Borderline variants of the Muckenhoupt-Wheeden inequality

Transcription

1 Borderline variants of the Muckenhoupt-Wheeden inequality Carlos Domingo-Salazar Universitat de Barcelona joint work with M Lacey and G Rey 3CJI Murcia, Spain September 10, 2015

2 The Hardy-Littlewood maximal operator Definition We consider the Hardy-Littlewood maximal operator defined by 1 Mf(x) = sup f(y) dy x Q Q Q

3 It is easy to see that M : L 1 (R n ) L 1 (R n ),

4 It is easy to see that M : L 1 (R n ) L 1 (R n ), but if we make the range a little bigger, then M : L 1 (R n ) L 1, (R n )

5 It is easy to see that M : L 1 (R n ) L 1 (R n ), but if we make the range a little bigger, then M : L 1 (R n ) L 1, (R n ) Now we add a weight (dx w(x)dx), and M : L 1 (w) L 1, (w) w A 1

6 It is easy to see that M : L 1 (R n ) L 1 (R n ), but if we make the range a little bigger, then M : L 1 (R n ) L 1, (R n ) Now we add a weight (dx w(x)dx), and M : L 1 (w) L 1, (w) w A 1 But if I want to have ANY weight w on the right-hand side, then what? For every weight w, M : L 1 (??) L 1, (w)

7 It is easy to see that M : L 1 (R n ) L 1 (R n ), but if we make the range a little bigger, then M : L 1 (R n ) L 1, (R n ) Now we add a weight (dx w(x)dx), and M : L 1 (w) L 1, (w) w A 1 But if I want to have ANY weight w on the right-hand side, then what? For every weight w, M : L 1 (??) L 1, (w)?? must be "larger" than w!

8 Fefferman-Stein inequality In 1971, C Fefferman and E M Stein, proved the following inequality: Theorem For every weight w, it holds that M : L 1 (Mw) L 1, (w) Or equivalently, λw(mf > λ) C f(x) Mw(x)dx, λ > 0 R n

9 Muckenhoupt-Wheeden conjecture The conjecture of B Muckenhoupt and R Wheeden was that "the same held for every Calderón-Zygmund operator T ": M : L 1 (Mw) L 1, (w) T : L 1 (Mw) L 1, (w) (FS inequality) (MW conjecture)

10 Muckenhoupt-Wheeden conjecture This was shown to be FALSE by M C Reguera and C Thiele in 2012 H : L 1 (Mw) L 1, (w)

11 Muckenhoupt-Wheeden conjecture This was shown to be FALSE by M C Reguera and C Thiele in 2012 H : L 1 (Mw) L 1, (w) Question How can we modify the maximal operator M M ϕ in the weight so that T : L 1 (M ϕ w) L 1, (w), for every w and Calderón-Zygmund operator T? We need to make M "larger"

12 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f(y) χ Q(y)dy x Q Q Q x Q R Q n

13 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f L1 (χ x Q Q Q / Q ) x Q Q

14 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f L1 (χ x Q Q Q / Q ) x Q We define Q M ϕ f(x) = sup f ϕ(l)(χq / Q ), x Q where ϕ is a Young function such as: ϕ(t) = t L 1 norm and M ϕ = M, ϕ(t) = t log t L log L norm,

15 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f L1 (χ x Q Q Q / Q ) x Q We define Q M ϕ f(x) = sup f ϕ(l)(χq / Q ), x Q where ϕ is a Young function such as: Larger ϕ ϕ(t) = t L 1 norm and M ϕ = M, ϕ(t) = t log t L log L norm,

16 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f L1 (χ x Q Q Q / Q ) x Q We define Q M ϕ f(x) = sup f ϕ(l)(χq / Q ), x Q where ϕ is a Young function such as: ϕ(t) = t L 1 norm and M ϕ = M, ϕ(t) = t log t L log L norm, Larger ϕ Larger operator M ϕ

