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Dyadic weights on $R^n$ and reverse Holder inequalities
Functional Analysis · Submitted 31 Jul 2014
abstract compiled · 2026-09-23 · no language-model review
The authors' main claim We prove that for any weight $\phi$ defined on $[0,1]^n$ that satisfies a reverse Holder inequality with exponent p > 1 and constant $c\ge1$ upon all dyadic subcubes of $[0,1]^n$, it's non increasing rearrangement satisfies a reverse Holder inequality with the same exponent and constant not more than $2^nc-2^n + 1$, upon all subintervals of $[0; 1]$ of the form $[0; t]$.
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The abstract, sentence by sentence
We prove that for any weight $\phi$ defined on $[0,1]^n$ that satisfies a reverse Holder inequality with exponent p > 1 and constant $c\ge1$ upon all dyadic subcubes of $[0,1]^n$, it's non increasing rearrangement satisfies a reverse Holder inequality with the same exponent and constant not more than $2^nc-2^n + 1$, upon all subintervals of $[0; 1]$ of the form $[0; t]$. This gives as a consequence, according to the results in [8], an interval $[p; p_0(p; c)) = I{p,c}$, such that for any $q \in I{p,c}$, we have that $\phi$ is in $L^q$.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | We prove that for any weight $\phi$ defined on $[0,1]^n$ that satisfies a reverse Holder inequality with exponent p > 1 and constant $c\ge1$ upon all dyadic subcubes of $[0,1]^n$, it's non increasing rearrangement satisfies a reverse Holder inequality with the same exponent and constant not more than $2^nc-2^n + 1$, upon all subintervals of $[0; 1]$ of the form $[0; t]$. | result 0.94 | 0.95 | 0.89 | asserted 1.00 |
| s1 | This gives as a consequence, according to the results in [8], an interval $[p; p_0(p; c)) = I{p,c}$, such that for any $q \in I{p,c}$, we have that $\phi$ is in $L^q$. | interpretation 0.80 | 0.40 | 0.13 | asserted 1.00 |
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The paper
Original abstract
We prove that for any weight $\phi$ defined on $[0,1]^n$ that satisfies a reverse Holder inequality with exponent p > 1 and constant $c\ge1$ upon all dyadic subcubes of $[0,1]^n$, it's non increasing rearrangement satisfies a reverse Holder inequality with the same exponent and constant not more than $2^nc-2^n + 1$, upon all subintervals of $[0; 1]$ of the form $[0; t]$. This gives as a consequence, according to the results in [8], an interval $[p; p_0(p; c)) = I{p,c}$, such that for any $q \in I{p,c}$, we have that $\phi$ is in $L^q$.
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How the literature uses it
2 arXiv papers cite this in 2 sentences
Papers citing it · 1 sentence without a confident label
Cites it as background 1 paper
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For numerous applications of monotonic rearrangements in variuos estimations see [ref] and references therein.
Monotonic rearrangements of functions with small mean oscillation · Introduction · 0.95
What it cites · 1 sentence without a confident label
Cites as background 1 paper
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All last results that are connected with $A_1$ dyadic weights $ϕ$ and the behavior of $ϕ^∗$ as an $A_1$-weight on $$, can be seen in [ref] .
The Bellman function of two variables for the dyadic maximal operator · Introduction · 0.93
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