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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

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    $\begingroup$ 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. $\endgroup$ – Enkidu Nov 30 '18 at 14:58
  • $\begingroup$ @Enkidu: I made an edit! You recommend inverse images of open sets? $\endgroup$ – Omojola Micheal Nov 30 '18 at 15:02
  • $\begingroup$ 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 $\endgroup$ – Enkidu Nov 30 '18 at 15:14
  • $\begingroup$ 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. $\endgroup$ – RRL Nov 30 '18 at 15:24
  • $\begingroup$ @RRL: Due to the fact that I already know the proof of the reverse triangular inequality, my question is: "is my proof wrong?" $\endgroup$ – Omojola Micheal Nov 30 '18 at 21:55
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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 $x,y \in X$

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

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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.

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

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

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