Duality of multiparameter Hardy spaces H p on spaces of homogeneous type

Duality of multiparameter Hardy spaces H p on spaces of homogeneous type Yongsheng Han, Ji Li, and Guozhen Lu Department of Mathematics Vanderbilt University Nashville, TN Internet Analysis Seminar 2012

Outline 1 Definitions and Tools 2 Sequence Spaces 3 Some Important Lemmas 4 Relating Sequence Spaces to H p and CMO p 5 Proof of Duality of s p and c p 6 More About the Proof of the Min-Max Inequality

Quasi-Metrics A quasi-metric ρ on a set X is a function from X X to [0, ) such that the following three conditions hold: 1 ρ(x, y) = 0 if and only if x = y. 2 ρ(x, y) = ρ(y, x) for all x, y X 3 There exists a constant A 1 such that for all x, y, z X, we have ρ(x, y) A[ρ(x, z) + ρ(z, y)]. Thus, a quasi-metric is like a metric, except that the normal triangle inequality is replaced by a triangle inequality that holds up to a multiplicative constant.

Spaces of Homogeneous Type Let 0 < θ 1. A space of homogeneous type is a collection (X, ρ, µ) θ, where X is a set, ρ is a quasi-metric on X, and µ is a nonnegative Borel regular measure on X. We require that for all r such that 0 < r < diam(x ), we have that µ(b(x, r)) r, where B(x, r) is the ball of radius r centered at x X. Lastly, we require that there is a constant C 0 > 0 such that for all x, x, y X, we have ρ(x, y) ρ(x, y) C 0 ρ(x, x ) θ [ρ(x, y) + ρ(x, y)] 1 θ. Note that if C 0 = 1 and θ = 1, this last condition is just the triangle inequality.

Slight Generalization Instead of assuming that µ(b(x, r)) r, we could have assumed that µ(b(x, r)) r d, where d > 0. However, in [MS79], Macias and Segovia proved that for d > 0, one can always find a quasi-metric ρ giving the same topology as ρ such that µ(b(x, r)) r, so throughout the paper the authors assume that d = 1.

Example From the definition, it is clear that R is a space of homogeneous type. By the result of [MS79], it is clear that R d is a space of homogeneous type for any natural number d.

Difficulties with spaces of Homogeneous Type The main difficulty of working on spaces of homogeneous type is that there are no translations or dilations, and no Fourier transform. Thus, the challenge is to find methods to prove results without using these basic tools. The paper in question relies heavily on Littlewood-Paley theory and a discrete Calderón reproducing formula for spaces of homogeneous type.

Dyadic Decomposition There exists a collection {Q k a X : k Z, a I k } of open subsets, called cubes, where I k is some index set, as well as constants C 1, C 2 > 0 such that 1 For each fixed k, we have µ(x \ a Qa k ) = 0, and Qa k Qb k = if a b, 2 For any a, b, k, l with l k, either Q l b Qk a or Q l b and Qk a are disjoint, 3 For each pair of k and a, there is a unique b such that Q k a Q l b, 4 diam(q k a ) C 1 (1/2) k 5 each Q k a contains some ball of radius C 2 (1/2) k.

More on Dyadic Decomposition Fix some large positive integer J. Let k Z and τ I k. We denote by N(k, τ) the number of cubes Qα k+j that are contained in Qτ k, and we denote such a cube by Qτ k,v, where 1 v N(k, τ).

Approximations to the Identity I A sequence {S k } k Z is said to be an approximation to the identity of order ɛ, where 0 < ɛ θ, if there is some constant C such that the kernel S k (x, y) of S k satisfies 1 2 kɛ S k (x, y) C (2 k + ρ(x, y)) 1+ɛ, 2 For ρ(x, x ) (1/2A)(2 k + ρ(x, y)), we have ( ρ(x, x S k (x, y) S k (x ) ) ɛ 2 kɛ, y) C 2 k + ρ(x, y) (2 k + ρ(x, y)) 1+ɛ 3 For ρ(y, y ) (1/2A)(2 k + ρ(x, y)), we have ( ρ(y, y S k (x, y) S k (x, y ) ) ɛ 2 kɛ ) C 2 k + ρ(x, y) (2 k + ρ(x, y)) 1+ɛ

Approximations to the Identity II 4 For ρ(x, x ), ρ(y, y ) (1/2A)(2 k + ρ(x, y)) we have C S k (x, y) S k (x, y ) S k (x, y) + S k (x, y ) ( ρ(x, x ) ) ɛ ( ρ(y, y ) ) ɛ 2 kɛ 2 k + ρ(x, y) 2 k + ρ(x, y) (2 k + ρ(x, y)) 1+ɛ 5 S k (x, y) dµ(y) = X X S k (x, y) dµ(x) = 1

Analogy As k increases, the kernels S k (x, y) become more concentrated around the diagonal x = y. The approximation to the identity S k (x, y) is analogous to the Poisson kernel P 2 k (x y) on the upper half plane.

