In this paper, the authors study the boundedness properties of a class of multilinear strongly singular integral operator with generalized kernels on product of weighted Lebesgue spaces and product of variable exponent Lebesgue spaces, respectively. Moreover, the types $ L^{\infty}\times \dots \times L^{\infty}\rightarrow BMO $ and $ BMO\times \dots \times BMO\rightarrow BMO $ endpoint estimates are also obtained.
Citation: Shuhui Yang, Yan Lin. Multilinear strongly singular integral operators with generalized kernels and applications[J]. AIMS Mathematics, 2021, 6(12): 13533-13551. doi: 10.3934/math.2021786
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In this paper, the authors study the boundedness properties of a class of multilinear strongly singular integral operator with generalized kernels on product of weighted Lebesgue spaces and product of variable exponent Lebesgue spaces, respectively. Moreover, the types $ L^{\infty}\times \dots \times L^{\infty}\rightarrow BMO $ and $ BMO\times \dots \times BMO\rightarrow BMO $ endpoint estimates are also obtained.
For the past few years, more and more attention has been paid to the topic of multilinear singular integrals. The multilinear Calderón-Zygmund theory originated in the works of Coifman and Meyer in [1,2,3]. Recently, this topic has been studied extensively by many scholars from different perspectives, see for example [4,5,6,7,8,9,10,11,12,13,14].
Lin in [15] introduced the multilinear strongly singular Calderón-Zygmund operator. The kernel does not need any size condition and is more singular near the diagonal than the standard case. It is defined as follows.
Definition 1.1. Suppose $ K(y_{0}, y_{1}, \dots, y_{m}) $ is a function defined away from the diagonal $ y_{0} = y_{1} = \dots = y_{m} $ in $ (\mathbb{R}^{n})^{m+1} $, where $ m\in \mathbb{N}_{+} $. For some $ \varepsilon > 0 $ and $ 0 < \alpha\leq1 $, it satisfies
$ \begin{align} |K(x, y_{1}, \dots, y_{m})-K(x', y_{1}, \dots, y_{m})| \leq \frac{C |x-x'|^{\varepsilon}}{(|x-y_{1}|+\cdots+|x-y_{m}|)^{mn+\varepsilon/\alpha}}, \end{align} $ | (1.1) |
whenever $ |x-x'|^{\alpha}\leq\frac{1}{2}\max_{1\leq j\leq m} |x-y_{j}| $.
$ T $ is an $ m $-linear operator with the kernel $ K $ given by the integral representation
$ \begin{align} T(f_{1}, \dots, f_{m})(x) = \int_{\mathbb{R}^{n}}\cdots\int_{\mathbb{R}^{n}} K(x, y_{1}, \dots, y_{m})\prod\limits_{j = 1}^{m}f_{j}(y_{j})dy_{1}\cdots dy_{m}, \end{align} $ | (1.2) |
where $ f_{j}\in C_c^\infty(\mathbb{R}^n)(j = 1, \dots, m) $ and $ x\notin \cap_{j = 1}^{m} \rm{supp} f_{j} $.
For some $ 1\leq r_{1}, \dots, r_{m} < \infty $ with $ 1/r = 1/r_{1}+\dots+1/r_{m} $, $ T $ is bounded from $ L^{r_{1}}\times \dots \times L^{r_{m}} $ to $ L^{r, \, \infty} $, and for some $ 1\leq l_{1}, \dots, l_{m} < \infty $ with $ 1/l = 1/l_{1}+\dots+1/l_{m} $, $ T $ is bounded from $ L^{l_{1}}\times \dots \times L^{l_{m}} $ to $ L^{q, \, \infty} $, where $ 0 < l/q\leq \alpha $. Then $ T $ is called an $ m $-linear strongly singular Calderón-Zygmund operator.
Lin in [15], Lin-Lu-Lu in [16] and Lin-Han in [17] established the sharp maximal estimates, the boundedness on product of weighted Lebesgue spaces and product of variable exponent Lebesgue spaces for the multilinear strongly singular Calderón-Zygmund operators, their multilinear commutators and multilinear iterated commutators, respectively. Moreover, the types $ L^{\infty}\times \dots \times L^{\infty}\rightarrow BMO $, $ BMO\times \dots \times BMO\rightarrow BMO $ and $ LMO\times \dots \times LMO\rightarrow LMO $ endpoint estimates were obtained. In [18], Spanne introduced the function space $ LMO $, which is a subspace of $ BMO $ space, equipped with semi-norm:
$ [f]_{LMO} = \sup\limits_{0 < r < 1}\frac{1+|\ln r|}{|B_r|}\int_{B_r}|f(x)-f_{B_r}|dx+\sup\limits_{r\geq1}\frac{1}{|B_r|}\int_{B_r}|f(x)-f_{B_r}|dx, $ |
where $ B_r $ denotes by the ball in $ \mathbb{R}^n $ with radius $ r $. Lin and Yan in [19] got the boundedness on the product of generalized Morrey spaces and weighted Morrey spaces for multilinear strongly singular Calderón-Zygmund operators, their multilinear commutators and multilinear iterated commutators, respectively. Other related results of this class of operators can be seen in [20] and so on.
In this paper, we will study a class of more general multilinear strongly singular integral operators $ T $ defined by (1.2). And instead of (1.1) the kernel $ K $ satisfies a weaker condition: there exist $ \varepsilon > 0 $, $ 0 < \alpha\leq1 $ and $ {p_0} > 1 $, such that for any $ k_1, \cdot\cdot\cdot, k_h\in \mathbb{N}_{+} $, $ h\in \{1, \dots, m\} $,
$ \Big(\int_{I_{i_1}}\int_{I_{i_2}}\cdot\cdot\cdot\int_{I_{i_h}}|K(x, y_{1}, \dots, y_{m})-K(x', y_{1}, \dots, y_{m})|^{p_0}dy_{i_1}dy_{i_2}\cdot\cdot\cdot dy_{i_h}\Big)^\frac{1}{p_0} $ |
$ \begin{align} \leq C|x-x'|^{\varepsilon-t{[n(m-\frac{h}{p_0})+{\frac{\varepsilon}{\alpha}}}]}\prod^h_{i = 1}2^{-{k_i}{[n(\frac{m}{h}-\frac{1}{p_0})+{\frac{\varepsilon}{h\alpha}}}]}, \end{align} $ | (1.3) |
where $ I_{i_s} = \{ y_{i_s}: {2^{k_s}|x-x'|^t\leq|y_{i_s}-x'|\leq2^{{k_s}+1}|x-x'|^t} \} $, $ s = 1, \cdots, h $, and $ t = 1 $ when $ |x-x'|\geq1 $, $ t = \alpha $ when $ |x-x'| < 1 $.
