V. CHOUSIONIS AND X. TOLSA

THE T THEOEM V. CHOUSIONIS AND X. TOLSA Introduction These are the notes of a short course given by X. Tolsa at the Universitat Autònoma de Barcelona between November and December of 202. The notes have been typed by V. Chousionis. We present a dyadic proof of the classical T theorem of David and Journé. We have preferred to state the T theorem in terms of the boundedness over characteristic functions of cubes rather than in terms of the typical conditions T, T BMO, which is the usual approach in the literature. Nevertheless, no attempt at originality is claimed. The notation used below is standard. We use the shortcuts L p or C for L p ( n ) or C ( n ), as well as other analogous terminology. As usual, the letter C denotes some constant that may change its value at different occurrences, and typically depends on some absolute constants and other fixed parameters.. Statement and reductions A kernel K : n n \ {(x, y) : x = y} is called Calderón-Zygmund standard, or simply standard, if there exist constants C, δ > 0 such that for all distinct x, y n and for all x such that x x x y, 2 () K(x, y) C x y n (2) K(x, y) K(x, y) + K(y, x) + K(y, x ) C x x δ. x y n+δ From now on K will always denote a standard kernel. Given a function f C0 we define Tf(x) = K(x, y)f(y)dy for x / supp(f). We then say that T is Calderón-Zygmund operator (CZO) with kernel K(, ). If f L p, for some < p <, we define the truncated singular integrals T ε associated with the standard kernel K by T ε (f)(x) = K(x, y)f(y)dy. x y >ε The operator T is said to be bounded in L 2 if there exists some constant C not depending on ε such that T ε (f) 2 C f 2

2 V. CHOUSIONIS AND X. TOLSA for all f L 2. The definition of the L 2 boundedness of T in terms of the truncated operators T ε is convenient because it avoids the delicate question of the existence of principal values for Tf. By T we denote the CZO associated with the kernel K(x, y) = K(y, x), furthermore it holds that, for f, g C 0, T ε f, g = f, T εg. The following theorem provides checkable criteria for the L 2 -boundedness of CZO s. Theorem (David-Journé). Let T be a CZO as above. Then T is bounded in L 2 if and only if there exists some constant C > 0 such that () sup ε>0 T ε χ Q 2 Cm(Q) /2 (2) sup ε>0 T ε χ Q 2 Cm(Q) /2 for all cubes Q. The first step consists of regularizing the kernel. We define a function ϕ :, ϕ C, 0 ϕ, such that ϕ = 0 in [ /2, /2] and ϕ = in \ [, ]. Given the function ϕ we regularize the kernel K at level ε by ( ) x y K ε (x, y) = K(x, y)ϕ. ε Notice that K ε (x, y) = K(x, y) when x y > ε and K ε (x, y) = 0 when x y ε. 2 It follows easily that K is also a standard kernel with constants not depending on ε. Next we consider the operators associated to K ε T ε f(x) = K ε (x, y)f(y)dy. Notice that, for K ε (x, y) = K(x, y)χ {r: r >ε} ( x y ), T ε f(x) T ε f(x) K ε (x, y) K ε (x, y) f(y) dy ε 2 < x y <ε C f(y) dy CMf(x) (ε/2) n x y <ε where M denotes the usual Hardy-Littlewood maximal operator. Since M is bounded in L p for all < p we deduce that T ε is bounded in L p, uniformly on ε, if and only if the operators T ε are bounded in L p uniformly on ε. In the same manner, for > 2ε, we can define the kernels ( x y K ε, (x, y) = ϕ ε and the corresponding operators T ε, f(x) = ) ( ϕ ( x y K ε, (x, y)f(y)dy. )) K(x, y),

