# Prove that $\Vert\cdot \Vert: X\to \Bbb{R},$ defined by $x\mapsto \Vert x \Vert,$ is continuous

I want to prove that $$\Vert\cdot \Vert: X\to \Bbb{R},$$ defined by $$x\mapsto \Vert x \Vert,$$ is continuous, where $$X$$ is a normed linear space.

Here's what I've tried.

TRIAL

Let $$\epsilon>0$$, we seek $$\delta$$ such that $$\Vert x_n-x \Vert<\delta,\;\forall\;n\geq N,$$ for some $$N$$ implies $$\Big|\Vert x_n\Vert-\Vert x \Vert \Big|<\epsilon,\;\forall\;n\geq N.$$ Then,

\begin{align}\Big|\Vert x_n\Vert-\Vert x \Vert \Big|\leq \Vert x_n-x \Vert <\delta \end{align} So, given any $$\epsilon>0,$$ take $$\delta=\epsilon.$$ Then, $$\forall\;n\geq N,\;\Big|\Vert x_n\Vert-\Vert x \Vert \Big|<\epsilon,\;\forall\;n\geq N.$$ Hence, we are done!

Please, I'm I right? If not, I need someone to help fine-tune the proof! Thanks

• You jneed to define what your $x_n$ is, also, I would recommend you prove it via preimages of open sets, since then it is a one liner. Nov 30, 2018 at 14:58
• @Enkidu: I made an edit! You recommend inverse images of open sets? Nov 30, 2018 at 15:02
• you are done, however, you are mixing up 2 different definitions of continuity, either you do sequence continuity or $\epsilon-\delta$, however, your approach looks like an incest of both of them (i.e. either say $x-y < \delta$ and imply $f(x)-f(y)\le \delta$, or $x_n\xrightarrow{\to \infty}x$ and imply $f(x_n)\xrightarrow{\to \infty} f(x)$) my approach would however be to prove it via: f is continuous if for every open set $U$, $f^{-1}(U)$ is open as well. Which is the most general and in my opinion useful definition for proofs, you might not know that definition, it is kind of abstract Nov 30, 2018 at 15:14
• The crux of the proof here is the reverse triangle inequality $|\|x\|-\|y\| | \leqslant \|x-y\|$. You take that as given making the proof of continuity trivial. If not see proof below.
– RRL
Nov 30, 2018 at 15:24
• @RRL: Due to the fact that I already know the proof of the reverse triangular inequality, my question is: "is my proof wrong?" Nov 30, 2018 at 21:55

Given a normed, linear space $$X$$, the norm $$\| \cdot \|$$ satisfies the triangle inequality

$$\|x + y\| \leqslant \|x \| + \|y\|$$

Hence,

$$\|x\| = \|y + (x-y) \| \leqslant \|y\| + \|x - y\|, \\ \|y\| = \|x - (x-y) \| \leqslant \|x\| + \|-1(x - y)\| = \|x\| + \|(x - y)\|,$$

The first inequality implies $$\|x\| - \|y \| \leqslant \|x-y\|$$ and the second implies $$\|x\| - \|y \| \geqslant -\|x-y\|$$

Thus,

$$| \, \|x\| - \|y\| \, | \leqslant \|x - y \|$$

This proves (uniform) continuity since for all $$x,y \in X$$

$$\|x - y\| < \delta (= \epsilon) \implies | \, \|x\| - \|y\| \, | < \epsilon$$

Attempt:

$$X$$ normed metric space, $$x_n, x \in X$$.

$$f(x):=||x||$$.

Let $$x_n \rightarrow x$$.

$$||x|| \le ||x-x_n|| +||x_n||;$$

$$||x_n|| \le ||x_n-x|| +||x|.$$

Hence $$|f(x_n)-f(x)| \le ||x-x_n||.$$

Let $$\epsilon >0$$ be given.

Since $$x_n \rightarrow x$$, there is a $$n_0$$ s.t. for $$n\ge n_0$$

$$||x-x_n|| \lt \epsilon$$, i.e.

$$|f(x)-f(x_n)| =$$

$$|(||x_n||-|x||)| \le ||x-x_n|| \lt \epsilon.$$