Bohr–Mollerup Theorem

Base case: $f(1) = 1 = 0!$ — given by hypothesis.

Inductive step: assume $f(k) = (k-1)!$. Applying the recurrence at $x = k$ gives $f(k+1) = k \cdot f(k) = k \cdot (k-1)! = k!$.

By mathematical induction, $f(n) = (n-1)!$ for every positive integer $n$.

The lesson: this proof uses only the first two conditions — $f$ is not determined at all between the integers.

Indeed, for $|a| < 1$ (e.g. the preset $a = 0.3$), consider the impostor $f_a(x) = \Gamma(x)(1 + a\sin 2\pi x)$.

If $|a| < 1$ then $1 + a\sin 2\pi x \ge 1 - |a| > 0$, so $f_a > 0$ and the codomain $(0,\infty)$ is respected (for $|a| \ge 1$ positivity can fail).

It matches $\Gamma$ exactly at every integer $n$ (since $\sin 2\pi n = 0$) while differing wildly between them.

The third condition (log-convexity) is what closes this gap.

Let $0 < t < 1$, $x, y > 0$, and $z = tx + (1-t)y$. Split the integrand of $\Gamma(z) = \int_0^\infty s^{z-1} e^{-s}\,ds$: $s^{z-1} e^{-s} = \left(s^{x-1} e^{-s}\right)^t \cdot \left(s^{y-1} e^{-s}\right)^{1-t}$ — checking exponents, $t(x-1) + (1-t)(y-1) = z - 1$ and the exponent of $e^{-s}$ is $t + (1-t) = 1$.

Set $F = \left(s^{x-1} e^{-s}\right)^t$, $G = \left(s^{y-1} e^{-s}\right)^{1-t}$ and apply Hölder with exponents $p = \frac1t$, $q = \frac{1}{1-t}$ (so $\frac1p + \frac1q = 1$): $\int_0^\infty FG \le \left(\int_0^\infty F^p\right)^{1/p} \left(\int_0^\infty G^q\right)^{1/q}$.

But $F^p = s^{x-1} e^{-s}$ and $G^q = s^{y-1} e^{-s}$, so the right side is $\Gamma(x)^t\,\Gamma(y)^{1-t}$. Hence $\Gamma(tx + (1-t)y) \le \Gamma(x)^t\,\Gamma(y)^{1-t}$; taking logs gives $\varphi(tx + (1-t)y) \le t\,\varphi(x) + (1-t)\,\varphi(y)$, so $\varphi = \ln\Gamma$ is convex.

Numerical check: the secant test above checks the very definition of convexity (the chord lies above the graph). The violation for $\ln\Gamma$ is below $10^{-9}$ — zero, agreeing with the proof.

Set $\varphi = \ln f$.

We use slope monotonicity of convex functions: for $a < b < c$, $\frac{\varphi(b) - \varphi(a)}{b-a} \le \frac{\varphi(c) - \varphi(b)}{c-b}$.

For $0 < x < 1$ and integer $n \ge 2$, applying this to $n-1 < n < n+x$ gives the lower bound $\varphi(n) - \varphi(n-1) \le \frac{\varphi(n+x) - \varphi(n)}{x}$ directly.

The upper bound does not follow directly: from $n < n+x < n+1$ we only get $\frac{\varphi(n+x) - \varphi(n)}{x} \le \frac{\varphi(n+1) - \varphi(n+x)}{1-x}$.

Use the weighted average of slopes with the same left endpoint: $\varphi(n+1) - \varphi(n) = x \cdot \frac{\varphi(n+x) - \varphi(n)}{x} + (1-x) \cdot \frac{\varphi(n+1) - \varphi(n+x)}{1-x}$, and since the second slope is at least the first, this weighted average is at least the first slope.

Hence $\frac{\varphi(n+x) - \varphi(n)}{x} \le \varphi(n+1) - \varphi(n)$.

