1
$\begingroup$

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 $\sigma(T) = \{0\}.$?

$\endgroup$
0
$\begingroup$

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 $\sigma(T) = \{0\}$

$\endgroup$
  • $\begingroup$ 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 $\lambda\neq 0$. $\endgroup$ – user112999 Dec 3 '13 at 6:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.