description/proof of that complex-conjugate-linear 'normed vectors space' isometry maps Cauchy sequence to Cauchy sequence
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of 'normed vectors space' isometry.
- The reader knows a definition of complex-conjugate-linear map.
- The reader knows a definition of Cauchy sequence on metric space.
Target Context
- The reader will have a description and a proof of the proposition that any complex-conjugate-linear 'normed vectors space' isometry maps any Cauchy sequence to a Cauchy sequence.
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
\(V_1\): \(\in \{\text{ the } F \text{ normed vectors spaces }\}\), with the metric induced by the norm
\(V_2\): \(\in \{\text{ the } F \text{ normed vectors spaces }\}\), with the metric induced by the norm
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the complex-conjugate-linear 'normed vectors space' isometries }\}\)
\(s\): \(: \mathbb{N} \to V_1\), \(\in \{\text{ the Cauchy sequences }\}\)
\(f \circ s\): \(: \mathbb{N} \to V_2\), \(\in \{\text{ the sequences }\}\)
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Statements:
\(f \circ s \in \{\text{ the Cauchy sequences }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(f \circ s\) satisfies the conditions to be a Cauchy sequence.
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N}\) such that for each \(m, n \in \mathbb{N}\) such that \(N \lt m, n\), \(dist (s (n), s (m)) = \Vert s (n) - s (m) \Vert \lt \epsilon\).
\(dist (f \circ s (n), f \circ s (m)) = \Vert f \circ s (n) - f \circ s (m) \Vert = \Vert f (s (n) - s (m)) \Vert\), because \(f\) is complex-conjugate-linear, \ = \Vert s (n) - s (m) \Vert\), because \(f\) is a 'normed vectors space' isometry, \(\lt \epsilon\).
So, \(f \circ s\) satisfies the conditions to be a Cauchy sequence.
