# What does it mean when the map $u \mapsto \lVert u \rVert$ is continuous with respect to the norm $\lVert u \rVert_1$

Given any norm $$\lVert \, \rVert$$ on a vector space of dimension $$n$$, for any basis $$(e_1, \dots, e_n)$$ of E, observe that for any vector $$x=x_1 e_1 + \dots x_n e_n$$, we have $$\lVert x \rVert = \lVert x_1 e_1 + \dots x_n e_n \rVert \leq C \lVert x \rVert_1\, ,$$ with $$C=\max_{1\leq i \leq n} \lVert e_i \rVert$$ and $$\lVert x \rVert_1 = \lVert x_1e_1+\dots +x_ne_n \rVert = \lvert x_1 \rvert + \dots + \lvert x_n \rvert$$
The above implies that $$\lvert \lVert u \rVert - \lVert v \rVert \rvert \leq \lVert u - v \rVert \leq C \lVert u-v \rVert_1 \, ,$$ which means that the map $$u \mapsto \lVert u \rVert$$ is continuous with respect to the norm $$\lVert \, \rVert_1$$

I think I have the intuition on how the inequality leads to the continuity, but need some further guidance on that. I used the $$\epsilon - \delta$$ definition of continuity and let $$f(u)=\lVert u \rVert$$ and $$\delta = \epsilon /C$$, so that when $$\lvert u - v \rvert \leq \delta$$, we will have $$\lVert u - v \rVert \leq \epsilon$$ and thereby the continuity.

What confused me is the last sentence, which says the map $$u \mapsto \lVert u \rVert$$ is continuous with respect to the norm $$\lVert \, \rVert_1$$. What does it mean when it says "continuous with respect to the p-norm"?

• Any two norm $||•||_1,||•||_2$ on a finite dimensional vector space is comparable i.e. we have positive numbers $a,b$ such that $||•||_1≤a||•||_2$ and $||•||_2≤b||•||_1$ – S.D. Oct 12 '18 at 16:21

In the $$\varepsilon-\delta$$ notation, this means: at any $$u$$, for every $$\varepsilon > 0$$, there exists some $$\delta(\varepsilon) > 0$$, such that whenever $$\|v-u\|_1 \le \delta$$ (note here is the $$1$$-norm), we have $$|(\|u\|-\|v\|)| \le \varepsilon$$. (here is the difference of your function value, i.e., in terms of $$\|\cdot\|$$.)
• Then is the $\epsilon - \delta$ definition only valid for 1-norm? – WeiShan Ng Oct 13 '18 at 5:51
• I think I get it....So from @UserS comment, if we can prove that the map $u \mapsto \lVert u \rVert$ is continuous with respect to 1-norm, then from the equivalence of norm, we can show that the map $u \mapsto \lVert u \rVert$ is continuous with respect to any p-norm. – WeiShan Ng Oct 13 '18 at 5:57