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Let $(a_n)$ be an enumeration of all complex numbers with rational real and imaginary parts which are not contained in the closed unit disk (i.e., $\{z\in\mathbb{Q}[i] \colon |z|>1\}$).

Let $(c_n)$ be a decreasing sequence of real numbers converging “rapidly” toward zero. (I'll say more on “rapidly” in a moment.)

Define $F(z) = \sum_{n=0}^{+\infty} \frac{c_n}{z-a_n}$ for all $z$ such that this series converges. Since by “rapidly” I mean at the very least that $\sum_{n=0}^{+\infty} c_n$ converges, $F$ is defined at least on the open unit disk $U := \{z \colon |z|<1\}$. Let $f(z) = F(z)$ on $U$. Because the series converges uniformly on every compact domain in $U$, the function $f$ is holomorphic on $U$.

The question: possibly discussing on the meaning of “rapidly”, is it possible that $f$ should have a holomorphic extension on some larger (connected) open set than $U$? (And, if it is possible, is there any relation between the values of that extension with those of $F$ for points outside the unit disk where it is defined?)

Possible assumptions for “rapidly” might be: •the series for $F$ (or possibly even all its derivatives) converges uniformly on the closed unit disk; •or perhaps: $c_n = o(n^{-\alpha})$ for $\alpha > \frac{3}{2}$ say (if my quick computation is correct, this implies that the series for $F$ converges outside a set of Lebesgue measure zero).

[The reason this question came up is that a student asked me for an example of a holomorphic function on the open unit disk which extends continuously to the closed unit disk but not to a larger domain of holomorphy, and my rather stupid reaction was to try to construct it in this way. (Instead, I should really have invoked the Ostrowski–Hadamard gap theorem.)

I asked this question years ago on sci.math.research and I suggested this problem to a number of people, none of whom was able to provide a satisfactory answer.]

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It is perfectly possible for the series to converge to $0$ uniformly in the closed unit disk without having zero coefficients. On the other hand, if the decay of the coefficients is really fast (like $e^{-n}$, say), then, if I remember it right, the function has the unit circle an the natural boundary. My memory is somewhat shaky but, since you are in France, you probably know Nikolay Nikolskii at the University of Bordeaux. He would know for sure :). –  fedja Aug 10 '11 at 17:39
    
Do you think the answer depends on which enumeration you choose? –  Igor Rivin Aug 10 '11 at 21:46
    
@fedja → Right, I should have been more careful about the possibility of converging to $0$ (I thought a mild condition like $c_n = o(n^-\alpha)$ for $\alpha>3/2$ would automatically preclude this, but it seems my reasoning was faulty, so I don't know); I'd love to see an example. If I don't get an answer here, I'll try bothering Nikolai as you suggest. @Igor Rivin → I guess I wasn't too clear on quantifiers. I would be asking: (given some meaning of “rapidly”) does there exist $(c_n)$ decreasing rapidly and does there exist some enumeration $(a_n)$ such that $f$ can be exended? –  Gro-Tsen Aug 10 '11 at 22:32
    
Look up Sibilev, R. V. A uniqueness theorem for Wolff-Denjoy series. (Russian. Russian summary) Algebra i Analiz 7 (1995), no. 1, 170--199; translation in St. Petersburg Math. J. 7 (1996), no. 1, 145–168 30B50 (30B40 47A15 47B38) He was Nikolski's student and this paper of his seems to be most relevant to your question. –  fedja Aug 11 '11 at 1:32

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