One-parameter Test Functions I Fix β, γ, r > 0. A function f defined on X is said to be a test function of type (β, γ) centered at x 0 X with width r if there is a constant C > 0 such that f satisfies the following conditions: 1 r γ f (x) C (r + ρ(x, x 0 )) 1+γ 2 If ρ(x, x ) 1 2A (r + ρ(x, x 0)) then ( ρ(x, x f (x) f (x ) ) β r γ ) C r + ρ(x, x 0 ) (r + ρ(x, x 0 )) 1+γ 3 X f (x) dµ(x) = 0.

One-parameter Test Functions II If f is such a test function, we say f G(x 0, r, β, γ) and we define its norm to be the infimum over the set of all C for which conditions 1 and 2 hold.

Two-Parameter Test Functions I For i = 1, 2, fix γ i, β i, r i > 0 and let (x 0, y 0 ) X X. We say that a function f defined on X X is a test function of type (β 1, β 2, γ 1, γ 2 ) centered at (x 0, y 0 ) and with widths (r 1, r 2 ) if there is a constant C > 0 such that 1 r γ 2 2 f (, y) G(x0,r 1,β 1,γ 1 ) C (r 2 + ρ(y, y 0 )) 1+γ, 2 2 For ρ(y, y ) 1 2A (r 2 + ρ(y, y 0 )) we have f (, y) f (, y ) G(x0,r 1,β 1,γ 1 ) ( ρ(y, y ) ) β2 r γ 2 2 C r 2 + ρ(y, y 0 ) (r 2 + ρ(y, y 0 )) 1+γ, 2 3 Condition 1 should hold with the roles of x and y reversed. 4 Condition 2 should hold with the roles of x and y reversed.

Two-Parameter Test Functions II We denote this space of test functions by G(x 0, y 0 ; r 1, r 2 ; β 1, β 2 ; γ 1, γ 2 ) and define the norm in this space to be the smallest C such that the above definition holds.

The Space G It is not difficult to see that no matter which x 0 and y 0 we choose, we get the same space of functions with equivalent norm. We choose some fixed (x 0, y 0 ) and let G(β 1, β 2 ; γ 1, γ 2 ) = G(x 0, y 0 ; 1, 1; β 1, β 2 ; γ 1, γ 2 ). If 0 < β 1, β 2, γ 1, γ 2 < θ, we define the space G(β 1, β 2 ; γ 1, γ 2 ) to be the completion of G(θ, θ; θ, θ) in G(β 1, β 2 ; γ 1, γ 2 ). Note that G(β 1, β 2 ; γ 1, γ 2 ) is a Banach space.

Littlewood-Paley We define the Littlewood-Paley operators D k as S k S k 1, where S is an approximation of the identity. We can now define the Littlewood-Paley-Stein square function by g(f )(x 1, x 2 ) = j= k= D j D k (f )(x 1, x 2 ) 2 It can be show that, for f L p for 1 < p <, we have g(f ) p f p. 1/2.

H p Spaces Let {S k } be an approximation to the identity of order θ. 1 Suppose that 1+θ < p 1 and 1 p 1 < β i, γ i < θ. Then we define the Hardy space H p (X X ) to be the set of all f ( G(β1, β 2, γ 1, γ 2 )) such that g(f ) L p (X X ) <. We define f H p (X X ) = g(f ) L p (X X )

The Carleson Measure Spaces CMO p Let β i, γ i, etc. be as before. The Carleson measure space CMO p (X X ) is the set of all f ( G(β 1, β 2, γ 1, γ 2 )) such that ( 1 sup D Ω µ(ω) 2 p 1 k1 D k2 (f )(x, y) 2 Ω χ Q k 1,v 1 Q k 1,v 1 τ 1 Q k 2,v 2 τ 2 Ω (x)χ k τ 1 Q 2,v 2 (y) dµ(x)dµ(y) τ 2 ) 1/2 is finite. The above expression defines the norm in CMO p.

Some Intuition The above definition can be though of as analogous to the single parameter case on R, where φ BMO if and only if u 2 y dx dy is a Carleson measure, where u is the harmonic extension of φ to the upper half plane.

