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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 $\kappa > 0$ ?2012-04-20
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    aha.. considering $ f(s)$ is positive and differentiable for each real number , so we avoid the solutions similar to $ |s(1-s)| $2012-04-20
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    False. Try $f(s) = \cos(2\pi s) + 2$.2012-04-20
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    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.2012-04-20
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    where can i find hamburguers theorem ??2012-04-20
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    Did you try an internet search?? Lots of hits come up. Here is one link: pages 127-129 at http://hh-mouvement.com.pagesperso-orange.fr/seminario0.pdf.2012-04-20
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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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