17 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f L1 (χ x Q Q Q / Q ) x Q We define Q M ϕ f(x) = sup f ϕ(l)(χq / Q ), x Q where ϕ is a Young function such as: ϕ(t) = t L 1 norm and M ϕ = M, ϕ(t) = t log t L log L norm, Larger ϕ Larger operator M ϕ Smaller space L 1 (M ϕ w)

18 Orlicz versions of the maximal operator The Hardy-Littlewood maximal operator can be written as 1 Mf(x) = sup f(y) dy = sup f L1 (χ x Q Q Q / Q ) x Q We define Q M ϕ f(x) = sup f ϕ(l)(χq / Q ), x Q where ϕ is a Young function such as: ϕ(t) = t L 1 norm and M ϕ = M, ϕ(t) = t log t L log L norm, Larger ϕ Larger operator M ϕ Smaller space L 1 (M ϕ w) More likely T : L 1 (M ϕ w) L 1, (w)

19 Current state of the problem Question What is the "least" Young function ϕ such that T : L 1 (M ϕ w) L 1, (w), for every w and Calderón-Zygmund operator T?

20 Current state of the problem Question What is the "least" Young function ϕ such that T : L 1 (M ϕ w) L 1, (w), for every w and Calderón-Zygmund operator T? ϕ(t) = t (Reguera 2011 / Reguera, Thiele 2012) FALSE

21 Current state of the problem Question What is the "least" Young function ϕ such that T : L 1 (M ϕ w) L 1, (w), for every w and Calderón-Zygmund operator T? ϕ(t) = t(log t) ɛ TRUE, ε > 0 (Pérez 1994 / Hytönen, Pérez 2015, with C = 1 ɛ )

22 Current state of the problem Question What is the "least" Young function ϕ such that T : L 1 (M ϕ w) L 1, (w), for every w and Calderón-Zygmund operator T? ϕ(t) = t log log t(log log log t) α TRUE, α > 1 (D-S, Lacey, Rey 2015, and C = 1 α 1 )

23 Current state of the problem Question What is the "least" Young function ϕ such that T : L 1 (M ϕ w) L 1, (w), for every w and Calderón-Zygmund operator T? ϕ(t) = o(t log log t) (Calderelli, Lerner, Ombrossi 2015) FALSE

24 Current state of the problem

25 Current state of the problem NEGATIVE RESULTS: With the Hilbert Transform POSITIVE RESULTS: With the reduction to sparse operators

26 Our contribution Theorem (D-S, Lacey, Rey 2015) Suppose the Young function ϕ satisfies c ϕ = k=1 1 ψ 1 (2 2k ) < Then, for all C-Z operator T, and any weight w, it holds that T : L 1 (M ϕ w) L 1, (w) with constant c ϕ That is, sup λw{t f > λ} c ϕ f(x) M ϕ w(x) dx λ>0 R n

27 Our contribution Theorem (D-S, Lacey, Rey 2015) Suppose the Young function ϕ satisfies c ϕ = k=1 1 ψ 1 (2 2k ) < Then, for all C-Z operator T, and any weight w, it holds that T : L 1 (M ϕ w) L 1, (w) with constant c ϕ That is, sup λw{t f > λ} c ϕ f(x) M ϕ w(x) dx λ>0 R n The function ψ is called the complementary function of ϕ, and whenever ϕ(t) = tl(t), with L a logarithmic part (log t, log log t, log log t(log log log t) α ), then essentially ψ 1 (t) L(t)

28 Our contribution Hence, for instance, when we have c ϕ = k=1 ϕ(t) = tl(t) = tlog log t(log log log t) α, 1 ψ 1 (2 2k ) 1 log log(2 2k )(log log log(2 2k )) α k=1 k=1 1 k(log k) α 1 α 1

