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A bounded map into yields a compact operator
Statement
Assume the Axiom of Choice. Let be a bounded open set and a bounded linear operator. Then is compact, where is the inclusion. No boundary regularity of is needed.
Facts & Assumptions
Given: the Axiom of Choice, a bounded open set , and a bounded linear operator , with the inclusion.
Zero-boundary Rellich theorem. is compactly embedded in : every sequence bounded in has a subsequence converging in . (Compactness of on bounded open sets, Compactly embedded normed spaces, The notation and the reserved zero-boundary symbol)
Bounded operators map bounded sequences to bounded sequences. If satisfies , then for the operator norm of A bounded linear operator between normed spaces. (A bounded linear operator between normed spaces)
Compactness is the sequential extraction criterion. A bounded operator is compact exactly when the image of every bounded sequence has a convergent subsequence. (Compact linear operator, Compactly embedded normed spaces)
Proof
Let be bounded in with . By [F2], is bounded in , so [F1] supplies a subsequence with in , that is, .
Since every bounded sequence in has an image under with a convergent subsequence, [F3] makes a compact operator. The inclusion is bounded because , and the Axiom of Choice is inherited through the Rellich theorem [F1].
Depends on
Used by
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Dependency tree · two levels
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Sources
- John K. Hunter, Notes on Partial Differential Equations, complete 242-page 2014 notes (standard reference, not scraped)