Bohr–Mollerup theorem

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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

$\,\Gamma (x)\,$ is the only function that satisfies $\,f(x+1)=xf(x)\,$ with $\,\log(f(x))\,$ 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

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

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

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

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:

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

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:

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

Now let $\,n\rightarrow \infty \,$ . The limit drives $\,{\frac {n+x}{n}}\rightarrow 1\,$ so the left side of the last inequality is driven to equal the right side. $\,{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\,$ is sandwiched in between. This can only mean that $\,\lim _{{n\rightarrow \infty }}{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\,$ is equal to $\,\Gamma (x)\,$ . In the context of this proof this means that $\,\lim _{{n\rightarrow \infty }}{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\,$ 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 $\,\lim _{{n\rightarrow \infty }}{\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\,$ exists. The problem is that our first double inequality

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

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

$\,{\begin{aligned}\Gamma (x+1)&=\lim _{{n\rightarrow \infty }}x\cdot \left({\frac {n^{x}n!}{(x+n)(x+n-1)\cdots (x+1)x}}\right){\frac {n}{n+x+1}}\\\Gamma (x)&=\left({\frac {1}{x}}\right)\Gamma (x+1)\end{aligned}}\,$

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

References

Hazewinkel, Michiel, ed. (2001), "Bohr–Mollerup theorem", Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4
Weisstein, Eric W., "Bohr–Mollerup Theorem", MathWorld.
Proof of Bohr–Mollerup theorem, PlanetMath.org.
Alternative proof of Bohr–Mollerup theorem, PlanetMath.org.
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)

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