# Prove spectrum of this operator

On the space $C[0,1]$ with the norm $||x|| = \max_{ 0 \leq t \leq 1} |x(t)|$, consider the linear operator $$Tx(t) = \int_{0}^t k(t,s)x(s)ds$$ where $k(t,s)$ is a jointly continuous function on $[0,1]\times [0,1]$. Show that $\sigma(T) = \{0\}.$

I can see ||T|| = M, where $M=\max_{s,t}|k(t,s)|$. Then $\sigma(T)$ is in the disk given by $|\lambda| \leq M$. But how to show $\sigma(T) = \{0\}.$?

You can prove by induction that $$\|T^n\| \leq \frac{M^n}{n!}$$ and then conclude that $$\lim_{n\to \infty} \|T^n\|^{1/n} = 0$$ Hence the spectral radius of $T$ is zero, whence $\sigma(T) = \{0\}$
• Thanks, very thoughtful. Can we see when the series $(T-\lambda I)^{-1}$ converges when $\lambda \neq 0$ converges? I can not see why it converses when $\lambda\neq 0$. – user112999 Dec 3 '13 at 6:45