TITLE: Show that $f \overset{T}{\rightarrow} \frac{1}{x} \int_0^x f(t) dt$ is Bounded, and is NOT Compact in $L^2(0, \infty)$
QUESTION [7 upvotes]: Can someone help me with this question?
Let $f \in X = L^2(0, \infty)$, and define
\begin{equation}
(Tf)(x) = \frac{1}{x} \int_0^x f(t) dt \ .
\end{equation}
Show that $T$ is Bounded, and is NOT Compact.
Thanks in advance.
Edit: For boundedness, I tried:
\begin{eqnarray}
\| f \| \leq 1 \Rightarrow \| Tf \|^2 &=& \int_0^{\infty} \lvert (Tf)(x) \rvert ^2 \ dx \\
&=& \int_0^{\infty} \lvert \frac{1}{x} \int_0^x f(t) \ dt \rvert ^2 \ dx \\
&\leq& \int_0^{\infty} \frac{1}{x^2} \int_0^x \lvert f(t) \rvert ^2 \ dt \ dx \\
&\leq& \int_0^{\infty} \frac{1}{x^2} \| f \| ^2 \ dx \\
&=& \| f \| ^2 \int_0^{\infty} \frac{1}{x^2} \ dx \\
\end{eqnarray}
But the problem is that $\int_0^{\infty} \frac{1}{x^2} \ dx$ is NOT finite.
To prove that $T$ is not compact, I think we should give a sequence of functions, $\{ f_n \}_{n=1}^{\infty}$, such that for every $n$, we have $\| f_n \| \leq 1$ and $\{ Tf_n \}$ has no Cauchy subsequence. But I can't find such sequence of functions.

REPLY [3 votes]: To show noncompactness, let $a> 0$ and set $f_a(x)= (a/2)^{-1/2}\chi_{(a/2,a)}.$ Then $f$ is a unit vector in $L^2.$ It is straightforward to see
$$T(f_a)(x) = \begin{cases} 0,\, 0 < x < a/2 \\
(a/2)^{1/2}/x,\, x> a.\end{cases}.$$
Suppose $a<b/2.$ Because $T(f_b) = 0$ on $(0,b/2),$
$$T(f_a)(x)-T(f_b)(x) = (a/2)^{1/2}/x, a<x<b/2.$$
Thus
$$\int_0^\infty|T(f_a)(x)-T(f_b)(x)|^2\, dx \ge \int_a^{b/2}|T(f_a)(x)|^2\, dx =\frac{1}{2} -\frac{a}{b}.$$
This is $\ge 1/4$ if $a/b\le 1/4.$ It's now easy: The unit vectors $f_{1/4^n}$ satisfy
$$\|T(f_{1/4^n}) - T(f_{1/4^m})\|_2 \ge \frac{1}{2}$$
if $m\ne n.$ Thus no subsequence of $T(f_{1/4^n})$ can be Cauchy. Hence $T$ is not compact.