A counterexample to a theorem of Bremermann on Shilov boundaries - revisited

Abstract

∂BD ⊂ ∂SD ⊂ ∂D. Notice that, in general, ∂BD ∂SD, e.g. for the domain D := {(z, w) ∈ C : 0 < |z| < 1, |w| < |z|− log |z|}. The algebra A(D) (resp. B(D) := the uniform closure in A(D) of O(D)) endowed with the supremum norm is a Banach algebra. A point a ∈ D is called a peak point for A(D) (resp. B(D)) if there is an f ∈ A(D) (resp. B(D)) with f(a) = 1 and |f(z)| < 1 for all z ∈ D {a}; f is called an associated peak function. It is known the peak points of A(D) (resp. B(D)) are dense in ∂SD (resp. ∂B(D))). Assume that the envelope of holomorphy D of D is univalent. Note that ∂SD ⊂ ∂SD and ∂BD ⊂ ∂BD. In the paper: [JP] M. Jarnicki, P. Pflug, A counterexample to a theorem of Bremermann on Shilov boundaries, Proc. Amer. Math. Soc. 143 (2015), 1675–1677,