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The Stolz angle is a condition used in Abel's Theorem:

$$|1-z|\leq M(1-|z|)$$

Q1) How do I intuitively remember (and understand this)?

Q2) In particular, is there a quick way to see that

$$(1-|z|)\leq M|1-z|$$

is the wrong condition?

Thanks for any help.

yoyostein
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1 Answers1

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Hmm, this is a slightly "soft" question...

A Stolz angle $S$ is also known as a "non-tangential approach region", the point being that if you approach $1$ along a curve $\gamma\subset S$ then $\gamma$ is not tangent to the unit circle at $1$.

Imagine a curve $\gamma$ in the unit disk that approaches $1$, but which is tangent to the unit circle at $1$. Draw a picture. Points on that curve close to $1$ are much closer to the boundary than they are to $1$, right? Tangential approach to the boundary says $1-|z|$ is much smaller than $|1-z|$.

So non-tangential approach says the opposite, that $1-|z|$ is not much smaller than $|1-z|$, which is to say $|1-z|\le M(1-|z|)$.

Or look at it this way: A lot of formulas are simpler for the upper half plane $y>0$. A non-tangential approach region (to the origin) in the upper half plane is defined by $$|x|<My.$$It's pretty clear I think that that defines an angle in the upper half plane, with vertex at $0$, and which is not tangent to the boundary.

Now, it's not hard to see that $|x|<My$ is equivalent to $(x^2+y^2)^{1/2}<M'y$, where $M'$ depends on $M$. And if you think about it a little, you see that (for points near the boundary point in either domain) the condition $(x^2+y^2)^{1/2}<M'y$ in the upper half plane corresponds roughly to $|1-z|<M'(1-|z|)$ in the disk.

David C. Ullrich
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    Thanks. This was very helpful. I can understand it better as the non-tangential approach now. – yoyostein Jul 17 '16 at 09:14
  • With $S:={z\in \mathbb{C} : \vert z \vert<1 \ \text{and} \ z = 1 - re^{-i\theta}, r>0, \theta \in [-\alpha,\alpha]}$ what would have happened if we have enabled $\alpha= \pm \frac \pi 2$ ? It seems that the Stolz domain would have been determined by a vertical tangent line to the point $1$. Hence maybe it means that every point of the convergence unit disk $z$ which also belongs to the Stolz domain would have approached $1$ "tangentially" ? In that case, I guess that the Abel's angular limit theorem does not hold for instance... – Maman May 12 '22 at 17:04
  • What is the homeomorphism you are referring when transferring the inequality from the upper half plane to the open unit disk? Using stereographic projection I find one such. Consider the function $f : \mathbb D^2 \longrightarrow \mathscr H$ defined by $$(x,y) \mapsto \left (\frac x {1-y}, \frac {\sqrt {1 - x^2 - y^2}} {1-y} \right ).$$ The inverse of $f$ is given by $$(x,y) \mapsto \left (\frac {2x} {1 + x^2 + y^2}, \frac {x^2 + y^2 -1} {1 + x^2 + y^2} \right ).$$ – Akiro Kurosawa Dec 16 '23 at 15:41