Meaning of "$f$ has a power series expansion around $p$"

Question

In Complex Analysis by Donald Marshall (page 29), there is an exercise problem that starts with "Suppose $f$ has a power series expansion at $0$ which converges in all of $\mathbb{C}$. " According to my professor, this means by definition there exists a power series $\sum a_nz^n$ which converges for every $z \in \mathbb{C}$ and is equal to $f$. I initially interpreted this sentence as follows: there exists $r>0$ and a power series $\sum a_nz^n$ such that $\sum a_nz^n$ converges for every $z\in\mathbb{C}$, and that $$f(z) = \sum a_nz^n$$ on $B(0,r)$. In other words, I thought the sentence meant $f$ is analytic at $0$ and its power series expansion around $0$ converges on $\mathbb{C}$. Then, I thought the fact that this series indeed converges to $f$ follows from the identity theorem. More generally, when one says "a function $f$ has a power series expansion around a point $p$, does this mean by definition that $f$ is equal to a power series $\sum a_n(z-p)^n$ on the disk $B(p,R)$ where $R=\frac{1}{\limsup_{n\to\infty} |a_n|^{1/n}}$, or does it mean $f$ is analytic at $p$? I think these are two different things which by means of the identity theorem become the same.

Answer 1

A necessary condition for $f : D \to\mathbb C$ having a power series expansion at $z_0$ is that the domain $D$ of $f$ contains an open neighborhood of $z_0$. I guess than the author only considers functions having an open domain (in which case the above condition is automatically satisfied), but that it not an essential point.

In the definition

$f$ has a power series expansion at $z_0$ if there exists $r > 0$ and a power series $\sum a_n(z-z_0)^n$ which converges in $B(z_0,r)$ and satisfies $f(z) = \sum a_n(z-z_0)^n$ for all $z \in B(z_0,r)$

one should add the requirement $B(z_0,r) \subset D$. Okay, this is implicit in "$f(z) = \sum a_n(z-z_0)^n$ for all $z \in B(z_0,r)$", but nevertheless one should explicitly mention it.

This definition does not require that $r$ is the radius of covergence of $\sum a_n(z-z_0)^n$ and in fact we cannot expect this. Even for functions which have a a power series expansion at all points of its domain $D$ (in which case $D$ must definitely be open) this is false in general. An example is a branch $\ln : D \to \mathbb C$ of the complex logarithm defined on the sliced plane $D = \mathbb C \setminus L$, where $L = \{ x \in \mathbb R \mid x \ge 0\}$. One can show

1. There is no contiuous extension of $\ln$ to a bigger open $D' \supset D$, thus $\ln$ is not the restricton of some continuuous $f : D' \to \mathbb C$ living on an open $D'$ (if that were the case, one could argue that the domain of $\ln$ was "artificially" made smaller).

2. $\ln$ has a power series expansion at all $z_0 \in D$, and the power series $\sum a_n(z-z_0)^n$ has radius of convergence $\lvert z_0 \rvert$.

Take for example $z_0 = 1 + i$. Then the power series at $z_0$ has radius of convergence $\sqrt 2$, but $B(z_0,\sqrt 2) \cap L \ne \emptyset$. This shows that $f(z) = \sum a_n(z-z_0)^n$ cannot be true for all $z \in B(z_0,\sqrt 2)$.

Therefore I would say that

Suppose $f$ has a power series expansion at $0$ which converges in all of $\mathbb{C}$

is a bit imprecise. However, it is clear that the intended meaning is that the power series has radius of convergence $\infty$ and converges to $f(z)$ for all $z \in B(0,\infty) = \mathbb C$.

Conclusion.

Given $r > 0$, one should understand the phrase

$f$ has a power series expansion $\sum a_n(z-z_0)^n$ at $z_0$ which converges in all of $B_r(z_0)$

in the sense that

1. $B_r(z_0) \subset D$.

2. $r$ is less or equal to the radius of convergence of the power series $\sum a_n(z-z_0)^n$.

3. $f(z) = \sum a_n(z-z_0)^n$ for all $z \in B(z_0,r)$.