# Gamma function and the somewhat extended version of Bohr-Mollerup theorem

The Gamma function $$\Gamma$$ is defined by $$\begin{equation*} \Gamma(x)=\int_{0}^\infty t^{x-1}e^{-t} \,\mathrm{d}t, \end{equation*}$$ for $$x>0$$. It satisfies the well-known functional equation $$\Gamma(x+1)=x\Gamma(x)\label{1}\tag{i}$$ for all $$x>0$$. Apparently, the definition of Gamma function can be extended to $$\mathbb{R}\setminus \{0,-1,-2,\dots\}$$ by using \eqref{1}. With this extended definition, the equation \eqref{1} will also hold for negative real numbers except at non-positive integers.

Now the question is:

Let $$A\subseteq \mathbb{R}$$ containing $$(0,\infty)$$, and let $$f:A\to\mathbb{R}$$. Suppose that you have shown that $$f(x)=\Gamma(x)$$ for all $$x>0$$ by using Bohr-Mollerup Theorem. However, if $$f(x+1)=xf(x)$$ for all $$x\in A$$, would it imply that $$f(x)=\Gamma(x)$$ hold for all $$x\in A\setminus \{0,-1,-2,\dots,\}$$?

The statement of the Bohr-Mollerup Theorem is:

Bohr-Mollerup Theorem: Gamma function is the only positive function $$f:(0,\infty)\to\mathbb{R}$$ that is logarithmically convex, $$f(1)=1$$ and $$f(x+1)=xf(x)$$ for all $$x>0$$.

But in Artin's "the Gamma function", he formulate (probably the translator) the Bohr-Mollerup Theorem a bit different:

B-M2 Theorem: If a function $$f(x)$$ satisfies the following three conditions, then it is identical in its domain of definition with the gamma function:

1. $$f(x+1)=xf(x)$$
2. The domain of definition of $$f(x)$$ contains all $$x>0$$, and is logarithmically convex for these x.
3. $$f(1)=1$$.

What Artin did was as follows (the way I understand it): It has been shown that $$\Gamma$$ satisfies these conditions, so the existence is OK. Let us assume $$f$$ that satisfies these conditions, and want to check its uniquenessness. First, assume $$x\in (0,1]$$. After some steps, we end up getting $$f(x)=\lim_{n\to\infty} \frac{n!n^x}{x(x+1)\cdots (x+n)}.$$ which shows that $$f$$ is unique on $$(0,1]$$. But the uniqueness of $$f$$ also holds on $$(1,2]$$ by condition (1). Keep doing like this, then $$f$$ is unique for whole $$(0,\infty)$$. From here, I am not really sure, how to conclude that $$f$$ is also unique on whole of its domain, when the domain is not explicitly known, which is why I asked the question.

$$\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}\newcommand\Z{\mathbb Z}$$The answer is yes. Indeed, the conditions $$f\colon A\to\R$$ and $$f(x+1)=xf(x)$$ for all $$x\in A$$ imply that $$x+1\in A$$ for any $$x\in A$$ and hence, by induction, $$x+k\in A$$ for all natural $$k$$. We also have $$(0,\infty)\subseteq A\subseteq\R$$ and $$f(x)=\Ga(x)$$ for all $$x>0$$.
Now take any $$y\in A\cap(-\infty,0]\setminus \{0,-1,-2,\dots,\}$$. It remains to show that $$f(y)=\Ga(y)$$. Take any natural $$n$$ such that $$y+n>0$$. Then $$\Ga(y+n)\ne0$$ and $$\begin{equation} f(y)\prod_{k=0}^{n-1}(y+k)=f(y+n)=\Ga(y+n)=\Ga(y)\prod_{k=0}^{n-1}(y+k). \tag{1} \end{equation}$$ So, $$\prod_{k=0}^{n-1}(y+k)\ne0$$ and $$f(y)=\Ga(y)$$, as desired. (The conclusion $$\prod_{k=0}^{n-1}(y+k)\ne0$$ also follows because $$y$$ is not an integer and hence $$y+k\notin\mathbb Z$$ for any integer $$k$$.)
• Could you please rephrase the statement what you mean by "... imply that $x+1\in A$ for any $x\in A$"? It sounds as if the set $A$ contains $x+1$ whenever it contains $x$. Also, why do we need $\Gamma(y+n)$ to be non-zero? As far as I can see, its requirement is not needed anywhere in your proof. Very thanks for your help! Feb 8 at 20:46
• @Mr.MathDoctor : (i) Your conditions $f\colon A\to\R$ and $f(x+1)=xf(x)$ for all $x\in A$ do imply that the set $A$ contains $x+1$ (as an element) whenever $A$ contains $x$ (as an element). Indeed, if $f(x+1)=xf(x)$ for all $x\in A$, then $f(x+1)$ must be defined for all $x\in A$. But $f$ is defined only on $A$. So, we must have $x+1\in A$ for any $x\in A$. (ii) To cancel the factor $\prod_{k=0}^{n-1}(y+k)$, we have to ensure that it is $\ne0$, which follows because $0\ne\Gamma(y+n)=\Gamma(y)\prod_{k=0}^{n-1}(y+k)$. Feb 8 at 21:10
• Thanks a lot for the elaboration. Makes better sense now. May I ask one more time, what's good knowing that $x+1\in A$ whenever $x\in A$? I mean, what can this information be used for? Feb 8 at 23:14
• @Mr.MathDoctor : This is needed in order to have $x+k\in A$ for all $x\in A$ and all natural $k$, which is in turn needed to have the first equality in (1). I have now added "and hence, by induction, $x+k\in A$ for all natural $k$". Feb 9 at 3:19