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:

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

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.