Remark 1.1. It is easy to see that the condition $ (1.1) $ implies the condition $ (1.3) $ as follows:
$ \begin{align*} &\Big(\int_{I_{i_1}}\int_{I_{i_2}}\cdot\cdot\cdot\int_{I_{i_h}}|K(x, y_{1}, \dots, y_{m})-K(x', y_{1}, \dots, y_{m})|^{p_0}dy_{i_1}dy_{i_2}\cdot\cdot\cdot dy_{i_h}\Big)^\frac{1}{p_0}\\ &\leq \Big(\int_{I_{i_1}}\int_{I_{i_2}}\cdot\cdot\cdot\int_{I_{i_h}}\Big(\frac{C |x-x'|^{\varepsilon}}{(|x'-y_{1}|+\cdots+|x'-y_{m}|)^{mn+\varepsilon/\alpha}}\Big)^{h\frac{p_0}{h}}dy_{i_1}dy_{i_2}\cdot\cdot\cdot dy_{i_h}\Big)^\frac{1}{p_0}\\ &\leq C|x-x'|^{\varepsilon}\Big(\int_{I_{i_1}}(\frac{1}{|x'-y_{i_1}|^{mn+\varepsilon/\alpha}})^{\frac{p_0}{h}}dy_{i_1}\Big)^\frac{1}{p_0}\cdots\Big(\int_{I_{i_h}}(\frac{1}{|x'-y_{i_h}|^{mn+\varepsilon/\alpha}})^{\frac{p_0}{h}}dy_{i_h}\Big)^\frac{1}{p_0}\\ &\leq C|x-x'|^{\varepsilon-t{[n(m-\frac{h}{p_0})+{\frac{\varepsilon}{\alpha}}}]}\prod^h_{i = 1}2^{-{k_i}{[n(\frac{m}{h}-\frac{1}{p_0})+{\frac{\varepsilon}{h\alpha}}}]}. \end{align*} $ |
Definition 1.2. Let $ T $ be an $ m $-linear operator with kernel $ K $ defined by (1.2). $ T $ is called an $ m $-linear strongly singular integral operator with generalized kernel if it satisfies the condition (1.3), and for some $ 1\leq r_{1}, \dots, r_{m} < \infty $ with $ 1/r = 1/r_{1}+\dots+1/r_{m} $, $ T $ is bounded from $ L^{r_{1}}\times \dots \times L^{r_{m}} $ to $ L^{r, \, \infty} $, and for some $ 1\leq l_{1}, \dots, l_{m} < \infty $ with $ 1/l = 1/l_{1}+\dots+1/l_{m} $, $ T $ is bounded from $ L^{l_{1}}\times \dots \times L^{l_{m}} $ to $ L^{q, \, \infty} $, where $ 0 < l/q\leq \alpha $.
In order to establish our main results, we need some notations as follows.
Definition 1.3. The Hardy-Littlewood maximal operator $ M $ is defined by
$ Mf(x) = \sup\limits_{B\ni x}\dfrac{1}{|B|}\int_B|f(y)|dy. $ |
Denote by $ M_{s}(f) = [M(|f|^{s})]^{1/s} $, where $ 0 < s < \infty $.
Definition 1.4. The sharp maximal function is defined by
$ M^{\sharp}(f)(x) = \sup\limits_{B\ni x}\dfrac{1}{|B|}\int_{B}|f(y)-f_{B}|dy\sim \sup\limits_{B\ni x} \inf\limits_{a\in \mathbb{C}}\dfrac{1}{|B|}\int_{B}|f(y)-a|dx, $ |
where the supremum is taken over all balls $ B $ containing $ x $ and $ f_{B} = \frac{1}{|B|}\int_{B}|f(x)|dx $.
Denote by $ M^{\sharp}_{s}(f) = [M^{\sharp}(|f|^{s})]^{1/s} $, where $ 0 < s < \infty $.
Definition 1.5. Suppose $ w $ is a non-negative measurable function. It belongs to Muckenhoupt class $ A_p $ with $ 1 < p < \infty $ if
$ \sup\limits_{Q}\biggl(\frac{1}{|Q|}\int_Q w(x)dx\biggr)\biggl(\frac{1}{|Q|}\int_Q w(x)^{1-p'}dx\biggr)^{p-1} < \infty, $ |
where $ 1/p+1/p' = 1 $.
When $ p = 1 $, we say that $ w $ belongs to $ A_1 $, if there exists a constant $ C > 0 $ such that for any cube $ Q $,
$ \frac{1}{|Q|}\int_Q w(y)dy\leq Cw(x), \quad a.e.\, x\in Q. $ |
Denote by $ A_{\infty} = \bigcup_{p\geq1}A_{p} $.
Definition 1.6. For the measurable function $ p(\cdot): \mathbb{R}^{n}\rightarrow [1, \infty) $, the variable exponent Lebesgue space $ L^{p(\cdot)}(\mathbb{R}^{n}) $ is defined by
$ L^{p(\cdot)}(\mathbb{R}^{n}) = \biggr\{f\, \, {\rm is\, \, measurable}: \int_{\mathbb{R}^{n}}\biggl(\frac{|f(x)|}{\lambda}\biggr)^{p(x)}dx < \infty\, {\rm for\, \, some\, \, constant}\, \lambda > 0\biggr\}. $ |
As well-known that the set $ L^{p(\cdot)}(\mathbb{R}^{n}) $ becomes a Banach space with respect to the norm
$ \|f\|_{L^{p(\cdot)}(\mathbb{R}^{n})} = \inf\biggl\{\lambda > 0: \int_{\mathbb{R}^{n}}\biggl(\frac{|f(x)|}{\lambda}\biggr)^{p(x)}dx\leq1\biggr\}. $ |
Denote by $ \mathcal{P}(\mathbb{R}^{n}) $ the set of all measurable functions $ p(\cdot): \mathbb{R}^{n}\rightarrow [1, \infty) $ such that
$ 1 < p_{-}: = {\rm essinf}_{x\in\mathbb{R}^{n}}p(x)\, \, {\rm and}\, \, p_{+}: = {\rm esssup}_{x\in\mathbb{R}^{n}}p(x) < \infty. $ |
Denote by $ \mathcal{B}(\mathbb{R}^{n}) $ the set of all $ p(\cdot)\in\mathcal{P}(\mathbb{R}^{n}) $ such that the Hardy-Littlewood maximal operator $ M $ is bounded on $ L^{p(\cdot)}(\mathbb{R}^{n}) $.
