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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 $z = R$? Thanksenter image description here

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  • $\begingroup$ If I understand correctly, shouldn't your first sentence be "... has convergence radius AT LEAST $R$?" $\endgroup$
    – D_S
    Commented Apr 27, 2021 at 4:57
  • $\begingroup$ @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. $\endgroup$
    – IGY
    Commented Apr 27, 2021 at 5:15
  • $\begingroup$ 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 $1$ $\endgroup$
    – D_S
    Commented Apr 27, 2021 at 12:57
  • $\begingroup$ 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. $\endgroup$ Commented Apr 29, 2021 at 17:13

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That is correct. An analytic function has at least one singularity on it's circle of convergence, in the sense that there is at least one point on it's circle of convergence in which the function cannot be analytically continued in an open ball around this point. You can find a discussion on this here.

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.

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