Bohr–Mollerup theorem

In mathematical analysis, the Bohr–Mollerup theorem is named after the Danish mathematicians Harald Bohr and Johannes Mollerup, who proved it. The theorem characterizes the gamma function, defined for x > 0 by

$\Gamma(x)=\int_0^\infty t^{x-1} e^{-t}\,dt$

as the only function f on the interval x > 0 that simultaneously has the three properties

• $f(1)=1, \,$ and
• $f(x+1)=xf(x) \text{ for } x>0, \,$ and
• f is logarithmically convex.

An elegant 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.

Proof

Statement of the Theorem

$\,\Gamma(z)\,$ is the only function that satisfies $\,f(z+1)=zf(z)\,$ with $\,\log(f(z))\,$ convex and also with $\,f(1)=1\,$.

Proof

Let $\,\Gamma(x)\,$ be a function with the assumed properties established above: $\,\Gamma(x+1)=x\Gamma(x)\,$ and $\,\log\left(\Gamma(x)\right)\,$ is convex, and $\,\Gamma(1)=1\,$. From the fact that $\,\Gamma(x+1)=x\Gamma(x)\,$ we can establish

$\, \begin{align} \Gamma(x+n)=(x+n-1)(x+n-2)(x+n-3)\ldots(x+1)x\Gamma(x) \end{align} \,$

The purpose of the stipulation that $\,\Gamma(1)=1\,$ forces the $\,\Gamma(x+1)=x\Gamma(x)\,$ property to duplicate the factorials of the integers so we can conclude now that $\,\Gamma(n)=(n-1)!\,$ if $\,n\in\mathbb{N}\,$ and if $\,\Gamma(x)\,$ exists at all. Because of our relation for $\,\Gamma(x+n)\,$, if we can fully understand $\,\Gamma(x)\,$ for $\,0<x\leq 1\,$ then we understand $\,\Gamma(x)\,$ for all values of $\,x\,$.

The slope of a line connecting two points $\,(x_1,\;f(x_1))\,$ and $\,(x_2,\;f(x_2))\,$ , call it $\,\mathcal{M}(x_1,x_2)\,$ is monotonically increasing for convex functions with $\,x_1<x_2\,$. Since we have stipulated $\,\log\left(\Gamma(x)\right)\,$ is convex we know

$\begin{align} \mathcal{M}(n-1,n)&\leq\mathcal{M}(n,n+x)\leq\mathcal{M}(n,n+1)\;\;\mathrm{when}\;0< x\leq 1\\ \frac{\log\left(\Gamma(n-1)\right)-\log\left(\Gamma(n)\right)}{(n-1)-n}&\leq \frac{\log\left(\Gamma(n)\right)-\log\left(\Gamma(n+x)\right)}{n-(n+x)}\leq \frac{\log\left(\Gamma(n)\right)-\log\left(\Gamma(n+1)\right)}{n-(n+1)}\\ \frac{\log\left((n-2)!\right)-\log\left((n-1)!\right)}{-1}&\leq \frac{\log\left(\Gamma(n+x)\right)-\log\left((n-1)!\right)}{x}\leq \frac{\log\left(n!\right)-\log\left((n-1)!\right)}{1}\\ -\log\left(\frac{(n-2)!}{(n-1)!}\right)&\leq \frac{\log\left(\Gamma(n+x)\right)-\log\left((n-1)!\right)}{x}\leq \log\left(\frac{n!}{(n-1)!}\right)\\ -\log\left(\frac{1}{(n-1)}\right)&\leq \frac{\log\left(\Gamma(n+x)\right)-\log\left((n-1)!\right)}{x}\leq \log\left(n\right)\\ x\cdot\log\left(n-1\right)+\log\left((n-1)!\right)&\leq \log\left(\Gamma(n+x)\right)\leq x\cdot\log\left(n\right)+\log\left((n-1)!\right)\\ \log\left((n-1)^x(n-1)!\right)&\leq \log\left(\Gamma(n+x)\right)\leq \log\left(n^x(n-1)!\right) \end{align}$

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. Now we recall that the function $\,\log()\,$ and $\,e^{()}\,$ are both monotonically increasing. Therefore if we exponentiate each term of the inequality, we will preserve the inequalities. Continuing:

