Take the 2-minute tour ×
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 $A\colon E\to E$ definied by $A(f)(x)= \int\limits_0^x f(t) dt$. I have to find the spectrum of $A$ in the cases $E=C[0,1]$ and $E=L_2[0,1]$. I have proved that $A$ has no eigenvalues, but I can't find full spectrum.

share|improve this question
1  
Right. What space are you working on? –  Jonas Teuwen Nov 21 '11 at 9:31
    
Sorry. I need to find spectrum in both $C[0,1]$ and $L_2[0,1]$. –  Philipp G. Sinicyn Nov 21 '11 at 9:34
6  
You can find an inverse of $A-a I$ by solving $g(x) = (\int^x f) - a f(x)$: differentiating gives an ode that you can solve explicitly, and then find an expression for $f$ even when the functions are not differentiable. –  Benoit Jubin Nov 21 '11 at 10:14

1 Answer 1

The operator $A$ is compact in both cases (when $E=L_2[0,1]$ it is actually more, it is Hilbert-Schmidt and when $E=C[0,1]$ the compactness follows from Arzela-Ascoli), so any nonzero $\lambda\in \sigma(A)$ is an eigenvalue. If you have proven that there no nonzero eigenvalues, since $\sigma(A)$ is nonempty, then $\sigma(A)=\{0\}$.

The other way of doing it is to use Gelfand's result for the spectral radius, i.e., the limit $\lim_{n\rightarrow \infty}\|A^n\|^{1/n}$ exists and is identical to $\sup_{\lambda\in\sigma(A)}|\lambda|$. For example, if $E=C[0,1]$, then it is not hard to show that $\|A^n\|\leq \frac{1}{n!}$ and $\lim_{n\rightarrow \infty}\frac{1}{(n!)^{1/n}}=0$, so that the spectral radius is zero and hence $\sigma(A)=\{0\}$.

share|improve this answer

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.