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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
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On an infinite-dimensional normed space, the identity operator is not compact
Statement
Let be a normed space that admits no ordered basis of finite length, and let be the identity map. Then is bounded, but it does not carry the closed unit ball of to a compact subset of . In that standard sense, the identity operator is not compact.
Facts & Assumptions
Given: A normed space with no ordered basis of finite length, its closed unit ball , and the identity map .
A bounded linear operator is a linear map satisfying one global norm bound (A bounded linear operator between normed spaces).
In this setting the closed unit ball is not compact (In an infinite-dimensional normed space the closed unit ball is not compact).
Proof
The identity map is linear and satisfies for every , so [L1] makes it a bounded linear operator.
One has . By [L2], that set is not compact. Therefore the identity operator does not send the closed unit ball to a compact subset of .
Remarks
- This item uses only the unit-ball criterion. It does not depend on a separate compact-operator definition item.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
6 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 (standard reference, not scraped)