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. – Enkidu Nov 30 '18 at 14:58
• @Enkidu: I made an edit! You recommend inverse images of open sets? – Omojola Micheal Nov 30 '18 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 – Enkidu Nov 30 '18 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 '18 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?" – Omojola Micheal Nov 30 '18 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.

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

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