The spectrum of the Volterra operator is $\{0\}$ I'm asked to prove that the spectrum of the Volterra operator $$V(f(x))= \int_0^x f(y)dy$$ given by $$ \sigma (V)= \{ \lambda \in \mathbb{C} | V-\lambda1 \text{ is not invertible} \}$$ contains only zero.
This needs to be done by first showing that $0 \in \sigma (V)$ and then that all other complex numbers cannot belong to the spectrum.
I don't see how to prove either one of those steps so any hints would really help.
 A: The kernel of $V$ is $K(x,t) = \mathsf 1_{(0,x)}(t)$ and
$$
\int_{[0,1]^2} |K(x,t)|^2\ \mathsf d(x\times t) = \int_0^1\int_0^x\ \mathsf dt \ \mathsf dx = \frac 12<\infty,
$$
so $V\in L^2([0,1]^2)$. It follows that $T$ is a Hilbert-Schmidt operator and therefore is compact. It follows then from the Fredholm alternative that if $\lambda\in\sigma(V)$, $\lambda\ne0$, that $\lambda$ is an eigenvalue of $V$. That is, $Vf(x)=\lambda f(x)$ for all $x\in[0,1]$. If $Vf=\lambda f$, then as $Vf\in C^0([0,1])$ (we can prove that $V$ is Hölder continuous using the Hölder inequality), we must have $f\in C^0([0,1])$. By the fundamental theorem of calculus, $Vf\in C^1([0,1])$, and so $f\in C^1([0,1])$. Differentiating both sides of the eigenvalue equation yields $f(x)= \lambda f'(x)$, and hence $f(x) = Ce^{\frac1\lambda x}$ for some $C\in\mathbb C$. Since $Vf(0) = 0$, we have $f(0)=0$, and therefore $C=0$. It follows that the spectral radius of $V$ is zero.
A: To prove $0$ is an element of $\sigma(T)$: notice that $Vf$ is injective and $0 \in \sigma(T)$ iff $V$ is not invertible. So if you can show that $Vf$ is not surjective then it can't be invertible anymore. So show that the image of $Vf$ is continuous. Since $L^2(\left[0,1 \right])$ contains also not continuous functions, $Vf$ is not surjective and thus not invertible.  
To show that any $\lambda \neq 0$ is not in the spectrum: notice $V- \lambda I$ is injective. If $V- \lambda I$ is surjective, it follows that $V-\lambda$ is invertible and $\lambda$ is not in the spectrum. If you pick a random $g \in C^1$ and try solve the equation $(V-\lambda)f=g$ to $f$ explicit. You see that there is always a solution if $\lambda \neq 0$. This link can help you to solve the equation: https://www.math24.net/linear-differential-equations-first-order/ . You have assumed $g$ is an element of $C^1$, so try to remove the argument by using that $C^1$ is dense in $L^2(\left[0,1 \right])$.
