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