Complex differentiable but not analytic on circle of convergence

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?

• This is a very subtle question. But what do you mean by complex differentiable at $1$, let alone by the rest of your sentence? The domain is only $\overline\Delta$, so I could take the difference quotient only for $z\in\Delta$ to consider the first question. But are you thinking there's some well-defined, continuous extension to a neighborhood of $z=1$ in $\mathbb C$? This seems highly unlikely. – Ted Shifrin May 26 '13 at 19:06
• Thank you Ted. I changed the wording of the question try to clarify some. I should have said that the function (or lack of) that I'm curious about should be defined on a neighborhood of the closed disk. – bryanj May 26 '13 at 19:20
• Well, probably you want it just in a neighborhood of $z=1$. If it's continuous on a neighborhood of all of $\overline\Delta$, then by the Cauchy integral formula you're going to have an analytic function on a neighborhood of $\Delta$. But trying to have a damped-out branch point at $1$ is not going to allow even a continuous extension to a neighborhood (functions like $\sqrt{1-z}$ or $(1-z)^2\log(1-z)$ won't work, although the latter will be complex differentiable at $z=1$ in the sense I mentioned above). – Ted Shifrin May 26 '13 at 19:41
• Thanks again Ted - I'm digesting your second comment. I think I don't have a full understanding of what you were getting at with the Cauchy integral formula comment. Could not one take any function $f$ analytic on $\overline{\Delta}$ and from it create another function $g$ as follows: $g(z) = f(z)$ when $|z| \le 1$, $g(z) = (2 - |z|) f(\frac{z}{|z|})$ when $1 < |z| < 2$, and $g(z) = 0$ when $|z| >= 2$? It seemed to me that this function should be analytic on $|z| < 1$ and continuous on $\mathbb{C}$ but not analytic anywhere outside the unit disk. – bryanj May 26 '13 at 22:24
• Ah, fair enough. So when we use your $g$ and apply the Cauchy integral formula on a circle of radius $1.1$ it will not reproduce $f$ inside $\Delta$, but it will define some analytic function. It is interesting. If you do this gluing in a smooth way (say, using bump functions), you'll get a real differentiable function on $S^1$ but it will (almost surely) not be complex differentiable other than from the interior. – Ted Shifrin May 26 '13 at 23:22

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 $1$.
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 $S$.
• Thanks! Just to clarify for myself: are you defining $(1-z)^{3/2}$ as $((1-z)^{1/2})^3$ using principal branch of log? – bryanj Jun 23 '13 at 15:47
• @bryanj Yes. It may be easier to first apply $1-z$ to $\mathbb C\setminus S$, which turns it to $\{re^{it}: |t|\le 2\pi/3\}$. Then $z^{3/2}$ is naturally $r^{3/2}e^{(3/2)it}$, which is nonpositive on the boundary. – ˈjuː.zɚ79365 Jun 23 '13 at 15:59