This paper will be organized as follows. The boundedness properties of multilinear strongly singular integral operators with generalized kernels on product of weighted Lebesgue spaces and product of variable exponent Lebesgue spaces will be obtained, respectively. Moreover, the types $ L^{\infty}\times \dots \times L^{\infty}\rightarrow BMO $ and $ BMO\times \dots \times BMO\rightarrow BMO $ endpoint estimates are also obtained as main results in Section 2. In Section 3, we will provide some necessary lemmas to prove the main results. Finally, in Section 4 we will give the proof details of the main results.
Theorem 2.1. Suppose $ T $ is an $ m $-linear strongly singular integral operator with generalized kernel and $ p_{0}'\geq\max \{r_{1}, \dots, r_{m}, l_{1}, \dots, l_{m}\} $, $ 1/p_{0} +1/p_{0}' = 1 $, where $ p_{0} $, $ r_{j} $ and $ l_{j} $ are given by Definition $ 1.2 $, $ j = 1, \dots, m $. For any $ p_{0}' < p_{1}, \dots, p_{m} < \infty $ with $ 1/p = 1/p_{1}+\cdots+1/p_{m} $, $ (w_{1}, \dots, w_{m})\in(A_{p_{1}/p_{0}'}, \dots, A_{p_{m}/p_{0}'}) $ and $ w = \prod_{j = 1}^{m}w_{j}^{p/{p_{j}}} $, there is
$ \|T(\vec{f})\|_{L^{p}(w)}\leq C\prod\limits_{j = 1}^{m}\|f_{j}\|_{L^{p_{j}}(w_{j})}. $ |
Theorem 2.2. Suppose $ T $ is an $ m $-linear strongly singular integral operator with generalized kernel and $ p_{0}'\geq max \{r_{1}, \dots, r_{m}, l_{1}, \dots, l_{m}\} $, $ 1/p_{0} +1/p_{0}' = 1 $, where $ p_{0} $, $ r_{j} $ and $ l_{j} $ are given by Definition $ 1.2 $, $ j = 1, \dots, m $. Suppose $ p(\cdot), p_{1}(\cdot), \dots, p_{m}(\cdot)\in\mathcal{B}(\mathbb{R}^{n}) $ and $ 1/p(\cdot) = 1/p_{1}(\cdot)+\cdots+1/p_{m}(\cdot) $. Let $ q_{0}^{j} $ be given by Lemma $ 3.5 $ for $ p_{j}(\cdot) $, $ j = 1, \dots, m $. If $ p_0'\leq \min_{1\leq j\leq m}q_{0}^{j} $, then
$ \|T(\vec{f})\|_{L^{p(\cdot)}(\mathbb{R}^{n})}\leq C\prod\limits_{j = 1}^{m}\|f_{j}\|_{L^{p_{j}(\cdot)}(\mathbb{R}^{n})}. $ |
Theorem 2.3. Suppose $ T $ is an $ m $-linear strongly singular integral operator with generalized kernel, $ p_{0} $, $ q $, $ r_{j} $, $ l_{j} $ are given by Definition $ 1.2 $, $ j = 1, \dots, m $ and $ q > 1 $, $ p_{0}'\geq \max \{r_{1}, \dots, r_{m}, l_{1}, \dots, l_{m}\} $, $ 1/p_{0} +1/p_{0}' = 1 $. Then
$ \|T(\vec{f})\|_{BMO}\leq C\prod\limits_{j = 1}^{m}\|f_{j}\|_{\infty}. $ |
Theorem 2.4. Suppose $ T $ is an $ m $-linear strongly singular integral operator with generalized kernel, $ p_{0} $, $ q $, $ r_{j} $, $ l_{j} $ are given by Definition $ 1.2 $, $ j = 1, \dots, m $ and $ q > 1 $, $ p_{0}'\geq \max \{r_{1}, \dots, r_{m}, l_{1}, \dots, l_{m}\} $, $ 1/p_{0} +1/p_{0}' = 1 $, and $ T(f_{1}, \cdots, f_{j-1}, 1, f_{j+1}, \cdots, f_{m}) = 0\, (j = 1, \dots, m) $. Then
$ \|T(\vec{f})\|_{BMO}\leq C\prod\limits_{j = 1}^{m}\|f_{j}\|_{BMO}. $ |
Lemma 3.1. [9,21] Let $ 0 < p < r < \infty $, then there exists a constant $ C = C_{p, r} > 0 $ such that for any measurable function $ f $ there has
$ |Q|^{-1/p}\|f\|_{L^{p}(Q)}\leq C|Q|^{-1/r}\|f\|_{L^{r, \infty}(Q)}. $ |
Lemma 3.2. Let $ T $ be an $ m $-linear strongly singular integral operator with generalized kernel and $ p_{0}'\geq \max \{r_{1}, \dots, r_{m}, l_{1}, \dots, l_{m}\} $, $ 1/p_{0} +1/p_{0}' = 1 $, where $ p_{0} $, $ r_{j} $ and $ l_{j} $ are given by Definition $ 1.2 $, $ j = 1, \dots, m $. If $ 0 < \delta < 1/m $, then
$ M^{\sharp}_{\delta}(T(\vec{f}))(x)\leq C\prod\limits_{j = 1}^{m}M_{p_0'}(f_{j})(x), $ |
for all $ m $-tuples $ \vec{f} = (f_{1}, \dots, f_{m}) $ of bounded measurable functions with compact support.
Proof of Lemma $ 3.2 $. Let us just consider the case $ m = 2 $ in order to simplify the proof. In fact, a similar procedure can be applied to all other situations.
Let $ f_1, f_2 $ be bounded measurable functions with compact support. Then for any ball $ B = B(x_{0}, r_{B}) $ containing $ x $ with $ r_{B} > 0 $, we will take into account two cases, respectively.
Case 1: $ r_{B}\geq \frac{1}{4} $.
