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Suppose I have (roughly speaking) a multivalued meromorphic function $f(z)$ on all of $\mathbb{C}$ that is single-valued and holomorphic on the open unit disc and has some branch points of finite order (but no poles) on the unit circle. Does the Taylor series always converge uniformly to $f(z)$ on the closed unit disc? This seems very likely but I do not think I have ever seen a proof.

I know an argument that works for the simplest possible case, namely the function $f(z)=\sqrt{1-z}$.

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The point is that it is not clear whether the Taylor series at $0$ of $f$ is even convergent on any point of the boundary. Or is it? –  Loïc Teyssier May 14 '13 at 20:17
    
This is by Abel's theorem. –  Alexandre Eremenko May 14 '13 at 21:21
    
Wouldn't uniform convergence on your disk imply that your function admits its derivative at the branch point? (By a "switch-sum-and-differentiation"-type argument.. Or am I wrong?) –  Tommaso Centeleghe May 14 '13 at 21:43
    
@Alexandre: how does this follow from Abel's theorem? That tells us that if we have a point $u$ on the unit circle and the Taylor series converges at $u$, then the Taylor series converges uniformly on the line segment $[0,u]$. It is not obvious that the hypothesis is satisfied, and even if it is, the conclusion is weaker than uniform convergence on the whole closed disk. –  Neil Strickland May 14 '13 at 21:47
    
@Tommaso: I don't think we can switch the sum and the derivative here. The usual argument for that involves a contour integral, and the contour would have to wind around the branch point, which would break the proof. –  Neil Strickland May 14 '13 at 21:50

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