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Is there any theorem or proof that if a function satisfy the functional equation $ f(1-s)=f(s)$ and $ f(s) >0$ for each real $s$ then $ f(s)= \xi(s)$ or $ f(s)= \operatorname{const}$?

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    Slightly tangential: Isn't the Riemann function denoted $\zeta(x)$ ("zeta"), not $\xi(x)$ ("xi")? Edit: Okay, I read http://mathworld.wolfram.com/Xi-Function.html. I learned something new today! Yay!2012-04-26

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If $f(s)$ is a solution then so is $f^2$ or $e^f$ or $H(f(s))$ for any positivity-preserving function $H$. The functional equation alone does not characterize the (completed) zeta function up to a finite number of parameters.

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The question is wrong.

In fact for the functional equation $f(1-s)=f(s)$, its general solution, according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1113.pdf, should be $f(s)=\Phi(s,1-s)$, where $\Phi(s,1-s)$ is any symmetric function of $s$ and $1-s$.

And also there are infintely many $\Phi(s,1-s)$ satisfly $\Phi(s,1-s)>0$ for each real $s$.

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Let $g(z):=f(s)=f((1/2)+iz)$. Then we have $f(1-s)=f((1/2)- iz)=g(-z)\tag{1}$ $f(s)=f((1/2)+ iz)=g(z)\tag{2}$

Using functional equation, we have:

$g(-z)=f(1-s)=f(s)=g(z)\tag{3}$.

Thus as long as $g(z)$ is an even function of $z$, $f(s)=g(-is-1/2)$ will satisfy the functional equation $f(s)=f(1-s)$.