# 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

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

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

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

Since every term in this series is positive, $\mathrm{\Gamma}$ is logarithmically convex. Furthermore, note that since each term is monotonically decreasing, $\mathrm{log}\mathrm{\Gamma}$ 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,

$${D}_{2}={\mathrm{\Delta}}_{2}(\mathrm{log}f;{x}_{0},{x}_{1},{x}_{0}+1)-{\mathrm{\Delta}}_{2}(\mathrm{log}\mathrm{\Gamma};{x}_{0},{x}_{1},{x}_{0}+1)$$ |

By periodicity, we have

$${D}_{2}={\mathrm{\Delta}}_{2}(\mathrm{log}f;{x}_{0}+n,{x}_{1}+n,{x}_{0}+n+1)-{\mathrm{\Delta}}_{2}(\mathrm{log}\mathrm{\Gamma};{x}_{0}+n,{x}_{1}+n,{x}_{0}+n+1)$$ |

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 |