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.


1 Answer 1


$\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}\newcommand\Z{\mathbb Z}$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 $k$.)

  • $\begingroup$ 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! $\endgroup$ Commented Feb 8, 2022 at 20:46
  • $\begingroup$ @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 $0\ne\Gamma(y+n)=\Gamma(y)\prod_{k=0}^{n-1}(y+k)$. $\endgroup$ Commented Feb 8, 2022 at 21:10
  • $\begingroup$ 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? $\endgroup$ Commented Feb 8, 2022 at 23:14
  • $\begingroup$ @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 $k$". $\endgroup$ Commented Feb 9, 2022 at 3:19
  • $\begingroup$ @Mr.MathDoctor : Are you now satisfied with this answer? $\endgroup$ Commented Feb 9, 2022 at 18:15

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