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Let $V$ be the vector space of all functions from $\mathbb R$ into $\mathbb R$ which are continuous. Let $T$ be the linear operator on $V$ defined by $$(Tf)(x) = \int_0^x f(t) dt$$ Prove that $T$ has no eigen values.

All those who are going to differentiate in the middle, please consider, there are functions which are continuous, but nowhere differentiable. Ex: Weirstrauss function. So, you can't differentiate anywhere in between.

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this might be a bit of an overkill, but you could use en.wikipedia.org/wiki/… –  mm-aops Jun 19 at 20:59
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It's true that $f$ need not be differentiable. But $Tf$ is automatically differentiable, and by the fundamental theorem of calculus, $(Tf)'=f$. –  mweiss Jun 19 at 21:00
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We have (by continuity of $f$) that $(Tf)'=f$. Therefore if $Tf=\lambda f$ with $\lambda\neq0$ we have that $f$ is differentiable also, and $\lambda f'=f$, which is a simple DE you hopefully can solve. But also $Tf(0)=0$. Hmm. –  Jyrki Lahtonen Jun 19 at 21:07

2 Answers 2

The only possible eigenvectors would have to be of the form $e^{\lambda x}$ but since $(Tf)(0)=0$, none of those work.

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Actually, you can differentiate $Tf(x)$ because integral of a continuous function is differentiable, and its derivative is the function, that's FTC.

Now if $Tf(x)=af(x)$ and $a\neq0$ then any eigenfunction $f(x)$ is also differentiable, and differentiating both sides gives $f(x)=af'(x)$. The general solution to this equation is $f(x)=Ce^{x/a}$, but since $f(0)=\frac1a Tf(0)=0$ we must have $C=0$ and hence $f(x)=0$.

It remains to consider the case $a=0$, but then $Tf(x)=0$ and differentiation gives $f(x)=0$ directly.

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