description/proof of that for finite-dimensional normed complex vectors space with canonical topology, norm map is continuous
Topics
About: vectors space
About: topological space
The table of contents of this article
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 complex vectors space.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional normed complex vectors space with the canonical topology, the norm map is continuous.
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:
\(V\): \(\in \{\text{ the } d \text{ -dimensional normed complex vectors spaces }\}\) with any norm, \(\Vert \bullet \Vert: V \to \mathbb{R}\), with the canonical topology
\(f'\): \(: V \to \mathbb{R}, v \mapsto \Vert v \Vert\)
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Statements:
\(f' \in \{\text{ the continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any basis, \(B = \{b_1, ..., b_d\}\), and see that for each \(v = \sum_{j \in \{1, ..., d\}} v^j b_j \in V\), \(\Vert v \Vert \le C (\sum_{j \in \{1, ..., d\}} \vert v^j \vert^2)^{1 / 2}\) where \(C\) does not depend on \(v\); Step 2: for each open ball around \(\Vert v \Vert\), \(B_{\Vert v \Vert, \epsilon} \subseteq \mathbb{R}\), take an open neighborhood of \(v\), \(U_v\), such that \(f' (U_v) \subseteq B_{\Vert v \Vert, \epsilon}\).
Step 1:
Let us take any basis, \(B = \{b_1, ..., b_d\} \subseteq V\).
Let \(v = \sum_{j \in \{1, ..., d\}} v^j b_j \in V\) be any.
\(\Vert v \Vert = \Vert \sum_{j \in \{1, ..., d\}} v^j b_j \Vert \le \sum_{j \in \{1, ..., d\}} \Vert v^j b_j \Vert = \sum_{j \in \{1, ..., d\}} \vert v^j \vert \Vert b_j \Vert\), by the definition of norm on real or complex vectors space. \(\le max (\{\Vert b_j \Vert\}) max (\{\vert v^j \vert\}) d\).
\(max (\{\vert v^j \vert\}) \le (\sum_{j \in \{1, ..., d\}} \vert v^j \vert^2)^{1 / 2}\).
So, \(\Vert v \Vert \le max (\{\Vert b_i \Vert\}) d (\sum_{j \in \{1, ..., d\}} \vert v^j \vert^2)^{1 / 2} = C (\sum_{j \in \{1, ..., d\}} \vert v^j \vert^2)^{1 / 2}\) where \(C := max (\{\Vert b_i \Vert\}) d\), which does not depend on \(v\).
Step 2:
Let \(f: V \to \mathbb{C}^d \to \mathbb{R}^{2 d}, v = \sum_{j \in \{1, ..., d\}} v^j b_j \mapsto (v^1 = w^1 + x^1 i, ..., v^d = w^d + x^d i) \mapsto (w^1, x^1, ..., w^d, x^d)\) be the map by which the canonical topology is defined.
For any open ball, \(B_{\Vert v \Vert, \epsilon} \subseteq \mathbb{R}\), let us take \(\delta = \epsilon / C\) and the open neighborhood of \(v\), \(U_v = \{p \in V \vert f (p) \in B_{f (v), \delta}\} \subseteq V\), which is indeed an open neighborhood of \(v\), because \(f (U_v) = B_{f (v), \delta} \subseteq \mathbb{R}^{2 d}\), open on \(\mathbb{R}^{2 d}\).
For each \(v' = \sum_{j \in \{1, ..., d\}} v'^j b_j \in U_v\), \(\Vert v' - v \Vert \le C (\sum_{j \in \{1, ..., d\}} \vert v'^j - v^j \vert^2)^{1 / 2} = C (\sum_{j \in \{1, ..., d\}} \vert w'^j + x'^j i - w^j - x^j i \vert^2)^{1 / 2} = C (\sum_{j \in \{1, ..., d\}} \vert (w'^j - w^j) + (x'^j - x^j) i \vert^2)^{1 / 2} = C (\sum_{j \in \{1, ..., d\}} (w'^j - w^j)^2 + (x'^j - x^j)^2)^{1 / 2} \lt C \delta = \epsilon\).
\(\vert \Vert v' \Vert - \Vert v \Vert \vert \le \Vert v' - v \Vert\), because \(\Vert v' \Vert = \Vert v' - v + v \Vert \le \Vert v' - v \Vert + \Vert v \Vert\), so, \(\Vert v' \Vert - \Vert v \Vert \le \Vert v' - v \Vert\), and likewise \(\Vert v \Vert - \Vert v' \Vert \le \Vert v' - v \Vert\).
So, \(\vert f' (v') - f' (v) \vert = \vert \Vert v' \Vert - \Vert v \Vert \vert \lt \epsilon\), which means that \(f' (U_x) \subseteq B_{\Vert v \Vert, \epsilon}\), which means that the map is continuous at any point, so, is continuous.
