# Finding eigenfunctions and eigenvalues

Let $K$ be the integral operator defined by

$$(Kf)(x)=\int_0^1 u(x)v(y)f(y) dy$$

for some continuous functions $u,v$ in the Hilbert space with inner product $\langle f,g \rangle = \int_0^1 f(x)^* g(x) dx$ on $(0,1)$. I want to find the eigenfunctions and eigenvalues corresponding to $K$. (this is problem 3.4 in http://www.mat.univie.ac.at/~gerald/ftp/book-fa/ )

The exercise is from a chapter about compact symmetric operators (which this operator is), but it only contains existence theorems.

If I could get some helpful hints on how to get started, I'd be thankful. (I have a suspicion this is easier than it looks)

You asked for a hint: Notice you can take $u(x)$ out of the integral!
Thanks! feels stupid So $u$ is an eigenfunction with eigenvalue $\int v(y)u(y)dy$. How do I know I have found all eigenfunctions (if I have!)? – Fredrik Meyer Jul 10 '11 at 5:55
Notice that you are in the following situation: you have a Hilbert space $H$ and two vectors $u$, $V\in H$, and your map looks like $T:x\in H\mapsto \langle x,u\rangle v\in H$. Can you see what the kernel is? Can you diagonalize it? Try to do this first with $H$ finite dimensional, maybe it helps. – Mariano Suárez-Alvarez Jul 10 '11 at 5:57
Seems like you wrote the map wrong. Isn't it $g \mapsto \langle v^*,g \rangle u$? From this we see that the kernel consists of all $g$ orthogonal to $v^*$. This should imply that $Ker(K)^\perp$ is one-dimensional, right? – Fredrik Meyer Jul 10 '11 at 6:10
My $u$ and $v$ were not intended to be yours: my example is quite generic. Yes, the kernel of my $T$ is the hyperplane orthogonal to $u$, so if you pick a basis for it, and add $u$, you get a basis of $H$ which diagonalizes $T$. – Mariano Suárez-Alvarez Jul 10 '11 at 6:22
In the book I'm using (referenced in the top post), it says every symmetric compact operator can be written as $Af=\sum_{j=1}^\infty \alpha_j \langle u_j,f \rangle u_j$ with $\alpha_j$ converging to $0$. From this I can conclude that the only non-zero eigenfunction is $u$. – Fredrik Meyer Jul 10 '11 at 20:18