I'm dealing with this problem:
Let $f(t) = (1-\frac{e^{it}}{2})^{-1} \in L^2(-\pi,\pi)$ and let $T$ be the Operator with integral Kernel $K(x,y)= f(x-y)$
I have to find eigenvalues and eigenvectors of $T$. Also I have a Hint which suggests me to write $f(t)$ in complex Fourier sum form. Obviously I noticed that \begin{equation} f(t) = \frac{1}{1-\frac{e^{it}}{2}} = \sum_{n \ge 0}\frac{e^{int}}{2^n} \end{equation}
And $2^{-n}$ should be my Fourier coefficients..
But How Should I proceed?
Sketch solution:
1) Show that $T : L^2(-\pi, \pi) \to L^2(-\pi, \pi)$ is the operator $$ T(u)(x) = \sum_{n \geq 0} \frac{e^{inx}}{2^n}\int_{-\pi}^\pi dy \ e^{-iny} u(y)$$
2) Every $u \in L^2( - \pi, \pi)$ can be written as a Fourier series, $$ u(y) = \sum_{k \in \mathbb Z} c_k e^{iky}.$$ Show that if $u$ has the above Fourier representation, then $$ T(u)(x) = 2\pi \sum_{n \geq 0} \frac{c_n }{2^n} e^{inx}.$$
3) Deduce that the eigenvectors are $e^{inx}$ (for $n \in \mathbb Z$), with corresponding eigenvalues are $2\pi / 2^n$ (if $n \geq 0$) or $0$ (if $n < 0$).