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.
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.
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
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 $0$.