Show that $f_n(z)=z^n$ converges locally uniformly on $|z|<1$ but not uniformly.
I am really confused. I saw a seemingly reliable reference stating that $z^n$ does not converge uniformly to $0$ on $\{|z|<1\}$ but it does converge uniformly on $\{|z|\le r,r<1\}$. What is the nuance between the two former sets? With $r_n<1,r_n\to 1$, shouldn't one conclude that $f_n(z)$ uniformly converges on $\{|z|<1\}$? And besides, this source tells me what I have to prove for Homework is disproved, which really makes it more frustrating.
I would appreciate any help.
We have pointwise convergence to $0$ for any $z \in \mathbb{C}$ with $|z| < 1$.
Uniform convergence requires that for any $\epsilon > 0$ there exists a positive integer $N$ independent of $z$ such that $n \geqslant N$ implies $|z^n - 0| < \epsilon$ for every $z \in \{z \in \mathbb{C} : |z| < 1\}.$ This is impossible when $z$ can be arbitrarily close to the boundary of that disc.
Given any positive integer $N$ take $z_n = (1 - 1/n)$ where $n > N$. Recall that $(1-1/n)^n \to e^{-1}$ and we must have $(1 - 1/n)^n > e^{-1}/2$ for all sufficiently large $n$. Hence, we can find some $n > N$ such that
$$|z^n -0|= |z|^n = (1-1/n)^n > e^{-1}/2,$$
in violation of uniform convergence.
With the definition in mind try to show yourself that convergence is uniform on any compact disc $\{z \in \mathbb{C}: |z| \leqslant r\}$ with $r < 1$.