Proving that the $\zeta$ function is convex on $(1,+\infty)$ is pretty simple if we use the derivative, but is there a proof without using derivative? I'm allowed to use just the definition of the convexity.
2026-04-05 17:35:37.1775410537
Convexity of the Riemann-zeta function without derivative
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hint to the answer : a finite sum of convex functions is still convex.
what about an infinite sum ? ( $\zeta(\sigma)$ is convex for $\sigma \in ]1;\infty[$ but it is concave for $\sigma \in [0;1[$ )