Bohr–Mollerup theorem

Template:Short description In mathematical analysis, the Bohr–Mollerup theorem is a theorem proved by the Danish mathematicians Harald Bohr and Johannes Mollerup. The theorem characterizes the gamma function, defined for x > 0 by

$\mathrm{\Gamma }(x)=\underset{0}{\overset{\mathrm{\infty }}{\int }}{t}^{x-1}{e}^{-t}\mathrm{d}t$

as the only positive function Template:Mvar, with domain on the interval x > 0, that simultaneously has the following three properties:

•  f (1) = 1, and
•  f (x + 1) = x f (x) for x > 0 and
• Template:Mvar is logarithmically convex.

A treatment of this theorem is in Artin's book The Gamma Function, which has been reprinted by the AMS in a collection of Artin's writings.

The theorem was first published in a textbook on complex analysis, as Bohr and Mollerup thought it had already been proved.

The theorem admits a far-reaching generalization to a wide variety of functions (that have convexity or concavity properties of any order).

Statement

Bohr–Mollerup Theorem.     Γ(x) is the only function that satisfies  f (x + 1) = x f (x) with log( f (x)) convex and also with  f (1) = 1.

Proof

Let Γ(x) be a function with the assumed properties established above: Γ(x + 1) = xΓ(x) and log(Γ(x)) is convex, and Γ(1) = 1. From Γ(x + 1) = xΓ(x) we can establish

$\mathrm{\Gamma }(x+n)=(x+n-1)(x+n-2)(x+n-3)\cdots (x+1)x\mathrm{\Gamma }(x)$

The purpose of the stipulation that Γ(1) = 1 forces the Γ(x + 1) = xΓ(x) property to duplicate the factorials of the integers so we can conclude now that Γ(n) = (n − 1)! if n ∈ N and if Γ(x) exists at all. Because of our relation for Γ(x + n), if we can fully understand Γ(x) for 0 < x ≤ 1 then we understand Γ(x) for all values of Template:Mvar.

For x₁, x₂, the slope S(x₁, x₂) of the line segment connecting the points (x₁, log(Γ (x₁))) and (x₂, log(Γ (x₂))) is monotonically increasing in each argument with x₁ < x₂ since we have stipulated that log(Γ(x)) is convex. Thus, we know that

$S(n-1,n)\le S(n,n+x)\le S(n,n+1)\text{for all }x\in (0,1].$

After simplifying using the various properties of the logarithm, and then exponentiating (which preserves the inequalities since the exponential function is monotonically increasing) we obtain

$(n-1{)}^x(n-1)!\le \mathrm{\Gamma }(n+x)\le n^x(n-1)!.$

From previous work this expands to

$(n-1{)}^x(n-1)!\le (x+n-1)(x+n-2)\cdots (x+1)x\mathrm{\Gamma }(x)\le n^x(n-1)!,$

and so

$\frac{(n-1{)}^x(n-1)!}{(x+n-1)(x+n-2)\cdots (x+1)x}\le \mathrm{\Gamma }(x)\le \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\left(\frac{n+x}{n}\right).$

The last line is a strong statement. In particular, it is true for all values of Template:Mvar. That is Γ(x) is not greater than the right hand side for any choice of Template:Mvar and likewise, Γ(x) is not less than the left hand side for any other choice of Template:Mvar. Each single inequality stands alone and may be interpreted as an independent statement. Because of this fact, we are free to choose different values of Template:Mvar for the RHS and the LHS. In particular, if we keep Template:Mvar for the RHS and choose n + 1 for the LHS we get:

$\begin{array}{rl}\frac{((n+1)-1{)}^x((n+1)-1)!}{(x+(n+1)-1)(x+(n+1)-2)\cdots (x+1)x}& \le \mathrm{\Gamma }(x)\le \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\left(\frac{n+x}{n}\right)\\ \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}& \le \mathrm{\Gamma }(x)\le \frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\left(\frac{n+x}{n}\right)\end{array}$

It is evident from this last line that a function is being sandwiched between two expressions, a common analysis technique to prove various things such as the existence of a limit, or convergence. Let n → ∞:

$\underset{n\to \mathrm{\infty }}{\lim }\frac{n+x}{n}=1$

so the left side of the last inequality is driven to equal the right side in the limit and

$\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$

is sandwiched in between. This can only mean that

$\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}=\mathrm{\Gamma }(x).$

In the context of this proof this means that

$\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$

has the three specified properties belonging to Γ(x). Also, the proof provides a specific expression for Γ(x), that must hold if the characterizing properties are true. Therefore, there is no other function defined on $x>0$ with all the properties assigned to Γ(x).

The remaining loose end is the question of proving that Γ(x) makes sense for all Template:Mvar where

$\underset{n\to \mathrm{\infty }}{\lim }\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}$

exists. The problem is that our first double inequality

$S(n-1,n)\le S(n+x,n)\le S(n+1,n)$

was constructed with the constraint 0 < x ≤ 1. If, say, x > 1 then the fact that Template:Mvar is monotonically increasing would make S(n + 1, n) < S(n + x, n), contradicting the inequality upon which the entire proof is constructed. However,

$\begin{array}{rl}\mathrm{\Gamma }(x+1)& =\underset{n\to \mathrm{\infty }}{\lim }x\cdot \left(\frac{n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}\right)\frac{n}{n+x+1}\\ \mathrm{\Gamma }(x)& =\left(\frac{1}{x}\right)\mathrm{\Gamma }(x+1)\end{array}$

which demonstrates how to bootstrap Γ(x) to all values of Template:Mvar where the limit is defined.

See also

• Wielandt theorem