analytic extension

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 $|z|<2$.

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I think this has to do with Laurent series but I could be wrong. I also thought of maybe showing $f$ has a removable singularity at 0, but I could not get it to work. –  john May 26 '12 at 14:09

Take the Laurent series for $f$ in the annulus and calculate the coefficients of negative powers using the polynomial approximation.

Conclude that the Laurent series is a power series.

A power series that converges in the annulus already converges in the disc.

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thank you for the hint. –  john May 26 '12 at 14:28
@john Do you know how to accept answers? –  AD. May 26 '12 at 16:52
"A power series that converges in the annulus already converges in the disc" Why is this true? –  john May 26 '12 at 19:53
A power series with powers of $z$ has a radius ofconvergence. It converges inside the radius, and it diverges outside (the cercle itself is a more delicate question). So, when you know it converges for points of radius 1.5, it automatically converges of all smaller ones. –  Phira May 26 '12 at 20:03
okay. Thank you –  john May 26 '12 at 20:20