Disclaimer: I posted this question seven days ago here on the Math.SE, with slightly different (however in an inessential way) comments. The question has been upvoted but no answer has been given, so I decided to repost it here to see if the members of this community can offer an answer.

Edit: I added a bit more information respect to the OP, answering (hopefully) the question posed by Pietro Majer in his comment. Precisely I have added what I mean for uniform convergence in this context (in the main question text) and why I have considered the limit of $f$ along general curves $\gamma$ (in the "Notes" section).

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 $z\in \Bbb{St}(M)$.
In this context, uniform convergence means that, for any $\varepsilon > 0$ the integer $m\in\Bbb N$ for which the relation $$ \left |f(z) -\sum_{n=0}^m a_n z^n\right|\triangleq|f(z)-f_m(z)|<\varepsilon $$ holds true is independent from the chosen 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 extensions of this theorem, I consulted the first paragraph of Hardy's paper [1], while for the summability theory based proof I read Knopp [2] since he offers a detailed analysis in §8, theorem 4 and theorem 5 p. 74, §52, theorem 1 pp. 391-392.
  • The problem was motivated by a research on power series on the unit disk: precisely, I am trying to use a result of Giovanni Ricci [4] with different hypothesis. He is able to estimate the remainder (or the limit of the remainder, as Ricci considers also series for which $f(1)$ does not exists but $\lim_{r\to 1^-} f(r)=s$ does) of a power series which cannot be analytically continued across the point $z=1$. The series I am studying have this property, but the results of Ricci are not directly applicable as he uses several more hypotheses which are not easily checked in my case.
    Then I tried to modify Ricci's method, itself also a modification of a proof method introduced by W.H. Young (see the references in [4], notably Titchmarsch's "Theory of functions") to prove Fatou's theorem for regular boundary points: and since it uses Cauchy's integral formula, I thougt that a judicious use of an integration path touching the boundary could be useful in determining the sought for modification, thus I started considering paths $\gamma$ with one endpoint in $z=1$. However, as pointed out again by Pietro Majer, it is not necessary to consider any curve since even the standard theorem holds irrespectively of any regularity hypothesis on $\gamma$: you can have $\gamma\big([0,1]\big)=\Bbb{St}(M)$.
  • Dealing with the task described in the previous point, I noticed the version of Fatou's theorem given by Privalov [3]: in my opinion, it is an interesting result since it lists between the necessary and sufficient conditions for the theorem to hold the fact that $f$ should be bounded in a non tangential conical region whose vertex is the relevant boundary point. During the reading of [3], I found a simple proof of the necessary part (I had to find a proof by myself, since Privalov considers this fact as obvious and thus does not give an explicit proof). Then I started to wonder if a stronger statement actually holds: and this thought has originated this question.


[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.

[4] Giovanni Ricci, "Sul resto delle serie di potenze alla periferia del cerchio di convergenza" (Italian) in Scritti Matematici in Onore di Filippo Sibirani, Bologna: Cesare Zuffi, pp. 233-242 (1957), MR0086864, Zbl 0077.28403

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    $\begingroup$ In R. Estrada, "Abel summability and angular convergence", SCIENTIA Series A: Mathematical Sciences 9 (2003), 45–51 power series $f(z)$ are constructed that converge on the open unit disc that are Abel summable at $1$ but $f(z)$ has no limit as $z \to 1$ within each Stolz angle at $1$ in the open unit disc. $\endgroup$
    – KConrad
    Jun 7, 2021 at 22:10
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    $\begingroup$ Why do you need to introduce the curves $\gamma$? What is the meaning of "uniform convergence" in the statement of Theorem AND of Question? $\endgroup$ Jun 8, 2021 at 5:37
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    $\begingroup$ Note that you do not need any notion of "convergence along curves" to state the theorem, since the thesis is just $f(z)\to s$ for $z\to1$, $z\in \mathbb{St}(M)\setminus\{1\}$. $\endgroup$ Jun 8, 2021 at 12:04
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    $\begingroup$ @PietroMajer do I really need the condition $|z-1|\le r$ and not simply $z\in\Bbb{St}(M)$? I do not understan why it is so, since when $\sum a_n$ is finite and converges to $s$ (classical Abel-Stolz theorem), the uniform convergence holds on the whole Stolz region: or do I simply misunderstand the statement of the theorem? $\endgroup$ Jun 8, 2021 at 13:12
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    $\begingroup$ that's true, forget about last comment :) $\endgroup$ Jun 8, 2021 at 13:25

2 Answers 2


In fact a trivial counterexample to the question, as now clarified, is just $a_n:=(-1)^n$ with $s:=\frac12$. In your notation, $a_n\to\frac12\;\bf (A)$, because the series $\sum_{n=0}^\infty (-z)^n$ converges on the open disk to $f(z):=\frac1{1+z}$, which is holomorphic at $z=1$. But the series does not converges uniformly on any Stolz angle, not even on $[1-\epsilon,1)$, since the remainder $f(z)-f_m(z)$ is $\frac{(-z)^{m+1}}{1+z}$, whose uniform norm on $[1-\epsilon,1)$ is at least $1/2$ .

  • $\begingroup$ Thank you very much: simple and clear as water. $\endgroup$ Jun 8, 2021 at 13:47

A simple counterexample is the following: consider $f(z)=e^{-z^3}$, restrict it to the right half-plane, and then take a composition $g=f\circ\phi$, where $\phi(z)=(1+z)/(1-z)$ is a conformal map of the unit disk onto the right half-plane which sends $[0,1)\to [1,\infty)$. Then $\lim_{r\to1-}g(z)=0$ but angular limits do not exist for Stolz angles of opening $>\pi/3$.

  • $\begingroup$ Thank you for your answer: I have to elaborate a bit on it. I'd like to figure out if the Stolz angles opening can be reduced to 0 or not. $\endgroup$ Jun 8, 2021 at 12:46
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    $\begingroup$ Yes, it can be reduced to 0. For this use the function $(\sin z)/z$ to begin with. $\endgroup$ Jun 9, 2021 at 4:32
  • $\begingroup$ In the above counter-example, why $g(z)$ does NOT possess angular limits for Stolz angle > $\pi/3$? $\endgroup$ Jul 6, 2021 at 12:20
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    $\begingroup$ @ Pierre MATSUMI: because the radial limit is $0$ but on some other segments $[a,1)$ the function tends to $\infty$ $\endgroup$ Jul 6, 2021 at 13:32
  • $\begingroup$ @AlexandreEremenko having had a look to the counterexamples you offered, am I right if I say that the way to produce them is to construct a holomorphic function with an essential singularity with a prescribed limit behavior along the rays (radial directions)? And if it is so, could I perhaps be able to construct an example of such kind with a countable set of rays along with the radial limit exists but not otherwise? $\endgroup$ Jul 17, 2021 at 7:17

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