Let $h$ be a continuous function with period $2\pi$. Define $T : L_2[−\pi, \pi] \to L_2[−\pi, \pi]$ by $(T f)(t) = \int \limits _{-\pi}^\pi h(t − s)f(s)ds$.
Let $\{\varphi_n(t) =\frac{1}{\sqrt{2\pi}} e^{int}\}_{n\in \mathbb Z}$ be the known trigonometric orthonormal basis for $L_2[−\pi, \pi]$.
Show that $\varphi_n$ is an eigenvector of the eigenvalue $\langle h(t), e^{int}\rangle$ for the linear operator $T$. Write $T(f)$ as $\sum \limits _{n \in \mathbb Z} \mu_n \langle f, \varphi_n\rangle \varphi_n$.
I have the first part but how do I write $T(f)$ as $\sum \limits_{n \in \mathbb Z} \mu_n \langle f, \varphi_n\rangle \varphi_n$?
Notice that you may expand $h$ as $h = \sum \limits _{n \in \Bbb Z} a_n \varphi_n$, so that
$$h (t-s) = \sum \limits _{n \in \Bbb Z} a_n \varphi_n (t-s) = \sum \limits _{n \in \Bbb Z} a_n \frac 1 {\sqrt {2 \pi}} \Bbb e ^{\textrm i n (t-s)} = \sum \limits _{n \in \Bbb Z} \sqrt {2 \pi} a_n \varphi_n (t) \overline {\varphi_n (s)} .$$
Inputting this in the integral that gives $T$ produces
$$\begin{align} (Tf) (t) =& \int \limits _{-\pi} ^\pi \sum \limits _{n \in \Bbb Z} \sqrt {2 \pi} a_n \varphi_n (t) \overline {\varphi_n (s)} f(s) \ \Bbb d s \\ =& \sum \limits _{n \in \Bbb Z} \sqrt {2 \pi} a_n \int \limits _{-\pi} ^\pi f(s) \overline {\varphi_n (s)} \ \Bbb d s \ \varphi_n (t) \\ =& \sum \limits _{n \in \Bbb Z} 2 \pi \sqrt {2 \pi} a_n \langle f, \varphi_n \rangle \ \varphi_n (t) \end{align}$$
and if you let $\mu_n = \sqrt {2\pi}^3 a_n$ and also drop the argument $t$, the above rewrites as
$$Tf = \sum \limits _{n \in \Bbb Z} \mu_n \langle f, \varphi_n \rangle \ \varphi_n .$$
(The $2\pi$ comes from the fact that $\langle f, g \rangle = \frac 1 {2\pi} \int \limits _{-\pi} ^\pi f(s) \overline {g(s)} \ \Bbb d s$.)