spectrum of convolution integral operator

Question

Let $A f(x)= \int_{-\pi}^{\pi} h(x-y) f(y) dy$ operator $L^2( {-\pi},{\pi})->L^2( {-\pi},{\pi}), h$ is continuous, periodic with period $2\pi$ and $h(x)=h(-x)$ on $ [ {-\pi},{\pi}] $.

How can I find the spectrum of $A$?

Answer 1

Take the Fourier transform. Then $\hat{Af}=\hat{f}\hat{h}$, and therefore we have turned convolution into multiplication.

Now, for the multiplication operator, the spectrum is the range of $h$ (since $h$ is cont. range=essential range).

As the Fourier transform is a unitary, we get that spectrum of $A$ is also the range of $h$.

Edit: Let $M_h$ be the multiplication operator on $L^2(-\pi,\pi)$, and $h$ is cont. Then spectrum of $M_h$ is the range of $h$.

Proof: $(M_h-\lambda I)f(x)=f(x)h(x)-\lambda f(x)=f(x)(h(x)-\lambda)$. So, if $\lambda$ does not belong to range of $h$ then $h(x)-\lambda$ is non zero for each $x$, and hence $(M_h-\lambda I)$ is invertible. The converse is also trivial from the above computation.