An example of a doubly connected domain which admits a quadrature identity

Abstract

In this paper we construct a doubly connected domain D 3 0 such that SSD f (z) do = Af (O) + Bf '(0) for any analytic and area integrable in D function f, which has a single-valued integral in D. l. Introduction. We first introduce the notation (see [1]). Let D be a bounded plane domain. By L' (D) we denote the set of single-valued analytic functions in D which are integrable in D with respect to the areal measure da, and by LI,, (D) the subset of L' (D) consisting of functions with single-valued integral. We say that D admits a quadrature identity (q.i.) relative to L' (D) (or Ll s(D)) if there exist a point z0 E D and complex numbers A, B such that (*) fDfda = Af '(zo) + Bf(z0) for every f E L (D) (or f E Ls &(D)). For a discussion of the background of this problem, see [1]. We note only that for a one point q.i., namely (* *) JD fda = Af (zo) there is no difference between Li and L'5. It can be shown [1, Theorem 7] that the validity of (**) for every f e Ll s(D) implies that D is simply connected and, hence, a disc centered at z0. In the present paper we show that the validity of (*) for allf in L' (D) does not imply that D is a simply connected domain. We prove the following THEOREM. There exists a bounded doubly connected domain D which admits a quadrature identity (*) for all f E L1 s (D). REMARKS. 1. It turns out that the validity of (*) for all f E L'(D) does imply that D is simply connected. This fact was proved by D. Aharonov and H. Shapiro [1, Theorem 4]. Such a domain D can be found explicitly. 2. Our theorem is closely related to a certain minimal-area problem considered in [2]. In fact, our example shows that the method in [2], as it stands now, is not sufficiently strong to conclude that a certain domain D is Received by the editors January 5, 1976. AMS (MOS) subject classifications (1970). Primary 30A80, 30A88. Copyright O 1977, American Mathematical Sowiet%