Bohr Mollerup Theorem

Math Calculus

Intro

The Gamma function $\Gamma (x)$ is a way to extend the factorial function, where $\Gamma (n+1)=n!$. This gives us two conditions defining $\Gamma (x)$:

$$\begin{gathered} \Gamma (1)=1 \\ \Gamma (x+1)=x\Gamma (x) \end{gathered}$$

However, by adding a third condition stating $\Gamma (x)$ is logarithimically convex ($\log \circ \Gamma$ is convex), we can prove that $\Gamma (x)$ is unique!

Proof

Let $G$ be a function with the properties above. Since $G(x+1)=xG(x)$, we can define any $G(x+n)$, where $n\in \mathbb{N}$ as:

$$G(x+n)=G(x)\prod _{i=0}^{n-1}(x+i)$$

This means that it is sufficient to define $G(x)$ on $x\in (0,1]$ for a unique $G(x)$.

Let $S(x_1,x_2)$ be defined as $\frac{\log (\Gamma (x_2))-\log (\Gamma (x_1))}{x_2-x_1}$. Observe that by log-concavity, for all $0<x\le 1$ and $n\in \mathbb{N}$:

$$\begin{gathered} S(n-1,n)\le S(n,n+x)\le S(n,n+1) \\ \log (G(n)))-\log (G(n-1))\le \frac{\log (G(n+x))-\log (G(n))}{x}\le \log (G(n+1))-\log (G(n)) \\ \log ((n-1)!)-\log ((n-2)!)\le \frac{\log (G(x+n))-\log ((n-1)!)}{x}\le \log (n!)-\log ((n-1)!) \\ \log (n-1)\le \frac{\log (\frac{G(x+n)}{(n-1)!})}{x}\le \log (n) \\ \log ((n-1{)}^x)\le \log (\frac{G(x+n)}{(n-1)!})\le \log (n^x) \end{gathered}$$

Raising to the $n$:

$$\begin{gathered} (n-1{)}^x\le \frac{G(x+n)}{(n-1)!}\le n^x \\ (n-1{)}^x(n-1)!\le G(x+n)\le n^x(n-1)! \end{gathered}$$

Using the above work to expand $G(x+n)$:

$$\begin{gathered} \frac{(n-1{)}^x(n-1)!}{{\prod }_{i=0}^{n-1}(x+i)}\le G(x)\le \frac{n^x(n-1)!}{{\prod }_{i=0}^{n-1}(x+i)} \\ \frac{(n-1{)}^x(n-1)!}{{\prod }_{i=0}^{n-1}(x+i)}\le G(x)\le \frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)}(\frac{n+x}{n}) \end{gathered}$$

Of course, taking the limit as $n$ goes to infinity on both sides by brute force will produce the value of $G(x)$, however I will present a more elegant solution. Notice we can take the inequalities separately, resulting in:

$$\begin{gathered} \frac{(n_1-1{)}^x(n_1-1)!}{{\prod }_{i=0}^{n_1-1}(x+i)}\le G(x) \\ G(x)\le \frac{n_2^x{n}_2!}{{\prod }_{i=0}^{n_2}(x+i)}(\frac{n_2+x}{n_2}) \end{gathered}$$

This shows that no matter $n_1$ and $n_2$, the equality still holds!

Now we can sub in $n_1=n+1$, $n_2=n$, to get:

$$\frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)}\le G(x)\le \frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)}(\frac{n+x}{n})$$

Taking a limit to infinity on both sides:

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)}\le G(x)\le \underset{n\to \infty }{\lim }\frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)}(\frac{n+x}{n}) \\ \underset{n\to \infty }{\lim }\frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)}\le G(x)\le \underset{n\to \infty }{\lim }\frac{n^{x}n!}{{\prod }_{i=0}^n(x+i)} \end{gathered}$$

Therefore there is only a singular function satisfying $G(x)$, as it is squeezed on $[0,1)$.

Exercise to the Reader

Prove that the definition:

$$\Gamma (n)={\int }_0^{\infty }{x}^{n-1}{e}^{-x}dx$$

is valid.