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Double annihilators give norm and weak-star closures
Statement
Let or . For a normed and a linear subspace , Consequently is weak-star closed if and only if , and weak-star dense in if and only if . For a linear subspace , the primal formula is .
Facts & Assumptions
Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.
From The weak-star topology from finite evaluations, with its stated hypotheses: Let or . For a normed with continuous dual from def-dual-space-of-a-normed-space, the weak-star topology is the initial topology of all evaluations into with its usual topology. At , a neighbourhood basis consists of where is finite and . For the set is all of . Finite intersections of inverse images of scalar open sets form the initial-topology basis; at the given point, finitely many disks can be refined using their smallest positive radius. Weak-star closure means closure in this topology, not merely sequential closure.
From Annihilator notation and the preannihilator, with its stated hypotheses: Let or . For a normed and arbitrary subsets , , define Here is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on , since linearity makes vanishing on equivalent to vanishing on its span. The preannihilator lies in , not in . Empty sets impose no conditions: and .
From Finite evaluations separate a functional from a dual subspace, with its stated hypotheses: Let or . Let be normed, a linear subspace, and with . Define . If satisfies , there is such that for all and .
From The annihilator detects the closure of a subspace, with its stated hypotheses: For every linear subspace ,
Proof
If , evaluation at vanishes on . Its kernel is weak-star closed by the definition of that topology. Intersecting these kernels shows .
Let lie outside the weak-star closure. Choose a basic neighbourhood disjoint from . Since , this neighbourhood cannot have . Its finite-coordinate map satisfies , since equality with would put that in .
Finite-evaluation separation supplies with . Hence . Together with step 1.1, this proves equality.
A set is closed exactly when it equals its closure, so the first equivalence follows in both directions. For density, if , the equality gives closure . Conversely if the closure is , any is annihilated by all of . The published primal formula with says , so .
The primal formula in the statement is exactly the published annihilator-closure identity, with the preannihilator notation unpacked. It also checks the extremes and : their closures and double annihilators are respectively and . For these coincide.
Depends on
Used by
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