Weighted inequalities for gradients on non-smooth domains ...

The key to our argument lies in defining the right maximal functi**on**. Let us assume that wehave a fixed finite linear combinati**on** f = ∑ Q λ Qφ (Q) .IfI ∈D,wedefineS(I) ≡{Q ∈D: Q ⊄ I}.It is useful to think of S(I) as the family of dyadic cubes that “surround” I. Ifx ∈ I, wedefineF (I,x) ≡∑Q:Q∈S(I)λ Q φ (Q) (x),and we do not define F (I,x)**on** **on** g ∗ (f) turns out to beF ∗ (x) ≡ sup |F (I)|.I:x∈ICorresp**on**ding to F ∗ (x) is a “maximal square functi**on**” adapted to g ∗ (f). For x ∈ I, wedefine⎛G(I,x) ≡ ⎝ ∑Q∈S(I)⎡∞∑|λ Q | 2 ⎣j=02 −j(2α−τ)⎤⎞ω(2 j Q) χ R j (Q)(x) ⎦⎠and we do not define G(I,x)**on** ofthe φ (Q) ’s requires us to surmount some n**on**-trivial technical obstacles.Lemma 1.2. For ω-a. e. x, |f(x)| ≤F ∗ (x).Proof. The inequality is obviously true Lebesgue almost everywhere. However, the **on**lyexcepti**on**al points lie **on** the faces of dyadic cubes, and these have ω-measure 0, because ω isdoubling. QED.Lemma 1.3. There is a c**on**stant C such that G ∗ (x) ≤ Cg ∗ (f)(x) almost everywhere.Proof. Let I ∈Dand x ∈ I. We need to show that G(I) ≤ Cg ∗ (f)(x), **on** above)(G(I)) 2 =Q:Q∈S(I)∑Q:Q∈S(I)j=0⎡∞∑|λ Q | 2 ⎣j=072 −j(2α−τ)1/2⎤ω(2 j Q) χ R j (Q)(x I ) ⎦ .,