Compact operator on $L^2[0,1]^2$

Let $$K\in L^2([0,1]\times[0,1])$$, and we define the operator $$T_k$$ on $$L^2[0,1]$$.

$$(T_kf)(x)=\int_{0}^{1}K(x,y).f(y).dy \quad \quad \forall f\in L^2[0,1]$$ How to prove that $$T_k$$ is a compact operator on $$L^2[0,1]$$. I thought of Ascoli's theorm, to prove that $$T_k(B_{L^2}(0,1))$$ is relatively compact. But we donc have the continuity !

• Out of curiosity? Do we even have that $T_kf$ is in $L^2$? With Hölder's Inequality I only get that it is in $L^1$. – Snake707 Jan 9 at 18:17
• @Snake707 cauchy schwarz and then calcul the square of both sides of inequality. – Anas BOUALII Jan 9 at 18:48

Hint: (The operator $$T_k$$ is called a Hilbert-Schmidt operator.) First show that for an orthonormal basis $$(e_n)_{n=1}^\infty$$ of $$L^2([0,1])$$, $$\sum_{n=1}^\infty \|T_k e_n\|^2 = \|K(\cdot,\cdot)\|^2_{L^2([0,1]^2)}<\infty$$ holds. Let $$P_N$$ be the orthogonal projection onto the subspace spanned by $$(e_n)_{n=1}^N$$. Finally prove that $$\lim_{N\to\infty}\|T_k - T_k P_N\|=0$$ and deduce the conclusion from the fact that $$T_kP_N$$ is a compact (finite rank) operator.