TITLE: Prove that this integral operator is compact
QUESTION [20 upvotes]: Let $X,Y=L^2(0,1)$, $k\in C^0([0,1]^2)$. Define
$$
K:X\to Y,\,\,\,\,\,Kf(x):=\int_0^1k(x,y)f(y)dy\,\,\,\,\forall\, f\in L^2(0,1).
$$
I have to show that $K$ is compact. 
My idea is to prove that $K$ is the limit of finite-rank operators. But I don't know exactly which kind of operators should I have to consider.

REPLY [5 votes]: Actually, if $k\in L^2(0,1)^2$, then you may consider $K$ as a limit of a sequence of finite rank operators. Since $L^2(0,1)$ is a separable Hilbert space, there exists a Hilbert basis denoted by $\{e_i\}_{i=1}^{\infty}$. Define $T_n (n\in\mathbb{N}^{+})$ as
$$
T_nf:=\sum\limits_{i=1}^{n}\langle f, e_i \rangle e_i,\quad \forall f\in L^{2}(0,1).
$$ 
Then, $K_n:=T_nKT_n$ is a finite rank operator. And $K_n$ is compact for each $n$. You need to show that $\lim\limits_{n\rightarrow \infty}||K_n-K||=0$. Then, you conclude that $K$ is compact, since the space of compact operators mapping $L^2(0,1)$ into $L^2(0,1)$ is closed.