proof of Bohr-Mollerup theorem

We prove this theorem in two stages: first, we establish that the gamma functionDlmfDlmfMathworldPlanetmath satisfies the given conditions and then we prove that these conditions uniquely determine a functionMathworldPlanetmath on (0,).

By its definition, Γ(x) is positive for positive x. Let x,y>0 and 0λ1.

logΓ(λx+(1-λ)y) = log0e-ttλx+(1-λ)y-1𝑑t
= log0(e-ttx-1)λ(e-tty-1)1-λ𝑑t
= λlogΓ(x)+(1-λ)logΓ(y)

The inequalityMathworldPlanetmath follows from Hölder’s inequality, where p=1λ and q=11-λ.

This proves that Γ 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, 0x1 and n.

n+x=(1-x)n+x(n+1) and by log-convexity of G, G(n+x)G(n)1-xG(n+1)x=G(n)1-xG(n)xnx=(n-1)!nx.

Similarly n+1=x(n+x)+(1-x)(n+1+x) gives n!G(n+x)(n+x)1-x.

Combining these two we get


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


Now these inequalities hold for every positive integer n and the terms on the left and right side have a common limit (limnanbn=1) so we find this determines G.

As a corollary we find another expression for Γ.

For 0x1,


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

Title proof of Bohr-Mollerup theorem
Canonical name ProofOfBohrMollerupTheorem
Date of creation 2013-03-22 13:18:14
Last modified on 2013-03-22 13:18:14
Owner lieven (1075)
Last modified by lieven (1075)
Numerical id 6
Author lieven (1075)
Entry type Proof
Classification msc 33B15
Defines Gauß’s product