751: For Finite-Dimensional Normed Real Vectors Space with Canonical Topology, Norm Map Is Continuous

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description/proof of that for finite-dimensional normed real vectors space with canonical topology, norm map is continuous

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Topics

About: vectors space
About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of norm on real or complex vectors space.
• The reader knows a definition of continuous map.
• The reader knows a definition of canonical topology for finite-dimensional real vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional normed real vectors space with the canonical topology, the norm map is continuous.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$V$: $\in \{\text{ the }d\text{ -dimensional normed real vectors spaces }\}$ with the norm, $\Vert \bullet \Vert :V\to \mathbb{R}$, with the canonical topology
$f^{\prime }$: $:V\to \mathbb{R},v\mapsto \Vert v\Vert$
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Statements:
$f^{\prime }\in \{\text{ the continuous maps }\}$
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2: Natural Language Description

For any $d$-dimensional normed real vectors space, $V$, with the norm, $\Vert \bullet \Vert :V\to \mathbb{R}$, with the canonical topology, the norm map, $f^{\prime }:V\to \mathbb{R},v\mapsto \Vert v\Vert$, is continuous.

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3: Proof

Let us take any basis, $B=\{b_1,...,b_d\}\subseteq V$.

Let $v=\sum _{j\in \{1,...,d\}}{v}^j{e}_j\in V$ be any.

$\Vert v\Vert =\Vert \sum _{j\in \{1,...,d\}}{v}^j{e}_j\Vert \le \sum _{j\in \{1,...,d\}}\Vert v^j{e}_j\Vert =\sum _{j\in \{1,...,d\}}|v^j|\Vert e_j\Vert$, by the definition of norm on vectors space. $\le max(\{\Vert e_j\Vert \})max(\{|v^j|\})d$.

$max(\{|v^j|\})\le (\sum _{j\in \{1,...,d\}}{v^j}^2{)}^{1/2}$.

So, $\Vert v\Vert \le max(\{\Vert e_i\Vert \})d(\sum _{j\in \{1,...,d\}}{v^j}^2{)}^{1/2}=C(\sum _{j\in \{1,...,d\}}{v^j}^2{)}^{1/2}$ where $C:=max(\{\Vert e_i\Vert \})d$, which does not depend on $v$.

Let $f:V\to {\mathbb{R}}^d,v=\sum _{j\in \{1,...,d\}}{v}^j{b}_j\mapsto (v^1,...,v^d)$ be the map by which the canonical topology is defined.

For any open ball, $B_{\Vert v\Vert ,ϵ}\subseteq \mathbb{R}$, let us take $\delta =ϵ/C$ and the open neighborhood of $v$, $U_v=\{p\in V|f(p)\in B_{f(v),\delta }\}\subseteq V$, which is indeed an open neighborhood of $v$, because $f(U_v)=B_{f(v),\delta }\subseteq {\mathbb{R}}^d$, open on ${\mathbb{R}}^d$.

For each $v^{\prime }=\sum _{j\in \{1,...,d\}}{v}^{\prime j}{b}_j\in U_v$, $\Vert v^{\prime }-v\Vert \le C(\sum _{j\in \{1,...,d\}}(v^{\prime j}-v^j{)}^2{)}^{1/2}<C\delta =ϵ$.

$|\Vert v^{\prime }\Vert -\Vert v\Vert |\le \Vert v^{\prime }-v\Vert$, because $\Vert v^{\prime }\Vert =\Vert v^{\prime }-v+v\Vert \le \Vert v^{\prime }-v\Vert +\Vert v\Vert$, so, $\Vert v^{\prime }\Vert -\Vert v\Vert \le \Vert v^{\prime }-v\Vert$, and likewise $\Vert v\Vert -\Vert v^{\prime }\Vert \le \Vert v^{\prime }-v\Vert$.

So, $|f^{\prime }(v^{\prime })-f^{\prime }(v)|=|\Vert v^{\prime }\Vert -\Vert v\Vert |<ϵ$, which means that $f^{\prime }(U_x)\subseteq B_{\Vert v\Vert -ϵ}$, which means that the map is continuous at any point, so, is continuous.

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References

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