Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question

1 Answer 1

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 $f\equiv 0$.

share|cite|improve this answer
Thanks, It is an elegant answer... – shane Oct 1 '12 at 8:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.