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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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    Possibly useful: http://mathoverflow.net/questions/23229/importance-of-log-convexity-of-the-gamma-function2012-05-25
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    Have you seen Luschny's work, by any chance?2012-06-16
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    After posting this question I came across his [page](http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html) on the gamma function. Very interesting stuff.2012-06-17

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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.

enter image description here

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    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.2012-05-25
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    @AD. Why is it incorrect?The graph is plotted on interval $(1,3)$.2012-05-25
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    I see, I thought it was $(0,3)$ - so the origin is $(1,0)$, in any case it is misleading.2012-05-25