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The BMO seminorm is unchanged by adding a constant

Example

For b∈Lloc1(Rn) and c∈C one has (b+c)Q=bQ+c for every cube Q and hence ∥b+c∥BMO=∥b∥BMO; so the BMO seminorm descends to the quotient BMO/C and is a norm there.

Facts & Assumptions

Given: b∈Lloc1(Rn), a constant c∈C and a cube Q, with the mean, the seminorm and the quotient of BMO seminorm and the quotient by constants.

[L1]

The mean is bQ=∣Q∣−1∫Qb, the seminorm is ∥b∥BMO=sup⁡Q∣Q∣−1∫Q∣b−bQ∣, and ∥b∥BMO=0 exactly when b is constant almost everywhere (BMO seminorm and the quotient by constants).

Verification

technique · direct
1.1L1algebra

Linearity of the integral over the cube Q gives (b+c)Q=∣Q∣−1∫Q(b+c)=∣Q∣−1∫Qb+∣Q∣−1∫Qc=bQ+c.

2.1step 1.1L1

By step 1.1, ∣(b+c)−(b+c)Q∣=∣b−bQ∣ pointwise on Q, so ∣Q∣−1∫Q∣(b+c)−(b+c)Q∣=∣Q∣−1∫Q∣b−bQ∣ for every cube; taking the supremum over all cubes gives ∥b+c∥BMO=∥b∥BMO, including the value +∞.

3.1step 2.1L1∎

The identity of step 2.1 shows that the seminorm is constant on each equivalence class modulo constants, so it descends to BMO(Rn)/C; on classes it is a norm because ∥[b]∥=0 holds exactly when b is almost everywhere constant by [L1], that is exactly for the zero class. No choice principle is used.

Depends on

Used by

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Dependency tree · two levels

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Sources