Gamma Function is Unique Extension of Factorial

Theorem

Let $f: \R_{>0} \to \R$ be a real function which is positive and continuous.

Let $\ln \mathop \circ f$ be convex on $\R_{>0}$.

Let $f$ satisfy the conditions:

$f \left({x + 1}\right) = \begin{cases} 1 & : x = 0 \\ x f \left({x}\right) & : x > 0 \end{cases}$

Then:

$\forall x \in \R_{>0}: f \left({x}\right) = \Gamma \left({x}\right)$

where $\Gamma \left({x}\right)$ is the Gamma function.

Proof

From the Gamma Function Extends Factorial:

$f \left({x + 1}\right) = \begin{cases} 1 & : x = 0 \\ x f \left({x}\right) & : x > 0 \end{cases}$

From Gamma Function is Continuous on Positive Reals, $\Gamma$ is positive and continuous on $\R_{>0}$.

From Log of Gamma Function is Convex on Positive Reals, $\ln \mathop \circ \Gamma$ is convex on $\R_{>0}$.

It remains to be shown that $\Gamma$ is the only real function which satisfies these conditions.

Let $s, t \in \R_{>0}$ such that $s \le t \le s + 1$.

Let $t = \alpha s + \beta \left({s + 1}\right)$ where $\alpha, \beta \in \R_{>0}, \alpha + \beta = 1$.

Then:

$t = \left({\alpha + \beta}\right) s + \beta = s + \beta$

and so:

$\beta = t - s$

Thus:

 $\displaystyle \ln \left({f \left({t}\right)}\right)$ $\le$ $\displaystyle \alpha \ln \left({f \left({s}\right)}\right) + \beta \ln \left({f \left({s + 1}\right)}\right)$ $\ln \mathop \circ f$ is convex on $\R_{>0}$ $\displaystyle \leadsto \ \$ $\displaystyle f \left({t}\right)$ $\le$ $\displaystyle \left({f \left({s}\right)}\right)^\alpha \left({f \left({s + 1}\right)}\right)^\beta$ $\displaystyle$ $=$ $\displaystyle \left({f \left({s}\right)}\right)^\alpha \left({s f \left({s}\right)}\right)^\beta$ $\displaystyle$ $=$ $\displaystyle s^\beta f \left({s}\right)$ as $\alpha + \beta = 1$ $(1):\quad$ $\displaystyle$ $=$ $\displaystyle s^{t - s} f \left({s}\right)$ as $\beta = t - s$

Because $s \le t \le s + 1$, it follows that $t - 1 \le s \le t$.

Substituting as appropriate in $(1)$:

 $\displaystyle f \left({s}\right)$ $\le$ $\displaystyle \left({t - 1}\right)^{s - t + 1} f \left({t - 1}\right)$ $(2):\quad$ $\displaystyle$ $=$ $\displaystyle \left({t - 1}\right)^{s - t} f \left({t}\right)$

Combining $(1)$ and $(2)$:

 $\displaystyle \left({t - 1}\right)^{t - s} f \left({s}\right)$ $\le$ $\displaystyle f \left({t}\right)$ $(3):\quad$ $\displaystyle$ $\le$ $\displaystyle s^{t - s} f \left({s}\right)$

Let $x \in \R$ such that $0 < x \le 1$.

Let $n \in \N$.

Let:

$s = n + 1$
$t = x + n + 1$

Then substituting in $(3)$:

 $\displaystyle \left({x + n}\right)^x f \left({n + 1}\right)$ $\le$ $\displaystyle f \left({x + n + 1}\right)$ $\displaystyle$ $\le$ $\displaystyle \left({n + 1}\right)^x f \left({n + 1}\right)$

Hence:

 $\displaystyle \left({x + n}\right)^x n!$ $\le$ $\displaystyle \left({x + n}\right) \left({x + n - 1}\right) \cdots x f \left({x}\right)$ $\displaystyle$ $\le$ $\displaystyle \left({n + 1}\right)^x n!$ $\displaystyle \leadsto \ \$ $\displaystyle \left({1 + \frac x n}\right)^x$ $\le$ $\displaystyle \frac {\left({x + n}\right) \left({x + n - 1}\right) \cdots x f \left({x}\right)} {n^x n!}$ $\displaystyle$ $\le$ $\displaystyle \left({1 + \frac 1 n}\right)^x$

Thus when $0 < x \le 1$:

$\displaystyle f \left({x}\right) = \lim_{n \mathop \to \infty} \frac {x \left({x + 1}\right) \cdots \left({x + n}\right)} {n^x n!}$

which is the Euler form of the Gamma function.

The general case is deduced by the use of:

$f \left({x + 1}\right) = x f \left({x}\right)$

$\blacksquare$