THE T THEOEM 3 Since > 2ε it follows that T ε, = T ε T therefore if T ε : L p L p is bounded for all ε > 0 then T ε, is bounded as well for all positive, ε such that > 2ε. Furthermore if T ε, is bounded in L 2 uniformly on ε and then T ε is bounded uniformly on ε. To see this let f L 2 with compact support and notice that the functions in L 2 with compact support are dense in L 2. Then T ε f(x) = K ε (x, y)f(y)dy + K ε (x, y)f(y)dy, x y x y > and the second term tends to 0 as. Therefore we can choose > 0 large enough such that T ε (f) 2 C f 2. Hence by density this operator uniquely extends to a bounded operator in L 2 with the same bounds. Now note that for f L 2 all x n, by Hölder s inequality, ( ) /2 ( ) /2 K ε (x, y) f(y) dy K ε (x, y) 2 dy f(y) 2 C, because K ε (x, y) 2 is integrable at infinity. Therefore by density we deduce that there exists a sequence (by passing to a subsequence if necessary) of f n L 2 with compact support such that f n f in L p and f n f a.e. Hence by the dominated convergence theorem, T ε (f n )(x) T ε (f)(x) and hence T ε is bounded in L 2 uniformly on ε. Therefore, if T ε χ Q 2 Cm(Q) /2 and T ε χ Q 2 Cm(Q) /2 for all cubes Q uniformly on ε, we also have that T ε, χ Q 2 C m(q) /2 and T ε, χ Q 2 C m(q) /2 for all cubes Q uniformly on ε and. The proof will be complete if we show that the above condition implies that T ε, : L 2 L 2 is bounded uniformly on ε and. Notice that T ε, : L p L p is bounded for p with norm depending on ε and. For f L, T ε, (f)(x) K ε, (x, y) f(y) dy f K ε, (x, y) dy C f ε< x y < C f ( ε ) n, x y ndy and analogously for the L bound. Hence it is enough to prove the following.

4 V. CHOUSIONIS AND X. TOLSA Theorem 2 (educed restatement of Theorem ). Let the CZO T : L p L p, bounded for some p, and associated with a kernel K such that K C 0 and K(x, y) = 0 whenever x y for some > 0. If furthermore () Tχ Q 2 C m(q) /2 and (2) T χ Q 2 C m(q) /2 for all cubes Q, then T L 2 L 2 C 2 with C 2 depending on C but independent of C 0 and. 2. The operators Q Let D denote the family of dyadic cubes of n. For Q D and f L loc we define { 0 if x / Q, Q f(x) = m P f m Q f if x P and P is a son of Q, where m Q f is the average of f on Q. The following proposition gathers several elementary but useful properties of the operators Q. Proposition 3. For Q, D and f L loc, () supp Q f Q, (2) Q is constant in every son of Q, (3) Q f = 0, (4) Q f, g = 0 whenever Q, (5) Q Q = Q. (6) Q : L 2 L 2 is bounded and Q = Q. For simplicity from now on we will denote 2 :=. Proposition 4. If f L 2 then f = Qf, the convergence is unconditional in L 2 and moreover f 2 = Qf 2. Proof. The essential step consists in showing that Q f 2 f 2. () To see this, let F D be finite and set g = f Q F Qf. Then g Q F Qf because by (4),(5) and (6) of Proposition 3: f Q F Q f, Q F Q f = Q F f, Q f Q F Q f, Q f = 0. Hence f 2 = g 2 + Q F Qf 2 and Q f 2 f 2. Q F Therefore since this holds for any finite subset F D, () follows.

