Gamma Function is Unique Extension of Factorial

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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:

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

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

Substituting as appropriate in $(1)$:

\(\ds \map f s\) \(\le\) \(\ds \paren {t - 1}^{s - t + 1} \map f {t - 1}\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \paren {t - 1}^{s - t} \map f t\)

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

\(\ds \paren {t - 1}^{t - s} \map f s\) \(\le\) \(\ds \map f t\)
\(\text {(3)}: \quad\) \(\ds \) \(\le\) \(\ds s^{t - s} \map f s\)

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 $(3)$:

\(\ds \paren {x + n}^x \map f {n + 1}\) \(\le\) \(\ds \map f {x + n + 1}\)
\(\ds \) \(\le\) \(\ds \paren {n + 1}^x \map f {n + 1}\)

Hence:

\(\ds \paren {x + n}^x n!\) \(\le\) \(\ds \paren {x + n} \paren {x + n - 1} \cdots x \map f x\)
\(\ds \) \(\le\) \(\ds \paren {n + 1}^x n!\)
\(\ds \leadsto \ \ \) \(\ds \paren {1 + \frac x n}^x\) \(\le\) \(\ds \frac {\paren {x + n} \paren {x + n - 1} \cdots x \map f x} {n^x n!}\)
\(\ds \) \(\le\) \(\ds \paren {1 + \frac 1 n}^x\)

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:

$\map f {x + 1} = x \map f x$

$\blacksquare$


Sources