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:
 * $\map f {x + 1} = \begin{cases}

1 & : x = 0 \\ x \map f x & : x > 0 \end{cases}$

Then:
 * $\forall x \in \R_{>0}: \map f x = \map \Gamma x$

where $\map \Gamma x$ is the Gamma function.

Proof
From the Gamma Function Extends Factorial:
 * $\map f {x + 1} = \begin{cases}

1 & : x = 0 \\ x \map f x & : 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 $t = \alpha s + \beta \paren {s + 1}$ where $\alpha, \beta \in \R_{>0}, \alpha + \beta = 1$.

Then:
 * $t = \paren {\alpha + \beta} s + \beta = s + \beta$

and so:
 * $\beta = t - s$

Thus:

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

Substituting as appropriate in $(1)$:

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

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 $(3)$:

Hence:

Thus when $0 < x \le 1$:
 * $\ds \map f x = \lim_{n \mathop \to \infty} \frac {x \paren {x + 1} \cdots \paren {x + n} } {n^x n!}$

which is the Euler form of the Gamma function.

The general case is deduced by the use of:
 * $\map f {x + 1} = x \map f x$