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I've been stuck forever with this old exam question:

Is there a holomorphic function $f : \mathbb{C} \to \mathbb{C}$ with $f(n) = f(-n) = \frac1{n^2}$ for all $n \in \mathbb{N}$?

The next question asks whether there is an entire function with $f(n) = f(-n) = \frac1{n^3}$, so I guess there is some difference.

What I did so far:

  • Trying to find a holomorphic function that meets the conditions. I came up with $f(z) = \frac{\sin^2(2\pi z)}{z^2}$, which is holomorphic but doesn't quite work. $f(z) = \frac{\sin^2(2\pi z + \frac{\pi}2)}{z^2}$ matches the points, but isn't holomorphic.

  • I read this related question, which uses the identity theorem. However, I don't see a way to use the identity theorem: the given points don't even have a limit point in $\mathbb{C}$.

  • As an alternative view on the problem, I tried to check if there is a harmonic function $u : \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ matching the points, which could be extended to a holomorphic function.

Sergjo
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    Maybe the answers to this question may help. – Bruno Krams Oct 06 '19 at 12:45
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    This looks related https://math.stackexchange.com/questions/626492/analytic-function-taking-prescribed-values/626598 [oops, Bruno's comment didn't load when I posted, and the link isnt the same] – Calvin Khor Oct 06 '19 at 12:59
  • Thank you, yes, that is indeed helpful! Apparently, the lecture stopped just before introducing the underlying theorems. I'm reading up on that now. – Sergjo Oct 06 '19 at 13:05

1 Answers1

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One can prescribe the values at infinitely many points as long as these don't have accumulation points (as per the questions linked in the comments). In the first case one can also give an explicit example: $$f(z)=\frac{1-\frac{\sin(\pi z)}{\pi z}}{z^2}.$$ For the second one, I don't think one can really get anything simpler than $$f(z)=2z\frac{\sin(2\pi z)}{2\pi}\sum_{n=1}^\infty\frac1{n^3(z^2-n^2)}$$ (which could be expressed in terms of the polygamma function).

EDIT: Actually, one can slightly modify the series for the second problem getting something more explicit: $$g(z)=\frac{\sin(2\pi z)}{\pi z}\sum_{n=1}^\infty\frac1{n(z^2-n^2)}=\frac{\sin(2\pi z)}{2\pi z^3}\Big(2\gamma +\frac{\Gamma'}{\Gamma}(1+z)+\frac{\Gamma'}{\Gamma}(1-z)\Big)$$ where $\gamma=-\Gamma'(1)=-\frac{\Gamma'}{\Gamma}(1)$ is Euler's constant. In fact the fact that this last expression give a good example could have been guessed directly since $\Gamma$ has simple poles at non-positive integers and no other pole or zero.

Sbt
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