Meaning of $f$ Extends to A Meromorphic Function

Question

What does it mean by saying that "$g$ extends to a meromorphic function $f$"?

Does it means there another function $f$ which has a function $g$ as a term, (i.e., $\zeta(s)$ is present in that function) and that function is holomorphic - means it is complex differentiable, i.e., has derivative on complex plane - except some point(s)?

Is above understanding correct?

The source of the above question is given below-

Answer 1

A "function", by definition, has three ingredients: a "domain of definition" $X$, a "domain of value" $Y$, and a rule that assigns to each $x\in X$ an element $y \in Y$.

When we say "$g$ extends to $f$", it basically means that $g$ is a priori defined only on a smaller set $X$ and $f$ is defined on a superset $X'$ containing $X$, such that the restriction of $f$ to $X$ coincides with $g$.

In your case, I guess $\Phi(s)$ is originally only defined for $\operatorname{Re}(s) > 1$, and then it extends to $\operatorname{Re}(s) > \frac{1}{2}$.

Answer 2

Let $A$ and $B$ be open subsets of $\mathbb C$, with $A\subset B$, and let $f\colon A\longrightarrow\mathbb C$ be a holomorphic function. We say that $f$ extends to a meromorphic function on $B$ if there is a meromorphic function $g\colon B\longrightarrow\mathbb C$ such that $g|_A=f$ (or, in other words, $(\forall z\in A):g(z)=f(z)$).

Note that, strictly speaking, the domain of $g$ doesn't have to be $B$. It is a subset $D_g$ of $B$ such that each $z_0\in B\setminus D_g$ is a pole of $g$.