# Prove that Volterra integral operator with deviating argument has no eigenvalue

Consider the space of continuous real-valued functions $$C[a,b]$$ and an integral kernel $$k \in C([a,b]\times[a,b])$$, where $$b-a<1$$. The operator $$K: L^2[a,b]\to L^2[a,b]$$, $$K f(x) =\int_{a}^{b-x+a} f(s) k(s,x) ds, \quad f\in L^2[a,b]$$ is Hilbert-Schmidt. Prove that the linear operator $$K$$ has no eigenvalues.

• Of course, $0$ can be an eigenvalue. For $\lambda\not= 0$, I think you can observe that an eigenfunction would have to be continuous, then solve by iteration and control the $L^{\infty}$ norm. – Christian Remling Jan 28 at 18:28