$\, \begin{align} (n-1)^x(n-1)!&\leq \Gamma(n+x)\leq n^x(n-1)!\\ (n-1)^x(n-1)!&\leq (x+n-1)(x+n-2)\ldots(x+1)x\Gamma(x)\leq n^x(n-1)!\\ \frac{(n-1)^x(n-1)!}{(x+n-1)(x+n-2)\ldots(x+1)x}\leq \Gamma(x)&\leq\frac{n^x(n-1)!}{(x+n-1)(x+n-2)\ldots(x+1)x}\\ \frac{(n-1)^x(n-1)!}{(x+n-1)(x+n-2)\ldots(x+1)x}&\leq \Gamma(x)\leq\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\left(\frac{n+x}{n}\right)\\ \end{align} \,$

The last line is a strong statement. In particular, it is true for all values of $\,n\,$. That is $\,\Gamma(x)\,$ is less than the right hand side for any choice of $\,n\,$ and likewise, $\,\Gamma(x)\,$ is greater than the left hand side for any other choice of $\,n\,$. 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 $\,n\,$ for the RHS and the LHS. In particular, if we keep $\,n\,$ for the RHS and choose $\,n+1\,$ for the LHS and get:

$\, \begin{align} \frac{((n+1)-1)^x((n+1)-1)!}{(x+(n+1)-1)(x+(n+1)-2)\ldots(x+1)x}&\leq \Gamma(x)\leq\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\left(\frac{n+x}{n}\right)\\ \frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}&\leq \Gamma(x)\leq\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\left(\frac{n+x}{n}\right)\\ \frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}&\leq \Gamma(x)\\ \Gamma(x)\left(\frac{n}{n+x}\right)&\leq\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x} \end{align} \,$

We can combine the last two lines into

$\, \begin{align} \Gamma(x)\left(\frac{n}{n+x}\right)\leq\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\leq\Gamma(x) \end{align} \,$

Now let $\,n\rightarrow\infty\,$. The limit drives $\,\frac{n}{n+x}\rightarrow 1\,$ so the left side of the last inequality is driven to equal the right side. $\,\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\,$ is sandwiched in between. This can only mean that $\,\lim_{n\rightarrow\infty}\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\,$ is equal to $\,\Gamma(x)\,$. In the context of this proof this means that $\,\lim_{n\rightarrow\infty}\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\,$ has the three specified properties belonging to $\,\Gamma(x)\,$. Also, the proof provides a specific expression for $\,\Gamma(x)\,$. And the final critical part of the proof is to remember that the limit of a sequence is unique. This means that for any choice of $\,x\in(0,1]\,$ only one possible number $\,\Gamma(x)\,$ can exist. Therefore there is no other function with all the properties assigned to $\,\Gamma(x)\,$. the assumptions of this theorem to

The remaining loose end is the question of proving that $\,\Gamma(x)\,$ makes sense for all $\,x\,$ where $\,\lim_{n\rightarrow\infty}\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\,$ exists. The problem is that our first double inequality

$\, \begin{align} \mathcal{M}(n-1,n)\leq\mathcal{M}(n+x,n)\leq\mathcal{M}(n+1,n) \end{align} \,$

was constructed with the constraint $\,0<x\leq 1\,$. If, say, $\,x>1\,$ then the fact that $\,\mathcal{M}\,$ is monotonically increasing would make $\,\mathcal{M}(n+1,n)<\mathcal{M}(n+x,n)\,$, contradicting the inequality upon which the entire proof is constructed. But notice

$\, \begin{align} \Gamma(x+1)&= \lim_{n\rightarrow\infty}x\cdot\left(\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}\right)\frac{n}{n+x+1}\\ \Gamma(x)&=\left(\frac{1}{x}\right)\Gamma(x+1) \end{align} \,$

which demonstrates how to bootstrap $\,\Gamma(x)\,$ to all values of $\,x\,$ where the limit is defined.

References

• Weisstein, Eric W., "Bohr–Mollerup Theorem" from MathWorld.
• Proof of Bohr–Mollerup theorem on PlanetMath
• Alternative proof of Bohr–Mollerup theorem on PlanetMath
• Artin, Emil (1964). The Gamma Function. Holt, Rinehart, Winston. 
• Rosen, Michael (2006). Exposition by Emil Artin: A Selection. American Mathematical Society. 
• Mollerup, J., Bohr, H. (1922). Lærebog i Kompleks Analyse vol. III, Copenhagen.  (Textbook in Complex Analysis)Template:Mathanalysis-