description/proof of for finite number of uniformly continuous maps from same metric space into \(1\)-dimensional Euclidean metric space, maximum or minimum 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.
- The reader admits the proposition that for any uniformly continuous map from any metric space into any Euclidean metric space, its norm map is uniformly continuous.
- The reader admits the proposition that any linear combination of uniformly continuous maps into any Euclidean metric space is uniformly continuous.
Target Context
- The reader will have a description and a proof of the proposition that for any finite number of uniformly continuous maps from any same metric space into the \(1\)-dimensional Euclidean metric space, the maximum or minimum 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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(\{f_1, ..., f_n\}\): \(f_j: M \to \mathbb{R} \in \{\text{ the uniformly continuous maps }\}\)
\(Max (\{f_1, ..., f_n\})\): \(: M \to \mathbb{R}, m \mapsto Max (\{f_1 (m), ..., f_n (m)\})\)
\(Min (\{f_1, ..., f_n\})\): \(: M \to \mathbb{R}, m \mapsto Min (\{f_1 (m), ..., f_n (m)\})\)
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Statements:
\(Max (\{f_1, ..., f_n\}) \in \{\text{ the uniformly continuous maps }\}\)
\(\land\)
\(Min (\{f_1, ..., f_n\}) \in \{\text{ the uniformly continuous maps }\}\)
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2: Proof
Whole Strategy: prove it inductively; Step 1: deal with the case that \(n = 1\); Step 2: when \(n = 2\), see that \(Max (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\) and \(Min (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\), and conclude the proposition for when \(n = 2\); Step 3: suppose that it holds when \(n = n' - 1\), and see that it holds when \(n = n'\).
Step 1:
It holds when \(n = 1\), because \(Max (\{f_1\}) = f_1\) and \(Min (\{f_1\}) = f_1\).
Step 2:
Let us suppose that \(n = 2\).
\(Max (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\), because for each \(m \in M\), \(f_1 (m) \le f_2 (m)\) or \(f_2 (m) \lt f_1 (m)\), and when \(f_1 (m) \le f_2 (m)\), \(Max (\{f_1, f_2\}) (m) = f_2 (m)\) while \(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) + \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) + f_2 (m) - f_1 (m)) = 1 / 2 (f_2 (m) + f_2 (m)) = 1 / 2 2 f_2 (m) = f_2 (m)\); and when \(f_2 (m) \lt f_1 (m)\), \(Max (\{f_1, f_2\}) (m) = f_1 (m)\) while \(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) + \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) + f_1 (m) - f_2 (m)) = 1 / 2 (f_1 (m) + f_1 (m)) = 1 / 2 2 f_1 (m) = f_1 (m)\); so, \(Max (\{f_1, f_2\}) (m) = 1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert) (m)\) anyway.
\(Min (\{f_1, f_2\}) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\), because for each \(m \in M\), \(f_1 (m) \le f_2 (m)\) or \(f_2 (m) \lt f_1 (m)\), and when \(f_1 (m) \le f_2 (m)\), \(Min (\{f_1, f_2\}) (m) = f_1 (m)\) while \(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) - \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) - f_2 (m) + f_1 (m)) = 1 / 2 (f_1 (m) + f_1 (m)) = 1 / 2 2 f_1 (m) = f_1 (m)\); and when \(f_2 (m) \lt f_1 (m)\), \(Min (\{f_1, f_2\}) (m) = f_2 (m)\) while \(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert) (m) = 1 / 2 (f_1 (m) + f_2 (m) - \vert f_1 (m) - f_2 (m) \vert) = 1 / 2 (f_1 (m) + f_2 (m) - f_1 (m) + f_2 (m)) = 1 / 2 (f_2 (m) + f_2 (m)) = 1 / 2 2 f_2 (m) = f_2 (m)\); so, \(Min (\{f_1, f_2\}) (m) = 1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert) (m)\) anyway.
\(1 / 2 (f_1 + f_2 + \vert f_1 - f_2 \vert)\) is uniformly continuous, by the proposition that for any uniformly continuous map from any metric space into any Euclidean metric space, its norm map is uniformly continuous and the proposition that any linear combination of uniformly continuous maps into any Euclidean metric space is uniformly continuous.
So, \(Max (\{f_1, f_2\})\) is uniformly continuous.
\(1 / 2 (f_1 + f_2 - \vert f_1 - f_2 \vert)\) is uniformly continuous, by the proposition that for any uniformly continuous map from any metric space into any Euclidean metric space, its norm map is uniformly continuous and the proposition that any linear combination of uniformly continuous maps into any Euclidean metric space is uniformly continuous.
So, \(Min (\{f_1, f_2\})\) is uniformly continuous.
Step 3:
Let us suppose that it holds when \(n = n' - 1\) where \(2 \le n'\).
Let us see that it holds when \(n = n'\).
\(Max (\{f_1, ..., f_{n'}\}) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\), because for each \(m \in M\), \(Max (\{f_1, ..., f_{n'}\}) (m) = f_j (m)\) for a \(j \in \{1, ..., n'\}\), and when \(j \in \{1, ..., n' - 1\}\), \(Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Max (\{f_j (m), f_{n'} (m)\}) = f_j (m)\); and when \(j = n'\), \(Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Max (\{f_l (m), f_{n'} (m)\})\) for a \(l \in \{1, ..., n' - 1\}\), \(= f_{n'} (m) = f_j (m)\); so, anyway, \(Max (\{f_1, ..., f_{n'}\}) (m) = Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m)\).
\(Max (\{f_1, ..., f_{n' - 1}\})\) is uniformly continuous, by the induction hypothesis, and \(Max (\{Max (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\) is uniformly continuous, by Step 2.
So, \(Max (\{f_1, ..., f_{n'}\})\) is uniformly continuous.
By the induction principle, \(Max (\{f_1, ..., f_n\})\) is uniformly continuous for each \(n \in \mathbb{N} \setminus \{0\}\).
\(Min (\{f_1, ..., f_{n'}\}) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\), because for each \(m \in M\), \(Min (\{f_1, ..., f_{n'}\}) (m) = f_j (m)\) for a \(j \in \{1, ..., n'\}\), and when \(j \in \{1, ..., n' - 1\}\), \(Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Min (\{f_j (m), f_{n'} (m)\}) = f_j (m)\); and when \(j = n'\), \(Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}) (m), f_{n'} (m)\}) = Min (\{f_l (m), f_{n'} (m)\})\) for a \(l \in \{1, ..., n' - 1\}\), \(= f_{n'} (m) = f_j (m)\); so, anyway, \(Min (\{f_1, ..., f_{n'}\}) (m) = Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\}) (m)\).
\(Min (\{f_1, ..., f_{n' - 1}\})\) is uniformly continuous, by the induction hypothesis, and \(Min (\{Min (\{f_1, ..., f_{n' - 1}\}), f_{n'}\})\) is uniformly continuous, by Step 2.
So, \(Min (\{f_1, ..., f_{n'}\})\) is uniformly continuous.
By the induction principle, \(Min (\{f_1, ..., f_n\})\) is uniformly continuous for each \(n \in \mathbb{N} \setminus \{0\}\).
