2011
DOI: 10.1016/j.jfa.2011.07.015
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Rosenthal inequalities in noncommutative symmetric spaces

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Cited by 59 publications
(70 citation statements)
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“…A version of Theorem 5.7 for the case where the Boyd indices satisfy the condition 2 < p E ≤ q E < ∞ was first obtained in [12,Theorem 6.3]. Similar line of result for martingale BMO-norms of sums of noncommuting independent sequences were also considered in [38,Theorem 5.3].…”
Section: If We Setmentioning
confidence: 81%
“…A version of Theorem 5.7 for the case where the Boyd indices satisfy the condition 2 < p E ≤ q E < ∞ was first obtained in [12,Theorem 6.3]. Similar line of result for martingale BMO-norms of sums of noncommuting independent sequences were also considered in [38,Theorem 5.3].…”
Section: If We Setmentioning
confidence: 81%
“…By the boundedness of martingale transforms on E(M) [, Proposition 4.9], there exists a constant κE such that for any given finite martingale x in E(M), κE1E‖‖n1εndxnE(M)xtrue∥E(M)κEE‖‖n1εndxnE(M)where (εn) denotes a Rademacher sequence on a given probability space. According to [, Theorem 4.3], we then have, (κE)1false(dxnfalse)true∥Efalse(scriptM;2cfalse)+Efalse(scriptM;2rfalse)xtrue∥E(M)cEfalse(dxnfalse)true∥Efalse(scriptM;2cfalse)Efalse(scriptM;2rfalse).Applying the noncommutative Stein inequality to the first inequality , we deduce that CE<...>…”
Section: Applications To Noncommutative Burkholder/rosenthal Inequalimentioning
confidence: 92%
“…Our strategy for the proof of Theorem is to use the corresponding Burkholder–Gundy alongside our Davis decomposition stated in Theorem . The following version of the Burkholder–Gundy inequalities is implicit in : Proposition Let 1<p<q< and E be a symmetric Banach function space with the Fatou property such that E Int (Lp,Lq). Then there exist positive constants CE and cE such that: (i)for every xE(M), the following inequality holds: inftrue∥xcscriptHEc+true∥xrscriptHErCExtrue∥E(M),where the infimum is taken over all xcscriptHEcfalse(scriptMfalse), and xrscriptHErfalse(scriptMfalse) such that x=xc+xr; (ii)for every xscriptHEcfalse(scriptMfalse)scriptHErfalse(scriptMfalse), the following inequality holds: xtrue∥E(M)cEmaxtrue∥xscriptH<...>…”
Section: Applications To Noncommutative Burkholder/rosenthal Inequalimentioning
confidence: 99%
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