description/proof of that norms on finite-dimensional real vectors space are equivalent
Topics
About: vectors space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
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.
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.
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.
Main Body
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:
\(\forall s_1, s_2 \in S (s_1 \sim s_2)\)
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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 \(s_1, s_2 \in S\), \(s_1 \sim s_2\).
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 \(0 \lt c_{k, 1}, c_{k, 2}\) 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 \(0 \lt c_{k, 1}, c_{k, 2}\) and \(c_{k, 1} s_0 (v) \le s_k (v) \le c_{k, 2} s_0 (v)\).
\(s_k (v) = s_k (\sum_{j \in \{1, ..., d\}} (v^j e_j)) \le \sum_{j \in \{1, ..., d\}} s_k (v^j e_j) = \sum_{j \in \{1, ..., d\}} (\vert v^j \vert s_k (e_j))\), by the definition of norm on vectors space, \(\le max (\{s_k (e_j)\}) max (\{\vert v^j \vert\}) d\).
\(max (\{\vert v^j\vert\}) \le (\sum_{j \in \{1, ..., d\}} {v^j}^2)^{1 / 2} = s_0 (v)\).
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 \(\{v \in V \vert s_0 (v) = 1\}\) is a compact subset of \(V\).
There is the canonical homeomorphism, \(f: V \to \mathbb{R}^d\), with respect to the basis. \(f (\{v \in V \vert s_0 (v) = 1\})\) 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, \(\{v \in V \vert s_0 (v) = 1\} = f^{-1} \circ f (\{v \in V \vert s_0 (v) = 1\})\) 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, \(c_{k, 1} \le s_k (v')\) for each \(v'\) such that \(s_0 (v') = 1\).
For any \(v \in V\) such that \(v \neq 0\), let us define \(v' := v / s_0 (v)\). Then, \(s_0 (v') = s_0 (v / s_0 (v)) = 1 / s_0 (v) s_0 (v) = 1\) and \(v = s_0 (v) v'\). \(c_{k, 1} s_0 (v) \le s_k (v') s_0 (v) = s_k (s_0 (v) v') = s_k (v)\).
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, \(s_1 \sim s_2\).
