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The Bohr-Mollerup theorem states that the Gamma function is the unique function that satisfies:

1) f(x+1) = x*f(x)

2) f(1) = 1

3) ln(f(x)) is convex

The Gamma function is meant to interpolate the factorial function, so I can see the importance of the first two properties. But why is log convexity important? How does it affect the Gamma function's applicability in other areas of mathematics?

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There is also Weilandt's characterisation of the gamma function, it is the unique meromorphic function which is bounded in vertical strips. –  teil May 2 '10 at 10:05
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From 1) (for natural x only) and 2) you can prove that ln(f(x)) is convex on the natural numbers. This doesn't explain why 3) is "important", but does explain why it's a natural property to ask for general x, and it's interesting that only one extension of the factorial keeps this property. –  Mark Meckes May 2 '10 at 13:33
    
Related to math.stackexchange.com/questions/1537/… –  Tom Copeland May 10 '12 at 1:30
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2 Answers

First, let me mention that log convexity of a function is implied by an analytic property, which appears to be more natural than log convexity itself. Namely, if $\mu$ is a Borel measure on $[0,\infty)$ such that the $r$th moment $$f(r)=\int_{0}^{\infty}z^r d\mu(z)$$ is finite for all $r$ in the interval $I\subset \mathbb R$, then $\log f$ is convex on $I$.

Log convexity can be effectively used in derivation of various inequalities involving the gamma function (particularly, two-sided estimates of products of gamma functions). It is linked with the notion of Schur convexity which is itself used in many applications.

An appetizer. Let $m=\max x_i$, $s=\sum x_i$, $x_i > 0$, $i = 1,\dots,n$, then $$[\Gamma(s/n)]^n\leq\prod\limits_{1}^{n}\Gamma (x_i)\leq \left[\Gamma\left(\frac{s-m}{n-1}\right)\right]^{n-1}\Gamma(m).\qquad\qquad\qquad (1)$$

(1) is trivial, of course, when all $x_i$ and $s/n$ are integers, but in general the bounds do not hold without assuming log convexity.

Edit added: a sketch of the proof. Let $f$ be a continuous positive function defined on an interval $I\subset \mathbb R$. One may show that the function $\phi(x)=\prod\limits_{i=1}^{n}f(x_i)$, $x\in I^n$ is Schur-convex on $I^n$ if and only if $\log f$ is convex on $I$. Thus the function $$\phi(x)=\prod\limits_{i=1}^n \Gamma(x_i),\quad x_i>0,\qquad \quad\qquad\qquad\qquad\qquad\qquad\quad (2)$$ is Schur-convex on $I^n=(0,\infty)^n$. Since $x_i\le m$, $i=1,\dots,n$, and $\sum x_i=s$, it is easy to check that $$x \prec \left(\frac{s-m}{n-1},\dots,\frac{s-m}{n-1},m\right).$$ The latter majorization and the fact that $\phi(x)$ defined by (2) is Schur-convex imply the upper bound (1). The lower bound follows from the standard majorization $x\succ (s/n,\dots,s/n)$.


Have a look at the recent short article by Marshall and Olkin concerning this and related inequalities for the gamma function.

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What is the finite Borel measure? Specifically, since $\Gamma(0)=\infty,$ doesn't this imply that $\mu$ is infinite? –  Victor Protsak May 3 '10 at 0:33
    
Thanks, Victor. I amended the statement to avoid ambiguity. –  Andrey Rekalo May 3 '10 at 17:29
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There is a lovely and detailed discussion of the uniqueness of the gamma function in Volume 2 of Markushevich's Theory of Functions of a Complex Variable. The essential points are as follows.

Any meromorphic solution $f$ to the functional equation $zf(z)=f(z+1)$, with $f(1)=1$, must have simple poles at all integers $m\le 0$ with residues $(-1)^m/m!$ Assuming that these are ALL the poles of $f$, Markushevich exhibits all meromorphic interpolations of the factorial function. It turns out that there is a whole zoo of them, and in some (not very satisfying) sense the usual gamma function is the simplest choice.

So ... The Bohr-Mollerup Theorem sheds some light on this situation. But maybe there are other "interesting" extensions of the factorial that are not log-convex.

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