# Compactness and spectrum of integral operator

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?

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When in doubt, invoke equicontiuity via Arzela-Ascoli. – Alex R. Jan 3 '13 at 23:27

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.
@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