proof of Bohr-Mollerup theorem

We prove this theorem in two stages: first, we establish that the gamma function satisfies the given conditions and then we prove that these conditions uniquely determine a function on $(0,\mathrm{\infty })$.

By its definition, $\mathrm{\Gamma }(x)$ is positive for positive $x$. Let $x,y>0$ and $0\le \lambda \le 1$.

$\log \mathrm{\Gamma }(\lambda x+(1-\lambda )y)$ $=$ $\log {\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-t}{t}^{\lambda x+(1-\lambda )y-1}𝑑t$

$=$ $\log {\displaystyle {\int }_0^{\mathrm{\infty }}}{(e^{-t}{t}^{x-1})}^{\lambda }{(e^{-t}{t}^{y-1})}^{1-\lambda }𝑑t$

$\le$ $\log ({({\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-t}{t}^{x-1}𝑑t)}^{\lambda }{({\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-t}{t}^{y-1}𝑑t)}^{1-\lambda })$

$=$ $\lambda \log \mathrm{\Gamma }(x)+(1-\lambda )\log \mathrm{\Gamma }(y)$

The inequality follows from Hölder’s inequality, where $p=\frac{1}{\lambda }$ and $q=\frac{1}{1-\lambda }$.

This proves that $\mathrm{\Gamma }$ is log-convex. Condition 2 follows from the definition by applying integration by parts. Condition 3 is a trivial verification from the definition.

Now we show that the 3 conditions uniquely determine a function. By condition 2, it suffices to show that the conditions uniquely determine a function on $(0,1)$.

Let $G$ be a function satisfying the 3 conditions, $0\le x\le 1$ and $n\in \mathbb{N}$.

$n+x=(1-x)n+x(n+1)$ and by log-convexity of $G$, $G(n+x)\le G{(n)}^{1-x}G{(n+1)}^x=G{(n)}^{1-x}G{(n)}^x{n}^x=(n-1)!n^x$.

Similarly $n+1=x(n+x)+(1-x)(n+1+x)$ gives $n!\le G(n+x){(n+x)}^{1-x}$.

Combining these two we get

$$n!{(n+x)}^{x-1}\le G(n+x)\le (n-1)!n^x$$

and by using condition 2 to express $G(n+x)$ in terms of $G(x)$ we find

$$a_n:=\frac{n!{(n+x)}^{x-1}}{x(x+1)\mathrm{\dots }(x+n-1)}\le G(x)\le \frac{(n-1)!n^x}{x(x+1)\mathrm{\dots }(x+n-1)}=:b_n.$$

Now these inequalities hold for every positive integer $n$ and the terms on the left and right side have a common limit (${lim}_{n\to \mathrm{\infty }}\frac{a_n}{b_n}=1$) so we find this determines $G$.

As a corollary we find another expression for $\mathrm{\Gamma }$.

For $0\le x\le 1$,

$$\mathrm{\Gamma }(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{n!n^x}{x(x+1)\mathrm{\dots }(x+n)}.$$

In fact, this equation, called Gauß’s product, goes for the whole complex plane minus the negative integers.

Title	proof of Bohr-Mollerup theorem
Canonical name	ProofOfBohrMollerupTheorem
Date of creation	2013-03-22 13:18:14
Last modified on	2013-03-22 13:18:14
Owner	lieven (1075)
Last modified by	lieven (1075)
Numerical id	6
Author	lieven (1075)
Entry type	Proof
Classification	msc 33B15
Defines	Gauß’s product