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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)

Thanks in advance.

1 Answers 1

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You asked for a hint: Notice you can take $u(x)$ out of the integral!

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    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!)?2011-07-10
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    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.2011-07-10
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    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?2011-07-10
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    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 $T$.2011-07-10
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    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 $u$.2011-07-10
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    @Fredrik: you can conclude that if you really insist, but you'd be wrong :) Any function in the kernel of your map is also an eigenfunction!2011-07-11
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    Of course :) More precisely then: the only eigenfunction with non-zero eigenvalue is $u$ and any function orthogonal to $v^*$ is an eigenfunction with eigenvalue $0$.2011-07-11
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    Well: if you *really* want to be precise, then **(i)** *any* non-zero multiple of $u$ is also an eigenfunction with non-zero eigenvalue, and **(ii)** any *non-zero* function orthogonal to $v^\*$ is an eigefunction with eigenvalue $0$ :)2011-07-11
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    :) Of course! Thanks!2011-07-11