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?

  • $\begingroup$ It means the same. $\endgroup$ – 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$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.