2018
DOI: 10.2140/apde.2018.11.2089
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On weak weighted estimates of the martingale transform and a dyadic shift

Abstract: We consider several weak type estimates for singular operators using the Bellman function approach. We disprove the A1 conjecture, which stayed open after Muckenhoupt-Wheeden's conjecture was disproved by Reguera-Thiele.By M w we will denote the dyadic maximal function of w, that is M w(x) = sup{ w J : J ∈ D, J ∋ x}. Then w ∈ A 1 with "norm" Q means that M w ≤ Q · w a. e. , and Q = [w] A 1 is the best constant in this inequality.Recall that a martingale transform is an operator given by T ε ϕ = J∈D ε J ∆ J ϕ .… Show more

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Cited by 9 publications
(6 citation statements)
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“…However, the fact that ϕ(t) ≃ t was disproved in [26], furthermore, in [18] it was established that ϕ(t) ≃ t log(e + t), and consequently the estimate in [19] is sharp.…”
Section: Introductionmentioning
confidence: 99%
“…However, the fact that ϕ(t) ≃ t was disproved in [26], furthermore, in [18] it was established that ϕ(t) ≃ t log(e + t), and consequently the estimate in [19] is sharp.…”
Section: Introductionmentioning
confidence: 99%
“…On the other hand, in [7], it was shown for the martingale transform (and explained how to transfer the result to the Hilbert transform case) that the dependence of [w] A 1 in the weighted weak type (1, 1) inequality cannot in general be made better than [w] A 1 log 1/5 (e + [w] A 1 ), thus disproving the weak Muckenhoupt-Wheeden conjecture. Later, in [8], the power of the logarithm was improved to 1/3 (this was again done for the martingale transform).…”
Section: Introductionmentioning
confidence: 99%
“…Summarizing the results in [5,7,8], if we denote by α H the best possible exponent for which the inequality…”
Section: Introductionmentioning
confidence: 99%
“…But the question was whether the logarithmic term can be dropped. The Bellman function construction in [16], [17] showed that linear estimate is false, in fact, in these preprints a sequence of weights w, was proved to exist such that [w] A 1 → ∞ but T : L 1 (w) → L 1,∞ (w) ≥ c[w] A 1 (log[w] A 1 ) 1/3 . Operator T was either a martingale transform, or the dyadic shift, or the Hilbert transform.…”
Section: Introductionmentioning
confidence: 99%