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A minimizing sequence in a convex set is Cauchy
Statement
Let be a nonempty convex subset of a real or complex inner-product space, let be a vector and put . If is a sequence with , then is a Cauchy sequence.
Facts & Assumptions
A subset is convex when for all and (Convex sets and continuous real-hyperplane separation in a normed space).
If then for every (Greatest lower bound (infimum)).
Every inner-product norm satisfies the parallelogram law (The parallelogram law).
The norm is induced by the pairing, in particular is the distance between and and (Real and complex inner-product spaces and their induced length).
Proof
Given: A nonempty convex set , a vector , the number and a sequence with .
Put ; for all the midpoint lies in by convexity, so because is a lower bound of the distances from to points of .
The parallelogram law applied to gives .
Given , convergence provides with for all , and then step 2.1 gives for all ; hence is Cauchy.
Depends on
Used by
Dependency tree · two levels
12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Theo Bühler and Dietmar Salamon, Functional Analysis, Theorem 1.44, pp.40–41 (standard reference, not scraped)
- Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Theorem 178 (standard reference, not scraped)