How to prove that eigenfunction of translationally invariant continuous operator $ K(t-t') $ is $ \exp(iwt) $?

Question

I was studying a book in computational neuroscience where I came across the following equations: $$ \int{W(t,t')e(t')dt'}=\lambda e(t) $$ and read that if $ W(t,t')=K(t-t') $ then the eigenfunction $ e(t) $ is a complex exponential. Could someone explain why?

Answer 1

If $T$ is a linear map, say $C^0(\Bbb{R})\to L^1_{loc}(\Bbb{R})$, such that for all $g,a$, $$T(g(.+a)) = (Tg)(.+a)$$ then every exponential $e^{st}, s\in \Bbb{C}$ is an eigenfunction.

If $\|T(g)\|_2\le C \|g\|_2$ then $T$ is said bounded in $L^2$ norm.

In that case let $H$ such that $T(e^{2i\pi y t}) =H(y)e^{2i\pi yt}$ for $y \in \Bbb{R}$. Then $T(g)= \mathcal{F}^{-1}(H.\mathcal{F}(g))$ (the Fourier transform).

It is hard to make it rigorous (this is the point of functional analysis) but you can think to the Fourier transform as diagonalizing $T$ (in a way similar to the diagonalization of a self-adjoint matrix).

If $H(y_1)=H(y_2)$ then $e^{2i\pi y_1 t}+e^{2i\pi y_2 t}$ is a non-exponential eigenfunction.