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I am looking for a possibly simple complex function $f(z)$ with the following properties: $$ f(-z)=-f(z);\quad \forall x\in\mathbb Z_+:\ f(x)=1. $$ Note that the above implies: $$ f(0)=0,\quad\forall x\in\mathbb Z_-:\ f(x)=-1. $$ The function $f(z)$ should be entire or have as few poles as possible (and they should be well determined).

I was not able to find a function with such properties in online sources. Though I am not sure that such function is known I consider this to be possible.

user
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  • What is a well determined pole? – José Carlos Santos Apr 22 '21 at 12:29
  • @JoséCarlosSantos Its position should be exactly known. Especially a residue of a product of $f(z)$ with another function at the pole can be computed. But an entire function would be the best choice. – user Apr 22 '21 at 12:57

2 Answers2

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Consider the function $g(z)=\sum_{n \ge 1}(-1)^n(\frac{1}{z-n}-\frac{1}{z+n}+\frac{2}{n})$

By construction, $g$ is even and meromorphic with poles at the non-zero integers since the series for $g$ converges absolutely and uniformly on compact sets avoiding the non zero integers

($|\frac{1}{z-n}-\frac{1}{z+n}+\frac{2}{n}|=|\frac{z}{n(z-n)}+\frac{z}{n(z+n)}| \le \frac{C_K}{n^2}, z \in K, n \ge n(K)$)

But now clearly $f(z)=\frac{g(z)\sin \pi z}{\pi}$ satisfies the required properties since near each $n$ non zero integer $\frac{\sin \pi z}{\pi}=(-1)^n(z-n)+(z-n)^2h_n(z)$, hence $f(\pm n)=\pm 1, n \ge 1$, while $f$ is clearly entire and $f$ is odd as a product of an even and an odd function

Conrad
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  • Thank you! Can the contour integral $I(a)=\int_{\Gamma(a)}\frac{g(z)}z dz$ along the square $a\to -a^\to -a\to a^\to a$ with $a=(k+1/2)(1+i)$ be evaluated as $k\in\mathbb Z$ goes to infinity? – user Apr 22 '21 at 13:49
  • $g$ has a double zero at the origin so no residue for $g(z)/z$ there, while the residues at $\pm n$ are opposite, so $g(z)/z$ has the same residues there and the integral above should be $2\pi i\sum_{1 \le n \le k} 2(-1)^n/n$ – Conrad Apr 22 '21 at 14:15
  • So it doesn't go to $0$... I should have of course mentioned this additional requirement. – user Apr 22 '21 at 14:35
  • if you ask about $f$ (the function you wanted, not the meromorphic auxiliary $g$) then the result is trivially true since $f$ is entire with a triple zero at the origin, hence $f/z$ is also entire so its integral on any closed curve is zero; $f/(z\sin \pi z)$ has indeed non zero integral on those squares – Conrad Apr 22 '21 at 14:37
  • No. I was interested in the function $\frac {f (z)}{z\sin (\pi z)} $, which is essentially your $g (z)/z $. – user Apr 22 '21 at 14:41
  • you can change $g$ by a constant so $g/z$ has whatever residue you want at zero and in particular, make the integral go to zero (something like $g+2 \log 2$ should work) – Conrad Apr 22 '21 at 14:47
  • Of course. But this is exactly what I wanted to avoid. But thanks to your answer I see now that this is impossible. I should have made an error in my derivation. – user Apr 22 '21 at 16:08
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In fact given a sequence $(z_n)$ of distinct complex numbers with $z_n\to\infty$ and any sequence $(w_n)$ there exists an entire function $f$ with $f(z_n)=w_n$; this is a standard result, see for example the second answer here.

When $z_n=n$ one can give a simple proof:

If $|w_n|\le 1$ there exists an entire function $f$ with $f(n)=w_n$ ($n\in\Bbb Z$).

Let $$s(z)=\begin{cases}\left(\frac{\sin(\pi z)}{\pi z}\right)^2,&(z\ne0), \\1,&(z=0).\end{cases}$$Let $$f(z)=\sum_{n\in\Bbb Z}w_ns(z-n);$$it's a routine exercise to verify that $f$ works.

Finally, note that if $w_{-n}=-w_n$ then the construction above gives an odd function: Since $s$ is even, the substitution $n\mapsto -n$ gives $$f(-z)=\sum w_ns(-z-n)=\sum w_n s(z+n)=\sum w_{-n}s(z-n)=-f(z).$$

(Not that it matters whether we got an odd function or not, since the oddness comes for free: If $g$ works except it's not odd let $f(z)=(g(z)-g(-z))/2$.)

Somewhat more intricate Exercise. Same as above, without the assumption that $|w_n|\le1$.

I thought the following outline worked; thinking about the details that's not so clear:

Show first that $s(z-n)\to0$ uniformly on compact sets, and also that for any $A>0$ there exists $\lambda\in(0,1)$ such that $$|s(t)|\le\lambda\quad(t\in[A,\infty)),$$ and deduce that there exist positive integers $k_n$ such that $f(z)=\sum w_n s(z-n)^{k_n}$ works.

David C. Ullrich
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