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On a semifinite measure space, a representing function is unique
Statement
Let be a semifinite measure space, let , and let be conjugate to . If satisfy then in . Equivalently, on a semifinite measure space a bounded functional on has at most one representing class.
Facts & Assumptions
Given: A semifinite measure space , an exponent , its conjugate exponent , and such that for every .
On a semifinite measure space, (The functional has norm ; for assume is semifinite).
For finite exponents, the norm is a genuine norm on the quotient space The norm descends to the quotient and makes a normed space for .
If , then almost everywhere, so almost everywhere (The essential supremum is attained as the least essential bound).
Proof
For every , the assumption gives [given, algebra] So is the zero functional on .
Applying [L1] to yields [L1, step 1.1]
If , [L2] says that zero norm means the zero class, so [L2, L3, step 2.1] in . If , then step 2.1 and [L3] give almost everywhere. In either case the representing class is unique.
Depends on
Used by
Dependency tree · two levels
15 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 B. Folland, Real Analysis, 2nd ed., Theorem 6.15 (standard reference, not scraped)
- John K. Hunter, Measure Theory, Proposition 7.13 and Theorem 7.14 (standard reference, not scraped)