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 !

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Snake707 cauchy schwarz and then calcul the square of both sides of inequality. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.