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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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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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    @HaraldHanche-Olsen Very interesting, thanks for your observation.2012-12-25