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Homeproof of Bohr-Mollerup theorem

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# 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.

## Mathematics Subject Classification

33B15*no label found*

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