Bloom Type Inequality: The Off-diagonal Case

DOI10.1007/S00025-023-01833-6arXiv1907.07292MaRDI QIDQ6322225

Author name not available (Why is that?)

Publication date: 16 July 2019

Abstract: In this paper, we establish a representation formula for fractional integrals. As a consequence, for two fractional integral operators

I_{l a m b d a_{1}}

and

I_{l a m b d a_{2}}

, we prove a Bloom type inequality �egin{align*} mbox{hbox to 8em{}}& hskip -8em left|�ig[I_{lambda_1}^1,�ig[b,I_{lambda_2}^2�ig]�ig] ight|_{L^{p_2}(L^{p_1})(mu_2^{p_2} imesmu_1^{p_1}) ightarrow L^{q_2}(L^{q_1})(sigma_2^{q_2} imessigma_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}( u)}, end{align*} where the indices satisfy

1 < p_{1} < q_{1} < i n f t y

,

1 < p_{2} < q_{2} < i n f t y

,

1 / q_{1} + 1 / {p_{1}}^{'} = l a m b d a_{1} / n

and

1 / q_{2} + 1 / {p_{2}}^{'} = l a m b d a_{2} / m

, the weights

m u_{1}, s i g m a_{1} i n A_{p_{1}, q_{1}} (m a t h b b R^{n})

,

m u_{2}, s i g m a_{2} i n A_{p_{2}, q_{2}} (m a t h b b R^{m})

and

u := m u_{1} s i g m a_{1}^{- 1} o t i m e s m u_{2} s i g m a_{2}^{- 1}

,

I_{l a m b d a_{1}}^{1}

stands for

I_{l a m b d a_{1}}

acting on the first variable and

I_{l a m b d a_{2}}^{2}

stands for

I_{l a m b d a_{2}}

acting on the second variable,

B M O_{m p r o d} (u)

is a weighted product

B M O

space and

L^{p_{2}} (L^{p_{1}}) (m u_{2}^{p_{2}} i m e s m u_{1}^{p_{1}})

and

L^{q_{2}} (L^{q_{1}}) (s i g m a_{2}^{q_{2}} i m e s s i g m a_{1}^{q_{1}})

are mixed-norm spaces.

Mathematics Subject Classification ID

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