Continuous but not compact operator on $L^2(0,\infty)$ Define the following operator on $L^2(0,\infty)$:
$$Tf(x)=\frac{1}{x} \int_0^xf(y)dy,\quad f\in L^2(0\infty).$$
I would like to see that it is continuous but not compact.
So, this is an integral operator with kernel $k(x,y)=\frac{1}{x}\mathbf1_{(0,x)}(y)$. The problem is that $k$ is not even in $L^2(0,\infty)^2$. Thus, the usual bound $\|Tf\|_2\leq \|k\|_2\cdot \|f\|_2$ does not work. Hence I am not even sure why the operator is well-defined.
I.e. why is $Tf$ even in $L^2(0,\infty)$? And how might we show continuity/non-compactness?
 A: Let $\def\norm#1{\left\|#1\right\|_{L^2}}f \in L^2(0,\infty)$. We have 
\begin{align*}
  \norm{Tf}^2 &= \def\abs#1{\left|#1\right|}\int_0^\infty \frac 1{x^2} \abs{\int_0^x f(y)\, dy}^2 \, dx\\
           &\le \int_0^\infty \frac 1{x^2}\left(\int_0^x y^{1/4}y^{-1/4}\abs{f(y)}\, dy\right)^2 \, dx \\
           &\le \int_0^\infty \frac 1{x^2} \left[\left( \int_0^x y^{-1/2}\, dy\right)^{1/2}\left(\int_0^x y^{1/2}\abs{f(y)}^2 \, dy\right)^{1/2}\right]^2 \, dx\\
   &= \int_0^\infty \frac {2x^{1/2}}{x^2} \int_0^x y^{1/2}\abs{f(y)}^2 \, dy\, dx\\
   &= \int_0^\infty \int_y^\infty \frac 2{x^{3/2}} \, dx\cdot  y^{1/2}\abs{f(y)}^2 \, dy\\
   &= \int_0^\infty \frac 4{y^{1/2}} \cdot y^{1/2}\abs{f(y)}^2\, dy\\
   &= 4 \norm f^2 
\end{align*}
Hence, $\norm{Tf}\le 2\norm f$, proving the continuity of $T$.
To see that $T$ is not compact, let $f_n := n^{1/2} \chi_{[0,1/n]}$, then $\norm{f_n} = 1$ and 
$$ \norm{Tf_n - Tf_m}^2 \ge \int_0^{1/m} (m^{1/2} - n^{1/2})^2\, dx = \left(1 - \left(\frac nm\right)^{1/2}\right)^2 $$
So $(Tf_n)$ does not have a convergent subsequence.
