Theorem: Bohr-Mollerup Theorem

Let $f: \R_{>0} \to \R_{>0}$ be a real function which is positive on $\openint 0 \to$.
Let $\ln \mathop \circ f$ be convex on $\R_{>0}$.
Let $f$ satisfy the conditions:
:<nowiki>$\map f {x + 1} = \begin{cases}
1 & : x = 0 \\
x \map f x & : x > 0 \end{cases}$</nowiki>
Then:
:$\forall x \in \R_{>0}: \map f x = \map \Gamma x$
where $\map \Gamma x$ is the Gamma function, defined in the Euler form.

Proof:
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:
{{begin-eqn}}
{{eqn | l = \map \ln {\map f t}
      | o = \le
      | r = \alpha \map \ln {\map f s} + \beta \map \ln {\map f {s + 1} }
      | c = $\ln \mathop \circ f$ is convex on $\R_{>0}$
}}
{{eqn | ll= \leadsto
      | l = \map f t
      | o = \le
      | r = \paren {\map f s}^\alpha \paren {\map f {s + 1} }^\beta
      | c = 
}}
{{eqn | r = \paren {\map f s}^\alpha \paren {s \map f s}^\beta
      | c = 
}}
{{eqn | r = s^\beta \map f s
      | c = as $\alpha + \beta = 1$
}}
{{eqn | n = 1
      | r = s^{t - s} \map f s
      | c = as $\beta = t - s$
}}
{{end-eqn}}
Because $s \le t \le s + 1$, it follows that $t - 1 \le s \le t$.
Substituting as appropriate in $(1)$:
{{begin-eqn}}
{{eqn | l = \map f s
      | o = \le
      | r = \paren {t - 1}^{s - t + 1} \map f {t - 1}
      | c = 
}}
{{eqn | n = 2
      | r = \paren {t - 1}^{s - t} \map f t
      | c = 
}}
{{end-eqn}}
Combining $(1)$ and $(2)$:
{{begin-eqn}}
{{eqn | l = \paren {t - 1}^{t - s} \map f s
      | o = \le
      | r = \map f t
      | c = 
}}
{{eqn | n = 3
      | o = \le
      | r = s^{t - s} \map f s
      | c = 
}}
{{end-eqn}}
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 $(3)$:
{{begin-eqn}}
{{eqn | l = \paren {x + n}^x \map f {n + 1}
      | o = \le
      | r = \map f {x + n + 1}
      | c = 
}}
{{eqn | o = \le
      | r = \paren {n + 1}^x \map f {n + 1}
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \paren {x + n}^x n!
      | o = \le
      | r = \paren {x + n} \paren {x + n - 1} \cdots x \map f x
      | c = 
}}
{{eqn | o = \le
      | r = \paren {n + 1}^x n!
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {1 + \frac x n}^x
      | o = \le
      | r = \frac {\paren {x + n} \paren {x + n - 1} \cdots x \map f x} {n^x n!}
      | c = 
}}
{{eqn | o = \le
      | r = \paren {1 + \frac 1 n}^x
      | c = 
}}
{{end-eqn}}
Since:
:$1 + \dfrac x n \to 1$
:$1 + \dfrac 1 n \to 1$
as $n \to \infty$, we have:
:$\paren {1 + \dfrac x n}^x \to 1$
:$\paren {1 + \dfrac 1 n}^x \to 1$
Therefore by Squeeze Theorem for Real Sequences:
:$\ds 1 = \lim_{n \mathop \to \infty} \frac {\paren {x + n} \paren {x + n - 1} \cdots x \map f x} {n^x n!}$
Thus when $0 < x \le 1$:
:$\ds \map f x = \lim_{n \mathop \to \infty} \frac {n^x n!} {x \paren {x + 1} \cdots \paren {x + n} } = \map \Gamma x$
by the Euler form of the Gamma function.
But then, by:
:$\map f {x + 1} = x \map f x$
combined with Gamma Difference Equation, we find that:
:$\map f x = \map \Gamma x$
for all $x > 0$.
{{qed}}
{{Namedfor|Harald August Bohr|name2 = Johannes Mollerup|cat = Bohr, H|cat2 = Mollerup}}