$T: L^2[0,1] \to L^2[0,1]$, $Tf(x)= \frac{1}{x}\int_{0}^x f(y)$, is a bounded but not compact operator. To show that the image of $T$ lies in $L^2$, and derive its bound, I tried the following: $$\|Tf\|_{2} = \left(\int|\int \frac{1}{x}f(y)dy|^2dx\right)^{\frac{1}{2}} \leq \int \sqrt{\int|\frac{1}{x}f(y)1_{[0,x]}(y)|^2dx}dy = \int_{0}^{1} \sqrt{ \frac{1}{y}-1 }|f(y)|.$$ Then probably using Holder's inequality, but this does not seem to work. Then to show it is not a compact operator, I want to construct a bounded sequence $f_n$ where $Tf_n$ does not have a convergent subsequence.
 A: Let us prove that $T$ is a bounded operator, and $\|T\| \leq 2$.
By density, it is enough to show that
$$
(1) \qquad \|Tf\|_2 \leq 2 \|f\|_2
\qquad
\forall f \in C^\infty_c((0,1)).
$$
Let $F(x) := \int_0^x f(y)\, dy$. Clearly, $F\in C^1$ and $F' = f$.
Integrating by parts and using Cauchy-Schwarz, we have that
$$
\begin{split}
\|Tf\|_2^2 & = \int_0^1\frac{1}{x^2} F(x)^2\, dx
= - F(1)^2 + 2 \int_0^1 f(x) \frac{1}{x}\, F(x)\, dx
\\ &\leq 2 \int_0^1 f(x) \ Tf(x)\, dx
\leq 2 \|f\|_2 \|Tf\|_2,
\end{split}
$$
so that (1) follows.
A: For the continuity of $T$, you can also check the link by mechanandroid.
For compactness, however, this link sends you to this question, where it is shown that $T : \mathcal{C} ([0,1]) \to \mathcal{C} ([0,1])$ is not compact, but $T : \mathcal{C} ([0,1]) \to \mathbb{L}^2 ([0,1])$ is compact. Of course, the functions used there are not suitable to show the non-compactness of $T : \mathbb{L}^2 ([0,1]) \to \mathbb{L}^2 ([0,1])$. Hence this question does not seem to be a duplicate.
Now, what would a suitable sequence of function $(f_n)$ look like? We want $\|f_n\|_{\mathbb{L}^2} \equiv 1$, an $T(f_n)$ as large as possible (so as to avoid convergence to $0$). The first thing is to avoid cancellations in the integral, since it makes $T(f_n)$ smaller. So let us look for non-negative $f_n$.
Then, we would like to put the most possible mass close to $0$; then $\int_0^x f_n (t) dt$ will be quite large for a small value of $x$, which makes $T(f_n)$ large. So, a good try is to take $f_n (t) := \sqrt{n} \mathbb{1}_{[0,1/n]} (t)$, which has unit norm. Then: 
$$T(f_n) (t) = \left \{ \begin{array}{ccc} \sqrt{n} & \text{if} & t \in [0,1/n] \\ 1/(\sqrt{n}t) & \text{if} & t \in [1/n,1] \end{array}\right. .$$
We compute $\|T(f_n)\|_{\mathbb{L}^2}^2 = 1+\int_{1/n}^1 1/(nt^2)dt = 2-1/n$.
In addition, $(T(f_n))_{n \geq 0}$ converges almost everywhere to $0$, so any limit point of this sequence must be $0$. Since the norm of $T(f_n)$ converges to $2$, this cannot happen, so $(T(f_n))_{n \geq 0}$ has no limit point.
