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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 !

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  • $\begingroup$ 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 $L^1$. $\endgroup$ – Snake707 Jan 9 at 18:17
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    $\begingroup$ @Snake707 cauchy schwarz and then calcul the square of both sides of inequality. $\endgroup$ – Anas BOUALII Jan 9 at 18:48
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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.

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