Write
$ f_{1} = f_{1}\chi_{32B}+f_{1}\chi_{(32B)^{c}}: = f_{1}^{1}+f_{1}^{2}, \quad f_{2} = f_{2}\chi_{32B}+f_{2}\chi_{(32B)^{c}}: = f_{2}^{1}+f_{2}^{2}. $ |
Choose a $ z_{0} \in 6B\setminus5B $ and take a $ c_{0} = T(f_{1}^{2}, f_{2}^{1})(z_{0})+T(f_{1}^{1}, f_{2}^{2})(z_{0})+T(f_{1}^{2}, f_{2}^{2})(z_{0}) $, then
$ \begin{eqnarray*} &&\biggl(\dfrac{1}{|B|}\int_{B}\bigl||T(f_{1}, f_{2})(z)|^{\delta}-|c_{0}|^{\delta}\bigr|dz\biggr)^{1/\delta}\\ &&\leq \biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(z)-c_{0}|^{\delta}dz\biggr)^{1/\delta}\\ &&\leq C\biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}^{1}, f_{2}^{1})(z)|^{\delta}dz\biggr)^{1/\delta}+C\biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{1})(z)-T(f_{1}^{2}, f_{2}^{1})(z_{0})|^{\delta}dz\biggr)^{1/\delta}\\ &&+C\biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}^{1}, f_{2}^{2})(z)-T(f_{1}^{1}, f_{2}^{2})(z_{0})|^{\delta}dz\biggr)^{1/\delta}+C\biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{2})(z)-T(f_{1}^{2}, f_{2}^{2})(z_{0})|^{\delta}dz\biggr)^{1/\delta}\\ &&: = \sum\limits_{j = 1}^{4}I_{j}. \end{eqnarray*} $ |
Notice that $ 0 < \delta < r < \infty $, where $ r $ is given as in Definition $ 1.2 $. By Lemma $ 3.1 $ and the boundedness of $ T $ from $ L^{r_{1}}\times L^{r_{2}} $ to $ L^{r, \, \infty} $, we have
$ \begin{eqnarray*} I_{1}&\leq& C|B|^{-1/\delta}\|T(f_{1}^{1}, f_{2}^{1})\|_{L^{\delta}(B)}\\ &\leq& C|B|^{-1/r}\|T(f_{1}^{1}, f_{2}^{1})\|_{L^{r, \infty}(B)}\\ &\leq& C\biggl(\dfrac{1}{|32B|}\int_{32B}|f_{1}(y_{1})|^{r_{1}}dy_{1}\biggr)^{\frac{1}{r_{1}}} \biggl(\dfrac{1}{|32B|}\int_{32B}|f_{2}(y_{2})|^{r_{2}}dy_{2}\biggr)^{\frac{1}{r_{2}}}\\ &\leq& CM_{r_{1}}(f_{1})(x)M_{r_{2}}(f_{2})(x)\\ &\leq& CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). \end{eqnarray*} $ |
For $ z\in B $ and $ y_{1}\in (32B)^{c} $, there are $ 2|z-z_{0}|\leq |y_{1}-z_{0}| $, $ 4r_{B}\leq |z-z_{0}|\leq 7r_{B} $ and $ |z-z_{0}|\geq1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} I_{2}&\leq& C\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{1})(z)-T(f_{1}^{2}, f_{2}^{1})(z_{0})|dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{2^k|z-z_{0}|\leq|y_{1}-z_{0}|\leq2^{k+1}|z-z_{0}|}\int_{32B} |K(z, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})||f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\Big(\int_{32B}\Big(\int_{2^k|z-z_{0}|\leq|y_{1}-z_{0}|\leq2^{k+1}|z-z_{0}|} |K(z, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|^{p_0}dy_{1}\Big)^{\frac{1}{p_0}}\\ &&\times\Big(\int_{2^k|z-z_{0}|\leq|y_{1}-z_{0}|\leq2^{k+1}|z-z_{0}|}|f_{1}(y_{1})|^{p_0'}dy_{1}\Big)^{\frac{1}{p_0'}} |f_{2}(y_{2})|dy_{2}\Big)dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}|B| 2^{-k(n(2-{\frac{1}{p_0}})+{\frac{\varepsilon}{\alpha}})}|z-z_{0}|^{\varepsilon-(n(2-{\frac{1}{p_0}})+{\frac{\varepsilon}{\alpha}})}(2^{k+1}|z-z_{0}|)^{\frac{n}{p_0'}}dz M_{p_0'}(f_{1})(x)M(f_{2})(x)\\ &\leq &C\sum^{\infty}_{k = 1}2^{-k(n+{\frac{\varepsilon}{\alpha}})}\int_{B} |z-z_{0}|^{-n-{\frac{\varepsilon}{\alpha}}+\varepsilon}dzM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq &C\sum^{\infty}_{k = 1}2^{-k(n+{\frac{\varepsilon}{\alpha}})} r_B^{\varepsilon-{\frac{\varepsilon}{\alpha}}}M_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq&CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). \end{eqnarray*} $ |
Similarly we can also get that
$ I_{3}\leq CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). $ |
For $ z\in B $ and $ y_{1}, y_{2}\in (32B)^{c} $, there are $ 2|z-z_{0}|\leq |y_{1}-z_{0}| $, $ 2|z-z_{0}|\leq |y_{2}-z_{0}| $, $ 4r_{B}\leq |z-z_{0}|\leq 7r_{B} $ and $ |z-z_{0}|\geq1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} I_{4}&\leq& C\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{2})(z)-T(f_{1}^{2}, f_{2}^{2})(z_{0})|dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\int_{2^{k_1}|z-z_{0}|\leq|y_{1}-z_{0}|\leq2^{{k_1}+1}|z-z_{0}|}\int_{2^{k_2}|z-z_{0}|\leq|y_{2}-z_{0}|\leq2^{{k_2}+1}|z-z_{0}|}|K(z, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|\\ &&\times|f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\Big(\int_{2^{k_1}|z-z_{0}|\leq|y_{1}-z_{0}|\leq2^{{k_1}+1}|z-z_{0}|}\int_{2^{k_2}|z-z_{0}|\leq|y_{2}-z_{0}|\leq2^{{k_2}+1}|z-z_{0}|}|K(z, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|^{p_0}\\ && dy_{2}dy_{1}\Big)^{\frac{1}{p_0}}\Big(\int_{2^{k_1}|z-z_{0}|\leq|y_{1}-z_{0}|\leq2^{{k_1}+1}|z-z_{0}|}|f_{1}(y_{1})|^{p_0'}dy_{1}\Big)^{\frac{1}{p_0'}}\Big(\int_{2^{k_2}|z-z_{0}|\leq|y_{2}-z_{0}|\leq2^{{k_2}+1}|z-z_{0}|}|f_{2}(y_{2})|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}}dz\\ &\leq&C\dfrac{1}{|B|}\int_{B}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}2^{-k_{i}(n(1-\frac{1}{p_0})+\frac{\varepsilon}{2\alpha})}|z-z_{0}|^{\varepsilon-(n(2-\frac{2}{p_0})+\frac{\varepsilon}{\alpha})}\prod^{2}_{i = 1}(2^{k_i+1}|z-z_{0}|)^{\frac{n}{p_0'}}dz M_{p_0'}(f_{1})(x)\\ &&\times M_{p_0'}(f_{2})(x)\\ &\leq&C\dfrac{1}{|B|}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1}\prod^{2}_{i = 1} 2^{\frac{-k_i\varepsilon}{2\alpha}}\int_{B}|z-z_{0}|^{\varepsilon-{\frac{\varepsilon}{\alpha}}}dzM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq&C\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1}\prod^{2}_{i = 1} 2^{\frac{-k_i\varepsilon}{2\alpha}}r_B^{\varepsilon-{\frac{\varepsilon}{\alpha}}}M_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq& CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). \end{eqnarray*} $ |
Case 2: $ 0 < r_{B} < \frac{1}{4} $.
