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It is well-known that the function $f(z)=\sum_{n=0}^\infty z^{n!}$ is analytic in the open unit disk and it can not be extended analytically to any proper open superset of the unit disk, i.e., the unit circle is the natural boundary of $f$.

Is there an example of a function, analytical in the unit disk, with "natural boundary" a part (of positive 1-dimensional measure) of the unit circle, which extends analytically to rest of $\mathbb C$.

I am not asking for a variation of $g(z)=\sqrt{z^2-1}$, which is defined in $\mathbb C\smallsetminus [-1,1]$, since the segment $[-1,1]$ is not a "natural boundary" of $g$, as $g$ can be analytically continued through this segment.

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    $\begingroup$ You can move the unit circle to the real line by a fractional linear transformation. There are somewhat contrived examples of functions which have the negative real axis as a natural boundary, but they also show up in the wild. For example, see Glasser, Abraham, and Lieb, "Analytic Properties of the Free Energy for the 'Ice' Models." J. Math. Phys. 13, 887 (1972). scitation.aip.org/content/aip/journal/jmp/13/6/10.1063/… $\endgroup$
    – Douglas Zare
    Commented Dec 15, 2013 at 1:02

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Every closed set $F$ on the unit circle is the set of singularities of some analytic function. Take a countable dense subset $z_k$ of $F$ and then choose positive $a_k$ so small that the series $$f(z)=\sum_k\frac{a_k}{z-z_k}$$ converges uniformly on compact subsets of $C\backslash F$.

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  • $\begingroup$ Is this the kind of Poincare simple poles? $\endgroup$
    – XL _At_Here_There
    Commented Jul 28, 2014 at 6:57
  • $\begingroup$ How to prove natural boundary of $f(z)=\sum_{n=0}^\infty z^{n!}$ is not countable dense set?Or it's uncountable? $\endgroup$
    – XL _At_Here_There
    Commented Jul 28, 2014 at 7:10
  • $\begingroup$ The set of singularities on the unit circle is closed (by definition). $\endgroup$
    – Alexandre Eremenko
    Commented Jul 28, 2014 at 21:21

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