Replicator equation

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

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