I am follow a line of proof and I am not sure about one part of the proof
The given assumption is that:
Given $f : \mathbb{R}^n \to \mathbb{R}^n$ where $f$ satisfies $\forall x(t),x'(t) \in \mathbb{R}^n$:
$\quad \|f(x(t)) - f(x'(t))\|_\infty \leq k\|x(t)-x'(t)\|_\infty, k \in [0, 1)$
The proof seems to be making the claim: Given a constant $x_o$
Since $f_i(x(t)) - f_i(x_o) \leq \|f(x(t)) - f(x_o)\|_\infty \leq k\|x(t)-x_o\|_\infty$, therefore
$-(f_i(x)-f_i(x_o)) \leq \min\limits_ik \|x-x_o\|_\infty$
or $-(f_i(x)-f_i(x_o)) \leq \min\limits_ik |x_i-{x_o}_i|$
where $i$ indexes the component of $f(x),x$ i.e. $x=\begin{bmatrix} x_1, \ldots, x_i, \ldots x_n, \end{bmatrix}$
I have been trying to justify this claim but can't exactly see why they were able to reach such conclusion. This is compounded with slight ambiguity in the notations which results in two different conclusions (as indicated by the 'or').
Expanding the proof out a bit, we know that
$\|f(x) - f(x_o)\|_\infty = \max\{|f_1(x)-f_1(x_o)|, \ldots, |f_n(x)-f_n(x_o)|\} \geq f_i(x)-f_i(x_o), \forall i \in \{1,\ldots,n\}$
So by assumption, $k\|x-x_o\|_\infty \geq f_i(x)-f_i(x_o) \implies -(f_i(x)-f_i(x_o)) \geq -k\|x-x_o\|_\infty$...Can anyone see how the min was introduced to reverse the inequality?
In your last line, the assumption is not that $$|x-x_0\|_\infty \geq f_i(x)-f_i(x_0) $$ but that $$|x-x_0\|_\infty \geq |f_i(x)-f_i(x_0)| $$ so the minus sign is irrelevant as there is no inequality that needs to be reversed.
Now, in the "claim", you say that $$-(f_i(x)-f_i(x_0)) \leq \min\limits_ik \|x-x_o\|_\infty.$$ This has two problems: first, that $k\|x-x_0\|_\infty$ does not depend on $i$, so it is not clear what the min is supposed to mean. Second, that in any case the inequality $$-(f_i(x)-f_i(x_0)) \leq \min\limits_ik |x_i-{x_0}_i|$$ is false in general. For instance let $n=2$ and take $$ f(x,y)=(y,x) $$ Then $$ - (f_1(0,1)-f_1(0,2))=1>0=\min\{|0-0|,|1-2|\}.$$