Compactness of a linear function

Question

Consider $T:L^2(0,1)\rightarrow L^2(0,1)$ such that $$ (Tu)(x)=\int_0^x u(t) dt \quad \forall x \in (0,1).$$ How could one prove that $T$ is compact?

Answer 1

Integral operators are where the original notion of compact came up through Arzela-Ascoli. Your integral operator maps bounded subsets of $L^2$ into equicontinuous sets of functions. This because, for $0 \le x \le x' \le 1$, $$ |(Lf)(x)-(Lf)(x')| \le \int_{x}^{x'}|f(t)|dt \\ \le \left(\int_{x}^{x'}|f(t)|^2dt\right)^{1/2}\sqrt{x'-x}. $$