# In search for a counterexample related to the Abel-Stolz theorem

Edit: cross-posted to the MathOverflow with some with some modifications in order to answer to questions posed in the comments. Now it has an accepted answer by Pietro Majer and a very interesting answer by Alexandre Eremenko.

Definition 1. A curve in the complex plane is the image $$\gamma([a,b])$$ of a segment $$[a,b]\subseteq\Bbb R$$ through a continuous non constant function $$\gamma:[a,b]\to\Bbb C$$: the points $$\gamma(a)$$ and $$\gamma(b)$$ are called endpoints of the curve. With abuse of notation, the curve is identified with its defining continuous function $$\gamma$$ (which is really simply a parametrization of the curve).
Definition 2. A complex sequence $$(a_n)_{n\in\Bbb N}$$ is said Abel-summable if, for $$x\in [0,1]$$, $$\lim_{x\to 1^-}\sum_{n=0}^\infty a_n x^n$$ is finite: if $$s\in\Bbb C$$ is the value of the limit, this is usually written as $$\sum_{n=0}^\infty a_n= s\;(\mathrm{A})$$.
Definition 3. A Stolz region $$\Bbb {St}(M)$$ in the unit disk $$\Bbb D\triangleq \{z\in\Bbb C: |z|^2\le1\}$$ is the set $$\Bbb {St}(M)\triangleq \big\{z\in \Bbb D : |1-z|\leq M(1-|z|)\big\} \quad M>1.$$ Be it noted that if $$M=1$$ then $$\Bbb {St}(M)=[0,1]$$, while if $$M<1$$ then $$\Bbb {St}(M)=\emptyset$$, therefore the condition $$M>1$$ is simply a non-triviality condition.
Theorem (Abel-Stolz). Let $$(a_n)_{n\in\Bbb N}\subset\Bbb C$$ be a complex sequence such that $$\sum_{n=0}^\infty a_n=s\in\Bbb C$$. Then the power series $$f(z)=\sum_{n=0}^\infty a_n z^n$$ converges to $$s$$ along every curve $$\gamma:[0,1]\to\Bbb C$$ with $$\gamma(0)=0$$ and $$\gamma(1)=1$$ and $$\gamma\subset\Bbb{St}(M)$$: moreover the convergence is uniform for every point $$z\in \Bbb{St}(M)$$.

My question:

do there exist a divergent but Abel summable sequence $$(a_n)_{n\in\Bbb N}\subset\Bbb C$$ for which $$f(z)$$ does not converge uniformly on the "deleted" Stolz region $$\Bbb{St}(M)\setminus\{ 1\}$$?

Some notes and why I am interested

• A (sketchy) proof of Abel-Stolz theorem can be found in the related wikipedia entry. For an historical survey on the various proofs and extension of this theorem see the first paragraph of Hardy's paper , while Knopp  offers a detailed analysis in §8, theorem 4 p.74, theorem 5 p. 74, §52, theorem 1 pp. 391-392.
• The problem was motivated by a research on Fatou's theorem: in particular, during the reading of  I found that any $$f(z)$$ such that $$\lim_{x\to 1}f(x)=s$$ converges to $$s$$ along every path $$\gamma\subset\Bbb{St}(M)$$ with endpoints $$0$$ and $$1$$ and is bounded on $$\Bbb {St}(M)$$ (I had to find a proof by myself, since Privalov consider this fact as obvious and thus does not give an explicit proof).

References

 Godfrey Harold Hardy, "Some theorems connected with Abel’s theorem on the continuity of power series". (English) Proceedings of the London Mathematical Society (2) 4, 247-265 (1906), JFM 37.0429.01.

 Konrad Knopp, Theory and Application of Infinite Series, Translated from the 2nd ed. and revised in accordance with the fourth by R. C. H. Young, London-Glasgow: Blackie & Son, 1951, XII+563, Zbl 0042.29203.

 Ivan Ivanovich Privalov, "Sur une généralisation du théorème de Fatou" (Russian, French abstract) Recueil Mathématique Moscou (Matematicheskiĭ Sbornik) 31, 232-235 (1923), JFM 49.0225.02.

• You can add $\sum_{n>0}\frac{z^{3^n}-z^{2\cdot 3^n}}{n}$ and $\sum_{n\geq0}(-1)^nz^n$. The resulting series is not summable, since the coefficients don't tend to zero. The sequence of coefficients is Abel summable, since first series has non-tangential limits at $z=1$ by Abel's theorem and the second is $\frac{1}{1+z}$. The second series has limits when $z\to 1$ regardless of how you approach $1$ from inside the unit disc, but the first one doesn't have non-non-tangential limit. So, the sum cannot have non-non-tangential limit either.
– plop
Jun 1 at 11:14
• @plop, it is true that the sum of functions you suggest does not converge in $z=1$, but both the functions you consider have non-tangential limits, so their sum does: and also I am asking for an example of power series for which the non-tangential limit exists (and this is implied by Abel summability) and it does not uniformly converge on the associated Stolz region. Jun 1 at 12:36
• I read "deleted" as the complement of the Stolz angle. I see now that you mean in(side) $\mathbb{St}(M)\setminus\{1\}$.
– plop
Jun 1 at 12:38
• @plop exactly: I've used the term "deleted" in the sense of the "deleted neighborhood" of singular integrals. It is really an abuse of notation, though it recalls quicly the idea of removing a point from the adherence of a given set. Jun 1 at 12:42