Denote by $ \tilde{B} = B(x_{0}, r_{B}^{\alpha}) $. Write
$ f_{1} = f_{1}\chi_{16\tilde{B}}+f_{1}\chi_{(16\tilde{B})^{c}}: = \tilde{f}_{1}^{1}+\tilde{f}_{1}^{2}, \quad f_{2} = f_{2}\chi_{16\tilde{B}}+f_{2}\chi_{(16\tilde{B})^{c}}: = \tilde{f}_{2}^{1}+\tilde{f}_{2}^{2}. $ |
Choose a $ \tilde{z}_{0} \in 3B\setminus2B $ and take a $ \tilde{c}_{0} = T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(\tilde{z}_{0})+T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{2})(\tilde{z}_{0}) +T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(\tilde{z}_{0}) $, then
$ \begin{eqnarray*} &&\biggl(\dfrac{1}{|B|}\int_{B}\bigl||T(f_{1}, f_{2})(z)|^{\delta}-|\tilde{c}_{0}|^{\delta}\bigr|dz\biggr)^{1/\delta}\\ &&\leq C\biggl(\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{1})(z)|^{\delta}dz\biggr)^{1/\delta}+C\biggl(\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(z)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(\tilde{z_{0}})|^{\delta}dz\biggr)^{1/\delta}\\ &&+C\biggl(\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{2})(z)-T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{2})(\tilde{z}_{0})|^{\delta}dz\biggr)^{1/\delta} +C\biggl(\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(z)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(\tilde{z}_{0})|^{\delta}dz\biggr)^{1/\delta}\\ &&: = \sum\limits_{j = 1}^{4}\tilde{I}_{j}. \end{eqnarray*} $ |
Notice that $ 0 < \delta < q < \infty $ and $ 0 < l/q\leq \alpha $, where $ l $ and $ q $ are given as in Definition 1.2. By Lemma 3.1, we have
$ \begin{eqnarray*} \tilde{I}_{1}&\leq& C|B|^{-1/\delta}\|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{1})\|_{L^{\delta}(B)}\\ &\leq& C|B|^{-1/q}\|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{1})\|_{L^{q, \infty}(B)}\\ &\leq& C|B|^{-1/q}|\tilde{B}|^{1/l} \biggl(\dfrac{1}{|16\tilde{B}|}\int_{16\tilde{B}}|f_{1}(y_{1})|^{l_{1}}dy_{1}\biggr)^{\frac{1}{l_{1}}} \biggl(\dfrac{1}{|16\tilde{B}|}\int_{16\tilde{B}}|f_{2}(y_{2})|^{l_{2}}dy_{2}\biggr)^{\frac{1}{l_{2}}}\\ &\leq& Cr_{B}^{n({\frac{\alpha}{l}}-{\frac{1}{q}})}M_{l_{1}}(f_{1})(x)M_{l_{2}}(f_{2})(x)\\ &\leq& CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). \end{eqnarray*} $ |
For $ z\in B $ and $ y_{1}\in (16\tilde{B})^{c} $, there are $ |z-\tilde{z}_{0}|^{\alpha}\leq (4r_{B})^{\alpha}\leq \dfrac{1}{2}|y_{1}-\tilde{z}_{0}| $, $ r_{B}\leq |z-\tilde{z}_{0}|\leq 4r_{B} $ and $ |z-\tilde{z}_{0}| < 1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} \tilde{I}_{2}&\leq& C\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(z)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(\tilde{z}_{0})|dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{2^k|z-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k+1}|z-\tilde{z}_{0}|^\alpha}\int_{16\tilde{B}} |K(z, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})||f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{16\tilde{B}}\Big(\int_{2^k|z-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k+1}|z-\tilde{z}_{0}|^\alpha} |K(z, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})|^{p_0}dy_{1}\Big)^{\frac{1}{p_0}}\\ &&\times\Big(\int_{2^k|z-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k+1}|z-\tilde{z}_{0}|^\alpha}|f_{1}(y_{1})|^{p_0'}dy_{1}\Big)^{\frac{1}{p_0'}} |f_{2}(y_{2})|dy_{2}dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}|\tilde{B}| 2^{-k(n(2-\frac{1}{p_0})+\frac{\varepsilon}{\alpha})}|z-\tilde{z}_{0}|^{\varepsilon-{\alpha}(n(2-\frac{1}{p_0})+\frac{\varepsilon}{\alpha})}(2^{k+1}|z-\tilde{z}_{0}|^\alpha)^{\frac{n}{p_0'}}dz M_{p_0'}(f_{1})(x)M(f_{2})(x)\\ &\leq &C\dfrac{1}{|B|}\sum^{\infty}_{k = 1}|\tilde{B}|2^{-k(n+\frac{\varepsilon}{\alpha})}\int_{B} |z-\tilde{z}_{0}|^{-{\alpha}n}dzM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq&CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). \end{eqnarray*} $ |
Similarly we can also get that
$ \tilde{I}_{3}\leq CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). $ |
