence compact, while if $K$ is replaced by $K_N(x,y)=\sum_{i=1}^N a_i(x)b_i(y)$, where $a_i\in L^q(U_1)$ and $b_i\in L^{p'}(U_2)$, then resulting operator is of rank $N$, and hence again compact. If $\|K_N-K\|\to 0$, then $K$ should be compact as the limit of compact operators.