THE T THEOEM 5 Now we order D = {Q, Q 2,... } and we have to prove that lim m m i= Q i f = f in L 2. We first see that the sequence is convergent because n 2 n Qi f = Qi f 2 0 i=m i=m as n since i Q i f 2 <. Let A = k N A k where A k are the classes of functions with compact support such that they are constant on some dyadic cube in D k and furthermore they have zero mean on each orthant. Then we notice that the result is true for the class of functions A and furthermore A is dense in L 2. Let ε > 0. Then for all f L 2, there exists a function g A such that f g < ε. We write m i= Q i f = m i= Q i (f g)+ m i= Q i g and we deduce using () that m Qi f f m m 2 Qi g g + g f + Qi (g f) i= i= i= ( m m ) /2 = Qi g g + g f + Qi (g f) 2 i= i= m Qi g g + 2 g f i= m 2ε + Qi g g, i= and the proof is complete after letting m. We will prove that 3. Proof of the T theorem Tf, g C f g, (2) for f = Q F Q f, g = Q F 2 Q g for finite F, F 2 D and a constant C not depending on f or g. Then by Proposition 4, (2) holds for all f, g L 2 and Theorem 2 follows, taking into account that T L 2 L2 < by assumption. We have Tf, g = T Q f, g. Q F, F 2 We then observe the following two easy facts: first, and, second, Q f 2 Q f 2 m(q) (3) T( Q f) C Q f. (4)

6 V. CHOUSIONIS AND X. TOLSA To see (3), suppose that Q = 2n i= P i where P i are the sons of Q and let c i = m Pi f m Q f. Then, Q f 2 = Q f 2 = and because Q f = maxc i, we get 2 n i= m(p i )c 2 i = m(q) 2 n 2 n i= 2 nm(q) Qf 2 Qf 2 m(q) Q f 2. Using assumption () of Theorem 2, (4) follows analogously. In the following we prove an auxiliary lemma which deals with the case when two cubes are far each other. Lemma 5. Let two functions ϕ Q, ψ be such that suppϕ Q Q, suppψ, ϕ Q, ψ L, ϕ Q = 0 and d(q, suppψ ) l(q). Then, { l(q) C δ Tϕ Q, ψ d(q,suppψ ϕ ) δ+n Q ψ (5) C ϕ Q ψ. Proof. Let x Q be the center of Q. Using that K is a standard kernel and ϕ Q = 0 we get TϕQ, ψ ( ) = K(x, y)ϕ Q (x)ψ (y)dx dy ( ) = (K(x, y) K(x Q, y))ϕ Q (x)ψ (y)dx dy x x Q C δ x Q, x y ϕ Q(x) ψ n+δ (y) dxdy y suppψ l(q) δ C d(q, suppψ ) ϕ Q δ+n ψ. To complete the proof of the lemma notice that for x Q and y, Therefore, x Q y x y + l(q) x y + d(q, suppψ ) 2 x y. x x Q δ x Q, x y ϕ Q(x) ψ n+δ (y) dxdy y suppψ Cl(Q) δ ϕ Q ψ C ϕ Q ψ. x Q y >l(q) c 2 i, x Q y n+δdy

x Q y >l(q) THE T THEOEM 7 The last inequality follows because x Q y >l(q) x Q d l(q) δ. This follows by y n+δ integrating on annuli, for instance, x Q y n+δdy x Q y n+δdy i=0 i=0 Cl(Q) δ. Now we do the following splitting: Q F, F 2 T Q, g = B(x Q,2 i+ l(q))\b(x Q,2 i l(q)) (l(q)2 i+ ) n (l(q)2 i ) n+δ = Q F, F 2, l(q) + = S + S 2. Q F, F 2, l(q)> 2n l(q) δ ( i=0 T Q, g 2 δ T Q, g We are going to bound S, the boundedness of S 2 follows analogously. We consider the following three cases for Q, D such that l(q) : () d( i P i, Q) l(q) γ γ where the P i s are the sons of, (2) d(q, ) l(q) γ γ, (3) d(q, i P i ) l(q) γ and Q, where γ = δ. 2(δ+n) We will use the notation (Q, ) () if Q, satisfy () and so on. We have S T Q f, g + T Q f, g + T Q f, g (Q,) () (Q,) (2) and we will bound each of the three previous terms separately. 3.. Estimates for (Q, ) (). First observe that T Q f, g Q f g ) i ( ) n/2 l(q). (6) To see this, let ψ = gχ 3Q c and ψ 2 = gχ 3Q.

