Volterra inverse differentiation has empty spectrum ($Af=i{f}^{\prime },D(A)=\{f\in H^1(0,1):f(0)=0\},\sigma (A)=\varnothing$)

The Volterra integration operator $J$ has spectrum $\{0\}$ and is injective. Hence $(\lambda I-i{J}^{-1}{)}^{-1}=J(\lambda J-iI{)}^{-1}$ 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