Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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
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

 
discard

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.