# 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 [1], while Knopp [2] 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 [3] 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

[1] 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.

[2] 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.

[3] 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