8 V. CHOUSIONIS AND X. TOLSA Since d(suppψ, Q) l(q), by Lemma 5, Hölder s inequality and (3) we obtain T Q f, ψ C Q f ψ C Q f g C Q f m(q) /2 g m() /2 ( ) n/2 l(q) = C Q f g. Furthermore, by (3), Hölder s inequality and (4) T Q f, ψ 2 C g T Q f Therefore (6) is proved. Lemma 6. We have (Q,) () 3Q C g m() /2 m(3q) /2 T Q f C g m() /2 m(q) /2 Q f ( ) n/2 l(q) = C Q f g. Q f g ( ) n/2 l(q) C f g. Proof. By Cauchy-Schwartz, ( ) n/2 l(q) Q f g Q f g (Q,) () :(Q,) () ( ) /2 Q f 2 Thus we only need to show that :(Q,) () f :(Q,) () ( ) 2 n/2 l(q) g ( ) n/2 l(q) :(Q,) () g ( ) 2 n/2 l(q) g /2 ( ) 2 n/2 l(q) /2 g. (7). /2

We start with the following trick: we rewrite ( ) n/2 l(q) g = :(Q,) () THE T THEOEM 9 :(Q,) () g ( ) n ε l(q) 2 ( ) ε l(q) 2 where ε is some small number to be chosen later. Now we apply Cauchy-Schwartz and we get ( ) 2 /2 n/2 l(q) g :(Q,) () ( ) n ε l(q) g 2 /2 ( ) ε l(q). :(Q,) () :(Q,) () For (Q, ) () we have that d( i P i, Q), where the P i s are the sons of. Now notice that for a fixed cube Q D, and k N, { D : (Q, ) () and = 2 k l(q)} C where C only depends on n, the dimension of the space. If Q is the dyadic cube which contains Q and has length 2 k l(q), then C is smaller than (3 n ) 2 which is the cardinality of the set of the neighbors of all neighbors of Q. Thus, :(Q,) () ( l(q) ) ε = k N :(Q,) (), =2 k l(q) ( l(q) Furthermore, by Fubini, ( ) n ε l(q) g 2 = g 2 D :(Q,) () ) ε = C 2 kε C. k N Q:(Q,) () ( ) n ε l(q). Hence in order to prove (7) and thus settle the proof, we only need to show that ( ) n ε l(q) C. (8) Q:(Q,) () To prove this we split again: ( ) n ε l(q) = Q:(Q,) () k N = k N Q:(Q,) (), l(q)=2 k 2 kε Q:(Q,) (), l(q)=2 k ( l(q) ) n ( ) ε l(q) ( ) n l(q).

0 V. CHOUSIONIS AND X. TOLSA Let d k = 2 2 kγ, and for A n denote Then for D fixed N dk (A) = {x : there is a A such that d(a, y) d k }. {Q D : (Q, ) (), l(q) = 2 k } N dk ( i P i ), where P i are the sons of. Therefore, ( ) n l(q) 2 kε = 2 kε k N Q:(Q,) (), l(q)=2 k k N k N C k N Q:(Q,) (), l(q)=2 k m(q) m() 2 kεm(n d k ( i P i )) m() 2 kεd k n n C k N 2 k(ε γ). Hence if we choose ε = γ/2 the proof of (8) is complete and we are done. Combining (6) and Lemma 6 we derive the desired estimate T Q f, g C f g. (Q,) () 3.2. Estimates for (Q, ) (2). In this subsection we are going to show that (Q,) (2) T Qf, g f g. We start with an auxiliary lemma. As usual if I is an index set, l 2 (I) = {(x i ) i I : x i and i I x2 i < }. Lemma 7 (Shur s lemma). Let I be some set of indices, and for each i I a number w i > 0. Suppose that, for some constant a 0, the matrix {T i,j } i,j I satisfies T i,j w j a w i for each i and j T i,j w i a w j for each j. i Then, the matrix {T i,j } i,j I defines a bounded operator in l 2 (I) with norm a. Proof. Let {x i } i I l 2 (I), and set y i = j I T i,jx j. We intend to show that i y i 2 a 2 i x i 2. So we write T i,j x j = ( T i,j /2 w /2 )( j Ti,j /2 w /2 j x j ),

