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.

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

Show that the operator $C: L^2([0,1]) \rightarrow L^2([0,1])$ defined by $$Cf(x) = \int_0^x\int_1^tf(s)dsdt$$ is compact and determine its spectrum.

Im not sure how to find the spectrum when we are in a function space. Do we solve $(C-\lambda I)f = 0$ or how can we do? If that is the case I can derivate two times and solve a differencial equation. This results in $$f(x) = a \sin(t\sqrt{\lambda}) +b \cos(t\sqrt{\lambda})$$ or $$ f(x) = a\exp{(t\sqrt{\lambda}})+ a\exp{(-t\sqrt{\lambda}})$$

are these solutions in $L^2([0,1])$ ? What conclusions should I get? Also is it compact since its volterra operator together with some integral operator?

share|cite|improve this question
When in doubt, invoke equicontiuity via Arzela-Ascoli. – Alex R. Jan 3 '13 at 23:27
up vote 4 down vote accepted

For compactness you can invoke a general theorem: Hilbert-Schmidt integral operators are compact. Indeed, by changing the order of integration your operator can be written as $Cf(x)=\int_0^1 k(x,y) f(y)\,dy$ with $k(x,y)=-\min(x,y)$. Thus, $C$ is compact and self-adjoint.

The idea to use the fact $(Cf)''=-f$ is sound. Regularity should be briefly discussed: we see that $Cf$ is always continuous; hence, $Cf=\lambda f$ implies that $f$ is continuous; from here we get that $Cf$ is in class $C^1$, etc... the conclusion is that eigenfunctions are $C^\infty$ smooth.

It is important to note the fact that not every every function $g$ with $\lambda g''=-g$ (i.e., a candidate for eigenfunction) is in the range of $C$. Indeed, $Cf(0)=0$ and $(Cf)'(1)=0$ by the definition of $Cf$. These boundary conditions will limit the set of eigenfunctions to a discrete (but still infinite) family.

share|cite|improve this answer
Many thanks! Can you please expand on one line how u change the order of integration, I still got a dubbel integral. – Johan Jan 5 '13 at 22:16
@Johan Yes you do get a double integral, but the inner variable is $t$ and $f(s)$ does not depend on $t$. So you pull $f(s)$ out of the inner integral as a constant, and evaluate. – user53153 Jan 5 '13 at 22:25

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.