spectrum of convolution integral operator 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 $A$?
 A: 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.
A: *

*By taking Fourier transforms, one gets $\mathcal{F}A\mathcal{F}^{-1} = M_{\hat{h}}$ as operators on $\ell^2(\mathbb{Z})$. Here $\hat{h}$ is a complex valued function defined on $\mathbb{Z}$ ("continuity" of these functions is a mute issue, since the topology of $\mathbb{Z}$ plays no role).  

*@voldemont proof applies, with a few modifications. For the specific case of the question, if $\lambda \ne 0$ is not in the range of $\hat{h}$ then $\hat{h}(n) - \lambda$ is non zero for each $n$, and, since $\hat{h}(n) - \lambda \rightarrow - \lambda \ne 0$ as $|n| \rightarrow \infty$, then $(\hat{h}(n) - \lambda)^{-1}$ is a bounded function of $n \in \mathbb{Z}$, so $M_{\hat{h}} - \lambda$ is invertible on $\ell^2(\mathbb{Z})$.
Note that $0 = \lim_{|n| \rightarrow{\infty}} \hat{h}(n) \in \overline{\hat{h}(\mathbb{Z})}$, but also $0 \in spec(M_{\hat{h}})$, to see why, let $\delta_n$ be the element of $\ell^2(\mathbb{Z})$ defined as $\delta_n(k)=1$ if $k=n, 0$ otherwise; then $M_{\hat{h}}(\delta_n) = \hat{h}(n)\ \delta_n \rightarrow 0$ in $\ell^2(\mathbb{Z})$, but $\|\delta_n\|_2 = 1 \nrightarrow 0$.
