Abstract. Let Ω be a Cartan domain of rank r and genus p and Bν, ν>p−1, the Berezin transform on Ω; the number Bνf(z) can be interpreted as a certain invariant-mean-value of a function f around z. We show that a Lebesgue integrable function satisfying f = Bνf = Bν+1f = ··· = Bν+rf, ν ≥ p,must be M-harmonic. In a sense, this result is reminiscent of Delsarte’s two-radius mean-value theorem for ordinary harmonic functions on the complex n-space Cn, but with the role of radius r played by the quantity 1/ν. Let Ω = G/K be an irreducible bounded symmetric (Cartan) domain in its Harish-Chandra realization (i.e. a circular convex domain in Cd centered at the origin), dm the Lebesgue measure on Ω normalized so that m(Ω) = 1, and denote by K(z,w) the Bergman kernel of Ω with respect to dm and by p, r its genus and rank, respectively. For ν ∈ R, it is known [FK1] that the integral c(ν) −1 ∫ Ω K(z,z)1−ν/p dm(z) is finite if and only if ν> p − 1; in that case, one can consider the weighted Bergman spaces A2 ν (Ω) of functions analytic on Ω and squareintegrable against the probability measure dµν: = c(ν)K(z,z) 1−ν/p dm(z). It can be shown that the point evaluations are continuous linear functionals on A 2 ν and the corresponding reproducing kernels are [FK1] Kν(z,w) =K(z,w) ν/p (1) The Berezin transform Bν on Ω is the integral operator defined b
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