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I would like to know the spectral radius of the operator $T_k$ from $C[0,1] \to C[0,1]$ : $$T_k x (t)= \int_0^1 k(t,s) x(s) ds$$ where $k(x,y)\colon [0,1]^2 \to \mathbb C$ is continuous.

And also although I know that $Tf(x)=\int_0^1f(x)dx$ is compact, I am not able to follow that $T_k$ is compact. Any hints and ideas ? Thanks!

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@DavideGiraudo : Since the map is from $C[0,1] \to C[0,1]$ and i know that the map $T(f)(s) =\int_0^s f(x)dx$ is compact , is there a reasoning by which i can immediatly say that $T_k$ is compact ? –  Theorem Jan 18 '13 at 12:42
    
An operator is compact only if it is the limit of finite rank operators. This is what Davide was suggesting. But you can also reason on the fact that $k$ is uniformly continuous on a compact.. –  uforoboa Jan 18 '13 at 12:45
    
I don't know. What you know is a special case; I'm not sure you can deal with the general one. But probably the proof of this case can be extended to the general one. Have you use Arzela-Ascoli theorem? –  Davide Giraudo Jan 18 '13 at 12:45
    
@uforoboa : yes , but can u tell me how does approximating $k$ as a limit of polynomials in $t,s$ have to do with the finite rank of $T_k$ . i think i am not understanding . –  Theorem Jan 18 '13 at 12:47
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@DavideGiraudo Are you sure that it is a limit of finite-rank operators? Now that I think about it, I would be sure if the integral were not dependent on $t$. But here it is... Do you see my point? How would you overcome it? –  uforoboa Jan 18 '13 at 13:11

1 Answer 1

To show compactness, we can use Arzela-Ascoli's theorem. Let $B$ the set of continuous functions on $[0,1]$ of norm $\leqslant 1$. We have to show equi-continuity of $T_k(B)$. It follows from the fact that $k$ is uniformly continuous on $[0,1]^2$.

For the spectral radius, show by induction that $$\lVert T_k^p\rVert\leqslant\frac{\lVert k\rVert_\infty^p}{p!},$$ the spectral radius formula and this thread.

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yes , i did it . Is $0$ spectrum of $T_k$ . it looks like that . –  Theorem Jan 18 '13 at 13:32

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