# Is there an analytic continuation at $z=R$?

Assume $$f(z)$$ is analytic in $$D_R(0)$$ (in other words, the corresponding power series $$\sum_{k=0}^{\infty}a_kz^k$$ has the convergence radius equal to R). Suppose we make an analytic continuation to a point $$z = r + i\ 0$$ with $$r < R$$ (see the diagram below) and consider the power series representation for f(z) centered at $$z = r$$. If it happens that the corresponding power series has the convergence radius exactly $$R − r$$ (the blue circle), In this case, can we claim that no analytic continuation exists at the point $$z = R$$? Thanks

• If I understand correctly, shouldn't your first sentence be "... has convergence radius AT LEAST $R$?"
– D_S
Commented Apr 27, 2021 at 4:57
• @D_S Thanks for the comment! I'm not quite sure about that, the original question I'm referencing says it's equal to R.
– IGY
Commented Apr 27, 2021 at 5:15
• Okay. My point is e.g. $f(z) = z^2$ is analytic in the disc $D_1(0)$, but its radius of convergence is more than $1$
– D_S
Commented Apr 27, 2021 at 12:57
• This is precisely a necessary and sufficient condition for a point $z$ on the boundary of the convergence disk to be a point of noncontinuability. In the linked Q&A the standard case $r=\frac{1}{2}$ and $R=1$ is shown, but the reference by Markushevich deals precisely with the the general $r$ (and $R$) case. Commented Apr 29, 2021 at 17:13

For the blue circle, this point cannot be any other point than $$z=R$$. Indeed, for all the other points we can simply take an open ball around the point contained in the big red circle, and the series of $$f$$ around $$0$$ as our analytic continuation.