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Consider the Lacunary function $$f(z)=\sum _{n=0}^{\infty}z^{2^{n}}=z+z^{2}+z^{4}+z^{8}+\cdots $$

which is analytic over the interior of the open unit disk. It is easy to show that $f(z)$ doesn't have limits at $2^n$-th roots of unity (see wikipedia), which form a dense subset of unit circle. Wikipedia claims that

hence by continuous extension every point on the unit circle must be a singularity of $f$.

Why does this hold? Suppose $f$ has a limit at one point of the unit circle, does it necessarily have limits everywhere in some neighborhood of it, and thus contradicts to the fact that the there is a dense subset of unit circle where $f$ doesn't have limits? (generally this is not sure, see Riemann Function)

I don't see how the linked post solve my question. I am asking about a very specific power series.

Alp
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I believe they are saying that the unit circle is a natural boundary for the function. If $f$ were analytic at any point on the unit circle, it would be analytic in a small disk about that point, but the disk would necessarily contain an $n$th root of unity, contradiction.

saulspatz
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  • Okay... but do limits exist there at all? I know analytic continuation will fail. – Alp Dec 05 '18 at 19:55
  • Even if the disk is a natural boundary for $f$, in general $f(\zeta)$ can exist for a given $\zeta$ on the unit disk without being analytic there. They are saying a little more: in the example analyzed, the series representing $f$ diverges at every point of the unit circle. – Daniele Tampieri Dec 05 '18 at 19:59
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    That I'm not sure about. Note that singularity is a point at which $f$ is not analytic. It's quite possible for a power series to converge at a boundary point, but the function not to have an analytic function at that point. See Pringsheim's Theorem In the special case in your question, I don't know. – saulspatz Dec 05 '18 at 20:00
  • @DanieleTampieri How does one see this? – saulspatz Dec 05 '18 at 20:01
  • There is an example of Wacław Sierpiński, published in 1916 on the "Rendiconti del Circolo Matematico di Palermo", titled "Sur une série potentielle qui, étant convergente en tout point de son cercle de convergence, représente sur ce cercle une fonction discontinue" where he constructs a series whose natural boundary is the unit circle, but converges at each of its point to a discontinuous function. – Daniele Tampieri Dec 05 '18 at 20:12
  • At present I do not remember a precise example involving only one point of non analyticity. However, again Sierpinski produced an example of a power series converging at one point of the unit circle by modifying an earlier example by Lusin. – Daniele Tampieri Dec 05 '18 at 20:15
  • The notes to chapter III of the book "An Introduction to Classical Complex Analysis" Burckel provide nice insight on the wild behavior of holomorphic function on the natural boundary of their domain of definition. – Daniele Tampieri Dec 05 '18 at 20:17
  • @DanieleTampieri Thanks, but what I meant was, how does one see in the present example that the series diverges at every point on the unit circle? – saulspatz Dec 05 '18 at 20:19
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    Some interesting remarks on Sierpinski's paper are given here – saulspatz Dec 05 '18 at 20:45
  • Let us continue this discussion in chat. – Daniele Tampieri Dec 05 '18 at 21:02
  • Have you got a clear answer @Alp? – Petra Axolotl Jun 10 '22 at 20:03