By Lawrence Zalcman

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**Sample text**

For R(X) x E X and let Then x Yn = is a peak point if and only if @O 2n Yn = ~ n=0 Proof. For convenience First, positive assume integer poles lie off we may suppose x = 0 . that the series above N X . and let We modify f converges. be a rational f S . off function a neighborhood in such a way that (1) IIflls2_<211 fllx (2) IIfllrN_< 211 fllrNnX Fix of a whose X -#6- Choose a compact set X c int K and boundary of (3) f K with smooth boundary so that is analytic on CK 0 [Izl _< 2 -N] • f(0) = ~ ~ K .

A peak point for Proof. R(X) ~ =~4n Theorem 7 •3. c x n {Iz-xl x Proof. = = 4n mn = = then (2 n ~/'Wnn)2 _<~ (~)(2n~ r] x E X (We h a v e . 12). Let < theorem. mn = ~ and ~n If Now apply Melnikov' s theorem. w n _< n4-n of Melnikov's is a peak point for dr = = . R(X) . We have ~o Clearly, value by = o ~(r~~ ( - ½ 7)=-~. 3 follows Although sufficient necessary. 2. 2 and 7 . 3 are t h e y are far f r o m A n = [Iz-3 -2-n-21 < n-12-n] and oO set X = A \ ~ that, but for An One s h o u l d continuous .

11) are of con- siderable interest in themselves. 1. Let K be analytic off > 0 . Proof. Then Let for the Our treatment follows [64] quite closely. be a compact set in the plane. K , f(~) = 0 , and If'(~)l _< 2~¥(K) f(z) = ~ A n z -n Let f Re f(z) < a , where . at Then the admissible ~ . n=l functior~ fl(z) = f(z)[f(z) fl(z) = (-Al/2~)z-1 + ... 2. Let - 2a] -I at a I! f ll,,. 1 If' (a)ip(p-~,(K)) Proof. 10, we may assume that Then I = to K . a 6 G(K) , and let the distance from . If' Thus as required.