I have tried to solve this problem. Please let me know if I need to do more. Of course, I know that there are some alternate way to prove the following operator is compact.
Let $T:L^{2}(0,1)\to L^{2}(0,1)$ be the linear operator defined by
$$ T(u)(x)=\int_{0}^{x}u(t)dt\;\; \forall x\in(0,1). $$ Prove that $T$ is compact.
Proof: Since $L^{2}(0,1)$ is a separable Hilbert space so we will prove that if $u_{n}\to 0\; weakly$ then $\Vert Tu_{n}\Vert\to 0 $ in norm. So assume that $u_{n}\to 0\; weakly$ then
$$\Vert Tu_{n}\Vert_{2}=(\int_{0}^{1}\vert \int_{0}^{x}u_{n}(t)dt \vert^{2} dx)^{1/2} $$ by the Minkowskis’inequality for integrals, $$ \leq\int_{0}^{x}(\int_{0}^{1}\vert u_{n}(t)\vert^{2}dx)^{1/2}dt=\int_{0}^{x} \vert u_{n}(t)\vert dt$$ Now, since $u_{n}\to 0\; weakly$ in $L^{2}(0,1),$ then
$$\int_{0}^{x} \vert u_{n}(t)\vert dt\to 0. $$ Therefore, $$\Vert Tu_{n}\Vert_{2}\to 0. $$