For $ z\in B $ and $ y_{1}, y_{2}\in (16\tilde{B})^{c} $, there are $ 2|z-\tilde{z}_{0}|^{\alpha}\leq|y_{1}-\tilde{z}_{0}| $, $ 2|z-\tilde{z}_{0}|^{\alpha}\leq|y_{2}-\tilde{z}_{0}| $, $ r_{B}\leq |z-\tilde{z}_{0}|\leq 4r_{B} $ and $ |z-\tilde{z}_{0}| < 1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} \tilde{I}_{4}&\leq& C\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(z)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(\tilde{z}_{0})|dz\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\Big(\int_{2^{k_1}|z-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k_1+1}|z-\tilde{z}_{0}|^\alpha}\int_{2^{k_2}|z-\tilde{z}_{0}|^\alpha\leq|y_{2}-\tilde{z}_{0}|\leq2^{k_2+1}|z-\tilde{z}_{0}|^\alpha}\\ &&|K(z, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})|^{p_0}dy_{2}dy_{1}\Big)^{\frac{1}{p_0}}\Big(\int_{2^{k_1}|z-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{{k_1}+1}|z-\tilde{z}_{0}|^\alpha}|f_{1}(y_{1})|^{p_0'}dy_{1}\Big)^{\frac{1}{p_0'}}\\ &&\times\Big(\int_{2^{k_2}|z-\tilde{z}_{0}|^\alpha\leq|y_{2}-\tilde{z}_{0}|\leq2^{{k_2}+1}|z-\tilde{z}_{0}|^\alpha }|f_{2}(y_{2})|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}}dz\\ &\leq&C\dfrac{1}{|B|}\int_{B}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}2^{-k_{i}(n(1-\frac{1}{p_0})+\frac{\varepsilon}{2\alpha})}|z-\tilde{z}_{0}|^{\varepsilon-\alpha(n(2-\frac{2}{p_0})+\frac{\varepsilon}{\alpha})}\prod^{2}_{i = 1}(2^{k_i+1}|z-\tilde{z}_{0}|^\alpha)^{\frac{n}{p_0'}}dz \\ &&\times M_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq&C\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1}\prod^{2}_{i = 1} 2^\frac{-k_i\varepsilon}{2\alpha}M_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x)\\ &\leq& CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x). \end{eqnarray*} $ |
Thus, according to the estimates in both cases, there is
$ \begin{eqnarray*} &M^{\sharp}_{\delta}(T(f_{1}, f_{2}))(x) \sim \sup\limits_{B\ni x} \inf\limits_{a\in \mathbb{C}}\biggl(\dfrac{1}{|B|}\int_{B}\bigl||T(f_{1}, f_{2})(z)|^{\delta}-a\bigr|dz\biggr)^{1/\delta}\\ &\leq CM_{p_0'}(f_{1})(x)M_{p_0'}(f_{2})(x), \end{eqnarray*} $ |
which completes the proof of the lemma.
Lemma 3.3. [9] Let $ 0 < p, \delta < \infty $ and $ w\in A_{\infty} $. Then there exists a constant $ C > 0 $ depending only on the $ A_{\infty} $ constant of $ w $ such that
$ \int_{\mathbb{R}^{n}}[M_{\delta}(f)(x)]^{p}w(x)dx\leq C\int_{\mathbb{R}^{n}}[M_{\delta}^{\sharp}(f)(x)]^{p}w(x)dx, $ |
for every function $ f $ such that the left-hand side is finite.
Lemma 3.4. [22] For $ (w_{1}, \dots, w_{m})\in(A_{p_{1}}, \dots, A_{p_{m}}) $ with $ 1\leq p_{1}, \dots, p_{m} < \infty $, and for $ 0 < \theta_{1}, \dots, \theta_{m} < 1 $ such that $ \theta_{1}+\cdots+\theta_{m} = 1 $, we have $ w_{1}^{\theta_{1}}\cdots w_{m}^{\theta_{m}}\in A_{\max\{p_{1}, \dots, p_{m}\}} $.
Lemma 3.5. [23] Let $ p(\cdot)\in\mathcal{P}(\mathbb{R}^{n}) $. Then $ M_{q_{0}} $ is bounded on $ L^{p(\cdot)}(\mathbb{R}^{n}) $ for some $ 1 < q_{0} < \infty $ if and only if $ M $ is bounded on $ L^{p(\cdot)}(\mathbb{R}^{n}). $
Lemma 3.6. [24] Given a family $ \mathcal{F} $ of ordered pairs of measurable functions, suppose for some fixed $ 0 < \tilde{p_{0} } < \infty $, every $ (f, g)\in \mathcal{F} $ and every $ w\in A_{1} $,
$ \int_{\mathbb{R}^{n}}|f(x)|^{\tilde{p_{0} }}w(x)dx\leq C_{0}\int_{\mathbb{R}^{n}}|g(x)|^{\tilde{p_{0} }}w(x)dx. $ |
Let $ p(\cdot)\in\mathcal{P}(\mathbb{R}^{n}) $ with $ \tilde{p_{0} }\leq p_{-} $. If $ \bigl(\frac{p(\cdot)}{\tilde{p_{0} }}\bigr)'\in\mathcal{B}(\mathbb{R}^{n}) $, then there exists a constant $ C > 0 $ such that for all $ (f, g)\in \mathcal{F} $, $ \|f\|_{L^{p(\cdot)}(\mathbb{R}^{n})}\leq C\|g\|_{L^{p(\cdot)}(\mathbb{R}^{n})} $.
Lemma 3.7. [25] Let $ p(\cdot), p_{1}(\cdot), \dots, p_{m}(\cdot)\in\mathcal{P}(\mathbb{R}^{n}) $ with $ 1/p(x) = 1/p_{1}(x)+\cdots+1/p_{m}(x) $. Then for any $ f_{j}\in L^{p_{j}(\cdot)}(\mathbb{R}^{n}) $, $ j = 1, \dots, m $, there has
$ \biggl\|\prod\limits_{j = 1}^{m}f_{j}\biggr\|_{L^{p(\cdot)}(\mathbb{R}^{n})}\leq 2^{m-1}\prod\limits_{j = 1}^{m}\|f_{j}\|_{L^{p_{j}(\cdot)}(\mathbb{R}^{n})}. $ |
Lemma 3.8. [23] Let $ p(\cdot)\in\mathcal{P}(\mathbb{R}^{n}) $. Then the following conditions are equivalent.
${\rm(i)} \ p(\cdot)\in\mathcal{B}(\mathbb{R}^{n}) $.
${\rm(ii)} \ p'(\cdot)\in\mathcal{B}(\mathbb{R}^{n}) $.
${\rm(iii)} \ \frac{p(\cdot)}{\tilde{p_{0} }}\in\mathcal{B}(\mathbb{R}^{n}) $ for some $ 1 < \tilde{p_{0} } < p_{-} $.
${\rm(iv)} \ \bigl(\frac{p(\cdot)}{\tilde{p_{0} }}\bigr)'\in\mathcal{B}(\mathbb{R}^{n}) $ for some $ 1 < \tilde{p_{0} } < p_{-} $.
