If we now sum over the I l ’s in a fixed S k ,weget∑H(l) ≤ C2 −kτ [ω(2 k I)] ∑ −1 ω(I l )l:I l ∈S k l:I l ∈S k≤ C2 −kτ [ω(2 k I)] −1 ω(2 k I)≤ C2 −kτ ,where the next-to-last line follows from ω’s doubling property. Summing over all k yields ∑ H(l) ≤C, and term (I) has been bounded.Handling (II). This **on**e’s easier.(II)=|≤∑Q:Q∈S(I)l(Q)>l(I)∑Q:Q∈S(I)l(Q)>l(I)λ Q(φ(Q) (x I ) − φ (Q) (x) ) |( ) β |xI − x| √ ∑ ∞ ( 2−jα|λ Q |ω(Q)l(Q)ω(2 j Q)j=0)(χ Rj (Q)(x I )+χ Rj (Q)(x)).Since we are c**on**sidering cubes Q that are larger than I, ifx I ∈ R j (Q) andx ∈ I, then x ∈ R j ′(Q)ong>forong> some j ′ such that |j − j ′ |≤C. Thereong>forong>e, our sum (II) is actually less than or equal toC∑Q:Q∈S(I)l(Q)>l(I)C**on**tinuing, this is bounded by:C=C∑Q:Q∈S(I)l(Q)>l(I)∞∑( ) β |xI − x| √ ∑∞ ( 2−jα|λ Q |ω(Q)l(Q)ω(2 j Q)( l(I)|λ Q |l(Q)∑k=1 Q:l(Q)=2 k l(I)⎛∞∑≤CG(I) ⎝k=1 Q:l(Q)=2 k l(I)j=0) β√ ∑ ∞ ( 2−jαω(Q)j=0|λ Q |2 −kβ√ ω(Q)∑ω(2 j Q)∞∑( 2−jαj=0j=0ω(2 j Q))χ Rj (Q)(x I ).)χ Rj (Q)(x I ))χ Rj (Q)(x I )⎞∞∑( ) 22 −2kβ −jτω(Q)ω(2 j χ Rj (Q)(x I ) ⎠Q)where the last line follows from the Cauchy-Schwartz inequality.We now need to show that the sec**on**d factor is bounded by a c**on**stant.We temporarily fix k and c**on**sider∑Q:l(Q)=2 k l(I)ω(Q)∞∑( 2−jτj=010ω(2 j Q)1/2)χ Rj (Q)(x I ). (1.3),

If χ Rj (Q)(x I ) ≠ 0 then Q ⊂ 2 k+j′ I ong>forong> some j ′ such that |j − j ′ |≤C. Also, ong>forong> such Q, ω(2 j Q)will be comparable to ω(2 k+j′ I), because ω is doubling. Thereong>forong>e, ong>forong> each j,∑( ) 2−jτω(2 j ≤ C2 −jτ .Q)ω(Q)Q:l(Q)=2 k l(I)χ Rj (Q) (x I )≠0Summing **on** j, we get that (1.3) is bounded by a c**on**stant. If we now multiply this by 2 −2kβ andsum **on** k, we get our result. Lemma 1.5 is proved. QED.If I is a dyadic cube, we define N(I) ≡ {I ∗ ∈ D : I ∗ ⊂ I, l(I ∗ ) = l(I)/2}, the “nextgenerati**on**” of cubes “begotten” by I.Lemma 1.6. If I ∗ ∈ N(I), then G(I) ≤ CG(I ∗ ).Proof. By Lemma 1.4, G(I) =G(I,x I ) ≤ CG(I,x I ∗). But G(I ∗ )=G(I ∗ ,x I ∗) ≥ G(I,x I ∗).QED.Lemma 1.7. There is a positive c**on**stant C such that, ong>forong> all I ∗ ∈ N(I) and all x ∈ I, G ∗ (x) ≥CG(I ∗ ).Proof. If x ∈ I ∗ , there is nothing to prove. So, let x ∈ I \ I ∗ ,andletL ⊂ I \ I ∗ be a dyadiccube such that x ∈ L and L is smaller than any of the cubes Q in the sum ∑ λφ (Q) defining f.Then:⎡⎤G(L) 2 =∑∞∑|λ J | 2 ⎣2 −j(2α−τ)ω(2 j J) χ R j (J)(x L ) ⎦≥J:J∈S(L)∑J:J∈S(I ∗ )j=0⎡∞∑|λ J | 2 ⎣j=02 −j(2α−τ)⎤ω(2 j J) χ R j (J)(x L ) ⎦ ,because the sec**on**d sum excludes the J’s c**on**tained in I ∗ . (Technically, the first sum excludes theJ’s c**on**tained in L, but, because L is so small, these c**on**tribute nothing to the sum.)The lemma will be proved **on**ce we show that, ong>forong> J ∈ S(I ∗ ),∞∑j=02 −j(2α−τ)ω(2 j J) χ R j (J)(x L ) ≥ C∞∑j=02 −j(2α−τ)ω(2 j J) χ R j (J)(x I ∗).But this follows from what are (by now) “the usual reas**on**s.” Simply note that, if J ∈ S(I ∗ ),then |x L − x J |≤C(d)|x I∗ − x J |. Thus, if 2−j(2α−τ)ω(2 j J)χ Rj (J)(x I ∗) ≠ 0, then 2−j′ (2α−τ)χω(2 j′ J) Rj ′ (J)(x L )will also be n**on**-zero, ong>forong> some j ′ such that |j − j ′ |≤C. The doubling property of ω ensures thatω(2 j′ J) ≤ Cω(2 j J), which finishes the proof of Lemma 1.7. QED.Lemma 1.8. There is a C such that, if I ∗ ∈ N(I), then |F (I) − F (I ∗ )|≤CG(I ∗ ).Proof.|F (I) − F (I ∗ )| = |F (I,x I ) − F (I ∗ ,x I ∗)|≤|F (I,x I ) − F (I,x I ∗)| + |F (I,x I ∗) − F (I ∗ ,x I ∗)|.11