The Bohr-Mollerup Theorem states that the gamma function is the unique function $f: (0, \infty )\rightarrow \mathbb{R}$ satisfying $f(1)=1,$ $f(x+1) = x f(x),$ and the condition that $\log f$ is convex. Logarithmically convex seems like a rather stringent imposition; what if $f$ is just convex? I'm not really sure how to devise a counterexample here, but I'm sure one must exist (otherwise the theorem would read prettier.) Anybody know anything about this?

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    $\begingroup$ Possibly useful: mathoverflow.net/questions/23229/… $\endgroup$ – Alex R. May 25 '12 at 8:18
  • $\begingroup$ Have you seen Luschny's work, by any chance? $\endgroup$ – J. M. is a poor mathematician Jun 16 '12 at 2:14
  • $\begingroup$ After posting this question I came across his page on the gamma function. Very interesting stuff. $\endgroup$ – user17794 Jun 17 '12 at 20:41

Construct a function $f$ with $f(x)=1,1 \leq x \leq 2$ with $f(x+1)=xf(x)$,then it is not difficult to find out that $f$ possess an increasing left derivative(not strictly increasing).So $f$ is a convex function. But $f$ is not a logarithmically convex function.

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  • $\begingroup$ The graph is incorrect. On $(0,1)$ we have $f(x)=1/x$. Also, it might help to consider other intervals: on $(3,4)$ we get $f(x)= (x-1)(x-2)$, on $(4,5)$ we get $f(x)= (x-1)(x-2)(x-3)$ etc. $\endgroup$ – AD. May 25 '12 at 11:22
  • $\begingroup$ @AD. Why is it incorrect?The graph is plotted on interval $(1,3)$. $\endgroup$ – zy_ May 25 '12 at 11:29
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    $\begingroup$ I see, I thought it was $(0,3)$ - so the origin is $(1,0)$, in any case it is misleading. $\endgroup$ – AD. May 25 '12 at 11:37

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