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Suppose that $f$ is analytic in the annulus $1<|z|<2$ and there exist a sequence of polynomials converging to $f$ uniformly on every compact subset of this annulus. Show $f$ has an analytic extension to all of the disc $|z|<2$.

It was answered on analytic extension.

Can anyone please show me how to calculate the coefficients of negative powers using the polynomial approximation???

J. W. Tanner
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ks1
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1 Answers1

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It is enough to have uniform polynomial approximation on a circle $C$ inside the annulus, say $|z|=c, 1<c<2$. Then if $||P_m-f||_{\infty}=\max_{|z|=c}|P_m(z)-f(z)| \to 0$ and $f(z)=\sum_{-\infty}^{\infty}a_nz^n$ we have by the triangle inequality

$2\pi|a_n|=|\int_Cf(z)z^{-n-1}dz| \le |\int_C(P_m(z)-f(z))z^{-n-1}dz|+|\int_CP_m(z)z^{-n-1}dz|\le$

$\le \int_C|P_m(z)-f(z)||z^{-n-1}dz| \le 2\pi c^{-n}||P_m-f||_{\infty} \to 0$ as $m \to \infty$

for all $n \le -1$ since then $-n-1 \ge 0$ and $P_m(z)z^{-n-1}$ is a polynomial too so has zero integral on $C$ by Cauchy

Conrad
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  • And this $||P_m-f||{\infty}=\max{|z|=c}|P_m(z)-f(z)| \to 0$ holds because of uniformly convergence ? – ks1 Apr 20 '20 at 16:32
  • that is the definition of uniform convergence as $|P_m-f|$ gets small uniformly in $z \in C$ which is equivalent to $\max|P_m-f|$ getting small- as noted it is enough to hold on a circle (or more generally a rectifiable Jordan curve), no need to hold on compact sets within the full annulus – Conrad Apr 20 '20 at 16:55
  • Thanks for clarification. – ks1 Apr 20 '20 at 17:10
  • happy to be of help – Conrad Apr 20 '20 at 17:18
  • sorry...but what about n= -infinity, is still an=0? – ks1 Apr 20 '20 at 23:21
  • $n$ is always finite - $\sum_{-\infty}^{\infty}$ means the usual sum where $n=0,1,-1,2,-2...$ – Conrad Apr 20 '20 at 23:23