# Can Bohr-Mollerup Theorem be extended to complex plane?

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.

• How do you define convex for a complex-valued function? Feb 7, 2014 at 8:36
• Not answering but just wondering : so we could find a function which has property (i) and (iii) but not log-convex which is not the Gamma function ? I'm curious, do we have such an example ? Feb 7, 2014 at 8:38
• I still dont't get your idea. In the complex plane one time differentiable=$C^\infty$=analytical. Feb 7, 2014 at 9:11
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.
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 $\lim_{z\to 0}z\Gamma(z)$.)