Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $|z|<2$.

share|improve this question
1  
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
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.