Bloom Type Inequality: The Off-Diagonal Case
نویسندگان
چکیده
In this paper, we establish a representation formula for fractional integrals. As consequence, two integral operators $I_{\lambda_1}$ and $I_{\lambda_2}$, prove 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 R^n)},[\mu_2]_{A_{p_2,q_2}(\mathbb R^m)} [\sigma_1]_{A_{p_1,q_1}(\mathbb R^n)},[\sigma_2]_{A_{p_2,q_2}(\mathbb R^m)}}} \|b\|_{\BMO_{\pro}(\nu)}, \end{align*} where the indices satisfy $1<p_1<q_1<\infty$, $1<p_2<q_2<\infty$, $1/q_1+1/p_1'=\lambda_1/n$ $1/q_2+1/p_2'=\lambda_2/m$, weights $\mu_1,\sigma_1 \in A_{p_1,q_1}(\mathbb R^n)$, $\mu_2,\sigma_2 A_{p_2,q_2}(\mathbb R^m)$ $\nu:=\mu_1\sigma_1^{-1}\otimes \mu_2\sigma_2^{-1}$, $I_{\lambda_1}^1$ stands acting on first variable $I_{\lambda_2}^2$ $I_{\lambda_2}$ second variable, $\BMO_{\rm{prod}}(\nu)$ is weighted product $\BMO$ space $L^{p_2}(L^{p_1})(\mu_2^{p_2}\times\mu_1^{p_1})$ $ L^{q_2}(L^{q_1})(\sigma_2^{q_2}\times\sigma_1^{q_1}) are mixed-norm spaces.
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ژورنال
عنوان ژورنال: Results in Mathematics
سال: 2023
ISSN: ['1420-9012', '1422-6383']
DOI: https://doi.org/10.1007/s00025-023-01833-6