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

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and chosen so that∑g(Q b )l(Q b ) −1 (u(X Q 1∣) − u(XQ 2 ))µ(T (Q b))∣Q b ∈G⎛≥ (1/2) ⎝ ∑ l(Q b ) −q |u(X Q 1 ) − u(XQ 2 )|q µ(T (Q b ) ⎠ . (3.2)Q b ∈G⎞1/qWe need to show that, <strong>for</strong> every such g, the left-hand side of (3.2) is not too big; i.e., that itis less than or equal to a constant timesWe define(∫1/p|f(s)| vdω(s)) p .∂ΩT (g)(s) ≡ ∑ Q b ∈Gg(Q b )µ(T (Q b ))l(Q b ) −1 (K(X Q 1 ,s) − K(XQ 2 ,s)),and notice that∑Q b ∈G∫g(Q b )l(Q b ) −1 (u(X Q 1 ) − u(XQ 2 ))µ(T (Q b)) =Recall that σ = v 1−p′ . The left-hand side of (3.2) will be<strong>for</strong> all g as we have defined, if(∫≤ C |f(s)| p vdω(s)∂Ω) 1/p∂Ωf(s) T (g)(s) dω(s).(∫|T (g)(s)| p′ σdω∂Ω) 1/p′⎛⎞≤ C ⎝ ∑ |g(Q)| q′ µ(T (Q b )) ⎠1/q ′ (3.3)Q∈G<strong>for</strong> all such g. It is this last inequality which we shall prove.WriteT (g)(s) = ∑ Q b ∈Gλ Q φ (Q) ,whereandφ (Q) (s) ≡ √ ω(Q b )(K(X Q 1 ,s) − K(XQ 2 ,s)),|λ Q |≤C |g(Q b)|µ(T (Q b ))l(Q b ) √ ω(Q b ) .24
The integral we need to estimate naturally breaks into two pieces. Let us recall the region wedenoted by R in Section 2:R≡{(x, y) : ψ(x) κ}. By our estimates onthe φ (Q) ’s,|Tg(x)| ≤C κ,Ω∑Q b ∈G|λ Q | √ ω(Q b )∞∑j=02 −jαω(2 j Q b ) χ R j (Q b )(x).2There is a C, independent of x ∈ℵ, such that−jαω(2 j Q b ) χ R j (Q b )(x) can be non-zero <strong>for</strong> at most Cmany j’s. For each of these j’s, 2 −j is essentially equal to l(Q b ), with the comparability constantsdepending on κ and Ω. Also, <strong>for</strong> such j, ω(2 j Q b ) is comparable to ω(∂Ω), which equals 1. There<strong>for</strong>e,<strong>for</strong> x ∈ℵ,∑|Tg(x)| ≤C κ,Ω |λ Q | √ ω(Q b )l(Q b ) α≤ C ∑ Q b ∈GQ b ∈G|g(Q)|µ(T (Q b ))l(Q b ) −1 l(Q b ) α ;which looks funny—but we have a good reason <strong>for</strong> not combining the exponents in the l(Q b )’s.We assume that we have⎛∫µ(T (Q b )) 1/q ⎜⎝∂Ω⎡⎣ω(Q b )⎤∞∑ 2 −jɛω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σ(x) dω(x) ⎠1/p ′ ≤ cl(Q b )ω(Q b )<strong>for</strong> every Q b ∈G. We may replace the integral on the left-hand side of this inequality by an integralover the smaller region ℵ. Doing so, we may rewrite the inequality (after a change in c) as:where ɛ>0 is small.Thus, <strong>for</strong> x ∈ℵ,|Tg(x)| ≤c ∑ Q b ∈G(µ(T (Q b )) 1/q ≤ cl(Q b ) −ɛ/2 l(Q b ) √ )ω(Q b )(∫,ℵ σdω) 1/p ′|g(Q)|µ(T (Q b ))l(Q b ) −1 l(Q b ) α= c ∑ |g(Q)|µ(T (Q b )) 1/q′ µ(T (Q b )) 1/q l(Q b ) −1 l(Q b ) αQ b ∈G⎡⎤≤ c ⎣ ∑ |g(Q)|µ(T (Q b )) 1/q′ l(Q b ) α−ɛ/2√ (∫ω(Q b ) ⎦ σdωQ b ∈Gℵ25) −1/p′.
Page 4: where ɛ>0 is a constant depending
Page 7 and 8: The key to our argument lies in def
Page 9 and 10: By the Cauchy-Schwarz inequality an
Page 11 and 12: If χ Rj (Q)(x I ) ≠ 0 then Q ⊂
Page 13 and 14: Thus, our problem has now reduced t
Page 15 and 16: Putting it all together, we get∫|
Page 17 and 18: x ∈ C(P i ) \ D,∑Q: Q⊂P jl(Q)
Page 19 and 20: comes almost (but not quite) <stron
Page 21 and 22: where, of course, R j (Q b )=2 j Q
Page 23: µ is a positive Borel measure on
Page 27 and 28: Comparing the sums term-by-term, we
Page 29 and 30: Thus, the last quantity is bounded
Page 31 and 32: The Dirichlet problem for</
Page 33 and 34: So, if M ν (h) is not identically

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