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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 ??

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