Davis-Garsia Inequalities for Hardy Martingales Paul F.X. M uller - PowerPoint PPT Presentation

Davis-Garsia Inequalities for Hardy Martingales Paul F.X. M¨ uller Johannes Kepler Universit¨ at Linz

Topics 1. Basic Examples 2. Maximal Functions 3. Davis Decomposition 4. Davis Garsia Inequalities

Complex analytic Hardy Spaces f ∈ L p ( T , X ) , T = { e iθ : | θ | ≤ π } , D = { z ∈ C : | z | < 1 } . The harmonic extension of f to the unit disk � π 1 − | z | 2 f ( z ) = 1 | z − e iα | 2 f ( e iα ) dα, z ∈ D . 2 π − π Define f ∈ H p ( T , X ) if f ∈ L p ( T , X ) and the harmonic extension of f is analytic in D .

Hardy Martingales H 1 ( T N , X ) T N the infinite torus-product with Haar measure d P . F k : T N → C is F k measurable iff x = ( x i ) ∞ F k ( x ) = F k ( x 1 , . . . , x k ) , i =1 Conditional expectation E k F = E ( F |F k ) is integration, � E k F ( x ) = T N F ( x 1 , . . . , x k , w ) d P ( w ) . An ( F k ) martingale F = ( F k ) is a Hardy martingale if ∈ H 1 ( T , X ) . y → F k ( x 1 , . . . , x k − 1 , y ) If ∆ F k ( x 1 , . . . , x k − 1 , y ) = m k ( x 1 , . . . , x k − 1 ) y then ( F k ) is called a simple Hardy martingale .

Example: Maurey’s embedding. Fix ǫ > 0 , w = ( w k ) ∈ T N . Put ϕ 1 ( w ) = ǫw 1 , and ϕ n ( w ) = ϕ n − 1 ( w ) + ǫ (1 − | ϕ n − 1 ( w ) | ) 2 w n . Then lim | ϕ n | = 1 and ϕ = lim ϕ n is uniformly dis- tributed over T . For any f ∈ H 1 ( T , X ) w ∈ T N F n ( w ) = f ( ϕ n ( w )) , is an integrable Hardy martingale with uniformly small increments � � sup E ( � F n � X ) = T � f � X dm and � ∆ F n � X ≤ 2 ǫ T � f � X dm. n ∈ N

Pointwise estimates for ∆ F n . Fix w ∈ T N , n ∈ N , z = ϕ n ( w ) , u = ϕ n − 1 ( w ) ∆ F n ( w ) = f ( ϕ n ( w )) − f ( ϕ n − 1 ( w )) . Cauchy integral formula � � ζ ζ � f ( z ) − f ( u ) = ζ − z − f ( ζ ) dm ( ζ ) . ζ − u T Triangle inequality | z − u | � � f ( z ) − f ( u ) � X ≤ T � f � X dm (1 − | u | )(1 − | z |

Example: Rudin Shapiro Martingales k =1 | c k | 2 ≤ 1 . Fix a complex sequence ( c n ) with � ∞ Define recursively: F 1 = G 1 = 1 and for w = ( w n ) ∈ T N F m ( w ) = F m − 1 ( w ) + G m − 1 ( w ) c m w m , G m ( w ) = G m − 1 ( w ) − F m − 1 ( w ) c m w m . Pythagoras for ( F m , G m ) and ( G m , − F m ) gives | F m +1 ( w ) | 2 + | G m +1 ( w ) | 2 = (1+ | c m +1 | 2 )( | F m ( w ) | 2 + | G m ( w ) | 2 ) . ( F n ) uniformly bounded and F m ( w ) − F m − 1 ( w ) = G m − 1 ( w ) c m w m ( F n ) is a simple Hardy martingale

The Origins I A. Pelczynski posed famous problems in “Banach Spaces of analytic functions and absolutely summing operators, (1977).” Does H 1 have an unconditional basis? Does there exist a subspace of L 1 /H 1 isomorphic to L 1 ? Does L 1 /H 1 have cotype 2? Are the spaces A ( D n ) and A ( D m ) not isomorphic when n � = m ? (Dimension Conjecture)

