In a book of functional analysis I encountered this problem
Find spectrum of $Af (x) = \int\limits_0^\pi \sum\limits_{n=1}^\infty 3^{-n} \cos(nx)\cos(nt) f(t)\,dt$ in $L_2[0,\pi]$.
To do this I just proved that the operator is self adjoint and thus its residual spectrum is empty. This sits well with me, but as this problem is the section of Fredholm theory, and I thought that it would be nice to solve it in another way.
For this I was thinking that I have to solve the following integral equation: $$f(x)-\lambda\int\limits_0^\pi \sum\limits_{n=1}^\infty 3^{-n} \cos(nx)\cos(nt) f(t)\,dt=g(x).$$ Substituting it on the operator to find all its eigen values and then argumenting that the residual spectrum is empty.
So to solve the equation, we consider the fact that this operator is self adjoint then we can solve the homogenous equation: $$f(x)-\lambda\int\limits_0^\pi \sum\limits_{n=1}^\infty 3^{-n} \cos(nx)\cos(nt) f(t)\,dt=0.$$ And here is where I am not sure on how to proceed, any help would be really appreciated.
For $n,m\ge 1$ we have $$\int\limits_0^\pi\cos(nt)\cos(mt)\,dt\\ ={1\over 2}\int\limits_0^\pi[\cos(n+m)t+\cos(n-m)t)]\,dt \\ =\begin{cases}{\pi\over 2}& n=m\\ 0 & n\neq m\end{cases}$$ Therefore for $f_n(x)=\sqrt{2\over \pi}\cos(nx)$ the collection $\{f_n\}_{n=1}^\infty $ forms an orthonormal set in $L^2(0,π).$ Moreover $Af_n={\pi\over 2}3^{-n}f_n.$ If $g\perp f_n$ for every $n$ then $Ag=0.$ In particular $A{\bf 1}=0.$ Thus $$\sigma(A)=\{{\pi\over 2}3^{-n}\}_{n=1}^\infty\cup\{0\}$$
Remark 1 The collection $\{f_n\}_{n=1}^\infty\cup\{\pi^{-1/2}\bf 1\} $ constitutes an orthonormal basis of $L^2(0,\pi).$ This follows from the Weierstrass theorem applied to $C[-1,1],$ as $\cos nt=P_n(\cos t),$ where $P_n$ is a polynomial of degree $n.$
Remark 2 Let $\{e_n\}_{n=1}^\infty $ be an orthonormal basis in the Hilbert space and $\{a_n\}_{n=1}^\infty$ be a bounded sequence. Then the operator defined by $$Ax=\sum_{n=1}^\infty a_n\langle x,e_n\rangle \,e_n$$ is hounded and $$\sigma(A)=\overline{\{a_n\, :\,n\in\mathbb{N}\}}$$ The operator $A$ is compact iff $a_n\to 0.$