29 Our contribution Hence, for instance, when we have c ϕ = k=1 ϕ(t) = tl(t) = tlog log t(log log log t) α, 1 ψ 1 (2 2k ) 1 log log(2 2k )(log log log(2 2k )) α k=1 k=1 Therefore, the theorem states that λw{t f > λ} 1 α 1 1 k(log k) α 1 α 1 R n f(x) M ϕ w(x) dx

30 Our contribution We also recover the sharp constant of Hytönen-Pérez s result, with ϕ(t) = t(log t) ɛ we have and hence c ϕ k=1 λw{t f > λ} 1 ɛ 1 (log 2 2k ) 1 ɛ 2 kɛ 1 ɛ, k=1 R n f(x) M ϕ w(x) dx

31 Our contribution How to prove it It is enough to show that, for every sparse operator S, λw{sf > λ} c ϕ R n f(x) M ϕ w(x) dx,

32 Our contribution How to prove it It is enough to show that, for every sparse operator S, λw{sf > λ} c ϕ R n f(x) M ϕ w(x) dx, where Sf(x) = Q S ( ) 1 f χ Q (x), Q Q and S is a family of dyadic cubes such that, for every Q S, Q S : Q Q Q Q 8

33 Our contribution How to prove it It is enough to show that, for every sparse operator S, λw{sf > λ} c ϕ R n f(x) M ϕ w(x) dx, where Sf(x) = Q S ( ) 1 f χ Q (x), Q Q and S is a family of dyadic cubes such that, for every Q S, Q S : Q Q In fact, by linearity, we reduce to (GOAL) Q Q 8 w{1 < Sf 2} c ϕ R n f(x) M ϕ w(x) dx

34 Our contribution How to prove it We split S into S k = { Q S : 2 k 1 < 1 } f 2 k Q Q

35 Our contribution How to prove it We split S into S k = { Q S : 2 k 1 < 1 } f 2 k Q Q By Fefferman-Stein, we can assume that S k = for k < 2, and, for every k 2, there is a finite number of layers S k = S k,0 S k,2 k: S k,0 S k,1 S k,2 k Figure : Layer decomposition of S k

36 Our contribution S k,0 S k,1 S k,2 k S k This gives a simpler description of the operator: Sf(x) = ( ) 1 f χ Q (x) Q Q S Q

37 Our contribution S k,0 S k,1 S k,2 k S k This gives a simpler description of the operator: Sf(x) = ( ) 1 f χ Q (x) Q Q S Q 2 k ( ) 1 = f χ Q (x) Q Q k=2 ν=0 Q S k,ν

38 Our contribution S k,0 S k,1 S k,2 k S k This gives a simpler description of the operator: Sf(x) = ( ) 1 f χ Q (x) Q Q S Q 2 k ( ) 1 = f χ Q (x) Q Q k=2 ν=0 Q S k,ν 2 k k=2 ν=0 Q S k,ν 2 k χ Q (x)

39 Our contribution S k,0 S k,1 S k,2 k S k This gives a simpler description of the operator: Sf(x) = ( ) 1 f χ Q (x) Q Q S Q 2 k ( ) 1 = f χ Q (x) Q Q = k=2 ν=0 Q S k,ν 2 k k=2 ν=0 k=2 Q S k,ν 2 k χ Q (x) k 2 2 k ν=0 Q S k,ν χ Q (x) overlapping

40 Our contribution The main lemma is the following: If S = k 2 S k, with S k f(x) = Q S k ( ) 1 f χ Q (x) Q Q

41 Our contribution The main lemma is the following: If S = k 2 S k, with S k f(x) = Q S k ( ) 1 f χ Q (x) Q Q Lemma For each k 2, if we denote E = {1 < Sf 2}, S k f(x)w(x)dx 2 k C w(e) + f(x) M E ψ 1 ϕ w(x)dx (2 2k ) R n Recall our goal was w{1 < Sf 2} c ϕ R n f(x) M ϕ w(x) dx

42 Thank you for your attention! Muchas Gracias!