By Kirkwood J.R.

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Now, for z' fixed, the function h(z', ·)is holomorphic on lznl ::;; on, with the possible exception of finitely many points, namely the k zeroes off lying over z'. By the classical one variable theorem of Riemann (see [Ahl], p. 2) equals h(z', zn) if (z', Zn) ¢E. Thus H = h on P(O, o) - E. 2) in the case just discussed. The main difficulty then involves proving that the integral indeed defines an extension of the given function. 1. 5. As in case n = 1, weaker growth conditions for h are sufficient for the existence of a holomorphic extension across thin sets.

Let r = r(w) = (lw 1 l, ... , lwnl). Then the power series Icvzv converges on the polydisc P(O, r). Moreover, the convergence is normal in the following sense: if K c P(O, r) is compact and e > 0 is arbitrary, there is a finite set A = A(K, e) c ~n, such that for all zEK. PROOF. Given K cc P(O, r), choose 0 < il < 1, such that K c P(O, ilr). lvl = ::5; lcvwvlillvl (LJ'=o A. it < ::5; Millvl for VE ~n. oo, the result follows. 16. 31). The convergence is normal inn. )(Q). 17. f(O) = oc! Ca. 32) PROOF.

Since E is nowhere dense, the extension H -if it exists-is determined uniquely by h. Therefore it is enough to construct a holomorphic extension of h to a neighborhood of an arbitrary point pEE. 1-fis zn-regular of some order k. 2) H(z', zn) = (2nir 1 I h(z', () d( Jl~l=~n Zn - ( 33 §3. Zero Sets of Holomorphic Functions clearly is defined and holomorphic on P(O, o). Now, for z' fixed, the function h(z', ·)is holomorphic on lznl ::;; on, with the possible exception of finitely many points, namely the k zeroes off lying over z'.

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