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???
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