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Let $Z\subseteq \mathbb{C}\setminus \overline{\mathbb{D}}$ be countable and discrete (here $\mathbb{D}$ stands for the unit disc).

Consider a function $f\colon \mathbb{D}\cup Z\to \mathbb{C}$ such that

1) $f\upharpoonright \overline{\mathbb{D}}$ is continuous

2) $f\upharpoonright \mathbb{D}$ is holomorphic

3) if $|z_0|=1$ and $z_n\to z_0$, $z_n\in Z$ then $(f(z_n)-f(z_0))/(z_n-z_0)\to f^\prime(z_0)$

Can $f$ be extended to a holomorphic function on some domain containing $Z$?

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How does the third condition work if $z_0$ is neither in $\mathbb D$ nor in $Z$? –  Henning Makholm Oct 22 '11 at 20:53
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Is the following a counterexample: $Z=\{2,3,4,\ldots\}$, $f(z)=0$ for $z\in \overline{\mathbb D}$ and $f(n)=n$ for $n\in Z$? –  Henning Makholm Oct 22 '11 at 20:55
    
@Henning: I was puzzled about that, too (I didn't have that slick a counterexample, though), but I thought I'd give a family of interesting examples (lacunary series) that can't be extended in any way to an analytic function beyond $\mathbb{D}$. –  t.b. Oct 22 '11 at 21:03
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2 Answers

up vote 3 down vote accepted

No.


The function $g(z) = 1+ 2z + \sum_{n=1}^{\infty} 2^{-n^2} z^{2^n}$ is holomorphic on the open disk $\mathbb{D}$ and infinitely often real differentiable in any point of the closed disk $\overline{\mathbb{D}}$ but cannot be analytically extended beyond $\overline{\mathbb{D}}$:

The radius of convergence is $1$. For $n \gt k$ we have $2^{nk} 2^{-n^2} \leq 2^{-n}$, hence the series and all of its derivatives converge uniformly on $\overline{\mathbb{D}}$, thus $g$ is indeed smooth in the real sense and holomorphic on $\mathbb{D}$. By Hadamard's theorem on lacunary series the function $g$ cannot be analytically continued beyond $\overline{\mathbb D}$.

[In fact, it is not difficult to show that $g$ is injective on $\mathbb{\overline{D}}$, so $g$ is even a diffeomorphism onto its image $g(\overline{\mathbb D})$ — but that's not needed here.]

I learned about this nice example from Remmert, Classical topics in complex function theory, Springer GTM 172, Chapter 11, §2.3 (note: I'm quoting from the German edition). The entire chapter is devoted to the behavior of power series on the boundary of convergence and gives theorems that provide positive and negative answers on whether a given power series can be extended or not.


Now, to get a counterexample, apply Whitney's theorem to extend $g$ to a smooth function $f$ on all of $\mathbb{C}$. Every restriction of $f$ to any set of the form $\overline{\mathbb{D}} \cup Z$ (I think that's what was intended) as in your question will provide a counterexample.

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As has already been mentioned, the answer is no. The identity theorem tells you that there is at most one way to extend $f$ analytically beyond the unit disk, which shows you that the answer must be negative (as per Henning's comment above).

Another fact to observe in this context is that there are functions that are analytic on the unit disk and extend smoothly to the boundary, but do not extend analytically beyond the boundary anywhere. (Take a Riemann map to a Jordan domain whose boundary is smooth but not analytic anywhere.)

However, even without any assumptions on derivatives, a continuous function as in your statement can be uniformly approximated arbitrarily closely by entire functions; this follows from Arakelyan's approximation theorem.

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No, you don't need Mergelyan. Arakelyan's theorem is enough here. (You use a little trick, first approximating the logarithm of your error function by an entire function.) –  mathstribble Mar 30 '13 at 16:06
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