Can Bohr-Mollerup Theorem be extended to complex plane?

Question

There is a famous theorem says about Gamma function:

Bohr-Mollerup Theorem

Let $f:(0,\infty)\rightarrow \mathbb{R}^+$ be a function satisfying below

(i) $f(x+1)=xf(x), \forall x\in (0,\infty)$

(ii) $f$ is a log-convex function

(iii) $f(1)=1$

Then, $f=\Gamma$ on its domain

This theorem only shows that the Gamma function deserves to be called the factorial function for $x\in\mathbb{R}^+$.

Is it possible to extend this idea to complex plane so Gamma function deserves to be the best factorial function?

Since $t^{z-1}e^{-t}$ can be decomposed into its real part and imaginary part using cosine&sine function, i guess it would show that the Gamma function on complex plane has some flexible property similar to that on real line.

Answer 1

A beautiful and simple characterization of the complex $\Gamma$-function exists. It is Wielandt's theorem where you just change condition (ii) "$f$ is a log-convex function" by "$f(x)$ is bounded in the strip $\{x\in\mathbb{C} \mid1\leq\Re(x)\leq2\}$". See Reinhold Remmert : Wielandt's Theorem About the Γ-Function.

Answer 2

As a function of complex variable, $\Gamma$ is meromorphic. There is an identity theorem: if two meromorphic functions on $\mathbb C$ agree on a set with a limit point in $\mathbb C$, then they agree everywhere in $\mathbb C$. In particular, two meromorphic functions that agree on $(0,\infty)$ agree everywhere. As a consequence, any modification of $\Gamma$ away on $\mathbb C\setminus (0,\infty)$ would lose the property of being meromorphic.

One can use the identity theorem to show that (i) continues to hold in the complex plane. Indeed, $\Gamma(z+1)$ and $z\Gamma(z)$ are two meromorphic functions that agree on $(0,\infty)$; as discussed above, this implies $\Gamma(z+1)=z\Gamma(z)$ for all $z\in\mathbb C$. (Minor detail: at $z=0$ one should interpret the product on the right as the limit $\lim_{z\to 0}z\Gamma(z) $.)