Abstract

Let (X, ~) be a connected Riemann domain over a Fr6chet space E, and let ~,~a(X) be the algebra of all holomorphic functions on X, with the compact-open topology. Let So(X) and Sb(X) respectively denote the continuous spectrum and the bounded spectrum of ~(X). If E is finite dimensional, then it is well known that Sb(X) coincides with So(X), and this set, endowed with the structure of a Riemann domain, can be identified with the envelope of holomorphy ~(X) of X. The question as to whether this result holds for infinite dimensional Riemann domains has been investigated by Alexander [11 Coeur~ [3], Hirschowitz [8], and Schottenloher [19, 20, 23]. By adapting a construction of Rossi [18], Alexander [1] endowed So(X) with the structure of a (not necessarily connected) Riemann domain, and proved that if ~c denotes the connected component of Sc(X) containing all point evaluations s with x~X, then the evaluation mapping e :X~Xr is a normal holomorphic extension of X, which is maximal among all normal holomorphic extensions of X. But in general ~ , is not the envelope of holomorphy of X, that is, the holomorphic extension e : X ~ c is not maximal among all holomorphic extensions of X. Indeed, Josefson [10] gave an example of a domain X in a nonseparable Banach space for which not every point of ~(X) defines a continuous complex homomorphism of g(X). However, Coeur6 [3] and Hirschowitz [8] showed that the complex homomorphisms defined by the points of ~(X) are at least bounded, and using this result, Schottenloher [19, 20] proved that t~(X)can always beidentified as a Riemann domain with a suitably structured subset Xb of Sb(X). Thl~(X)us ~( ) slways equals Xb, but all questions concerning the nontrivial relations between c(X), S~(X), J~c, and )~b remained unanswered. In a recent paper, Schottenloher [23] proved that if E is a separable Fr6chet space with the bounded approximation property, then the sets Sc(X), 2~, ~ , and ~(X) all coincide, but the problem as to whether these sets coincide or not with S~(X) remained open. In this note this problem is solved in the affirmative under

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