description/proof of for uniformly continuous map from metric space into Euclidean metric space, its norm map is uniformly continuous
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean metric space.
- The reader knows a definition of uniformly continuous map between metric spaces.
Target Context
- The reader will have a description and a proof of the proposition that for any uniformly continuous map from any metric space into any Euclidean metric space, its norm map is uniformly 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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean metric space }\)
\(f\): \(: M \to \mathbb{R}^d \in \{\text{ the uniformly continuous maps }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(\Vert f \Vert\): \(: M \to \mathbb{R}, m \mapsto \Vert f (m) \Vert\)
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Statements:
\(\Vert f \Vert \in \{\text{ the uniformly continuous maps }\}\)
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2: Note
When \(d = 1\), \(\Vert f \Vert = \vert f \vert\), the absolute map.
3: Proof
Whole Strategy: Step 1: see that \(\vert \Vert f \Vert (m') - \Vert f \Vert (m) \vert \le \Vert f (m') - f (m) \Vert\); Step 2: see that \(\Vert f \Vert (B_{m, \delta}) \subseteq B_{\Vert f \Vert (m), \epsilon}\) independent of \(m\).
Step 1:
Let \(m, m' \in M\) be any.
\(\Vert f (m') \Vert = \Vert f (m') - f (m) + f (m) \Vert \le \Vert f (m') - f (m) \Vert + \Vert f (m) \Vert\), by the property of norm.
So, \(\Vert f (m') \Vert - \Vert f (m) \Vert \le \Vert f (m') - f (m) \Vert\).
Symmetrically, \(\Vert f (m) \Vert - \Vert f (m') \Vert \le \Vert f (m) - f (m') \Vert = \Vert f (m') - f (m) \Vert\).
As \(\vert \Vert f (m') \Vert - \Vert f (m) \Vert \vert = \Vert f (m') \Vert - \Vert f (m) \Vert \text{ or } \Vert f (m) \Vert - \Vert f (m') \Vert\), \(\vert \Vert f (m') \Vert - \Vert f (m) \Vert \vert \le \Vert f (m') - f (m) \Vert\).
So, \(\vert \Vert f \Vert (m') - \Vert f \Vert (m) \vert \le \Vert f (m') - f (m) \Vert\).
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and for each \(m \in M\), \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\), because \(f\) is uniquely continuous.
Let \(m \in M\) be any.
Let \(m' \in B_{m, \delta}\) be any.
\(\vert \Vert f \Vert (m') - \Vert f \Vert (m) \vert \le \Vert f (m') - f (m) \Vert \lt \epsilon\), by Step 1.
That means that \(\Vert f \Vert (B_{m, \delta}) \subseteq B_{\Vert f \Vert (m), \epsilon}\).
As \(\delta\) is determined independent of \(m\), \(\Vert f \Vert\) is uniformly continuous.
