Welcome to MathOverflow

Question

0

I ask about this claim: let $f$ be an entire function satisfying $f(s)=u(s)f(a-s)$. Assume that $s$ and $a-s$ are not zeroes of $f$ and $f (bar)(a-s)=f(s)$ in a region $D$ ($f(bar)$ is the conjugate of $f$). Then the module of $f(s)/f(bar)(a-s)$ is equal to $1$, implying that the module of $u(s)$ is also $1$. The question is: Does this result implies that in fact the function $u(s)$ is constant. Thank you in advance.

edited Nov 6 at 17:03

asked Nov 4 at 9:02

RH
105●1●9

1

I do not know what you mean by the conjugate of $f$. I see two options: either it is $s\mapsto \overline{f(s)}$ or it is $\bar f$ with $\bar f(s)=\overline{f(\bar s)}$. In the first case, $f$ must be holomorphic and antiholomorphic, hence constant. In the second case, $f(s)=e^{is}$ and $u(s)=e^{-2is}$ with $a=0$ is a counterexample. – anton Nov 4 at 10:14

Yes, it is the first case. Thank you very much. – RH Nov 4 at 11:43

Please use TeX in your questions. – Alexandre Eremenko Nov 4 at 13:07

1

Where I can find it – RH Nov 4 at 13:13

See the "How to write math" box in the right-lower corner of this page. – András Bátkai Nov 5 at 22:07

quid · Accepted Answer · 2012-11-06 16:35:50Z UTC

1

Technically, the answer is yes $u(s)=1$.

However, for a quite boring reason (as hinted at in the comment of Xogn Ambandl):

If $\overline{f(a-s)}=f(s)$, then $2 \ \Re{f}(s)$ would be holomorphic in that region, as it is equal to $f(s)+\overline{f(s)}= f(s) + f(a-s)$ a sum of holomorphic functions. Yet, then as a real-valued holomorphic function $\Re{f}(s)$ is constant. And, so $f(s)$ is constant.

Thus, the only functions $f$ fulfilling your assumptions are constant functions, for which what you ask about is clear.

edited Nov 6 at 16:35

answered Nov 6 at 16:27

quid
13.1k●31●51

But I speak about the modulus of u(s). See this link for a similar answer: mathoverflow.net/questions/85351/… Thanks. – RH Nov 9 at 15:02

What do you mean by you speak about the modulus of u(s), the function u(s) is always equal to 1, what more do you expect to say about it? I tried to explain that the only function f satisfying your condition is a constant function. So f(u)=f(v) for any u,v whatsoever and therefore of course your u(s) is always 1 because in fact f(s) = f(a-s) for f that satisfies the other condition f(s) = f(bar)(a-s). – quid Nov 9 at 17:55

Yes, this is the case. Thank you. – RH Nov 9 at 18:19

1

You are welcome. – quid Nov 9 at 19:59

Functional equation and constant functions

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.