By Lawrence Zalcman

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**Additional resources for Analytic Capacity and Rational Approximation**

**Sample text**

T® ask is: Melnikov's a complete is a p e a k point answer. w h e n is t h e o r e m allows 0 for R(X) . 1. Let X be of type (L) . 0 Then is a peak OO point for R(X) if and only if (rn/Xn) = . n=l Proof. Let I = CX n [0, I] , I n = I n A n , Jn = I n A n . First we show that ]I x-1 dx diverges with ~ (rn/Xn) . Indeed, we have O0 CO ~ rn ~ 2rn n=l q -< X~rn n=l Hence, n~l -< =i Jn dx ~ dx x : I x " for the sufficiency it is enough to show that is a peak point if [ x -I dx = ~ . Suppose, 0 then, that I the integral diverges.

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. 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 .

C. Curtis Then if take . x = 0 . < rn])/r n > e > O c . 1. 9. Then be [iI$]. -53- for large k n kk+l _< rn < ~k , where , of course, we have depends upon n . 9 twice, we obtain ;kk+l e _< r n g _< y(CX CI [ I z l < rn} ) < c y(ox n {Izl < ~k+21)+ oy(cx n {~k+2 < Izl < ~k]) < c}k+2 + c2 Yk ( )") + 02Yk+l ( ~") where " yn(k)" = y(CX CI [kn+l_< Izl _< } n ] ) . Hence , ~k+l o • (~- ~) _< Yk(~) + yk+l(~) • It follows that @o a-n yn(X ) = n=O By the remark following the proof of Melnikov's theorem we are done.