proof of Bohr-Mollerup theorem

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

$$\mathrm{\Gamma }(x)=\frac{1}{x}{e}^{-\gamma x}\prod _{n=1}^{\mathrm{\infty }}\frac{n}{x+n}{e}^{-x/n}$$

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

$$\log \mathrm{\Gamma }(x)=-\log x-\gamma x-\sum _{n=1}^{\mathrm{\infty }}\log \left(\frac{n}{x+n}\right)-\frac{x}{n}$$

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

$$\frac{d^2}{d{x}^2}\log \mathrm{\Gamma }(x)=\frac{1}{x^2}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(x+n)}^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(x+n)}^2}$$

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

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

Next, let $f$ satisfy the hypotheses of the Bohr-Mollerup theorem. Consider the function $g$ defined as $e^{g(x)}=f(x)/\mathrm{\Gamma }(x)$. 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 $g(x_0)\ne g(x_1)$. 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 . Let $D_2$ denote the second divided difference of $g$:

$$D_2={\mathrm{\Delta }}_2(g;x_0,x_1,x_0+1)=-\frac{g(x_0)}{x_0-x_1}+\frac{g(x_1)}{(x_1-x_0)(x_1-x_0-1)}-\frac{g(x_0+1)}{x_0-x_1+1}$$

By our assumptions, . By linearity,

$$D_2={\mathrm{\Delta }}_2(\log f;x_0,x_1,x_0+1)-{\mathrm{\Delta }}_2(\log \mathrm{\Gamma };x_0,x_1,x_0+1)$$

By periodicity, we have

$$D_2={\mathrm{\Delta }}_2(\log f;x_0+n,x_1+n,x_0+n+1)-{\mathrm{\Delta }}_2(\log \mathrm{\Gamma };x_0+n,x_1+n,x_0+n+1)$$

for every integer $n>0$. However,

As $n\to \mathrm{\infty }$, 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 $n$,

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

Title	proof of Bohr-Mollerup theorem
Canonical name	ProofOfBohrMollerupTheorem
Date of creation	2013-03-22 14:53:39
Last modified on	2013-03-22 14:53:39
Owner	Andrea Ambrosio (7332)
Last modified by	Andrea Ambrosio (7332)
Numerical id	18
Author	Andrea Ambrosio (7332)
Entry type	Proof
Classification	msc 33B15