Classical Calderón Reproducing Formula f (x) = 0 (ψ t ψ t f )(x) dt t where ψ t (x) = t n ψ(x/t). Also, we assume that ψ is real, 2 dt radial, and well behaved, and that ( ψ(tx)) = 1 for all 0 t x 0. This is not difficult to prove by using the Fourier transform.

Discreet Calderón Reproducing Formula Let R denote an arbitrary dyadic rectangle of the form Q k 1,v 1 τ 1 Q k 2,v 2 τ 2. Let the notation be the same as above. For each R, choose a point (x, y ) R. Then there are families of linear operators { D k } and {D k } such that f (x, y) = R = R µ(r) D k1 Dk2 (x, y, x, y )D k1 D k2 (f )(x, y ) µ(r)d k1 D k2 (x, y, x, y )D k1 D k2 (f )(x, y ),

The Main Result The dual space of H p is CMO p if 2 2 + θ = p 0 < p 1. (Recall that θ is a parameter measuring the regularity of the space of Homogeneous type. The higher θ, the more regular the space is

Sequence Spaces The sequence space s p is defined to be the space of all complex valued sequences {λ k Q 1,v 1 } such that τ 1 Q k 2,v 2 τ 2 [ ( λ s p = λr χ R ( ) ) ] 1/2 2 Here χ R = µ(r) 1/2 χ R. R <. L p

Sequence Spaces The sequence space c p is the space of all sequences {t k Q 1,v 1 } such that τ 1 Q k 2,v 2 τ 2 t c p = sup Ω (µ(ω) 1 (2/p) 1/2 ( t R χ(x, y)) dµ(x)dµ(y)) 2 <. Ω R Ω

Notice the Similarity [ ( λ s p = λr χ R ( ) ) ] 1/2 2 <. R L p 1/2 f H p = D j D k (f )(x 1, x 2 ) 2 j= k= 1/2.

Notice the Similarity t c p = sup Ω ( 1 µ(ω) 2 p 1 1/2 ( t R χ(x, y)) dµ(x)dµ(y)) 2. Ω R Ω f CMO p = ( 1 sup Ω µ(ω) 2 p 1 D k1 D k2 (f )(x, y) 2 χ R (x, y) dµ(x)dµ(y) Ω R Ω ) 1/2

Duality The authors show that the dual of s p is c p. To do this, they use inequalities involving the Hardy-Littlewood maximal function and also use the duality properties of certain weighted l 2 spaces related to both s p and c p.

Min-Max for Hardy Spaces I Let all notation be the same as in definition 2.6. Let {P k } be another approximation to the identity of order θ, and let E k denote the corresponding Littlewood-Paley operators. Suppose that 1 1+θ < p 1 and 1 p 1 < β i, γ i < θ. Then there is a constant C > 0 such that for all f ( G(β1, β 2, γ 1, γ 2 )), we have

Min-Max for Hardy Spaces II ( k 1 = k 2 = τ 1 I k1 τ 2 I k2 N(k 1,τ 1 ) v 1 =1 N(k 2,τ 2 ) v 2 =1 sup z Q k 1,v 1 τ 1,w Q k 2,v 2 τ 2 D k1 D k2 (f )(z, w) 2 χ k Q 1,v 1 (x)χ k τ 1 Q 2,v 2 (y) τ 2 ( N(k 1,τ 1 ) N(k 2,τ 2 ) C k 1 = k 2 = τ 1 I k1 τ 2 I k2 v 1 =1 v 2 =1 ) 1/2 L p inf z Q k 1,v 1 τ 1,w Q k 2,v 2 τ 2 E k1 E k2 (f )(z, w) 2 χ k Q 1,v 1 (x)χ k τ 1 Q 2,v 2 (y) τ 2 ) 1/2 where the integration for the L p norm is respect to dµ(x)dµ(y). L p

Min-Max for Hardy Spaces III This lemma is very useful for several reasons. It lets one show that the definition of H p given does not depend on the choice of approximation to the identity. To see this, note that the left hand side is greater than the H p norm with operators D k, and the right side is less than C times the H p norm with operators E k. Also, the lemma in question allows one to approximate the integration defining the L p norm of the g function by a sort of Riemann sum. The lemma states that it does not matter whether we take the smallest or largest possible Riemann sum, we still get an equivalent norm. Of course, the sums in the lemma are infinite. As the coefficient k gets larger, we need to sum over smaller and smaller dyadic boxes Qτ k,v. This lemma allows us to relate Hardy space functions to sequences, which is crucial in proving duality.

