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On another post a user asked for a proof that if a map is linear then it is continuous. "Every linear mapping on a finite dimensional space is continuous"

I have pasted an image of the answer which I am having trouble understanding. In the last line he replaces $\sum_i \lvert \alpha_i\rvert$ with $|x|$, so it must either be equal or less for the conclusion to follow.

Question Why is this sum of scalars equal or less than the norm?

Thanks

enter image description here

john fowles
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  • There was a mistake in that proof and it was pointed out in the comments. It is not true that $\sum |\alpha_i| \leq ||x||$ for any norm when $x=(\alpha_1,...,\alpha_n)$. For example this is not true for the usual norm on $\mathbb R^{n}$. – Kavi Rama Murthy Jul 19 '18 at 05:23
  • Could we fix it by writing $\sum |\alpha_i| \leq q||x||$, for some $q \in \mathbb{R}$ and letting $\delta =\dfrac{\epsilon}{mq}$ – john fowles Jul 19 '18 at 06:12
  • Yes, that is how you can fix that proof. – Kavi Rama Murthy Jul 19 '18 at 06:13

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Norms on finite-dimensional spaces are equivalent. You can check that $||x||_0 := \sum_i |\alpha_i|$ defines a norm on $X$ (in fixing that basis for $X$). Since the given norm $||\cdot||$ is another norm on $X$, a finite-dimensional space, it then follows that $||\cdot||_0$ and $||\cdot||$ are equivalent, i.e. there exists some constant $C>0$ such that $||\cdot||_0 \leq C ||\cdot||$. Adjusting your $\delta$ to be instead $\varepsilon / (CM)$ gets what you want

You can read something such as this post to learn more about this property of norm equivalence in finite-dimensional spaces

Lucas
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  • So you're letting each $\alpha_i$ in $\sum_i |\alpha_i|$ represent basis vectors that span $X$? – john fowles Jul 19 '18 at 02:16
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    No I am using the original post's notation (e.g. in the attached picture), where the $\alpha_i$ are the coordinates of $x \in X$ relative to the fixed basis $e_1, \ldots, e_n$ of $X$, i.e. $x = \sum_i \alpha_i e_i$ or $x = (\alpha_1, \ldots, \alpha_n)$. – Lucas Jul 19 '18 at 14:00
  • ah, I missed that. Thanks! Also by the $\delta$ $\epsilon$ definition to show continuity wouldn't we have to show $|x-a|< \delta$ gives $|T(x)-T(a)| < \epsilon$? But in their proof they just show $|x|< \delta$ implies $|T(x)|< \epsilon$. Does the $|x-a|$ case follow trivially from proving the $|x|$ case? Because showing $|x|$ shows continuity at $0$, or am I missing something. Thanks again. – john fowles Jul 19 '18 at 20:37
  • no problem ^_^. Actually there is an important property of continuity of linear maps used there: Any linear map $T$ between normed spaces is continuous if and only if it is continuous at $0$ (e.g. see this post). So showing that $||x||<\delta$ implies $||Tx||<\varepsilon$ is showing continuity at $0$, which as noted is sufficient for continuity everywhere (e.g. that $||x-a||<\delta$ implies $||Tx-Ta||<\varepsilon$). – Lucas Jul 19 '18 at 23:48