description/proof of linear combination of uniformly continuous maps into Euclidean metric space 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 any linear combination of uniformly continuous maps into any Euclidean metric space 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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean metric space }\)
\(\{f_1, ..., f_n\}\): \(f_j: M \to \mathbb{R}^d \in \{\text{ the uniformly continuous maps }\}\)
\(\{r_1, ..., r_n\}\): \(\subseteq \mathbb{R}\)
\(r_1 f_1 + ... + r_n f_n\): \(: M \to \mathbb{R}^d\)
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Statements:
\(r_1 f_1 + ... + r_n f_n \in \{\text{ the uniformly continuous maps }\}\)
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2: Proof
Whole Strategy: Step 1: for each \(\epsilon\), take for each \(j\) such that \(r_j \neq 0\), \(\delta_j\) such that \(f_j (B_{m, \delta_j}) \subseteq B_{f_j (m), \epsilon / (n \vert r_j \vert)}\) and \(\delta\) as the minimum of such \(\delta_j\) s, and see that \((r_1 f_1 + ... + r_n f_n) (B_{m, \delta}) \subseteq B_{(r_1 f_1 + ... + r_n f_n) (m), \epsilon}\).
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
For each \(j\) such that \(r_j \neq 0\), there is a \(\delta_j \in \mathbb{R}\) such that \(0 \lt \delta_j\) and for each \(m \in M\), \(f_j (B_{m, \delta_j}) \subseteq B_{f_j (m), \epsilon / (n \vert r_j \vert)}\), because \(f_j\) is uniformly continuous.
When not all of \(\{r_1, ..., r_n\}\) are zeros, let \(\delta \in \mathbb{R}\) be the minimum of such \(\delta_j\) s, then, \(0 \lt \delta\).
Otherwise, let \(\delta \in \mathbb{R}\) be \(1\), then, \(0 \lt \delta\).
Let \(m \in M\) be any.
Let \(m' \in B_{m, \delta}\) be any.
\(dist ((r_1 f_1 + ... + r_n f_n) (m'), (r_1 f_1 + ... + r_n f_n) (m)) = \Vert (r_1 f_1 + ... + r_n f_n) (m') - (r_1 f_1 + ... + r_n f_n) (m) \Vert = \Vert r_1 f_1 (m') + ... + r_n f_n (m') - r_1 f_1 (m) - ... - r_n f_n (m) \Vert = \Vert r_1 (f_1 (m') - f_1 (m)) + ... + r_n (f_n (m') - f_n (m)) \Vert \le \Vert r_1 (f_1 (m') - f_1 (m)) \Vert + ... + \Vert r_n (f_n (m') - f_n (m)) \Vert\), as a property of norm, \(= \vert r_1 \vert \Vert f_1 (m') - f_1 (m) \Vert + ... + \vert r_n \vert \Vert f_n (m') - f_n (m) \Vert\).
For each \(j\) such that \(r_j \neq 0\), \(m' \in B_{m, \delta} \subseteq B_{m, \delta_j}\), so, \(f_j (m') \in B_{f_j (m), \epsilon / (n \vert r_j \vert)}\), so, \(\vert r_j \vert \Vert f_j (m') - f_j (m) \Vert \lt \vert r_j \vert \epsilon / (n \vert r_j \vert) = \epsilon / n\).
For each \(j\) such that \(r_j = 0\), \(\vert r_j \vert \Vert f_j (m') - f_j (m) \Vert = 0 \lt \epsilon / n\).
So, \(dist ((r_1 f_1 + ... + r_n f_n) (m'), (r_1 f_1 + ... + r_n f_n) (m)) \lt \epsilon / n + ... + \epsilon / n = \epsilon\).
That means that \((r_1 f_1 + ... + r_n f_n) (B_{m, \delta}) \subseteq B_{(r_1 f_1 + ... + r_n f_n) (m), \epsilon}\).
As \(\delta\) is determined independent of \(m\), \(r_1 f_1 + ... + r_n f_n\) is uniformly continuous.