Min-Max for CMO p I A very important theorem is the Min-Max comparison principle for CMO p. For ease of notation, let R denote an arbitrary dyadic rectangle Q k 1,v 1 τ 1 Q k 2,v 2 τ 2. If 2/(2 + θ) < p 1, then there is some constant C > 0 such that ( ) 1/2 1 sup µ(r) sup D Ω µ(ω) 2 p 1 k1 D k2 (f )(x, y) 2 R Ω (x,y) R ( C sup Ω 1 µ(ω) 2 p 1 µ(r) inf D k 1 D k2 (f )(x, y) 2 (x,y) R R Ω where R = Q k 1,ν 1 τ 1 Q k 2,ν 2 τ 2. ) 1/2

Min-Max for CMO p II In fact, one can use different approximations to the identity on the left and right side of the inequality, which allows us to see that the CMO spaces are well defined. As before, another reason this theorem is important is because it allows us to consider a sort of Riemann sum instead of integration in the definition of CMO. This allows us to relate the CMO spaces to sequences, which again is key to proving the duality.

Overview of Proof of Min-Max for CMO 1 Choose (x, y ) to be the point where D k1 D k2 (f )(x, y) is minimized in R. 2 Apply D k1 D k2 to both sides of the Discrete Calderón Reproducing Formula. 3 We obtain an expression that can be used to relate the left side of the inequality in the Min-Max Theorem to something resembling the right side. 4 Change the order of summation and carefully consider the various geometric quantities involved to prove the Min-Max Theorem.

The Lifting Operator Define the lifting operator { } S D (f ) = µ(q k 1,ν 1 τ 1 ) 1/2 µ(q k 2,ν 2 τ 2 ) 1/2 D k1 D k2 (f )(y 1, y 2 ), which maps elements of ( G(β 1, β 2 ; γ 1, γ 2 )) into sequences. Here y j is the center of Q k j,ν j τ j. Also, 0 < β j, γ j < θ.

The Projection Operator For any sequence, we define the projection operator by T D(s)(x 1, x 2 ) = k 1 = k 2 = τ 1 I k1 τ 2 I k2 N(k 1,τ 1 ) ν 1 =1 N(k 2,τ 2 ) ν 2 =1 s Q k 1,ν 1 τ 1 Q k 2,ν 2 τ 2 µ(q k 1,ν 1 τ 1 ) 1/2 µ(q k 2,ν 2 τ 2 ) 1/2 Dk1 Dk2 (x 1, x 2, y 1, y 2 ) The D k are the kernels of the operators associated with the D k by the Calderón reproducing formula.

Relating H p to the s p Let 1 1 + θ < p 1. For any f Hp (X X ), Also, for any s s p, S D (f ) s p C f H p. T D(s) H p C s s p. In addition, T D S D equals the identity on H p.

Relating CMO p to c p Let 2 2 + θ < p 1. For any f CMOp, Also, for any t c p, S D (f ) s p C f CMO p. T D(t) CMO p C t c p. In addition, T D S D equals the identity on CMO p.

A Word About the Proofs The proofs of these two statements are similar to the proof of the Min-Max inequalities. Also, note that the adjoint of S D is T D and the adjoint of T D is S D. The fact that (H p ) = CMO p follows fairly easily from the above statements about the lifting and projection operators, and from the fact that (s p ) = c p.

Overview of the Proof of Duality of s p and c p First, we need to show that s R t R s s p t c p. R Define the { sets Ω k = (x 1, x 2 ) : ( R ( s R χ R (x 1, x 2 )) 2) } 1/2 > 2 k } B k = {R : µ(ω k R) > 1 2 µ(r), µ(ω k+1 R) 1 2 µ(r) { } Ω k = (x, y) : M s (χ Ωk ) > 1 2 where M s is the strong maximal function.

Duality of s p and c p s R t R p/2 s R 2 p/2 1/p t R 2 R k R Bk R Bk µ( Ω k ) 1 p 2 p/2 p/2 1/p s R 2 1 t k R Bk µ( Ω k ) 2 R 2 p 1 R Ω k p/2 1/p µ( Ω k ) 1 p 2 s R 2 t c p k R Bk

1 2 R B k s R 2 R B k s R 2 µ(r) 1 1 2 µ(r) R B k s R 2 µ(r) 1 µ( Ω k \ Ω k+1 R) Ω k \Ω k+1 R B k ( s R χ R (x)) 2 dµ(x) 2 2(k+1) µ( Ω k \ Ω k+1 ) C2 2k µ(ω k ) Substitute back into the previous slide to see that 1 s t c p R t R C 2 kp µ( Ω k ) C 2 kp µ(ω k ) C s s p R k k Thus L C t c p.

