Let be $I=[0,1]$ the unit interval and define the operator $$(Af)(x)=\int_0^x f(t)dt $$ with domain $C_0^{\infty}(I)\subseteq L^2(I)$. I want to show the following:
$A$ is bounded and compact;
find the spectrum of $A$;
Regarding proving $A$ is bounded, I computed, by Fubini-Tonelli, $$\| Af\|_2^2=\int_0^1 \left| \int_0^xf(t)dt \right|^2 dx\leq\int_0^1 dx \iint_{[0,x]^2}|f(t)f(s)|dtds$$
but I don't know how to manage a subsequent upper bound. To prove compactness, I have to prove that $A$ sends a weakly convergent sequence in a strongly convergent sequence, but how?
In the second point, if I search for an eigenvector $f$ with eigenvalue $\lambda$, the condition is
$$\int_0^xf(t)dt=\lambda f(x) $$
for every $x\in I$. If I could differentiate both sides, I'd find $\lambda f^{\prime}(x) = f(x)$ with the condition $f(0)=0$, that doesn't have non zero solutions. What does this means?
