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}$?
Functional equation and Riemann function $ \xi(s) $
1
$\begingroup$
functional-equations
-
0Did you mean to say $f(s) = \kappa \cdot \xi(s)$ for some $\kappa > 0$ ? – 2012-04-20
-
0aha.. considering $ f(s)$ is positive and differentiable for each real number , so we avoid the solutions similar to $ |s(1-s)| $ – 2012-04-20
-
3False. Try $f(s) = \cos(2\pi s) + 2$. – 2012-04-20
-
5Or, 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
-
0where can i find hamburguers theorem ?? – 2012-04-20
-
0Did 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
-
0Slightly 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