THE T THEOEM and then by Cauchy-Schwartz, ( ) ( y i 2 T i,j w j j I j I T i,j w j x j 2 ) a w i j I T i,j w j x j 2. Summing on i and interchanging the sums, we get y i 2 a T i,j w i w j x j 2 a 2 i j i j x j 2. For Q, (2) we have that d(q, ) l(q) hence by Lemma 5 and Cauchy- Schwartz we have T Q f, g C l(q)δ d(q, ) Qf δ+n g C l(q)δ m() /2 d(q, ) δ+nm(q)/2 Q f g. We denote D(Q, ) = l(q) + + d(q, ). It then follows that l(q) δ d(q, ) C l(q)δ/2 δ/2. (9) n+δ D(Q, ) n+δ To prove (9) we consider two cases. If d(q, ) then D(Q, ) 3d(Q, ) and furthermore l(q) δ l(q) δ/2 δ/2 hence (9) follows. If d(q, ) then D(Q, ) 3 and recalling that for (Q, ) (2), d(q, ) l(q) γ γ, as γ = l(q) δ d(q, ) n+δ δ. Therefore, 2(n+δ) l(q) δ l(q) γ(n+δ) = l(q)δ γ(n+δ) γ(n+δ) ( γ)(n+δ) (n+δ) C l(q)δ/2 δ/2 D(Q, ) n+δ, T Q f, g C l(q)δ/2 δ/2 D(Q, ) δ+n m(q)/2 m() /2 Q f g. (0) Our goal now is to apply Schur s Lemma. To this end, consider the matrix (T Q, ) (Q,) (2) defined by T Q, = l(q)δ/2 δ/2 D(Q, ) δ+n m(q)/2 m() /2.

2 V. CHOUSIONIS AND X. TOLSA Applying Cauchy-Schwartz in l 2 (I), where I is the index set associated to the set {(Q, ) (2)} we get, T Q, Q f g (Q,) (2) (Q,) (2) g (T Q, Q f ) 2 (Q,) (2) /2 (T Q, Q f ) 2. (Q,) (2) /2 g 2 /2 () Hence we only need to show that (T Q, Q f ) 2 C Q f 2. (2) (Q,) (2) (Q,) (2) For Q D let w Q = m(q) /2, by Schur s lemma in order to prove (2) it is enough to show that :(Q,) (2) T Q, w Cw Q (3) and T Q, w Q Cw. (4) Q:(Q,) (2) For (3), :(Q,) (2) T Q, w k N :=2 k l(q) = l(q) n/2 k N l(q) δ/2 2 k 2l(Q) δ δ/2 l(q) n/2 n D(Q, ) δ+n 2 k δ 2 l(q) δ :=2 k l(q) n D(Q, ) n+δ.

THE T THEOEM 3 Let x Q be the center of Q. Then for x, x Q x + C(l(Q) + + d(q, )) = CD(Q, ) and we obtain n D(Q, ) = n+δ :=2 k l(q) :=2 k D(Q, ) n+δdx l(q) C dx :=2 k ( x Q x + ) n+δ l(q) C ( x Q x + 2 k l(q)) n+δdx ( = C dx x x Q 2 k l(q) ( x Q x + 2 k l(q)) n+δ ) + ( x Q x + 2 k l(q)) dx n+δ For I, I x x Q 2 k l(q) :=2 k l(q) = C(I + I 2 ). x x Q >2 k l(q) (2 k l(q)) n+δ dx (2k l(q)) n (2 k l(q)) n+δ = (2k l(q)) δ, and for I 2 as usual after splitting the set x Q x > 2 k l(q) in annuli and integrating we get I 2 x x Q >2 k l(q) x Q x dx C n+δ (2 k l(q)). δ Therefore, n D(Q, ) C n+δ (2 k l(q)), δ and by (3.2), :(Q,) (2) T Q, w Cl(Q) n/2 k N 2 k δ 2 Cl(Q) n/2. The proof of (4) follows in a similar manner and it is omitted. 3.3. Estimates for (Q, ) (3). In this subsection we are going to show that T Qf, g f g. To this end we need the following discrete version of the famous embedding theorem of Carleson. Theorem 8. [Carleson s Embedding Theorem] Let σ be a adon measure on d. Let D be the dyadic lattice from d and let {a Q } be a family of non negative numbers. Suppose that for every cube D we have a Q c 2 σ(). (5) :Q