The Origins II Hardy martingales gave rise to the operators by which Maurey proved that H 1 has an unconditional basis; and to the isomorphic invariants by which Bourgain proved the dimension conjecture, that L 1 /H 1 has co- type 2 and that L 1 embeds into L 1 /H 1 . Pisier’s L 1 /H 1 valued Riesz products form a Hardy martingale that is strongly intertwined with Bourgain’s solutions and played an important role for the work of Garling, Tomczak-Jaegermann, W. Davis on Hardy martingale cotype and complex uniformly convexity of Banach spaces.

Maximal Functions estimate For any X valued Hardy martingale F = ( F k ), and any 0 < α ≤ 1 , ( � F k − 1 � α X ) is a non- negative submartingale and E (sup � F k � ) ≤ e sup E ( � F k � ) . k ∈ N k ∈ N

Davies Decomposition (PFXM) A Hardy martingale F = ( F k ) can be decomposed into Hardy martingales as F = G + B such that � ∆ G k � X ≤ C � F k − 1 � X , and ∞ � E ( � ∆ B k � X ) ≤ C E ( � F � X ) . k =1 Lemma If h ∈ H 1 0 ( T , X ) , z ∈ X there exists g ∈ H ∞ 0 ( T , X ) with � g ( ζ ) � X ≤ C 0 � z � X , ζ ∈ T and � z � X + 1 � � T � z + h � X dm. T � h − g � X dm ≤ 8

Proof of Lemma (Sketch) . Let { z t : t > 0 } denote complex Brownian Motion started at 0 ∈ D , and τ = inf { t > 0 : | z t | > 1 } . Define ρ = inf { t < τ : � h ( z t ) � X > C 0 � z � X } , A = { ρ < τ } . • By choice of ρ, � h ( z ρ ) � ≤ C 0 � z � X . • Doob’s projection generates the analytic function g ( ζ ) = E ( h ( z ρ ) | z τ = ζ ) , ζ ∈ T , and also the testing function which yields lower esti- mates for � T � z + h � X dm : p ( ζ ) = 1 q = e ln(1 − p )+ iH ln(1 − p ) , 2 E (1 A | z τ = ζ ) , 1 = p + | q | .

Basic Steps. • � T � z + h � X pdm ≥ (1 / 2 − 1 / (2 C 0 )) E ( � h ( z τ )1 A � X • � T � z + h � X | q | dm ≥ � x � X (1 − 3 P ( A )) • 1 = p + | q | . � T � z + h � X dm ≥ � x � X + (1 / 2 − δ ( C 0 )) E ( � h ( z τ )1 A � X • � T � h − g � X dm ≤ 2 E ( � h ( z τ )1 A � X ) Summing up: � � T � z + h � X dm ≥ � x � X + δ T � h − g � X dm

Sketch of Proof. Fix x ∈ T k − 1 . Put h ( y ) = ∆ F k ( x, y ) and z = F k − 1 ( x ) . Lemma yields a bounded analytic g with � � � z � X +1 / 8 T � h − g � X dm ≤ T � z + h � X dm ; � g ( ζ ) � X ≤ C 0 � z � X . Define ∆ G k ( x, y ) = g ( y ) , ∆ B k ( x, y ) = h ( y ) − g ( y ) . Then � F k − 1 � X + 1 / 8 E k − 1 ( � ∆ B k � X ) ≤ E k − 1 ( � F k � X ) . Integrate and take the sum, � E ( � ∆ B k � X ) ≤ 4 sup E ( � F k � X ) .

