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.

  • $\begingroup$ How do you define convex for a complex-valued function? $\endgroup$ Feb 7, 2014 at 8:36
  • $\begingroup$ 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 ? $\endgroup$
    – Rivten
    Feb 7, 2014 at 8:38
  • $\begingroup$ @Rivten: See e.g. luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html $\endgroup$ Feb 7, 2014 at 8:47
  • $\begingroup$ @Martin Of course i'm not saying the complex Gamma function would have some inequality property. I just want to know how much the Gamma function is smooth*(informally speaking) while the real part of it is very *smooth. $\endgroup$
    – John. p
    Feb 7, 2014 at 8:59
  • $\begingroup$ I still dont't get your idea. In the complex plane one time differentiable=$C^\infty$=analytical. $\endgroup$ Feb 7, 2014 at 9:11

2 Answers 2


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) $.)


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