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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.

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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 '12 at 10:14
Yes, it is the first case. Thank you very much. – China-Hong Kong Nov 4 '12 at 11:43
Please use TeX in your questions. – Alexandre Eremenko Nov 4 '12 at 13:07
Where I can find it – China-Hong Kong Nov 4 '12 at 13:13
See the "How to write math" box in the right-lower corner of this page. – András Bátkai Nov 5 '12 at 22:07
up vote 1 down vote accepted

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.

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But I speak about the modulus of u(s). See this link for a similar answer:… Thanks. – China-Hong Kong Nov 9 '12 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 '12 at 17:55
Yes, this is the case. Thank you. – China-Hong Kong Nov 9 '12 at 18:19
You are welcome. – quid Nov 9 '12 at 19:59

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