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By Vandebril R., Van Barel M., Golub G.

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Example text

2 So we set A k = Rk ∩ j |ϕsj | ≤ 2C14 λ , and then µ(Ak ) ≥ µ(Rk )/2. The constant αk is chosen so that for ϕk = αk χAk we have Qk f wk dµ. Then we obtain |αk | ≤ 1 µ(Ak ) 2 ≤ µ(Rk ) Qk |f | dµ ≤ 1 2 Rk |f | dµ ≤ 2 µ(Rk ) Qk 2 µ(Rk ) |f | dµ 1/p p 1 2 Rk |f | dµ (this calculation also applies to k = 1). Thus, |ϕk | + |ϕsj | ≤ (2C14 + C15 ) λ. 5) follows. ϕk dµ = ≤ C15 λ 134 X. 6) also holds. 8) Qi |f |p dµ 1/p C µ(2Qi )1/p ≤ λp−1 Qi |f |p dµ. 6). Suppose now that the collection of cubes {Qi }i is not finite.

1,p 5. 34-40]. We will introduce the atomic spaces Hatb (µ), and we will prove 1,∞ 1,∞ (µ) and that the dual of Hatb (µ) is RBMO(µ) that they coincide with Hatb simultaneously. For a fixed ρ > 1 and p ∈ (1, ∞), a function b ∈ L1loc (µ) is called a p-atomic block if 1. there exists some cube R such that supp(b) ⊂ R, BMO and H 1 for non doubling measures 2. 119 b dµ = 0, 3. there are functions aj supported in cubes Qj ⊂ R and numbers λj ∈ R such that b = j∞=1 λj aj , and aj −1 ≤ µ(ρQj )1/p−1 KQ .

In a similar way, it can be proved that if the 144 X. 8) holds for any pair of cubes Q ⊂ R with KQ,R not too large, then it holds for any pair of cubes Q ⊂ R. 4) taken only over doubling cubes Q ⊂ R such that x ∈ Q and KQ,R ≤ P0 . Then, by the preceding lemma it follows that M (f ) ≈ M (f ). 4) M ([b, T ]f )(x) ≤ Cp b ∗ (Mp,(9/8) f (x) + Mp,(3/2) Tf (x) + T∗ f (x)), where, for η > 1, Mp,(η) is the non centered maximal operator 1 µ(ηQ) Mp,(η) f (x) = sup Q x and T∗ is defined as Q |f |p dµ 1/p , T∗ f (x) = sup |Tε f (x)|.

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