Understanding the proof to continuity of norms in normed spaces

Question

Following the proof outlined here Why are norms continuous?,

this is what I expected to be shown: $\forall \epsilon >0,\, \exists \delta>0,\, \mid\mid x-y \mid\mid<\delta \, \Rightarrow $ $\mid\mid f(x)-f(y) \mid\mid< \epsilon\,\,$ where $f$ represents the norm function.

Instead, the given proof aims to show $\mid f(x)-f(y)\mid <\epsilon.$

Why instead of using the (arbitrary) norm as a metric for measuring the distance between $f(x)$ and $f(y)$, the absolute value is used? Perhaps I am missing something about the definition of continuity in a normed spaced, here? Thanks for any help.

Note that $f$ is real-valued, being a norm. We are showing $f$ is continuous using the absolute value norm in the codomain. — Minus One-Twelfth, Mar 10 '19 at 06:05

score 1 · Answer 1 · answered Mar 10 '19 at 06:14

The norm is a function $||\cdot||_X:X\to [0,\infty)$. Hence if you define $f(x)=||x||_X$, then $f(x)$ is a real positive number. In $\mathbb{R}$, the norm is the usual euclidean norm which is simply the absolute value, so $||f(x)-f(y)||_\mathbb{R}=\big|||x||-||y||\big|$ :) If you have any questions, let me know !

score 1 · Accepted Answer · answered Mar 10 '19 at 06:16

To make this crystal clear, let us recall the definition of continuity in metric spaces:

$$f: A \to B$$ is uniformly continuous if $$\forall \epsilon >0,\, \exists \delta>0,\, d_A(x,y)<\delta \, \Rightarrow d_B(f(x),f(y)) < \epsilon$$

Since in your case $f: (E, \| \, \|) \to (\mathbb R, | \, |)$ you have $$d_A(x,y)=\| x -y\|\\ d_B(a,b)=|a-b|$$

Understanding the proof to continuity of norms in normed spaces

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