What does it mean to say that a function of an arbitrary norm is "continuous with respect to the 1-norm"? I am trying to understand a proof of the equivalence of all norms. Many of the proofs start by showing that two arbitrary norms $|| x ||_p$ and $|| x ||_q$ are equivalent to $||x||_1$ and thus the equivalence relation
$$ \alpha || x ||_p \le || x ||_q \le \beta || x ||_p $$
is transitive. Finally with the transitivity of $|| x ||_p$ and $|| x ||_q$ through $||x||_1$ is established, one only needs to show that $||x||_1$ is equivalent to some other arbitrary norm, say $||x||_k$.
This is done by first constructing the inequality
$$ \alpha \le ||v||_k \le \beta, \forall v \text{ with} ||v||_1 = 1 $$
and finding some bounds $\alpha, \beta$ on $||v||_k$.

This process is described succinctly here

Though I don't exactly understand what it means when it says "Finally we show that $f(x)\dots$ is continuous with respect to the $||.||_1$ norm".
I think I understand the concept geometrically. For instance, the $||x||_1$ norm is the unit circle/ball, and if we "draw" another norm enclosing this (by selecting $\alpha, \beta$ so that it does enclose), we are basically creating the first inequality stated.


Could someone explain the final step of this proof visually/geometrically with some light rigorous math information to back it up?

Assume that $\|\cdot\|_{\beta}\leq c\|\cdot\|_{1}$, for the function $f:(X,\|\cdot\|_{1})\rightarrow{\bf{R}}$, $f:x\rightarrow\|x\|_{\beta}$, if $x_{n}\rightarrow x$ in $\|\cdot\|_{1}$, then $\|x_{n}-x\|_{1}\rightarrow 0$, apply Squeeze Theorem to the inequality $\|x_{n}-x\|_{\beta}\leq c\|x_{n}-x\|_{1}$, we get $\|x_{n}-x\|_{\beta}\rightarrow 0$, by triangle inequality, we have $\|x_{n}\|_{\beta}\rightarrow\|x\|_{\beta}$, so $f(x_{n})\rightarrow f(x)$, the continuity of $f$ is then established.
