Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am solving the following question : If $k:[0,1]^2\to \mathbb C$ is continuous and $T_k : C[0,1] \to C[0,1]$ such that $$(T_kx)(t)=\int_0^t k(t,s)x(s) ds$$ Define $k_n: [0,1]^2\to \mathbb C$ inductively , ie $k_1=k$ and

$$k_{n+1}=\int k_n(t,r).k(r,s)dr$$ for $0\le s\le t$

I showed that $$(T^n_kx)(t)= \int_0^t k_n(t,s)x(s) ds$$

from that taking $|x(s)|\le 1 $ , i can easily see that $\|T_k^n\| \le\|k_n\|_{\infty}$

(am i right for the approximation of $\|T_k^n\|$ ?

next i approximate $$|k_n(t,s)| \le \frac{\|k\|_{\infty}^n}{(n-1)!}|t-s|^{n-1}$$

Now according to the definition of spectral radius $r(T) = \lim_{n \to \infty} \|T^n\|^{1/n}$

But we have $\|T_k^n\| \le\|k_n\|_{\infty}$ , so there eixsts $M \in \mathbb N$ such that $\|T_k^n\| \le M$ which implies the spectral radius is $0$ , but i am confused that the norm of $k_n$ may depend on $n$ , which seems to be the case . How do i go further from here .

And if the radius of the spectrum is $0$ it follows that the specrum is $0$ , but the whole reasoning fails if sup norm of $k_n$ depends on $n$ . How do i go further ? And is $T_k$ compact ??

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.