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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"?

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  • $\begingroup$ 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 $||•||_2≤b||•||_1$ $\endgroup$ – S.D. Oct 12 '18 at 16:21
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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 $\|\cdot\|$.)

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  • $\begingroup$ Then is the $\epsilon - \delta$ definition only valid for 1-norm? $\endgroup$ – WeiShan Ng Oct 13 '18 at 5:51
  • $\begingroup$ 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. $\endgroup$ – WeiShan Ng Oct 13 '18 at 5:57

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