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O depend on G . (K~]). ~s -456. MELNIKOV' S THEOPd~4 We are now p r e p a r e d proceeding with the proof, set some notation. and £n = [Izl For z 6 C from z = 2 -n] of • If p(z,S) S . 1 to prove Melnikov's Finally, = inf Before it will be convenient let S c ¢ theorem. A n = [2 -n-1 ~ we shall write to Izl ~ 2 -n] CS = 2 \ [Iz-x I :x E S] , the distance d(S) = sup [Iz-gl :z,C 6 S] is S . (Melnikov). Let y(CX fl {2 -n-1 _< Iz-xl _< 2 - n ] ) . 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.

Since ~(~) it is clear that any condition which implies that is a peak point for ~) D . R(D) implies x ~(D~ ; o ReR(D), x 6 D is a peak point for and hence a regular point for the Dirichlet problem on Thus theory. 3 strengthen known results from potential -578. APPLICATIONS Melnikov's t h e o r e m gives us useful in the simplest a compact set a sequence (nontrivlal) X of cases. constructed open discs information Consider, as follows. [IZ-Xnl < rn] , where (i) I > Xl > x2 > ... - ~ 0 (2) x I + rI < i ; (3) Xn+ I + rn+ I < x n - r n ; for all n .

D (Kn)]k-i IZn lk k[d(Kn) ]k-i k--2 Y(Kn ) oo ->T~ [l-e~ 8-g~ ]k k--2 Y(Kn) 2n = Cl T ~ >- Cl Y(Kn) " Hence, by (23) and (17), M (2~) M f(0) = (I/9) ~ ~n mn(O) >- (C1/9) ~ 2n Y(Kn) > OI/9 ' n=N which is (***) . n=N This completes the proof. It is worth noticing that Melnikov's theorem can be given a formulation slightly more general than that presented above. Indeed, let 0 < k < 1 y(CX n {~n+l ~ iz_x I ~ kn)). and set yn(k) = Then the proof of Melnikov's theorem shows that X n=O x -n yn(x) = ~, is also necessary and sufficient for point for R(X) .