Eigenvalues of a kernel integral operator

Question

I need to find the eigenvalues of the following operator $T: L^2(0,2\pi) \to L^2(0,2\pi)$ defined as $Tf = \int_0^{2\pi} sin(x+y)f(y)dy$.

I know that this is a Hilbert-Schmidt operator, so it is compact, yet I cannot find any eigenvalues when I apply the condition $\int_0^{2\pi} \sin(x+y)f(y)dy = \lambda f(x)$, since I obtain that $f$ is the zero function, which means that there are no eigenvectors. But a compact operator must have a singular-value decomposition, so there must be a basis of eigenvectors!

Can anybody point out where I am doing wrong? Thank you very much.

Answer 1

Hint:

Take Fourier transform from both sides.

Answer 2

Suppose $\lambda f(x)=\int_0^{2\pi}\sin(x+y)f(y)dy=\sin x\int_0^{2\pi}\cos yf(y)dy+\cos x\int_0^{2\pi}\sin y f(d)dy$ then $$f(x)=a\sin x+b\cos x.$$ But $\lambda a\sin x+\lambda b\cos x=\lambda f(x)=\lambda(\sin x \int_0^{2\pi}(a\sin y+b\cos y)\cos ydy+\cos x \int_0^{2\pi}(a\sin y+b\cos y)\sin ydy)=\lambda \pi b\sin x+\lambda\pi a \cos x.$ for non zero $f$, the only possibly is that $\lambda=0$.

Note that if an operator is not one-one then zero is an eigenvalue and the kernel is the corresponding eigenspace.