What does it mean to say that a function of an arbitrary norm is "continuous with respect to the 1-norm"?

Question

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 $||v||_k$.

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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.

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

Answer 1

Just write down the definition of continuity of $f$: for any vector $x$ and $\epsilon > 0$, there exists $\delta > 0$ such that $|\|y\|_\beta - \|x\|_\beta| < \epsilon$ for any $y$ satisfying $\|y - x\|_1 \le \delta$.

Geometrically, continuity of $f$ at $x=0$ implies that for any $\epsilon > 0$, we can fit a small $\|\cdot\|_1$-ball inside a $\|\cdot\|_\beta$-ball of radius $\epsilon$.

Answer 2

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.