Functional equation and constant functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:43:44Z http://mathoverflow.net/feeds/question/111437 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111437/functional-equation-and-constant-functions Functional equation and constant functions RH 2012-11-04T09:02:40Z 2012-11-06T17:03:33Z <p>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.</p> http://mathoverflow.net/questions/111437/functional-equation-and-constant-functions/111662#111662 Answer by quid for Functional equation and constant functions quid 2012-11-06T16:27:42Z 2012-11-06T16:35:50Z <p>Technically, the answer is yes $u(s)=1$. </p> <p>However, for a quite boring reason (as hinted at in the comment of Xogn Ambandl):</p> <p>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 <em>real-valued</em> holomorphic function $\Re{f}(s)$ is constant. And, so $f(s)$ is constant. </p> <p>Thus, the only functions $f$ fulfilling your assumptions are constant functions, for which what you ask about is clear.</p>