For $f\in L^2(0,\infty),$ define $(Tf)(x)=x^{-1}\int\_0^x f(s)ds,$ for $x\in(0,\infty),$ then from hardy's inequality, $T\in B(L^2),$ my question is how to show that $T$ is not compact?



Show a linear operator is not compact
For a natural number $j$ let $f\_j$ be the indicator function of the interval $[0,1/j]$ times the square root of $j$.
Then the $L^2$-norm of $f\_j$ is one.
A simple calculation shows that one has 
$||Tf\_i-Tf\_j||^2\ge\int\_0^{1/j}(\sqrt i-\sqrt j)^2dx= (1-\sqrt{i/j})^2$ for $i\le j$, which implies that no subsequence can be Cauchy.


It is natural to look for an example around $x=0$ since that is where the kernel of $T$ fails to be $L^2$.


