TITLE: Example of a noncompact operator on $L^2([0,1])$
QUESTION [6 upvotes]: For $f \in L^2([0,1])$, define operator $Tf: x \mapsto \frac{1}{x}\int_0^x f(y)dy$. Show that $T$ is not a compact operator on $L^2([0,1])$ and that $T$ is bounded. 
For the second part, I can show $T$ is bounded by looking at $\|Tf\|_2$ and rewriting it by integration by parts and then apply the Cauchy-Schwartz inequality. However, I was not able to find a bounded sequence of $L^2$ functions so that its image under $T$ is not precompact in $L^2$. 
Any help is tremendously appreciated.

REPLY [2 votes]: Another approach using the same sequence $f_n = \sqrt{n}\chi_{[0, 1/n]}$. 
Here it is shown that $(f_n)_n$ converges to $0$ weakly:

$\|f_n\|_2 = 1, \forall n\in\mathbb{N}$ so $(f_n)_n$ is bounded
for $g \in L^\infty[0,1] \cap L^2[0,1]$ we have 
$$|\langle f_n, g\rangle| \le \| g \|_\infty \cdot \sqrt{n} \int_0^{1/n}\! dx = \frac{\| g \|_\infty}{\sqrt{n}} \to 0$$
and $L^\infty[0,1] \cap L^2[0,1]$ is dense in $L^2[0,1]$.

If $T$ were compact, it would map weakly convergent sequences to strongly convergent sequences so we would have $Tf_n \to 0$ strongly.
But $\|Tf_n\|_2 = \sqrt{2-\frac1n}$ which doesn't converge to $0$.