Pictures of some generalized analytical continuations

[Caleb]

THANK YOU, CALEB!

I AM AWARE OF EVERYTHING IN YOUR POST. IT MAY BE CONTINUOUS AT A BOUNDARY POINT! IT MAY EVEN BE COMPLEX DIFFERENTIABLE!

BUT!!!!!!!!!!!!!!!!!!!!!! IT IS NOT COMPLEX DIFFERENTIABLE ON AN OPEN DOMAIN!

Funny you mention \(1/n^2\). Let me take the function:

\[
f(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{(n+1)^2}\\
\]

Then MY FORMULA STILL WORKS!!!!!

My formula doesn't care that \(1/2^n\) goes to zero fast; All I care is that \(G(n) \to 0\) and:

\[
\sum_{n=0}^\infty |G(n)| < \infty\\
\]

Then:

\[
f_G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} G(n)\\
\]

Satisfies the reflection formula. Let's make it even more complex. Let's say \(m(n) : \mathbb{N} \to \mathbb{N}\), so instead of \(n \mapsto n\), we have \(n \mapsto m(n)\). So we write:

\[
f_G^m(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{m(n)}} G(n)\\
\]

THIS SATISFIES THE SAME REFLECTION FORMULA!

---

The fact you are noticing values in which this object is finite/differentiable along \(|z| = 1\), is entirely expected. At every value \(|z| = 1\) such that \(z \neq e^{\pi i/ n}e^{2\pi i \frac{k}{n}}\) for all \(n \ge 1\) and \(0 \le k \le n\)-- this object is differentiable.

The point, and maybe, I'll chop it up to you not understanding some subtleties. Just because \(f(z)\) is complex differentiable at \(z_0\), does not mean it is holomorphic at \(z_0\). For \(f(z)\) to be holomorphic at \(z_0\), we need \(f(z)\) to be complex differentiable for \(|z-z_0| < \delta\) for some \(\delta> 0\). This is a SUPER SUBTLE DIFFERENCE!

What you are detailing is that \(f(z_0)\) and \(f'(z_0)\) exist! Which, all the power to you! That's correct as fuck! But HOLOMORPHY, is a stricter requirement. So when I say it is NOWHERE HOLOMORPHIC on \(|z| = 1\), I am correct. BUT IT ALLOWS, for \(f(z_0)\) and \(f'(z_0)\) to still exist!

I am entirely on board with you, Caleb! No where in this post am I trying to disagree. I am just clarifying terminology. This function is NOWHERE HOLOMORPHIC for \(|z| = 1\). It can be pointwise differentiable! BUT NOT HOLOMORPHIC! At least, assuming that there are dense poles about \(z^n = -1\)....

Yes, I agree with what you have here, thank you for clarifying the terminology. Of course the function is nowhere holomorphic at the boundary, the boundary has a dense set of singularity. 

But, if we want to study function beyond natural boundary, I think there are some regular notions we must let go of. In particular, I thinking about what happens when we reduce the definition of holomorphic from "F is holomorphic on an open subset" TO "F is holomorphic on an open subset minus a set of measure zero".

My point is just that the functions we are dealing with here only have a measure 0 set of bad points-- the rest of the points are well defined. So, I'm interested if your extension will recover any familar theorems when we weaken the definition of holomorphic in the way I have suggested.

Also, I don't mean to disagree with the work you have done. I'm not imply anything you have is wrong. I'm just suggesting that we look at some details of coherence on the unit circle to see if a given extension is 'good'. I.e., we might be able to classify 'bad' extensions by their bad behaviour on the unit circle. My cursory look at generalized analytical continuation suggests they do something like this-- they look for matching boundary values as one condition for continuation. I think complex differentiablity might be another condition to look for, to shift out bad extensions from the good ones

Lets call my defintion of holomorphic weakly holomorphic. We might wonder if there are any functions that have a natural boundary but a weakly holmorphic extension. Is the natural extension of c(x) weakly holomorphic? Weakly holomorphic seems like a pretty strong condition, so I'm thinking that showing c(x) satisfies it would be a pretty strong step towards full uniqueness.

EDIT: Sorry-- I now realize "holomorphic except on a set of measure zero" isn't a coherent idea. I think there is something there-- but thats not the right idea. I'll have to think about it a bit more to figure out what I really want to say. Anyway, I still think that having such good convergence on the boundary (convergence except on a set of measure 0) definitely means something, and it means that a lot of analysis can be done with a sufficent weakening of definitions