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:
\(\ds \map \ln {\map f t}\) | \(\le\) | \(\ds \alpha \map \ln {\map f s} + \beta \map \ln {\map f {s + 1} }\) | $\ln \mathop \circ f$ 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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.6$