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The question is as follows:

Show that the linear operator $(Ax)(t)= \int_{0}^{1} \frac{\sin(ts)}{\mid t-s \mid^{\frac{1}{3}}} x(s) ds$ for $A : L_2[0,1] \to L_2[0,1] $ is compact.

$\textbf{An idea:}$

For to prove that $A$ is compact, it is enough to show that its kernel $k(t,s) = \frac{\sin(ts)}{\mid t-s \mid^{\frac{1}{3}}} $ is continuous and is in $L_2[0,1] \times L_2[0,1]$, i.e we have to show that $\int_{0}^{1} \int_{0}^{1} \left| \frac{\sin(ts)}{\mid t-s \mid^{\frac{1}{3}}} \right|^2 ds dt \leq +\infty $.

For to show this, we have \begin{align} \int_{0}^{1} \int_{0}^{1} \left| \frac{\sin(ts)}{\mid t-s \mid^{\frac{1}{3}}} \right|^2 ds dt &\leq \int_{0}^{1} \int_{0}^{1} \frac{1}{\mid t-s \mid^{\frac{2}{3}}} ds dt \\& = \int_{0}^{1} \left( \int_{0}^{t} \frac{1}{( t-s )^{\frac{2}{3}}} + \int_{t}^{1} \frac{1}{( s-t )^{\frac{2}{3}}} ds \right) dt \\&= \int_{0}^{1} \left( -3(t-s)^{frac{1}{3}}\mid_{0}^{t} + 3(s-t)^{frac{1}{3}}\mid_{t}^{1} \right) dt \\& = \int_{0}^{1} \left( -3 t^{\frac{1}{3}} + 3(1 -t)^{\frac{1}{3}} \right) dt \\&= -\frac{9}{4} t^{\frac{4}{3}}\mid_{0}^{1} - \frac{9}{4} (1-t)^{\frac{4}{3}}\mid_{0}^{1} = - \frac{9}{4} + \frac{9}{4} =0 < +\infty \end{align} This proves the claim.

Please let me know if I am wrong or if we need to show something else for to ensure that the mentioned integral operator is compact?

Thanks!

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  • $\begingroup$ I am rather dubious about the fact that the integral of a non-negative and non-constant function over $(0,1)^2$ turns out to be $\leq 0$. $\endgroup$ – Jack D'Aurizio Dec 10 '17 at 17:38
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    $\begingroup$ Anyway, $$\iint_{(0,1)^2}\frac{ds\,dt}{|s-t|^{2/3}}=\frac{9}{2}<+\infty.$$ $\endgroup$ – Jack D'Aurizio Dec 10 '17 at 17:41
  • $\begingroup$ @JackD'Aurizio Many thanks! Can you please let me know if my claim (i.e. If $k(t,s) \in L_2[0,1] \times L_2[0,1]$ then our integral operator is compact) is true? $\endgroup$ – user510716 Dec 10 '17 at 18:08
  • $\begingroup$ @JackD'Aurizio But I checked it twice! It gave me zero!! $\endgroup$ – user510716 Dec 13 '17 at 5:16
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We can simplify the argument: For any $t\in [0,1],$

$$\int_{0}^{1} \frac{1}{\mid t-s \mid^{\frac{2}{3}}} ds\le \int_{t-1}^{t+1} \frac{1}{\mid t-s \mid^{\frac{2}{3}}} ds = \int_{-1}^{1} \frac{1}{\mid s \mid^{\frac{2}{3}}} ds <\infty.$$


Added later: Just a comment on why the operator $A$ is compact. If we are given an infinite matrix $M=(a_{ij})$ such that $\sum |a_{ij}|^2 < \infty$ (i.e., $M$ is Hilbert-Schmidt), we can define an operator $T:l^2 \to l^2$ by $T(x)=Mx^T.$ I don't think it's too difficult to show $T$ is compact: You take a sequence $(x_n)$ in the unit ball, and by diagonalization, show there is a subsequence $x_{n_k}$ that converges in each coordinate. The condition on $M$ then shows $Tx_{n_k}$ converges in $l^2.$

I think the result for $k\in L^2[0,1]^2$ follows from this: Take an orthonormal basis $\phi_j$ of $L^2[0,1].$ Then $\phi_i\phi_j$ is an orthonormal basis $L^2[0,1]^2.$ Write $k$ as $\sum_{ij} a_{ij}\phi_i\phi_j,$ and write functions $x\in L^2[0,1]$ as $x= \sum_j b_j \phi_j.$ The operator $A$ now corresponds to the matrix situation above, and compactness of $A$ follows.

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  • $\begingroup$ Many thanks! Can you please let me know if my claim (i.e. If $k(t,s) \in L_2[0,1] \times L_2[0,1]$ then our integral operator is compact) is true? $\endgroup$ – user510716 Dec 10 '17 at 18:08
  • $\begingroup$ en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_integral_operator $\endgroup$ – Diesirae92 Dec 12 '17 at 20:24
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    $\begingroup$ @Diesirae92 Thanks! You mean that I have to prove that $A$ is self-adjoint operator? Or What I have done above is enough? What about when that $K(t,s) \in C[0,1] \times C[0,1]?$ $\endgroup$ – user510716 Dec 12 '17 at 20:49
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    $\begingroup$ nope, what you did is enough. In the case of continuous functions you characterise compactness with Ascoli Arzelà. Compactness there is more difficult to get then in Lebesgue spaces. Probabily there is non compact, but I am fust guessing (tipically you build a sequence of bump functions escaaping to $\infty$) $\endgroup$ – Diesirae92 Dec 12 '17 at 21:33

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