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Prove that: $x \mapsto \|x\|$ is a continuous mapping of $(X,\|.\|) \rightarrow \Bbb R$

What does it mean for a function from a normed spaced to a normed space to be continuous?

I know that any normed spaced has an associated metric space by $$(X, \| . \|) \rightarrow (X, d(x,y) = \|x - y\|)$$

So is this asking to show continuity for the function $$T: (X, d_1(x,y)) = \| x-y \|_X \rightarrow (\Bbb R, d_2(x,y) = \| x - y\|_{\Bbb R}) \text{ by } x \mapsto \|x\|?$$

If so, then it's simply showing that $$(\forall \epsilon \gt 0)(\exists \delta \gt 0) (\|x-y\|_X \lt \delta \Rightarrow \|\|x\|_X - \|y\|_X\|_{\Bbb R} \lt \epsilon)$$

Which is true by $|\|x\| - \|y\| | \le \|x - y \|$.

Is this a correction assumption or does continuity between two normed spaced mean something else?

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  • $\begingroup$ It means the same. $\endgroup$ – Shahab Aug 8 at 14:47
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Your interpretation is correct, but there is no need to use the notation $\lVert\cdot\rVert_{\mathbb R}$ here. The distance in $\mathbb R$ is the usual one, and therefore continuity at $x$ means that$$(\forall\varepsilon>0)(\exists\delta>0):\lVert y-x\rVert<\delta\implies\bigl\lvert\lVert y\rVert-\lVert x\rVert\bigr\rvert<\varepsilon.$$

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Just take $\varepsilon=\delta$ and apply the triangle inequality:

Let $x\in X,\varepsilon>0$ and $y\in X$ such that $0<||x||-||y||<\varepsilon$.

then: $$||x||=||x-y+y||\le||x-y||+||y||$$ $$||y||=||y-x+x||\le||x-y||+||x||$$

So you can conclude that $||x||-||y||\le ||x-y||=\delta=\varepsilon $

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Let $f(x)= ||x||$;

$f: X \rightarrow \mathbb{R};$

$f$ is continuous at $y \in X$:

Given $\epsilon >0$ there is a $\delta >0$ s.t.

$||x-y|| < \delta$ implies $|f(x)-f(y)| < \epsilon$.

$|(||x||-||y||)| \le ||x-y||$ (reverse triangle inequality);

Choosing $\delta =\epsilon$ :

$|f(x)-f(y)| =|(||x||-||y||)| \le$

$ ||x-y|| <\delta=\epsilon$.

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