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Let operator $A: L^2(0,2\pi) \rightarrow L^2(0,2\pi)$ be given by $$(Au)(x)=\sin x \int_{0}^{2\pi} u(y)\cos y \, dy$$ for $u\in L^2(0,2\pi), x\in [0,2\pi]$. What are eigenvalues of $A$?

Thanks for help.

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Unless I'm missing something here, the only eigenfunctions must be multiples of u(x) = sin x. Substitute that into the equation and there is only one possible eigenvalue: zero. –  Simon S Jan 19 '12 at 18:34
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There are more eigenfunctions: $sin(nx)$ are also eigenfunctions to eigenvalue 0 for all $n\in\mathbb{N}$, so are $\cos(nx)$ for $n\geq 2$ and $1$. –  Fabian Jan 19 '12 at 18:40
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You're exactly right. What I should have said the only possible eigenfunction corresponding with a nonzero eigenvalue must be sin(x). But even that function has eigenvalue = 0. Now, are there other eigenfunctions for that eigenvalue? Yes. Must drink more coffee. –  Simon S Jan 19 '12 at 18:48

2 Answers 2

up vote 1 down vote accepted

Suppose we have a Fourier expansion for an eigenfunction $u(x)=a_0+\sum\limits_{k=1}^\infty \left(a_k\sin(kx)+b_k\cos(kx)\right)$. Then $\int_0^{2\pi} u(t)\cos(t)\,dt=b_1\pi$

So in order to have $A[u] = \lambda u$ we would need $b_1\pi\sin(x)=\lambda u(x)$. So $\lambda u(x)$ has nothing but $\sin(x)$ appearing in its expansion. In particular, there is no $\cos(x)$ term and thus $b_1=0$. Therefore, we must have $\lambda u(x)=0$ so $\lambda=0$ if $u(x)\not=0$. So (among functions with a Fourier expansion), $0$ is the only possible eigenvalue (with any non-zero function having a Fourier expansion with $b_1=0$ as eigenfunctions).

Edit: Well, there you go. I feel very silly. Wikipedia says:

Theorem. If $f \in L^2([−\pi, \pi])$, then the Fourier series converges to $f$ in $L^2([−\pi, \pi])$.

So I guess every such $u \in L^2([0,2\pi])$ has a convergent Fourier series. Thus my proof works for all functions in question.

I really don't know anything about Fourier series! :P

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I suppose that your assumption that Fourier series of $u$ is convergent to $u$ is not essential, because $u$ is in particular integrable and also $u(y) \cosy$ and hence Fourier series of $u(y) cos y$ can be integrated term by term, even if its Fourier series is not convergent. –  Alex Jan 19 '12 at 19:28
    
@Alex thanks for the comment. I went to look at the Wikipedia article and found the relevant theorem about convergence which lays all these issues to rest. –  Bill Cook Jan 19 '12 at 19:38

It may be worth adding that some of the funny behavior of this operator (explicated in Bill Cook's answer) is due to the fact that, in $L^2[0,1]$, say, the integral computes the projection to the one-dimensional space of scalar multiples of $\cos(y)$. Then the coefficient is used to multiply the function $\sin(x)$. So this operator is a rank-one operator of an explicit sort.

Edit: Indeed, as Yemon Choi notes, since $\sin$ and $\cos$ are orthogonal, "of course" the square of the operator is $0$: the operator projects everything to multiples of $\sin$, then "rotates by 90 degrees" to $\cos$, so repeating the projection immediately gives $0$.

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Square-zero rank one, no less –  user16299 Jan 19 '12 at 19:29

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