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.

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    $\begingroup$ 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. $\endgroup$ Nov 21, 2011 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 $\sigma(A)=\{0\}$.

  • $\begingroup$ Thank you @Nonliapunov But how can we see that $A$ is compact? $\endgroup$ Aug 7, 2016 at 7:01
  • $\begingroup$ @MorganeMaPh With the Arzelá-Ascoli theorem, i.e. on an answer to this question. $\endgroup$ Oct 27, 2019 at 22:34

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