Assumptions: Let $A \subseteq \mathbb C$. Let $f_n,f: A \to \mathbb C$, for all $n \ge 1$. As to whether or not $\{f_n\}_{n=1}^{\infty}$ converges uniformly or pointwise to $f$ (on $A$):
Definitions:
Complex pointwise convergence is true:
For any $(\varepsilon,z) \in (0,\infty) \times A$, there exists $N_{(\varepsilon,z)} > 0$ s.t. $|f_n(z)-f(z)| < \varepsilon$ whenever $n > N_{(\varepsilon,z)}$.
Complex uniform convergence is true:
For any $\varepsilon > 0$, there exists $N_{\varepsilon} > 0$ s.t. $|f_n(z)-f(z)| < \varepsilon$ whenever $n > N_{\varepsilon}$ and $z \in A$.
Complex uniform convergence is false:
There exists $\varepsilon > 0$ s.t. for all $N > 0$, there exists $n_{\varepsilon,N} > N$ and $z_{\varepsilon,N} \in A$ s.t. $|f_n(z)-f(z)| \ge \varepsilon$.
Questions:
- What I understand about these definitions in particular for the $|z|$:
For (real uniform continuity and) real uniform convergence, what we want is our ($\delta$ and) $N$ to not depend on $x$.
For complex uniform convergence (and I guess complex uniform continuity), well...what we want in disproving uniform convergence seems to be showing that our $N$ (and I guess $\delta$) depends on $|z|$, eg here (for $\{\frac{1}{z^2+\frac 1 n}\}_{n=1}^{\infty}$), but technically $N$ doesn't depend on $z$ given its dependence on $|z|$.
Do I understand this correctly?
- 1.1. I mean, are there examples where disproving uniform convergence really shows dependence on really $z$ (and not just $|z|$)? I have a feeling it's really gonna be just $|z|$ because $|z|$ is what we get when we do $|f_n(z)-f(z)|$.
- Does $N$'s dependence, eg here (for $\{\frac{1}{z^2+\frac 1 n}\}_{n=1}^{\infty}$), on $|z|$ even though we (might) have independence of $z$ given $|z|$ mean that $N$ still 'depends' on $z$ anyway?
I guess yes because well...$|z|$ isn't (necessarily) fixed so for different moduli, we have different (choices for) $z$.
