Hi! I am Max.

Question:
We know that
if a series converges absolutely at magnitude $r$,
then it converges at every $z$ such that $|z|=r$.
Is the converse true?
My answer: abs_converse.pdf

$\hat{f}(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi ix\xi} dx = \int_{-\infty}^\infty f(-x) e^{-2\pi i(-x)\xi} dx = \pm \int_{-\infty}^\infty f(x) e^{-2\pi ix(-\xi)} dx = \pm \hat{f}(-\xi)$ This proves the $\implies$ direction.
Similar argument but with Fourier inversion formula proves the $\impliedby$ direction.

Denote $B_a(z) = \frac{ a-z }{ 1-\overline{a}z }$ and recall SS ch 1 ex 7.
Definition: the trivial automorphism of $\mathbb{D}$ is $\operatorname{id}_\mathbb{D}$.
Definition: a fixed point of a function $f: X \to Y$ is an $x \in X$ such that $f(x) = x$.

Claim: Let $f \in \operatorname{Aut}(\mathbb{D})$. If $f$ is nontrivial, then $f$ has at most one fixed point in $\mathbb{D}$.
Proof:
We prove the contrapositive; assume $f$ has two distinct fixed points $z_0,z_1 \in \mathbb{D}$.
The function $g = B_{z_0} \circ f \circ B_{z_0}$ fixes $0$ and $B_{z_0}(z_1)$
so Schwarz lemma gives $g = (z \mapsto \lambda z)$ for some $\lambda \in \partial\mathbb{D}$,
and we must have $\lambda = 1$ since $g$ fixes $B_{z_0}(z_1)$,
hence $g = \operatorname{id}_\mathbb{D}$
hence $f = B_{z_0} \circ g \circ B_{z_0} = \operatorname{id}_\mathbb{D}$.
QED.

Examples:
If $f = (z \mapsto \lambda z)$ with $1 \neq \lambda \in \partial\mathbb{D}$, then $0$ is the unique fixed point of $f$ because $f$ is a nontrivial rotation.
If $f = B_a$, with $a \in \mathbb{D}$, then $f$ has a unique fixed point in $\mathbb{D}$ because the equation $B_a(z) = z$ has a unique solution in $\mathbb{D}$.
If $f = -B_a$, with $0 \neq a \in \mathbb{D}$, then $f$ has no fixed points in $\mathbb{D}$; its fixed points are $\pm \frac{ a }{ |a| } \in \partial\mathbb{D}$, since those are the solutions of $-B_a(z) = z$.

Hyperbolic geometry: every nontrivial isometry of the hyperbolic plane has at most one fixed point; this fails for the euclidean plane because of reflections.

Fixed point in $\overline{\mathbb{D}}$: the Brouwer fixed point theorem implies that every $f \in \operatorname{Aut}(\mathbb{D})$ has a fixed point in $\overline{\mathbb{D}}$ (once $f$ has been extended holomorphically to an open superset of $\overline{\mathbb{D}}$). It would be interesting to see a complex-analytic proof, especially since I don't know how to prove the Brouwer fixed point theorem.

Unresolved question: which $f \in \operatorname{Aut}(\mathbb{D})$ have no fixed point in $\mathbb{D}$?
Maybe they are precisely those $f$ which have two fixed points in $\partial\mathbb{D}$.

Thoughts: The function $f = \lambda B_a$ is nonconstant whenever $a \notin \partial\mathbb{D}, \lambda\neq0$. It's meromorphic on $\hat{\mathbb{C}}$. Let $a \neq 0$; then the quadratic formula shows that $f$ has exactly two fixed points with multiplicity. They may be equal: let $\lambda \in \partial\mathbb{D}$; then $f$ has a double fixed point iff $\frac{ |1+\lambda| }{ 2|a| } = 1$, and the double fixed point is $\frac{ 1+\lambda }{ 2\overline{a} }$. (Note that $\arg \lambda = 2 \arg (1+\lambda)$; this has a straightedge-and-compass proof.) Let $\lambda \neq -1,1$; those $a$ solving the equation form a circle centered at $0$ with radius $\tfrac{1}{2} | 1+\lambda | < 1$.
From this we can see that for any $a \in \mathbb{D}$ there exists a unique rotation $(z \mapsto \lambda z)$ such that $\lambda B_a$ has a double fixed point on the unit circle (and no other fixed points).
Rouché's theorem might give a little more info.

Solution: Just realized that, whenever $\lambda \in \partial\mathbb{D}$ and $0 \neq a \notin \partial\mathbb{D}$, the fixed points of $\lambda B_a$ are $u \left(1 \pm \sqrt{ 1 - |u|^{-2} } \right)$ where $u = \frac{ 1+\lambda }{ 2\overline{a} }$.
This completely resolves the unresolved question; I might use this to write up a cleaner version of this post at some point.

1.pdf
2.pdf
5.pdf
6.pdf
7.pdf
8.pdf
9.pdf
10.pdf