758: Norms on Finite-Dimensional Real Vectors Space Are Equivalent

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description/proof of that norms on finite-dimensional real vectors space are equivalent

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Topics

About: vectors 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 equivalence relation on norms on vectors space.
• The reader knows a definition of canonical topology for finite-dimensional real vectors space.
• The reader admits the proposition that for any finite-dimensional normed real vectors space with the canonical topology, the norm map is continuous.
• The reader admits the Heine-Borel theorem: any subset of any Euclidean topological space is compact if and only if it is closed and bounded.
• The reader admits the proposition that the image of any continuous map from any compact topological space to the $\mathbb{R}$ Euclidean topological space has the minimum and the maximum.

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Target Context

• The reader will have a description and a proof of the proposition that any norms on any finite-dimensional real vectors space are equivalent.

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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:
$V$: $\in \{\text{ the real vectors spaces }\}$
$S$: $=\{\text{ the norms on }V\}$
$\sim$: $\subseteq S\times S$, $=\text{ the equivalence relation on }S$
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Statements:
$\mathrm{\forall }{s}_1,s_2\in S(s_1\sim s_2)$
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2: Natural Language Description

For any real vectors space, $V$, the set of the norms on $V$, $S$, and the equivalence relation on $S$, $\sim :\subseteq S\times S$, for each $s_1,s_2\in S$, $s_1\sim s_2$.

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3: Proof

Whole Strategy: Step 1: take any basis for $V$ and define the Euclidean norm, $s_0$, with respect to the basis; Step 2: find for each $k=1,2$, $c_{k,1},c_{k,2}\in \mathbb{R}$ such that $0<c_{k,1},c_{k,2}$ and for each $v\in V$, $c_{k,1}{s}_0(v)\le s_k(v)\le c_{k,2}{s}_0(v)$; Step 3: conclude that $s_1\sim s_2$.

Step 1:

Let us take any basis of $V$, $\{e_1,...,e_d\}$. Any vector, $v\in V$, is $v=\sum _{j\in \{1,...,d\}}(v^j{e}_j)$.

There is the Euclidean norm with respect to the basis, $s_0\in S:v\mapsto (\sum _{j\in \{1,...,d\}}{v^j}^2{)}^{1/2}$.

$s_0$ is indeed a norm, because it is an Euclidean norm.

Step 2:

Let us find for $k=1,2$ some $c_{k,1},c_{k,2}\in \mathbb{R}$ such that $0<c_{k,1},c_{k,2}$ and $c_{k,1}{s}_0(v)\le s_k(v)\le c_{k,2}{s}_0(v)$.

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

$max(\{|v^j|\})\le (\sum _{j\in \{1,...,d\}}{v^j}^2{)}^{1/2}=s_0(v)$.

So, $s_k(v)\le max(\{s_k(e_j)\})d{s}_0(v)=c_{k-2}{s}_0(v)$ where $c_{k,2}=max(\{s_k(e_i)\})d$, which does not depend on $v$.

Let us regard $V$ as the canonical topological space, by the definition of canonical topology for finite-dimensional real vectors space.

The norm map from the canonical topological space, $V$, $s_k:V\to \mathbb{R}$, is continuous, by the proposition that for any finite-dimensional normed real vectors space with the canonical topology, the norm map is continuous.

Let us see that $\{v\in V|s_0(v)=1\}$ is a compact subset of $V$.

There is the canonical homeomorphism, $f:V\to {\mathbb{R}}^d$, with respect to the basis. $f(\{v\in V|s_0(v)=1\})$ is obviously closed and bounded, and so, is a compact subset of ${\mathbb{R}}^d$, by the Heine-Borel theorem: any subset of any Euclidean topological space is compact if and only if it is closed and bounded. So, $\{v\in V|s_0(v)=1\}=f^{-1}\circ f(\{v\in V|s_0(v)=1\})$ is compact on $V$.

So, its image under the norm map, $s_k$, has the minimum, $c_{k,1}$, by the proposition that the image of any continuous map from any compact topological space to the $\mathbb{R}$ Euclidean topological space has the minimum and the maximum, while $c_{k,1}$ is positive, because any norm is positive definite.

So, $c_{k,1}\le s_k(v^{\prime })$ for each $v^{\prime }$ such that $s_0(v^{\prime })=1$.

For any $v\in V$ such that $v\ne 0$, let us define $v^{\prime }:=v/s_0(v)$. Then, $s_0(v^{\prime })=s_0(v/s_0(v))=1/s_0(v)s_0(v)=1$ and $v=s_0(v)v^{\prime }$. $c_{k,1}{s}_0(v)\le s_k(v^{\prime })s_0(v)=s_k(s_0(v)v^{\prime })=s_k(v)$.

For $v=0\in V$, $c_{k,1}{s}_0(v)=0\le 0=s_k(v)$.

So, $c_{k,1}{s}_0(v)\le s_k(v)\le c_{k,2}{s}_0(v)$.

So, $s_1\sim s_0$ and $s_2\sim s_0$.

Step 3:

As $\sim$ is an equivalence relation, $s_1\sim s_2$.

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References

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