# proof of Bohr-Mollerup theorem

To show that the gamma function is logarithmically convex, we can examine the product representation:

 $\Gamma(x)={1\over x}e^{-\gamma x}\prod_{n=1}^{\infty}{n\over x+n}e^{-x/n}$

Since this product converges absolutely for $x>0$, we can take the logarithm term-by-term to obtain

 $\log\Gamma(x)=-\log x-\gamma x-\sum_{n=1}^{\infty}\log\left({n\over x+n}\right% )-{x\over n}$

It is justified to differentiate this series twice because the series of derivatives is absolutely and uniformly convergent.

 ${d^{2}\over dx^{2}}\log\Gamma(x)={1\over x^{2}}+\sum_{n=1}^{\infty}{1\over(x+n% )^{2}}=\sum_{n=0}^{\infty}{1\over(x+n)^{2}}$

Since every term in this series is positive, $\Gamma$ is logarithmically convex. Furthermore, note that since each term is monotonically decreasing, $\log\Gamma$ is a decreasing function of $x$. If $x>m$ for some integer $m$, then we can bound the series term-by-term to obtain

 ${d^{2}\over dx^{2}}\log\Gamma(x)<\sum_{n=0}^{\infty}{1\over(m+n)^{2}}=\sum_{n=% m}^{\infty}{1\over n^{2}}$

Therefore, as $x\to\infty$, $d^{2}\Gamma/dx^{2}\to 0$.

Next, let $f$ satisfy the hypotheses of the Bohr-Mollerup theorem. Consider the function $g$ defined as $e^{g(x)}=f(x)/\Gamma(x)$. By hypothesis 3, $g(1)=0$. By hypothesis 2, $e^{g(x+1)}=e^{g(x)}$, so $g(x+1)=g(x)$. In other , $g$ is periodic.

Suppose that $g$ is not constant. Then there must exist points $x_{0}$ and $x_{1}$ on the real axis such that $g(x_{0})\neq g(x_{1})$. Suppose that $g(x_{1})>g(x_{0})$ for definiteness. Since $g$ is periodic with period 1, we may assume without loss of generality that $x_{0}. Let $D_{2}$ denote the second divided difference of $g$:

 $D_{2}=\Delta_{2}(g;x_{0},x_{1},x_{0}+1)=-{g(x_{0})\over x_{0}-x_{1}}+{g(x_{1})% \over(x_{1}-x_{0})(x_{1}-x_{0}-1)}-{g(x_{0}+1)\over x_{0}-x_{1}+1}$

By our assumptions, $D_{2}<0$. By linearity,

 $D_{2}=\Delta_{2}(\log f;x_{0},x_{1},x_{0}+1)-\Delta_{2}(\log\Gamma;x_{0},x_{1}% ,x_{0}+1)$

By periodicity, we have

 $D_{2}=\Delta_{2}(\log f;x_{0}+n,x_{1}+n,x_{0}+n+1)-\Delta_{2}(\log\Gamma;x_{0}% +n,x_{1}+n,x_{0}+n+1)$

for every integer $n>0$. However,

 $|\Delta_{2}(\log\Gamma;x_{0}+n,x_{1}+n,x_{0}+n+1)|<\max_{x_{0}+n\leq x\leq x_{% 0}+1}{d^{2}\over dx^{2}}\log\Gamma(x)$

As $n\to\infty$, the right hand side approaches zero. Hence, by choosing $n$ sufficiently large, we can make the left-hand side smaller than $|D_{2}|/2$. For such an $n$,

 $\Delta_{2}(\log f;x_{0}+n,x_{1}+n,x_{0}+n+1)<0$

However, this contradicts hypothesis 1. Therefore, $g$ must be constant. Since $g(0)=0$, $g(x)=0$ for all $x$, which implies that $e^{0}=f(x)/\Gamma(x)$. In other words, $f(x)=\Gamma(x)$ as desired.

Title proof of Bohr-Mollerup theorem ProofOfBohrMollerupTheorem 2013-03-22 14:53:39 2013-03-22 14:53:39 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 18 Andrea Ambrosio (7332) Proof msc 33B15