I want to show that for each $f\in\mathcal{C}([0,\pi])$, there exists a unique twice continuously differentiable solution to the Neumann problem on an interval $$ -\frac{d^2u}{dx^2}+u=f$$ on $(0,\pi)$ with boundary conditions $\frac{du}{dx}(0)=\frac{du}{dx}(\pi)=0$. I also want to show that the map that sends $f\mapsto u$ extends to a compact self-adjoint operator on $L^2(0,\pi)$.
It was suggested that I use the fact that $c_k\cos(kx)$ is an orthonormal basis of $L^2(0,\pi)$ for normalizing coefficients $c_k$ but I'm unsure how to proceed. Ideally, I would use Hilbert space theory to solve this.
Thanks!
So here is the general approach. The calculations are omitted, mostly due to my laziness, but this also enables you to complete them on your own.
The basis you have is more than orthonormal - it consists of eigenvectors of your operator (which we call $L$). Call this basis $\varphi_1,\varphi_2,\ldots$. Any function $u$ in your space can be written as$$u=\sum u_i\varphi_i,$$ and since every $\varphi_i$ is an eigenfunction with eigenvalue $\lambda_i$, you obtain$$L(u)=\sum\lambda_iu_i\varphi_i.$$All you have left to do is solve the (very easy) system$$\lambda_iu_i=f_i,\quad i=1,2,\ldots,$$where the $f_i$'s are given by $f=\sum f_i\varphi_i$.