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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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Did you mean to say $f(s) = \kappa \cdot \xi(s)$ for some $\kappa > 0$ ? – Sasha Apr 20 '12 at 18:55
aha.. considering $ f(s)$ is positive and differentiable for each real number , so we avoid the solutions similar to $ |s(1-s)| $ – Jose Garcia Apr 20 '12 at 19:22
False. Try $f(s) = \cos(2\pi s) + 2$. – KCd Apr 20 '12 at 20:17
Or, more simply, $f(s) = (s(1-s))^2 + 1$. You should look up Hamburger's theorem if you're trying to find conditions under which the completed zeta-function is characterized up to a scaling factor. – KCd Apr 20 '12 at 20:23
where can i find hamburguers theorem ?? – Jose Garcia Apr 20 '12 at 20:49

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, 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:


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

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