Where do the following sequences converge pointwise? Do they converge uniformly on this domain? (a)$\frac{nz^n}{1}$ (b) $\frac{z^n}{n}$ (c) $ \frac{1}{1+nz}$
These sequences were asked about in these questions: Uniform convergence of $\frac{1}{1+nz}$, Where do the following sequences converge pointwise and uniformly? and $\sum_{n=1}^\infty\frac{z^n}{n}$ does not converge uniformly on $\mathbb{D}$. , but I have other questions. ((c) and (d) are here.)
Firstly, is this supposed to be unambiguous? I mean 'this domain' refer to the answers we got for pointwise convergence? Or are 7.22(a) and (b) missing specified 'domains', which I guess would be what (c) has? Anyhoo, I'm just gonna assume 'this domain' refer to the answers we got for pointwise convergence for (a) and (b).
(b1)
Pointwise convergence: I think $\lim \frac{z^n}{n} = 0 \iff |z| \le 1$? Pf: $$\lim |\frac{z^n}{n}| = \lim \frac{|z^n|}{n} = \lim \frac{|z|^n}{n} = 0 \iff |z| \le 1. \therefore, \lim \frac{z^n}{n} = 0 \iff |z| \le 1$$ QED
(b2)
Uniform convergence: I think $\lim \frac{z^n}{n} \stackrel{u}{=} 0 \iff$ also $|z| \le 1$ by Exer 7.19(a)? (*)
(a1)
Pointwise convergence: I think $\lim n z^n = 0 \iff |z| < 1$ as proved here, but is my reasoning below right?
Pf:
$$\lim |n z^n| = \lim n |z|^n = 0 \iff |z| \le 1. \therefore, \lim n |z|^n = 0 \iff |z| \le 1$$
QED
(a2)
Uniform convergence: $\lim n z^n \stackrel{u}{=} 0 \iff |z| \le r < 1$ as proved here, but is my reasoning below right?
Pf:
$|n z^n| = n |z|^n \le n r^n \to 0$ Conclude with Exer 7.19(a). (*)
QED
(a3)
Uniform convergence: $\lim n z^n \stackrel{u}{\ne 0} \iff |z| < 1?$
How exactly would we prove this?
- Just say that pointwise convergence to $0$ (or any $\mathbb C$-constant for that matter) doesn't hold on $|z|=1$?
I think I remember reading somewhere that uniform convergence is for closed sets or something so if it holds in $A$ then it should hold in $\overline{A} := A \cup \partial A$. I don't think this is in the text, or maybe I overlooked it. Might be assumed to be known from elementary analysis. I might be confusing this with uniform continuity.
- How would we show this by definition? I tried finding $\varepsilon$ s.t. $\forall N > 0, \exists z \in G: n > N \wedge n|z|^n \ge \varepsilon$ to no avail. Hints?
(*)
Exer 7.19(a)
Suppose $G \subseteq \mathbb C$ and $f_n: G \to \mathbb C, n \ge 1$. Then $\lim a_n =0, a_n \in \mathbb R, |fn(z)| \le a_n \ \forall z \in G \implies f_n \stackrel{u}{\to} 0 \ $ in G.
I don't agree with any of the proofs for the pointwise convergence. If you are allowed to claim e.g. $ \lim_{n\to\infty} \frac{|z^n|}{n} = 0 \iff |z|\leq 1$ then there is no point to the question at all. So you should justify this with a full argument. The fact that this is sloppy is also important when you find out that in the proof of $(a1)$, when you claim
$$\lim_{n\to\infty} n z^n=0 \iff |z|\leq 1 $$ you also claim in particular that $n\to 0$ as $n\to\infty$ by choosing $|z| = 1$. Curiously you copied correctly the answer from the other MSE post (strict inequality important.)
For the proofs of uniform convergence, the answers are right. But I don't like what you wrote because you did not explain how the exercise was used. This exercise talks about a sequence $a_n$. What is this sequence?
Finally, for $(a3)$, I have no idea what you are doing. It seems you use $\iff$ to not mean "if and only if", because $$\lim n z^n \stackrel{u}{\ne 0} \iff |z| < 1?$$ is $[A]$ redundant, since we already characterised the criteria for uniform convergence in $(a2)$ and $[B]$ not consistent with $(a2)$, if I understand you correctly. So the fact that you think $(a3)$ is needed means you don't understand your argument for $(a2)$, and I suggest you write it out in full, both "if" and "only if" directions. I also suggest you use less sentences made up purely of mathematical symbols.