description/proof of that linear map from finite-dimensional normed vectors space into normed vectors space is continuous
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of normed vectors space.
- The reader knows a definition of dimension of vectors space.
- The reader knows a definition of continuous, normed vectors spaces map.
- The reader admits the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm.
- The reader admits the proposition that any norms on any finite-dimensional real vectors space are equivalent.
- The reader admits the proposition that any norms on any finite-dimensional complex vectors space are equivalent.
Target Context
- The reader will have a description and a proof of the proposition that any linear map from any finite-dimensional normed vectors space into any normed vectors space 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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(d\): \(\in \mathbb{N}\)
\(V_1\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with any norm
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any norm
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
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Statements:
\(f \in \{\text{ the continuous maps }\}\)
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2: Note
\(f\) is also continuous as the map between the topological spaces induced by the norms, by the proposition that for any map between any normed vectors spaces, the map is continuous if and only if the map is continuous as the map between the topological spaces induced by the metrics induced by the norms.
3: Proof
Whole Strategy: Step 1: take any basis, \(\{b_1, ..., b_d\}\), for \(V_1\), and see that \(: V_1 \to \mathbb{R}, v = v^j b_j \mapsto \sum_{j \in \{1, ..., d\}} \vert v^j \vert\) is a norm; Step 2: take a constant, \(c \in \mathbb{R}\), such that \(0 \lt c\) and \(c \sum_{j \in \{1, ..., d\}} \vert v^j \vert \le \Vert v \Vert\); Step 3: see that for each \(v' \in B_{v, \delta}\), \(c \sum_{j \in \{1, ..., d\}} \vert v'^j - v^j \vert \lt \delta\); Step 4: take any \(B_{f (v), \epsilon}\), and find a \(\delta\) such that \(f (B_{v, \delta}) \subseteq B_{f (v), \epsilon}\).
Step 1:
Let us take any basis, \(\{b_1, ..., b_d\}\), for \(V_1\).
\(: V_1 \to \mathbb{R}, v = v^j b_j \mapsto \sum_{j \in \{1, ..., d\}} \vert v^j \vert\) is a norm on \(V_1\), by the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm.
Step 2:
By the proposition that any norms on any finite-dimensional real vectors space are equivalent or the proposition that any norms on any finite-dimensional complex vectors space are equivalent, there is a constant, \(c \in \mathbb{R}\), such that \(0 \lt c\) and for each \(v = v^j b_j \in V_1\), \(c \sum_{j \in \{1, ..., d\}} \vert v^j \vert \le \Vert v \Vert\).
Step 3:
Let \(v = v^j b_j \in V_1\) be any.
Let \(B_{v, \delta} \subseteq V_1\) be the \(\delta\)-'open ball' around \(v\).
For each \(v' = v'^j b_j \in B_{v, \delta}\), \(c \sum_{j \in \{1, ..., d\}} \vert v'^j - v^j \vert \lt \delta\), because \(c \sum_{j \in \{1, ..., d\}} \vert v'^j - v^j \vert \le \Vert v' - v \Vert \lt \delta\).
Step 4:
Let \(B_{f (v), \epsilon} \subseteq V_2\) be the \(\epsilon\)-'open ball' around \(f (v)\).
Let \(v' = v'^j b_j \in B_{v, \delta}\) be any.
\(\Vert f (v') - f (v) \Vert = \Vert f (v'^j b_j) - f (v^j b_j) \Vert = \Vert v'^j f (b_j) - v^j f (b_j) \Vert = \Vert (v'^j - v^j) f (b_j) \Vert \le \vert v'^j - v^j \vert \Vert f (b_j) \Vert\).
Let \(M := max (\{\Vert f (b_j) \Vert \vert j \in \{1, ..., d\}\})\).
\(\Vert f (v') - f (v) \Vert \le \vert v'^j - v^j \vert \Vert f (b_j) \Vert \le \sum_{j \in \{1, ..., d\}} \vert v'^j - v^j \vert M\).
But \(\sum_{j \in \{1, ..., d\}} \vert v'^j - v^j \vert M \lt \delta / c M\), so, \(\Vert f (v') - f (v) \Vert \lt \delta / c M\).
So, let us take \(\delta = c \epsilon / M\).
Then, \(\Vert f (v') - f (v) \Vert \lt \epsilon\).
That means that \(f (B_{v, \delta}) \subseteq B_{f (v), \epsilon}\).
That means that \(lim_v f = f (v)\).
So, \(f\) is continuous.
