Let $T: C([a,b])\rightarrow C([a,b])$ be the Volterra Integral Operator, where $T(\phi)(t)=\int_a^tk(t,s)\phi(s)ds$.
I have already seen that this operator is compact using the Ascoli-Arzela Theorem. Now I am trying to show that $\sigma(T)=\{0\}$.
We know that $0\in \sigma(T)$ because the operator is compact and the space has infinite dimensions but I am having some trouble showing that it is the only one. So what I am thinking is that we can try and show that $T-\lambda I$ is invertible for $\lambda\neq 0$, and $T-\lambda I=-(I-\frac{T}{\lambda})$ so maybe try using the neumann series to see that this is invertible in the banach algebra $L(C([a,b]))$, but I am not being able to do it.
I can't really show that $\sum||T^n||$ has to converge , so any help is aprecciated. Thanks in advance.
Best Answer
Since $V$ is a compact operator, you have $\sigma(V) = \{0\} \cup \sigma_p(V)$. Id est, every element of the spectrum that is different from zero must be an eigenvalue.
Assume $\lambda \neq 0$ is an eigenvalue. We have $Vf = \lambda f$. Since $f$ is continuous we have that $Vf$ is differentiable, so $f$ is differentiable too. Then, you can derive the expression to obtain $\frac{1}{\lambda}f = f', f(0) = 0$, and the Picard theorem of uniqueness of solution of ODEs give you the contradiction.