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Let $V$ be the vector space of all continuous functions from $\mathbb{R}$ into $\mathbb{R}$ and let $T\colon V \rightarrow V$ be a linear map defined by $T(f)(x)=\int^{x}_{0}f(t)dt$. How can we prove $T$ has no eigenvalue?

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up vote 7 down vote accepted

If $\lambda$ was an eigenvalue, and $f$ an eigenvector for $\lambda$, then for each $x\in \Bbb R$, we would have $$\int_0^xf(t)dt=\lambda f(x).$$ If $\lambda$ was equal to $0$, we would have $f\equiv 0$, which is not allowed. So $\lambda\neq 0$, and since $f$ is continuous, $f$ is $C^1$, as a primitive of a continuous function. So we have $f(x)=\lambda f'(x)$ for all $x$ and $f(0)=0$. This gives $$\left(\frac 1{\lambda}f(x)-f'(x)\right)e^{-\frac x{\lambda}}=0,$$ hence $f(x)=Ce^{\frac x{\lambda}}$. This gives that $f(0)=C=0$, hence $f\equiv 0$.

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Thanks, It is an elegant answer... –  shane Oct 1 '12 at 8:45
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