description/proof of that for uniformly convergent sequence of value-bounded maps into normed vectors space, convergence is value-bounded
Topics
About: vectors space
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of value-bounded map from set into seminormed vectors space.
- The reader knows a definition of metric induced by norm on real or complex vectors space.
- The reader knows a definition of uniformly convergent sequence of maps from set into metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any uniformly convergent sequence of any value-bounded maps into any normed vectors space with the induced metric, the convergence is value-bounded.
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:
\(S\): \(\in \{\text{ the sets }\}\)
\(V\): \(\in \{\text{ the normed vectors spaces }\}\), with the induced metric
\(J\): \(\subseteq \mathbb{N}\)
\(s\): \(: J \to \{g: S \to M\}\), \(\in \{\text{ the uniformly convergent sequences }\}\), such that \(\forall j \in J (s (j) \in \{\text{ the value-bounded maps }\})\)
\(f\): \(= lim s\), \(: S \to M\)
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Statements:
\(f \in \{\text{ the value-bounded maps }\}\)
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2: Proof
Whole Strategy: Step 1: let any value-bound of \(s (j)\) be \(L_j\), take any \(\epsilon\) and \(N\) such that for each \(N \lt j\), for each \(p \in S\), \(\Vert f (p) - s (j) (p) \Vert \lt \epsilon\); Step 2: see that \(\Vert f (p) \Vert \le \Vert f (p) - s (j) (p) \Vert + \Vert s (j) (p) \Vert \lt \epsilon + L_j\).
Step 1:
For each \(j \in J\), there is an \(L_j \in \mathbb{R}\) such that for each \(p \in S\), \(\Vert s (j) (p) \Vert \lt L_j\), because \(s (j)\) is value-bounded.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in J\) such that for each \(j \in J\) such that \(N \lt j\), for each \(p \in S\), \(\Vert f (p) - s (j) (p) \Vert \lt \epsilon\), because \(f\) is the uniform convergence of \(s\).
Step 2:
For any such \(j\), for each \(p \in S\), \(\Vert f (p) \Vert = \Vert f (p) - s (j) (p) + s (j) (p) \Vert \le \Vert f (p) - s (j) (p) \Vert + \Vert s (j) (p) \Vert \lt \epsilon + L_j\).
So, for any \(\epsilon\) and any such \(j\), \(\epsilon + L_j\) is an upper bound of \(\Vert f (p) \Vert\).
So, \(f\) is value-bounded.
