Power series in $\mathbb{R}$ and $\mathbb{C}$

Question

I do have some general problems understanding power series. So I have a few questions:

(1) - The radius of convergence of a power series $\sum_{k=1}^{\infty} a_k(x-x_0)^k$ can be calculated with $R = \dfrac{1}{\text{lim sup}_{k \rightarrow \infty} \sqrt[k]{\mid a_k \mid}}$. Okay. Is that actually only for $x_0 = 0$?

(2) - The formula in (1) shows, that the series converges absolute for $\mid x - x_0 \mid < R$ and diverges for $\mid x - x_0 \mid > R$. That means that the domain of convergence is ALWAYS a circle? In complex analysis I learned that you can find a holomorphic function for every given domain. Lets consider the domain $D$ not circular. Then there is a function $f$ which is holomorphic on that domain $D$ such that there exist no function $F$ which is holomorphic in a bigger domain $G$ with $D \subset G$ and $F(z) = f(z)$ for every $z \in D$. So there is no analytic conitnuation for $f$. Since $f$ is holomorphic there is a power series for $f$. The formula from (1) says, that the radius of convergence is a circle, and the power series diverges for a bigger radius. How is that possible? We considered that $D$ is not circular.

(3) - Please check, that I didnt get that wrong. We defined $C^\infty (D)$. Is $f \in C^\infty (D)$ then $f$ is infinitely often continuously differentiable. That doesnt mean that $f$ is analytic right? See $e^{-\frac{1}{x^2}}$. If $f$ wants to be analytic, $f$ needs to have a power series in every point $x_0 \in D$. And therefore $f$ needs to be in $C^\infty(D)$. Is the following true?

$f \in C^\infty (D)$ and $f$ has a power series in every $x_0 \in D \Rightarrow f$ analytic $\Rightarrow f$ infinitely often continuously differentiable in every $x_0 \in D$, which just means, that $f \in C^\infty (D)$.

Is that right?

(4) - When $f$ has a local power series, $f$ is analytic. When $f$ is analytic, does that imply that $f$ has a local power series? I know when $f$ is analytic, then $f$ has a convergent taylor series. But the taylor series doesnt have to converge to $f$. Does the taylor series always converge to $f$ when $f$ is analytic?

Answer 1

No, it's not only for $x_0=0$. The series will converge in the interior of a circle of radius $R$ centered at $x_0$ (exceptions: if $\limsup_{k\to\infty}\sqrt[k]{|a_k|}=0$, then the convergence is on the whole complex plane; if $\limsup_{k\to\infty}\sqrt[k]{|a_k|}=\infty$ the convergence is only for $x=x_0$). The series may or may not converge on points on the boundary of the circle.

If $f$ is a holomorphic function on a domain $D$ (connected open set in the complex plane), then the Taylor series for $f$ at $x_0$ will have as radius of convergence the supremum of the real numbers $r>0$ such that the circle centered at $x_0$ with radius $r$ is contained in $D$ (this is not an elementary result, but it follows from Cauchy’s integral formula).

Let's now agree that analytic means having a local power series at any point, with positive (or infinite) radius of convergence.

If a complex function on $D$ has derivative at any point, it's analytic (Goursat’s theorem), in particular it is of class $C^{\infty}$. The converse is obvious.

For functions defined on an open interval of the real line, the situation is different. A function may be of class $C^{\infty}$ without being analytic. The classical example is indeed $$ f(x)=\begin{cases} 0 & x=0\\ e^{-1/x^2} & x\ne0 \end{cases} $$ which is of class $C^{\infty}$, but its Taylor series at $x_0=0$ doesn't have positive radius of convergence.

The converse is true, however: any analytic function is of class $C^{\infty}$. Suppose $f$ is (real) analytic at $x_0$, with Taylor series $$ \sum_{n\ge0}a_n(x-x_0)^n \tag{*} $$ Then the series $$ \sum_{n\ge1}na_n(x-x_0)^{n-1} \tag{**} $$ has the same radius of convergence as (*) and the sum of (**) is the derivative of (*) in the (open) interval of convergence.