Classical Analysis – Gamma Function and Extended Bohr-Mollerup Theorem

ca.classical-analysis-and-odesgamma functionspecial functions

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.

Best Answer

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

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