Together, $\varphi(n) - \varphi(n-1) \le \frac{\varphi(n+x) - \varphi(n)}{x} \le \varphi(n+1) - \varphi(n)$.

From (i) and (ii) (problem 1), $f(n) = (n-1)!$, so $\varphi(n) - \varphi(n-1) = \ln(n-1)$ and $\varphi(n+1) - \varphi(n) = \ln n$. Multiplying by $x$ gives the key inequality $x\ln(n-1) \le \varphi(n+x) - \varphi(n) \le x\ln n$.

Exponentiating gives $(n-1)^x \le \frac{f(n+x)}{f(n)} \le n^x$, i.e. $(n-1)^x (n-1)! \le f(n+x) \le n^x (n-1)!$. Iterating (ii) $n$ times gives $f(n+x) = x(x+1)\cdots(x+n-1)\,f(x) = f(x)\prod_{k=0}^{n-1}(x+k)$; dividing through yields $\frac{(n-1)^x (n-1)!}{\prod_{k=0}^{n-1}(x+k)} \le f(x) \le \frac{n^x (n-1)!}{\prod_{k=0}^{n-1}(x+k)}$.

The ratio of the upper to the lower bound is $\left(\frac{n}{n-1}\right)^x \to 1$ as $n \to \infty$. Writing $L_n$ and $U_n$ for the bounds, $L_n \le f(x) \le U_n$ and $U_n - L_n = L_n\left(\left(\frac{n}{n-1}\right)^x - 1\right) \le f(x)\left(\left(\frac{n}{n-1}\right)^x - 1\right) \to 0$; by the squeeze theorem both bounds converge to the same limit $G(x)$, and $f(x) = G(x)$ for $0 < x < 1$.

Connecting the Gauss limit to Euler's integral.

In the truncated integral $I_n(x) = \int_0^n t^{x-1}\left(1-\frac{t}{n}\right)^n\,dt$, substitute $s = \frac{t}{n}$ to get $I_n(x) = n^x\int_0^1 s^{x-1}(1-s)^n\,ds$.

Repeated integration by parts gives $\int_0^1 s^{x-1}(1-s)^k\,ds = \frac{k}{x}\int_0^1 s^{x}(1-s)^{k-1}\,ds$ (boundary terms vanish for $x > 0$), so $\int_0^1 s^{x-1}(1-s)^n\,ds = \frac{n!}{x(x+1)\cdots(x+n)}$ — hence $I_n(x) = \frac{n^x n!}{x(x+1)\cdots(x+n)} = \frac{n}{n+x}\,U_n$, where $U_n$ is the upper squeeze bound above.

Now $n \to \infty$: for each $t$, $\left(1-\frac{t}{n}\right)^n \to e^{-t}$, so the integrand converges pointwise to $t^{x-1}e^{-t}$.

Domination comes from $\ln(1-u) \le -u$: for $0 \le t \le n$, $\left(1-\frac{t}{n}\right)^n = e^{n\ln(1-t/n)} \le e^{-t}$, giving the integrable dominating function $t^{x-1}e^{-t}$ on $(0,\infty)$ (for $x > 0$).

By dominated convergence, $I_n(x) \to \int_0^\infty t^{x-1}e^{-t}\,dt = \Gamma(x)$.

The two limits agree: from $I_n(x) = \frac{n}{n+x}\,U_n$, since $\frac{n}{n+x} \to 1$ and $U_n \to G(x)$ (squeeze), $I_n(x) \to G(x)$; but also $I_n(x) \to \Gamma(x)$. Hence $G(x) = \Gamma(x)$ for $0 < x < 1$ — Gauss's product formula equals Euler's integral, fully proved.

Finally extend by (ii): writing $x > 0$ as $x = y + m$ with $0 < y \le 1$ and integer $m \ge 0$, $f(x) = y(y+1)\cdots(y+m-1)\,f(y)$, and since $\Gamma$ satisfies the same recurrence, the agreement on $(0,1]$ extends to all $x > 0$: $f = \Gamma$.