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Bloom Type Inequality: The Off-diagonal Case (1907.07292v1)

Published 17 Jul 2019 in math.CA

Abstract: In this paper, we establish a representation formula for fractional integrals. As a consequence, for two fractional integral operators $I_{\lambda_1}$ and $I_{\lambda_2}$, we prove a Bloom type inequality \begin{align*} \mbox{\hbox to 8em{}}& \hskip -8em \left|\big[I_{\lambda_1}1,\big[b,I_{\lambda_2}2\big]\big] \right|{L{p_2}(L{p_1})(\mu_2{p_2}\times\mu_1{p_1})\rightarrow L{q_2}(L{q_1})(\sigma_2{q_2}\times\sigma_1{q_1})} % \ %& \lesssim{\substack{[\mu_1]{A{p_1,q_1}(\mathbb Rn)},[\mu_2]{A{p_2,q_2}(\mathbb Rm)} \ [\sigma_1]{A{p_1,q_1}(\mathbb Rn)},[\sigma_2]{A{p_2,q_2}(\mathbb Rm)}}} |b|{\BMO{\pro}(\nu)}, \end{align*} where the indices satisfy $1<p_1<q_1<\infty$, $1<p_2<q_2<\infty$, $1/q_1+1/p_1'=\lambda_1/n$ and $1/q_2+1/p_2'=\lambda_2/m$, the weights $\mu_1,\sigma_1 \in A_{p_1,q_1}(\mathbb Rn)$, $\mu_2,\sigma_2 \in A_{p_2,q_2}(\mathbb Rm)$ and $\nu:=\mu_1\sigma_1{-1}\otimes \mu_2\sigma_2{-1}$, $I_{\lambda_1}1$ stands for $I_{\lambda_1}$ acting on the first variable and $I_{\lambda_2}2$ stands for $I_{\lambda_2}$ acting on the second variable, $\BMO_{\rm{prod}}(\nu)$ is a weighted product $\BMO$ space and $L{p_2}(L{p_1})(\mu_2{p_2}\times\mu_1{p_1})$ and $ L{q_2}(L{q_1})(\sigma_2{q_2}\times\sigma_1{q_1}) $ are mixed-norm spaces.

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