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Mazur theorem: weak and norm closure agree for convex sets
Statement
For a convex subset of a normed space , its norm closure equals its weak closure. Here the weak topology is generated by the basic neighbourhoods of
for finitely many and some .
Facts & Assumptions
Given: A convex set and .
Disjoint convex sets, one open, are strictly separated by a nonzero continuous functional (Separation of disjoint convex sets when one is open).
Proof
Every is norm-continuous, so every displayed basic weak neighbourhood is norm-open. Thus every weak-open set is norm-open, and .
If , both closures are empty. Otherwise is nonempty; if , choose with . This latter set is open and convex, so [F1] yields with for , .
Taking suprema over gives . The weak neighbourhood therefore misses , hence misses .
Thus is not in the weak closure whenever it is not in , proving the reverse inclusion and equality.
Depends on
Used by
Dependency tree · two levels
10 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
- Gerald Teschl, Topics in Real and Functional Analysis, Theorem 5.11 (standard reference, not scraped)