0

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.

glS
  • 7,963
Someone
  • 5,059

2 Answers2

2

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.

javi1996
  • 1,005
  • Yeah I aprecciate the answer but I am trying to do this in a context of functional analysis . – Someone Jun 03 '20 at 08:33
  • 1
    @Something This answer is essnetially Functional Analytic. The main point used is the fact that non-zero points of the spectrum are eigen values. To find eigen values you have to solve a simple DE. – Kavi Rama Murthy Jun 03 '20 at 08:47
  • 1
    You cannot treat this in a completely theorical way, because you are handling a concrete operator in a concrete function space, and you have to deal with the properties of primitives and continuous functions in a way or another. Following your argument, if you want to see that $(\lambda I - V)$ is not invertible, if this operator were suprajective you could express every continuous function f as $f = \lambda f - Vf$, so $(\lambda+1)f = Vf,$ but $Vf$ is a C1 function, so $f$ would be $C1$ too and you have a contradiction. As you see, you need to use theproperties of your space. – javi1996 Jun 03 '20 at 10:38
1

Also, in addition to the other good suggestions, you can compute (yes, using the specifics of the situation, as other comments and answer note!) that $\lim_n |T^n|^{1/n}=0$. Then invoke the theorem that that limit is a bound for the spectrum.

paul garrett
  • 55,317