Other Direction of Duality Now we have shown that c p (s p ), and we need to show the other direction. Define s i R = 1 when R = R i and s i r = 0 otherwise. Then s i R s p = 1. We can write s = {s R } = i s R i s i R i. For L (s p ), define a sequence t by t Ri = L(s i ). Then L(s) = R s Rt R. Need to show that t c p L.

l 2 (µ) Spaces For fixed Ω, define µ(r) = µ(r) µ(ω). Define s l 2 (µ) = ( R Ω Then (l 2 (µ)) = l 2 (µ). s R 2 µ(r) µ(ω) p 2 1 ) 1/2 This will allow us to estimate norms of sequences in c p by using l 2 (µ) duality.

l 2 (µ) Spaces I ( ) 1/2 1 t µ(ω) p 2 1 R 2 = µ(r) 1/2 t R l 2 (µ) R Ω = sup ( t s: s l 2 (µ) 1 R µ(r) 1/2 µ(r) )s R R Ω µ(ω) p 2 1 ( ) sup s: s l 2 (µ) 1 L χ R Ω (R) µ(r)1/2 s R µ(ω) p 2 1 sup L s: s l 2 (µ) 1 χ R Ω(R) µ(r)1/2 s R µ(ω) p 2 1 s p

l 2 (µ) Spaces II But by Hölder s Inequality, this is less than or equal to So t c p L. sup L s: s l 2 (µ) 1 ( R Ω ) 1/2 s R 2 µ(r) µ(ω) p 2 1 sup L s l 2 (µ) L. s: s l 2 (µ) 1

Some Words About the Min-Max Inequality The Min-Max Inequalities for H p and for CMO p are crucial to the paper. Recall: If 2/(2 + θ) < p 1, then there is some constant C > 0 such that sup Ω ( C sup Ω 1 µ(ω) 2 p 1 ( 1 µ(ω) 2 p 1 R Ω ) 1/2 µ(r) sup D k1 D k2 (f )(x, y) 2 (x,y) R µ(r) inf D k 1 D k2 (f )(x, y) 2 (x,y) R R Ω where R = Q k 1,ν 1 τ 1 Q k 2,ν 2 τ 2. ) 1/2

Almost Orthogonality Almost orthogonality of D j Dk : D j Dk (x, y) C2 k j ε 2 min(k,j)θ (2 min(k,j) + ρ(x, y)) 1+θ where 0 < ε < θ Basically, D j Dk (x, y) is small if k and j are far apart, or if x and y are far apart. Now we apply D k1 D k2 to both sides of the Calderon Reproducing Formula, where x and y are chosen such that D k1 D k2 (f )(x, y ) is minimized in R. Use the almost orthogonality to get an estimate.

Some Notation R = Q k 1,ν 1 τ 1 Q k 2,ν 2 τ 2, S R = sup (x,y) R D k1 D k2 (f )(x, y) 2 T R = inf (x,y) R D k1 D k2 (f )(x, y) 2 v(r, R ) and r(r, R ) and P(R, R ) are geometric quantities depending on µ and the rectangles R and R.

By using the estimate obtained earlier and analysis of the geometric quantities involved, we find that the left side of the Min-Max inequality, which is is bounded by 1 µ(ω) 2 p 1 1 µ(ω) 2 p 1 R Ω µ(r)s R R Ω R v(r, R )r(r, R )P(R, R )T R. To finish the proof, we want to show that this is bounded by sup Ω 1 µ(ω ) 2 p 1 R Ω µ(r)t R where Ω ranges over all open sets with finite measure.

Let Ω 0 = R Ω 3A2 R. For each R, we consider smallest j, k 0 such that 3A 2 2 j Q 1 3A2 2 k Q 2 intersects 3A2 R. Here R = Q 1 Q 2. We divide the R into different classes based on the values of j and k. We switch orders of summation. Divide the R into even more classes based on what fraction of their area lies in Ω 0. We estimate the geometric quantities v, r, and P for the various classes. Eventually, after analyzing many different cases, we get the estimate we want.

Yongsheng Han, Ji Li, and Guozhen Lu. Duality of multiparameter Hardy spaces Hp on spaces of homogeneous type. Ann. Sc. Norm. Super. Pisa Cl. Sci. (5), 9(4):645 685, 2010. Roberto A. Macías and Carlos Segovia. Lipschitz functions on spaces of homogeneous type. Adv. in Math., 33(3):257 270, 1979.