Lemma 3.9. [24] If $ p(\cdot)\in\mathcal{P}(\mathbb{R}^{n}) $, then $ C^{\infty}_{c}(\mathbb{R}^{n}) $ is dense in $ L^{p(\cdot)}(\mathbb{R}^{n}) $.
Lemma 3.10. [26] Let $ f $ be a function in $ BMO $. Suppose $ 1\leq p < \infty $, $ x\in \mathbb{R}^{n} $, and $ r_{1}, r_{2} > 0 $. Then
$ \biggl(\frac{1}{|B(x, r_{1})|}\int_{B(x, \, r_{1})}|f(y)-f_{B(x, \, r_{2})}|^{p} dy\biggr)^{1/p}\leq C\biggl(1+\bigl|\ln\frac{r_{1}}{r_{2}}\bigr|\biggr)\|f\|_{ BMO}, $ |
where $ C > 0 $ is independent of $ f $, $ x $, $ r_{1} $ and $ r_{2} $.
Proof of Theorem $ 2.1 $. From the fact that $ w_{j}\in A_{p_{j}/p_{0}'} $ and $ p_{j} > p_{0}' $ ($ j = 1, \dots, m $), it follows that the Hardy-Littlewood maximal operator $ M $ is bounded on $ L^{p_{j}/p_{0}'}(w_{j}) $. By Lemma 3.4, we obtain $ w\in A_{\max\{p_{1}/p_{0}', \dots, p_{m}/p_{0}'\}}\subset A_{\infty} $.
Take a $ \delta $ such that $ 0 < \delta < 1/m $, then by Lemma 3.2 and Lemma 3.3, we have
$ \begin{eqnarray*} &&\|T(\vec{f})\|_{L^{p}(w)}\leq\|M_{\delta}(T(\vec{f}))\|_{L^{p}(w)}\leq C\|M_{\delta}^{\sharp}(T(\vec{f}))\|_{L^{p}(w)}\\ &&\leq C\biggl\|\prod\limits_{j = 1}^{m}M_{p_{0}'}(f_{j})\biggr\|_{L^{p}(w)}\leq C\prod\limits_{j = 1}^{m}\|M_{p_{0}'}(f_{j})\|_{L^{p_{j}}(w_{j})}\\ && = C\prod\limits_{j = 1}^{m}\|M(|f_{j}|^{p_{0}'})\|_{L^{p_{j}/p_{0}'}(w_{j})}^{1/p_{0}'}\leq C\prod\limits_{j = 1}^{m}\bigl\||f_{j}|^{p_{0}'}\bigr\|_{L^{p_{j}/p_{0}'}(w_{j})}^{1/p_{0}'}\\ && = C\prod\limits_{j = 1}^{m}\|f_{j}\|_{L^{p_{j}}(w_{j})}. \end{eqnarray*} $ |
This completes the proof of the theorem.
Proof of Theorem $ 2.2 $. Since $ p(\cdot)\in\mathcal{B}(\mathbb{R}^{n}) $, then by Lemma 3.8, there exists a $ \tilde{p_{0} } $ such that $ 1 < \tilde{p_{0} } < p_{-} $ and $ \bigl(\frac{p(\cdot)}{\tilde{p_{0} }}\bigr)'\in\mathcal{B}(\mathbb{R}^{n}) $. Take a $ \delta $ such that $ 0 < \delta < 1/m $. For any $ w\in A_{1} $, by Lemma 3.2 and Lemma 3.3, we can get that
$ \begin{eqnarray*} \int_{\mathbb{R}^{n}}|T(\vec{f})(x)|^{\tilde{p_{0} }}w(x)dx &\leq&\int_{\mathbb{R}^{n}}[M_{\delta}(T(\vec{f}))(x)]^{\tilde{p_{0} }}w(x)dx\\ &\leq& C\int_{\mathbb{R}^{n}}[M_{\delta}^{\sharp}(T(\vec{f}))(x)]^{\tilde{p_{0} }}w(x)dx\\ &\leq& C\int_{\mathbb{R}^{n}}\biggl[\prod\limits_{j = 1}^{m}M_{p_0'}(f_{j})(x)\biggr]^{\tilde{p_{0} }}w(x)dx\\ &\leq& C\int_{\mathbb{R}^{n}}\biggl[\prod\limits_{j = 1}^{m}M_{q_{0}^{j}}(f_{j})(x)\biggr]^{\tilde{p_{0} }}w(x)dx \end{eqnarray*} $ |
holds for all $ m $-tuples $ \vec{f} = (f_{1}, \dots, f_{m}) $ of bounded measurable functions with compact support.
By Lemma 3.9, we can get that $ C^{\infty}_{c}(\mathbb{R}^{n}) $ is dense in $ L^{p_j(\cdot)}(\mathbb{R}^{n})(j = 1, 2, \cdots, m) $. For $ f_j\in C^{\infty}_{c}(\mathbb{R}^{n}) $, applying Lemma 3.6 to the pair $ \bigl(T(\vec{f}), \prod_{j = 1}^{m}M_{q_{0}^{j}}(f_{j})\bigr) $, we have
$ \|T(\vec{f})\|_{L^{p(\cdot)}(\mathbb{R}^{n})}\leq C\biggl\|\prod\limits_{j = 1}^{m}M_{q_{0}^{j}}(f_{j})\biggr\|_{L^{p(\cdot)}(\mathbb{R}^{n})}. $ |
Then by Lemma 3.5 and Lemma 3.7, we have
$ \|T(\vec{f})\|_{L^{p(\cdot)}(\mathbb{R}^{n})}\leq C\prod\limits_{j = 1}^{m}\|M_{q_{0}^{j}}(f_{j})\|_{L^{p_{j}(\cdot)}(\mathbb{R}^{n})} \leq C\prod\limits_{j = 1}^{m}\|f_{j}\|_{L^{p_{j}(\cdot)}(\mathbb{R}^{n})}. $ |
Since the denseness of $ C^{\infty}_{c}(\mathbb{R}^{n}) $ in $ L^{p_j(\cdot)}(\mathbb{R}^{n})(j = 1, 2, \cdots, m) $, this completes the proof of the theorem.
Proof of Theorem $ 2.3 $. To simplify the proof, we only give the proof when $ m = 2 $ since their similarities.
Let $ f_1, f_2\in L^{\infty} $, then for any ball $ B = B(x_{0}, r_{B}) $ with $ r_{B} > 0 $, we will think about two cases, respectively.
Case 1: $ r_{B}\geq \frac{1}{4} $.
