# Bohr–Mollerup theorem

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

${\displaystyle \Gamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}\,dt}$

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

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

### Proof

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

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

The slope of a line connecting two points ${\displaystyle \,(x_{1},\;f(x_{1}))\,}$ and ${\displaystyle \,(x_{2},\;f(x_{2}))\,}$, call it ${\displaystyle \,{\mathcal {M}}(x_{1},x_{2})\,}$ is monotonically increasing for convex functions with ${\displaystyle \,x_{1}. Since we have stipulated ${\displaystyle \,\log \left(\Gamma (x)\right)\,}$ is convex we know

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

{\displaystyle \,{\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 ${\displaystyle \,n\,}$. That is ${\displaystyle \,\Gamma (x)\,}$ is not greater than the right hand side for any choice of ${\displaystyle \,n\,}$ and likewise, ${\displaystyle \,\Gamma (x)\,}$ is not less than the left hand side for any other choice of ${\displaystyle \,n\,}$. 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 ${\displaystyle \,n\,}$ for the RHS and the LHS. In particular, if we keep ${\displaystyle \,n\,}$ for the RHS and choose ${\displaystyle \,n+1\,}$ for the LHS and get:

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

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

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

{\displaystyle \,{\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 ${\displaystyle \,\Gamma (x)\,}$ to all values of ${\displaystyle \,x\,}$ where the limit is defined.

## References

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