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Let $X=(C([0,1]), \Vert \cdot \Vert_{\infty})$. Determine the spectrum of $$ \begin{split} M \colon & X \to X\\ & u(t) \mapsto \int_0^t h(s)u(s)ds \end{split} $$ where $h \in C([0,1])$ is fixed.

First of all, I have proved the operator is compact (by Ascoli-Arzelà). Hence $0 \in \sigma(M)$.

Now how can I find the other elements of $\sigma(M)$? I know that they are eigenvalues and they are either finite, either a sequence (converging to $0$), since $M$ is compact.

Here's what I've tried: let $\lambda \ne 0$ and eigenvalue $g(\cdot)$ a corresponding eigenvector. So I have $$ \lambda g(x) = \int_0^x h(s)g(s)ds $$ hence $g \in C^1$ and $$ \begin{cases} \lambda g'(x) = h(x)g(x)\\ g(0)=0 \end{cases} $$ Solving this Cauchy problem, I get $$ g(x)=\exp{\left( \frac{1}{\lambda}\int_0^x h(s)ds\right) } -1. $$ Now I don't manage to finish: I should find some conditions on $\lambda$, but can't see how...

Thanks in advance.

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up vote 2 down vote accepted

Solving this Cauchy problem actually gives that $$\frac d{dx}\left(g(x)\exp\left(-\frac 1\lambda\int_0^xg(t)h(t)dt\right)\right)=0$$ so $g(x)\exp\left(-\frac 1\lambda\int_0^xg(t)h(t)dt\right)$ is constant on $[0,1]$. As $g(0)=0$, this constant is $0$ so $g$ is identically $0$ and $\lambda$ is not an eigenvalue of $M$.

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Here is a reason why this should not be surprising: The operator in question is a bit like a triangular matrix $G$, with “entries” $G_{xt}=g(t)$ if $t<x$, $G_{xt}=0$ otherwise. Stretching the analogy, you expect to find the eigenvalues on the diagonal; but the diagonal isn't “really there”, or more precisely, the diagonal in $[0,1]\times[0,1]$ has measure zero. (None of this is a substitute for actual proof, of course.) – Harald Hanche-Olsen Dec 25 '12 at 11:41
Hi Davide, thanks for your answer. I'm sorry, but I think there's a typo: it should be $\exp\left(\frac{-1}{\lambda}\int_0^x h(t)dt \right)$ (the integrand is not $g(x)h(x)$). Do you agree? Indeed, I was almost at the end but I made a mistake (I should have written a multiplicative constant, instead of an additive one). Thanks a lot for your help. – Romeo Dec 25 '12 at 13:31
@HaraldHanche-Olsen Very interesting, thanks for your observation. – Romeo Dec 25 '12 at 13:32

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