# 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 $(0,\infty)$.

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

 $\displaystyle\log\Gamma(\lambda x+(1-\lambda)y)$ $\displaystyle=$ $\displaystyle\log\int_{0}^{\infty}e^{-t}t^{\lambda x+(1-\lambda)y-1}dt$ $\displaystyle=$ $\displaystyle\log\int_{0}^{\infty}(e^{-t}t^{x-1})^{\lambda}(e^{-t}t^{y-1})^{1-% \lambda}dt$ $\displaystyle\leq$ $\displaystyle\log((\int_{0}^{\infty}e^{-t}t^{x-1}dt)^{\lambda}(\int_{0}^{% \infty}e^{-t}t^{y-1}dt)^{1-\lambda})$ $\displaystyle=$ $\displaystyle\lambda\log\Gamma(x)+(1-\lambda)\log\Gamma(y)$

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

This proves that $\Gamma$ 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, $0\leq x\leq 1$ and $n\in{\mathbb{N}}$.

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

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

Combining these two we get

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

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

 $a_{n}:=\frac{n!(n+x)^{x-1}}{x(x+1)\dots(x+n-1)}\leq G(x)\leq\frac{(n-1)!n^{x}}% {x(x+1)\dots(x+n-1)}=:b_{n}.$

Now these inequalities hold for every positive integer $n$ and the terms on the left and right side have a common limit ($\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=1$) so we find this determines $G$.

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

For $0\leq x\leq 1$,

 $\Gamma(x)=\lim_{n\rightarrow\infty}\frac{n!n^{x}}{x(x+1)\dots(x+n)}.$

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

Title proof of Bohr-Mollerup theorem ProofOfBohrMollerupTheorem 2013-03-22 13:18:14 2013-03-22 13:18:14 lieven (1075) lieven (1075) 6 lieven (1075) Proof msc 33B15 Gauß’s product