The Davis decomposition yields vector valued Davis and Garsia inequalities for Hardy martingales. At this point a hypothesis on the Banach space X is necessary : Let q ≥ 2 . A Banach space X satisfies the hypothesis H ( q ) , if for each M ≥ 1 there exists δ = δ ( M ) > 0 such that for any x ∈ X with � x � = 1 and g ∈ H ∞ 0 ( T , X ) with � g � ∞ ≤ M, � � T � g � q X dm ) 1 /q . T � z + g � X dm ≥ (1 + δ (1) Remarks: • Condition (1) is required for uniformly bounded analytic functions g, and δ = δ ( M ) > 0 is allowed to depend on the uniform estimates � g � ∞ ≤ M. • If X = C , the hypothesis “ H ( q )” hold true with q = 2 .

Theorem 1 Let q ≥ 2 . Let X be a Banach satisfying H ( q ) . There exists M > 0 δ q > 0 such that for any h ∈ H 1 0 ( T , X ) and z ∈ X there exists g ∈ H ∞ 0 ( T , X ) satisfying � g ( ζ ) � X ≤ M � z � X , ζ ∈ T , and � 1 /q + 1 � � � � � z � q T � g � q T � z + h � X dm ≥ X + δ q X dm T � h − g � X dm. 16 The Davis decomposition and hypothesis H ( q ) com- bined give a decomposition of a Hardy martingale F into Hardy martingales such that F = G + B and ∞ ∞ ( E k − 1 � ∆ G k � q X )) 1 /q + E ( � � E ( � ∆ B k � X ) ≤ A q E ( � F � X ) . k =1 k =1

Non- Linear teleskoping: Let 1 ≤ q ≤ ∞ , 1 /p + 1 /q . If k ) 1 /q ≤ E M k E ( M q k − 1 + v q for 1 ≤ k ≤ n, (2) then n v q k ) 1 /q ≤ 2( E M n ) 1 /q ( E max k ≤ n M k ) 1 /p � E ( (3) k =1 (All random variables are non-negative, integrable)

Let X be complex Banach space. Assume that for every X valued Hardy martingale ( F k ) we have: • (X has ARNP) If sup E � F k � < ∞ then ( F k ) converges a.e. •• (Hardy martingale cotype q ) There exists q < ∞ such that ( E � ∆ k F � X ) q ) 1 /q ≤ C sup � ( E � F k � X . k k • • • (AUMD) For every choice of signs ± � ± ∆ k F � X ≤ C sup E � E � F k � X . k AUMD and ARNP for a Banach space are already de- termined by testing simple Hardy martingales. This re- duction is open for non trivial Hardy martingale Cotype

Proof of Non- Linear teleskoping: P = q = 2 For 0 ≤ s ≤ 1 , and A, B ≥ 0 , Bs ≤ s 2 A + ( A 2 + B 2 ) 1 / 2 − A. (4)

Let 0 ≤ ǫ ≤ 1 . Choose bounded functions 0 ≤ s k ≤ ǫ k ≤ ǫ 2 to linearize the square function. k =1 s 2 with � n k ) 1 / 2 − M k − 1 v k s k ≤ s 2 k M k − 1 + ( M 2 k − 1 + v 2 (5) Integrate k ) 1 / 2 − E M k − 1 . E ( v k s k ) ≤ E ( s 2 k M k − 1 ) + E ( M 2 k − 1 + v 2 Use hypothesis for E ( M 2 k − 1 + v 2 k ) 1 / 2 . E ( v k s k ) ≤ E ( s 2 k M k − 1 ) + E M k − E M k − 1 − E w k .

Sum over k ≤ n n n n s 2 � � � E ( v k s k ) + E w k ≤ E M n + E ( k M k − 1 ) (6) k =1 k =1 k =1 ≤ E M n + ǫ 2 E max k ≤ n M k − 1 Since � n k =1 s 2 k ≤ ǫ 2 , n n k ) 1 / 2 + v 2 E w k ≤ E M n + ǫ 2 E max � � ǫ E ( k ≤ n M k − 1 . k =1 k =1 Divide by 0 < ǫ ≤ 1 , with ǫ 2 = ( E M n )( E max k ≤ n M k ) − 1 .