Prove that $Af(x) = \frac 1x \int\limits_0^x f(t) dt$ isn't compact in $L_2[0,1]$ I need to prove that operator $\displaystyle Af(x) = \frac 1x\int\limits_0^x\ f(t) dt$ isn't compact in $L_2[0,1]$.
I have tried to calculate spectrum of $A$, but failed.
 A: First method. 
We use the following result:

Let $H$ a Hilbert space and $T\colon H\to H$ a linear bounded operator. Then $T$ is compact if and only if for all sequence $\{x_n\}$ weakly convergent to $0$ the sequence $\{\lVert Tx_n\rVert\}$ converges to $0$.

Consider $f_n(x):=\sqrt n\mathbb 1_{\left[0,\frac 1n\right]}(x)$. This sequence is bounded and for all polynomial $P$ we have $\lim_{n\to\infty}\int_{\left[0,1\right]}f_nPd\lambda x=0$, an the sequence $\{f_n\}$ converges weakly to $0$. 
 But 
$$\lVert Af_n\rVert^2=\int_0^{\frac 1n}\frac 1ndx+\int_{\frac 1n}^1\frac 1{nx^2}dx\geq\frac 1n\left[-\frac 1x\right]_{1/n}^1=1-\frac 1n $$
which shows that the sequence $\{Af_n\}$ doesn't converges strongly to $0$, therefore $A$ is not compact.
Second method.
We use the fact that in a separable Hilbert space, a compact self-adjoint operator has a countable spectrum.
The fact that $A$ is self-adjoint follow from an easy computation. Now, we will look at the spectrum of $A$. We first look at the non-zero eigenvalues of $A$.
If $\lambda$ is an eigenvalue of $A$, let $f\neq 0$ and eigenvector. Then, by dominated convergence and using the integral formula, $f$ is continuous, then thanks to this formula, differentiable and $f(x)=\lambda f(x)+x\lambda f'(x)$ hence $\frac{\lambda-1}{\lambda x}f(x)+f'(x)=0$ and $\exp\left(1-\frac 1{\lambda}\ln x\right)f(x)$ is constant: $f(x)=Cx^{\frac 1{\lambda}-1}$. We need $f$ to be in $L_2$, hence $2-\frac 2{\lambda}<1$ and $1<\frac 2{\lambda}$, therefore $0<\lambda <2$. Conversely, the previous computations show that for $0<\lambda<2$, $x^{\frac 1{\lambda}-1}$ is an eigenvector. Hence the spectrum contains $(0,2)$ and cannot be countable.