4 V. CHOUSIONIS AND X. TOLSA Then, every family of non negative numbers {w Q } satisfies w Q a Q c 2 sup w Q dσ(x). (6) Q x In particular, if f L 2 (σ), f σ,q 2 a Q c c 2 f 2 L 2 (σ), (7) where f σ,q = f dσ/σ(q) and c is an absolute constant. Q Proof. To prove (6), consider the characteristic function defined by χ(q, t) = if 0 < t < w Q, and 0 otherwise. Then, χ(q, t) a Q dt = χ(q, t) a Q dt. (8) For each t > 0, let w Q a Q = 0 Ω t = : w Q >t Notice that if χ(q, t) =, then Q Ω t, and thus χ(q, t) a Q Q. 0 :Q Ω t a Q For m, let I m D be the subfamily of the cubes Q Ω t such that l(q) 2 m, and let J m I m be the subfamily of maximal cubes from I m. Then we have a Q = lim a Q = lim a Q. m m :Q Ω t Q I m Q By the assumption (5), for all m, a Q c 2 and so J m Q J m J m σ() c 2 σ(ω t ), χ(q, t) a Q c 2 σ(ω t ). Since Ω t coincides with {x d : w (x) > t}, where w (x) = sup Q x w Q, by (8) we obtain w Q a Q c 2 σ(ω t ) = c 2 w (x) dσ(x). 0 To prove the estimate (7), we take w Q = f σ,q 2, and then from (6) we deduce f σ,q 2 a Q c 2 M σ,d f 2 L 2 (σ),

THE T THEOEM 5 where M σ,d is the Hardy-Littlewood maximal dyadic operator with respect to σ. From the L 2 (σ) boundedness (with absolute constants) of this operator, we obtain (7). We now observe that if Q is the son of that contains Q, g is constant in Q and g Q = m Q g m g := c,q (g). Therefore we write, Hence, g = χ \Q g + c,q (g)χ Q = χ \Q g + c,q (g) c,q (g)χ c Q. T Q f, g + T Q f, χ \Q g = A + A 2 + A 3. T Q f, c,q gχ c Q g + T Q f, c,q g We will finish the proof of Theorem 2 by showing that A i C f g for i =, 2, 3. The following lemma deals with A. Lemma 9. We have A = Proof. If ψ = χ \Q g then Q f, T (χ \Q g) C f g. d(q, suppψ ) l(q) γ γ. easoning as in the proof of (0) we get that, T Q f, χ \Q g C l(q)δ/2 δ/2 D(Q, ) δ+n m(q)/2 m() /2 Q f 2 g 2, and following the same proof as in case (2) we get that, Q f, T (χ \Q g) C f g. We know turn our attention to A 2. Lemma 0. We have A 2 = c,q (g) T Q f, χ c Q C f g. (9)

