# $x \mapsto \|x\|$ is a continuous mapping of $(X,\|.\|) \rightarrow \Bbb R$?

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?

• It means the same. – Shahab Aug 8 at 14:47

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$$(\forall\varepsilon>0)(\exists\delta>0):\lVert y-x\rVert<\delta\implies\bigl\lvert\lVert y\rVert-\lVert x\rVert\bigr\rvert<\varepsilon.$$

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 $$||x||-||y||\le ||x-y||=\delta=\varepsilon$$

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 $$\delta =\epsilon$$ :

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

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