Bohr–Mollerup theorem

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


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

as the only positive function $f$, 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
 * $f$ 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


 * $$\Gamma(x+n)=(x+n-1)(x+n-2)(x+n-3)\cdots(x+1)x\Gamma(x)$$

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 $x$.

For $x_{1}$, $x_{2}$, the slope $S(x_{1}, x_{2})$ of the line segment connecting the points $(x_{1}, log(Γ (x_{1})))$ and $(x_{2}, log(Γ (x_{2})))$ is monotonically increasing in each argument with $x_{1} < x_{2}$ since we have stipulated that $log(Γ(x))$ is convex. Thus, we know that
 * $$S(n-1,n) \leq S(n,n+x)\leq S(n,n+1)\quad\text{for all }x\in(0,1].$$

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
 * $$(n-1)^x(n-1)! \leq \Gamma(n+x)\leq n^x(n-1)!.$$

From previous work this expands to
 * $$(n-1)^x(n-1)! \leq (x+n-1)(x+n-2)\cdots(x+1)x\Gamma(x)\leq n^x(n-1)!,$$

and so
 * $$\frac{(n-1)^x(n-1)!}{(x+n-1)(x+n-2)\cdots(x+1)x} \leq \Gamma(x) \leq \frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\left(\frac{n+x}{n}\right).$$

The last line is a strong statement. In particular, it is true for all values of $n$. That is $Γ(x)$ is not greater than the right hand side for any choice of $n$ and likewise, $Γ(x)$ is not less 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 we get:


 * $$\begin{align}

\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^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\left(\frac{n+x}{n}\right)\\ \frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}&\leq \Gamma(x)\leq\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}\left(\frac{n+x}{n}\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. Let $n → ∞$:


 * $$\lim_{n\to\infty} \frac{n+x}{n} = 1$$

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


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

is sandwiched in between. This can only mean that


 * $$\lim_{n\to\infty}\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x} = \Gamma (x).$$

In the context of this proof this means that


 * $$\lim_{n\to\infty}\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}$$

has the three specified properties belonging to $Γ(x)$. Also, the proof provides a specific expression for $Γ(x)$. 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 $0 < x ≤ 1$ only one possible number $Γ(x)$ can exist. Therefore, there is no other function with all the properties assigned to $Γ(x)$.

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


 * $$\lim_{n\to\infty}\frac{n^xn!}{(x+n)(x+n-1)\cdots(x+1)x}$$

exists. The problem is that our first double inequality


 * $$S(n-1,n)\leq S(n+x,n)\leq S(n+1,n)$$

was constructed with the constraint $0 < x ≤ 1$. If, say, $x > 1$ then the fact that $S$ 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{align}

\Gamma(x+1)&= \lim_{n\to\infty}x\cdot\left(\frac{n^xn!}{(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{align}$$

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