I'm trying to get a better handle on behavior of complex power series on the boundary of their maximal disk of convergence.
I'm reading Bak-Newman's Complex Analysis, Chapter 18.1.
A regular point $z_0$ on the circle bounding the maximal disk of convergence is defined as one where the function in question can be continued analytically to some open neighborhood of $z_0$. My understanding of analytic at a point in this book is that it is always used to indicate differentiability on some neighborhood of the point, so that being analytic at a point is equivalent to being analytic in some open disk around a point.
(By differentiable at a point $z_0$ I mean that $\lim_{z \to z_0}\frac{f(z)-f(z_0)}{z – z_0}$ exists, or that equivalently the function $\mathbb{R}^2 \to \mathbb{R}^2$ is differentiable and the Cauchy-Riemann equations hold.)
This section of the book also defines a singularity on the circle of convergence to be a point which is not a regular point. I'm trying to figure out how this definition of singularity relates to the notion of isolated singularity with which I'm already (more or less) comfortable.
My question is this: Is it possible to be differentiable at a point on the circle of convergence, but not analytic at that point?
More or less, I'm trying to figure out if there is a function which is defined in some open neighborhood of the closed unit disk, analytic in the open unit disk, complex differentiable at $z = 1$, but not differentiable at each of a sequence of real $x_n > 1$ which converge to $1$.
Can this happen?
Best Answer
Yes. Let $S$ be the set $\{1+re^{it}: r\ge 0, |t|\le \pi/3\}$ (angle/sector/cone or what you call it). In the domain $\mathbb C\setminus S$ the function $(1-z)^{3/2}$ has a single valued branch with $(1-0)^{3/2}=1$. On the boundary of $S$ this branch takes real, nonpositive values. Let $$f(z)=\begin{cases} (1-z)^{3/2} \quad & z\notin S \\ -|1-z|^{3/2} & z\in S \end{cases}$$ The function $f$ is continuous on $\mathbb C$, analytic in the open unit disk, complex differentiable at $1$ (with $f'(1)=0$), but is not holomorphic in any neighborhood of $1$, since it is real-valued in $S$.