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!