6 V. CHOUSIONIS AND X. TOLSA Proof. Using (3) of Proposition 3 we have that for x c Q and y Q denoting the center of Q, K(x, y Q ) Q f(y)dy = 0, since K is a standard kernel, T Q f(x) = (K(x, y) K(x, y Q )) Q f(y)dy K(x, y) K(x, y Q ) Q f(y) dy Q C Q y y Q δ x y Q n+δ Qf(y) dy l(q) δ C Q f x y Q n+δ. (20) Therefore, as x y Q d( c Q, Q) l(q)γ γ for x c Q, T Q f, χ c Q C Q f C Q f c Q l(q) δ x y Q n+δdx x y Q l(q) γ γ l(q) δ C Q f l(q) γδ ( γ)δ ( ) δ( γ) l(q) C Q f m(q) /2, l(q) δ x y Q n+δdx where we used integration on annuli and Hölder s inequality. Furthermore by (3), c,q (g) g g m() /2. Thus, c,q (g) T Q f, χ c Q C Q f g ( ) δ( γ) ( ) /2 l(q) m(q), m()

and so, for ε = δ( γ) c,q (g) T Q f, χ c Q C C D g D Q f g Q: Q:Q THE T THEOEM 7 ( ) δ( γ) ( ) /2 l(q) m(q) m() Q f ( ) δ( γ) ( ) /2 l(q) m(q) m() ( ) /2 C g 2 ( Q f D D Q:Q = C g ( ( ) ε/2 ( l(q) l(q) Q f D Q:Q ( [ ( ) ][ ε l(q) C g Q f 2 ( l(q) Q:Q ) δ( γ) ( m(q) m() ) ε/2 ( m(q) m() ( l(q) ) /2 ) 2 ) /2 ) 2 /2 ) ]) ε /2 m(q), m() where we used twice Cauchy-Schwartz. For a cube D let D k () = {Q D : Q, l(q) = 2 k }. We then notice that, ( ) ε l(q) m(q) m() = ( ) ε l(q) m(q) m() Q:Q k N = k N k () 2 kε k () Therefore by the previous two estimates and Fubini, ( c,q (g) T Q f, χ c Q C g D: Q m(q) m() = 2 kε C. k N D Q:Q ( = C g Q f 2 C f g. The last inequality follows, because as previously ( ) ε l(q) = ( l(q) k N : Q, =2 k l(q) ( ) ) ε /2 l(q) Q f 2 : Q ( ) ) ε /2 l(q) ) ε = ( ) k C. k N 2 ε /2

8 V. CHOUSIONIS AND X. TOLSA Given a cube Q D let ( i ) i N be the increasing sequence of cubes in D which strictly contain Q. ecall that iq denotes the son of i which contains Q and notice that Q = Q, i+ Q = i and since g = F 2 f and F 2 is a finite set of cubes, there exists some i 0 and some F 2 F 2 such that Hence, by Proposition 3, and m i0 g = 0. Furthermore, Q c,q (g) = i N suppg i0 = F2. i0 g = F 2 g = 0 c i,qg = i N(m iq g m i g) By the previous observations this sum is telescopic and since m i0 = 0 we get c,q (g) = m Q g. Q Now we can estimate: T Q f, c,q (g) = (Q,): Q Furthermore, c,q (g) T Q f, = (Q,): Q = (Q,): Q = S 3 S 4. (Q,): Q = c,q (g) T Q f, ( ) c,q (g) T Q f, Q = m Q (g) Q f, T. c,q (g) T Q f, m Q (g) Q f, T (Q,): Q, (Q,)/ (3) (Q,): Q, (Q,)/ (3) c,q (g) T Q f, c,q (g) T Q f, Therefore A 3 S 3 + S 4, and in order to finish the proof we only need to bound the terms S 3 and S 4. Lemma. For S 4 as in (2) we have S 4 C f g. (2)

