Volterra inverse differentiation has empty spectrum
= Volterra inverse differentiation has empty spectrum
{c}
{title2=$Af=if',\quad D(A)=\{f\in H^1(0,1):f(0)=0\},\quad\sigma(A)=\varnothing$}
The <Volterra integration operator> $J$ has spectrum $\{0\}$ and is injective. Hence $(\lambda I-iJ^{-1})^{-1}=J(\lambda J-iI)^{-1}$ exists and is bounded for every complex $\lambda$. Empty spectrum is possible for closed unbounded operators, unlike bounded operators on a nonzero complex Banach space.
// Target: analysis.bigb