Wielandt's theorem, spectral sets and Banach algebras (Q389544)

Let \(A\) be a unital complex Banach algebra. A closed subset \(S\) of \(\mathbb{C}\) containing the spectrum \(\sigma(a)\) of \(a\in A\) is a spectral set of \(a\) if \(\| f(a)\|\leq\sup_{z\in S}| f(z)|\) for each rational function \(f\) which does not have poles in \(S\). The element \(a\) is a von Neumann element if \(\sigma(a)\) is a spectral set of \(a\). Let \(a,b\in A\) be von Neumann elements. Suppose that \(\sigma(b)\subset K\), where \(K\) is the image by some Möbius transformation of either the open or the closed unit disk. Then the authors show that \(\sigma(a+b)\subset \sigma(a)+K\). Further, \(\sigma(ab)\subset\sigma(a) K\), in the case where at least one of the elements is invertible.