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The BMO seminorm is unchanged by adding a constant
Example
For and one has for every cube and hence ; so the BMO seminorm descends to the quotient and is a norm there.
Facts & Assumptions
Given: , a constant and a cube , with the mean, the seminorm and the quotient of BMO seminorm and the quotient by constants.
The mean is , the seminorm is , and exactly when is constant almost everywhere (BMO seminorm and the quotient by constants).
Verification
Linearity of the integral over the cube gives .
By step 1.1, pointwise on , so for every cube; taking the supremum over all cubes gives , including the value .
The identity of step 2.1 shows that the seminorm is constant on each equivalence class modulo constants, so it descends to ; on classes it is a norm because holds exactly when is almost everywhere constant by [L1], that is exactly for the zero class. No choice principle is used.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
5 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
- Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)