THE T THEOEM 9 Proof. We have, S 4 (Q,): Q, d(q,) l(q) γ γ (Q,): Q, d(q,) l(q) γ γ c,q (g) T Q f, c,q (g) T Q f. We write T Q f = (T Q f)χ 3Q + (T Q f)χ (3Q) c, as in (20) we have that for x (3Q) c and y Q being the center of Q, l(q) δ T Q f(x) C Q f x y Q n+δ, By integrating on annuli and using Hölder s inequality we conclude that l(q) δ (T Q f)χ (3Q) c C Q f x y Q n+δdx C Qf m(q) /2. Again by Hölder s inequality and (4), x y Q >l(q) (T Q f)χ 3Q C T Q f m(q) /2 C Q f m(q) /2. On the other hand, by (3), c,q (g) g g m() /2. So by all the previous estimates and Cauchy-Schwartz, S 4 C ( m(q) g Q f m() C ( 2)/2 ( g D = C f g. ) /2 Q f 2 ) /2 Finally we need to bound S 3. By (5) and (6) of Proposition 3, S 3 = m Q g Q f, T = m Q g f, Q (T ) = Q F m Q g f, Q (T ) = f, Q F m Q g Q (T ).

20 V. CHOUSIONIS AND X. TOLSA Notice that the functions g Q = m Q g Q (T ) are orthogonal, and thus S 3 f, m Q g Q (T ) f 2 m Q g Q (T ) Q F Q F = f 2 Q F m Q g 2 Q (T ) 2. The following lemma settles the case for S 3. Lemma 2. We have m Q g 2 Q T 2 C g 2. (22) Proof. By Theorem 8 it suffices to prove that Q T 2 Cm() (23) Q, for all D. For Q, D, Q and then Q ((T m (T ))χ ) = 0. (24) To see this let Q be the son of Q which contains. Then, for h = (T m (T ))χ h = T m()m (T ) = 0, hence for x Q h(x) = m Q h m Q h = 0. Now by Proposition 4, since (T m (T ))χ L 2, using the fact that Q (m Q (T )) = 0 and (24), we get Q T 2 = Q ((T m (T ))χ ) 2,Q,Q = Q ((T m (T ))χ ) 2 = (T m (T ))χ 2. Thus in order to complete the proof it is enough to show that for all D, T m (T ) 2 C. (25) m() This is equivalent to saying that T BMOd 2, the dyadic BMO space. Lemma 3. If T χ 2 2 Cm(), T m (T ) 2 C. m() 2

Proof. We have THE T THEOEM 2 T m (T ) = T χ 2 m (T χ 2 ) + T χ (2) c m (T χ (2) c) We first deal with A: = A + B. A 2 = T χ 2 m (T χ 2 ) 2 T χ 2 2 + m() m (T χ 2 ) 2 Cm() + m() m (T χ 2 ) 2. But by Hölder s inequality, ( ) 2 m (T χ 2 ) 2 T χ 2 m() ( ( ) ) /2 2 T χ 2 2 m() /2 m() Hence, C m() T χ 2 2 C m() = C. m() A 2 Cm(). Now we will estimate B 2. The first step is to show that for all x, x 2, T χ (2) c(x ) T χ (2) c(x 2 ) C. (26) To see this, T χ (2) c(x ) T χ (2) c(x 2 ) K(y, x ) K(y, x 2 ) dy (2) c x x 2 C (2) x c y n+δdy C δ (2) x c y n+δdy C δ δ = C, where in the last inequality we used integration in annuli as usual.

22 V. CHOUSIONIS AND X. TOLSA Now, B 2 = T χ (2) c(x ) 2 T χ (2) c(x 2 )dx 2 dx m() = T χ (2) c(x )dx 2 2 T χ (2) c(x 2 )dx 2 dx m() m() 2 = m() 2 (T χ (2) c(x ) T χ (2) c(x 2 ))dx 2 dx ( T χ m() 2 (2) c(x ) T χ (2) c(x 2 ) )m()dx 2 Cm(). For the last two inequalities we used Hölder s inequality and (26). This finishes the proof of Lemma 3 and the proof of Lemma 2 as well. Hence the proof of Theorem 2 is completed. emark 4. Expressions of the form Π h g = m Q g Q h are called paraproducts of g associated to h. In our case Πg = m Qg Q (T ) is the paraproduct associated to T (). Using Lemma 2 and orthogonality it follows that Πg C g.