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