[1] pointwise convergence is uniform convergence under some condition
[2] A course in metric geometry - Burago, Burago and Ivanov
Question : I want to understand intuitively or visually the proof of Theorem 8.5.4 in [2], which is in the below and which seems to be a little technical proof containing net-notion (cf. We already considered a continuous case (?) in [1]).
More geometrically, I want to know how pointwise convergence bacomes uniform convergence.
Note that there exists a Riemannian metric $\mathbb{R}^2$ with $G= \mathbb{Z}^2$-action. Hence from this action, we have a $G$-invariant metric on the set $G$.
But without this action, we can define a $G$-invariant metric $d$ on $G$ : For instance, $$f: G \rightarrow \mathbb{R},\ f(x,y)=|x| -1+ \sum_{i=0}^{|y|}\ \frac{1}{2^i} $$ We set $$ d((x,y),(a,b))=f((x,y)-(a,b))$$
We set $F ((x,y)) = \lim_{n\in\mathbb{N}, \ n\rightarrow \infty}\ \frac{f(n(x,y)) }{ n} $ so that we have a semi-norm $\|\ \|$ on $ \mathbb{R}^2$ which is not a norm and which is an extension of $F$.
Here when we choose $f$ suitably so that $\|\ \|$ is in fact a norm, then prove that given $ \varepsilon$ there is $R$ s.t. $\|(x,y) \| >R $ implies $|\frac{f ((x,y)) }{\| (x,y) \| } - 1| < \varepsilon $ (cf Theorem 8.5.4)
The Proof in the book :
i) $v=(x,y)$. Note that $f(nv)\leq nf(v)$ That is we have $$ \|\ \|\leq f \leq C\cdot |\ |$$ for some $C>0$ where $|\ |$ is Euclidean norm.
ii) Assume that $B$ is $\|\ \|$-unit ball
Define $S \subset \mathbb{Q}G$ to be finite $\varepsilon$-net in $B$
For $v=(x,y)\in S$, there is a natural number $M'$ s.t. $M'v \in G$
Hence there is $M''$ s.t. $Mv=M'' M'v$ and $ | \frac{f(Mv) }{M''} -\| M'v\|| < \varepsilon$ for all $v\in S$, since $S$ is a finite set
Hence $$ f(Mv)\leq M\varepsilon +M\|v\| \leq M(1+\varepsilon ) $$
For any $w\in G $, there is $v$ s.t. $$ \| \frac{w}{\| w\|} -v \| < \varepsilon $$
When $k\leq \| w\|< k+1$, then $$ \| w-\| w\| v \| < \|w\|\varepsilon ,\ \| w-kv\| \leq \|w\|\varepsilon +1 $$
Hence \begin{align*} f(Mw) &\leq f(kMv ) + f(kMv - Mw) \\&< k(1+ \varepsilon ) M + C\| kv-w\|_E M \\&< k(1+\varepsilon ) M + CC_2 \|kv-w\| M \\&< k(1+ \varepsilon )M +CC_2 M\bigg\{ \|w \| \varepsilon +1 \bigg\} \end{align*}
Hence there is $k_0$ s.t. $\| w\|>k_0$ implies $|\frac{f(Mw)}{\| Mw\|} -1|<\varepsilon$
iii) Here when $G'$ is set of $M$-divisible vectors, then $G'$ is a net in $G$
Proof : $ (x,y)=(Mq_1+r_1,Mq_2+r_2)$
iv) When $F(v)> Mk+2M$, then there is $w$ s.t. $\|v-w\| <M$
Hence $\|w\|>Mk$ and $\frac{k}{k-1}<1+\varepsilon$ so that $$ \frac{f(v)}{\|v\|} < \frac{f(w) +f(v-w)}{ \| w\| - M} < 1+2\varepsilon + \frac{cM}{\| w\|-M} $$