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Inspired of course by http://mathoverflow.net/questions/20688/whats-a-natural-candidate-for-an-analytic-function-that-interpolates-the-tower-f I am minded to ask what looks to me like a more natural question: given a sequence $a_1,a_2,a_3,\ldots$ of complex numbers, is there always a holomorphic function $f$ defined on the entire complex plane, with $f(n)=a_n$ for $n=1,2,3,\ldots$? No idea what the answer is myself, but wouldn't surprise me if it were well-known and even easy.

flag	asked Apr 8 2010 at 12:06 Kevin Buzzard 22.8k●2●70●137

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Sure. More generally, for any Stein space $X$ and discrete set $S$ in $X$ and effective divisor $D$ supported on $S$, the surjective map of coherent sheaves $O_X \rightarrow O_D$ has coherent kernel and so induces a surjection on global sections. So by description of $O_D(X)$ via discreteness of $S$, there exists holomorphic $f$ on $X$ whose germ at each point of $S$ has whatever "initial part of Taylor expansion" we wish. In dimension 1 can play similar game with meromorphic $f$ holomorphic outside $S$ and Laurent tails at $S$ (generalizing Mittag-Leffler theorem) – BCnrd Apr 8 2010 at 12:24

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Typo correction above: $D$ isn't a divisor when $X$ has dimension $> 1$. I meant it to be a 0-dimensional analytic space structure on $S$ (of which there are zillions of choices as "multiplicity" grows). Presumably this intent was clear. – BCnrd Apr 8 2010 at 15:03

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This question was asked twice before: mathoverflow.net/questions/2944, mathoverflow.net/questions/7328/… – Jonas Meyer Apr 8 2010 at 15:47

@Jonas: But now with a completely different solution. :) – BCnrd Apr 8 2010 at 18:24

As Brian Conrad said, the important property is that $\mathbb{Z}\subset \mathbb{C}$ is a discrete subset. You can take any collection of isolated points $\Omega\subset \mathbb{C}$ and define an entire function with any values you want at the points of $\Omega$. My favorite consequence: If you define the sum of divisors function for the Gaussian integers, there is an entire function that outputs the sum of the divisors of the input when the input is of the form $a+bi$, $a, b\in \mathbb{Z}$. – Matt Apr 8 2010 at 23:58

Harald Hanche-Olsen · Answer 1 · 2010-04-08 19:17:01Z UTC

This is Exercise 6, Page 26, of Knopp's Problem Book in the Theory of Functions, Volume 2: For any sequence of complex numbers $z_n$ with no finite limit point, and for any sequence of complex numbers $w_n$, there is an entire function mapping $z_n$ to $w_n$. The proof goes like this: Use the Weierstrass Factor Theorem to construct a function $W$ with simple zeros at the $z_n$. Use the Mittag-Leffler theorem to construct a function $M$ with simple poles at the $z_n$ with residues $\frac{w_n}{W'(z_n)}$. Then the function $W\cdot M$ does the job.

Harald Hanche-Olsen · Answer 2 · 2010-04-08 12:27:11Z UTC

Probably well-known. Easy? I'd venture to guess that an expression like $$\sum_{n=1}^\infty b_n\frac{e^{c_n(z-n)}}{(n-1)!}\prod_{k=1}^{n-1}(z-k)$$ can be made to work. You'll have to pick the $b_n$ successively to make the $N$'th partial sum equal to $a_N$ , and real constants $c_n$ large enough to obtain uniform convergence to the left of any fixed vertical line. E.g., so that the $n$'th term has absolute value less than $2^{-n}$ when $\operatorname{Re} z<n/2$ .

Does there exist a holomorphic function which takes given values on the positive integers?

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