Write
$ f_{1} = f_{1}\chi_{32B}+f_{1}\chi_{(32B)^{c}}: = f_{1}^{1}+f_{1}^{2}, \quad f_{2} = f_{2}\chi_{32B}+f_{2}\chi_{(32B)^{c}}: = f_{2}^{1}+f_{2}^{2}. $ |
Choose a $ z_{0} \in 6B\setminus5B $ and take a $ c_{0} = T(f_{1}^{2}, f_{2}^{1})(z_{0})+T(f_{1}^{1}, f_{2}^{2})(z_{0})+T(f_{1}^{2}, f_{2}^{2})(z_{0}) $, then
$ \begin{eqnarray*} &&\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(x)-c_{0}|dx\\ &&\leq \dfrac{1}{|B|}\int_{B}|T(f_{1}^{1}, f_{2}^{1})(x)|dx+\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{1})(x)-T(f_{1}^{2}, f_{2}^{1})(z_{0})|dx\\ &&\quad+\dfrac{1}{|B|}\int_{B}|T(f_{1}^{1}, f_{2}^{2})(x)-T(f_{1}^{1}, f_{2}^{2})(z_{0})|dx+\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{2})(x)-T(f_{1}^{2}, f_{2}^{2})(z_{0})|dx\\ &&: = \sum\limits_{j = 1}^{4}J_{j}. \end{eqnarray*} $ |
Take $ p_{1}, \dots, p_{m} $ such that $ \max\{p_0', m\} < p_{1}, \dots, p_{m} < \infty $. Let $ 1/p = 1/p_{1}+\cdots+1/p_{m} $, then $ 1 < p < \infty $. By Theorem 2.1, we obtain that $ T $ is bounded from $ L^{p_{1}}\times \cdots \times L^{p_{m}} $ into $ L^{p} $.
By Hölder's inequality and the $ L^{p_{1}}\times L^{p_{2}}\rightarrow L^{p} $ boundedness of $ T $, we have
$ \begin{eqnarray*} J_{1} &\leq& \biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}^{1}, f_{2}^{1})(x)|^{p}dx\biggr)^{\frac{1}{p}}\\ &\leq& C|B|^{-1/p}\|f_{1}^{1}\|_{L^{p_{1}}}\|f_{2}^{1}\|_{L^{p_{2}}}\\ &\leq& C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. \end{eqnarray*} $ |
For $ x\in B $ and $ y_{1}\in (32B)^{c} $, there are $ 2|x-z_{0}|\leq |y_{1}-z_{0}| $, $ 4r_{B}\leq |x-z_{0}|\leq 7r_{B} $ and $ |x-z_{0}|\geq1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} J_{2} &\leq& \dfrac{1}{|B|}\int_{B} \int_{(32B)^{c}}\int_{32B} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})||f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dx\\ &\leq &\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{2^k|x-z_{0}|\leq|y_{1}-z_{0}|\leq2^{k+1}|x-z_{0}|}\int_{32B} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|dy_{2}dy_{1}dx\\ &\leq &\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{32B}\Big(\int_{2^k|x-z_{0}|\leq|y_{1}-z_{0}|\leq2^{k+1}|x-z_{0}|} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|^{p_0}dy_{1}\Big)^{\frac{1}{p_0}}\\ &&\times(2^{k+1}|x-z_{0}|)^{\frac{n}{p_0'}}dy_{2}dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq &C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}|B| 2^{-k(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}|x-z_{0}|^{\varepsilon-(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}(2^{k+1}|x-z_{0}|)^{\frac{n}{p_0'}}dx \\ &\leq &C r_B^{-\frac{\varepsilon}{\alpha}+\varepsilon}\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq &C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. \end{eqnarray*} $ |
Similarly we can also get that
$ J_{3}\leq C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. $ |
For $ x\in B $ and $ y_{1}, y_{2}\in (32B)^{c} $, there are $ 2|x-z_{0}|\leq |y_{1}-z_{0}| $, $ 2|x-z_{0}|\leq |y_{2}-z_{0}| $, $ 4r_{B}\leq |x-z_{0}|\leq 7r_{B} $ and $ |x-z_{0}|\geq1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} J_{4} &\leq& \dfrac{1}{|B|}\int_{B} \int_{(32B)^{c}}\int_{(32B)^{c}} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})||f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dx\\ &\leq &\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\Big(\int_{2^{k_1}|x-z_{0}|\leq|y_{1}-z_{0}|\leq2^{{k_1}+1}|x-z_{0}|}\int_{2^{k_2}|x-z_{0}|\leq|y_{2}-z_{0}|\leq2^{{k_2}+1}|x-z_{0}|}\\ && |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|^{p_0}dy_{2}dy_{1}\Big)^{\frac{1}{p_0}}\prod^{2}_{i = 1}(2^{k_i+1}|x-z_{0}|)^{\frac{n}{p_0}}dx\\ &\leq&C\dfrac{1}{|B|}\int_{B}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}2^{-k_{i}(n(1-{\frac{1}{p_0}})+\frac{\varepsilon}{2\alpha})}|x-z_{0}|^{\varepsilon-(n(2-{\frac{2}{p_0}})+\frac{\varepsilon}{\alpha})}\prod^{2}_{i = 1}(2^{k_i+1}|x-z_{0}|)^{\frac{n}{p_0}} dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq&C\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1}\prod^{2}_{i = 1} 2^\frac{-k_i\varepsilon}{2\alpha}\dfrac{1}{|B|}\int_{B}|x-z_{0}|^{\varepsilon-\frac{\varepsilon}{\alpha}}dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq& C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. \end{eqnarray*} $ |
Case 2: $ 0 < r_{B} < \frac{1}{4} $.
Denote by $ \tilde{B} = B(x_{0}, r_{B}^{\alpha}) $. Write
$ f_{1} = f_{1}\chi_{16\tilde{B}}+f_{1}\chi_{(16\tilde{B})^{c}}: = \tilde{f}_{1}^{1}+\tilde{f}_{1}^{2}, \quad f_{2} = f_{2}\chi_{16\tilde{B}}+f_{2}\chi_{(16\tilde{B})^{c}}: = \tilde{f}_{2}^{1}+\tilde{f}_{2}^{2}. $ |
Choose a $ \tilde{z}_{0} \in 3B\setminus2B $ and take a $ \tilde{c}_{0} = T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(\tilde{z}_{0})+T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{2})(\tilde{z}_{0}) +T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(\tilde{z}_{0}) $, then
$ \begin{eqnarray*} &&\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(x)-\tilde{c}_{0}|dx\\ &&\leq \dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{1})(x)|dx +\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(x)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{1})(\tilde{z}_{0})|dx\\ &&\quad+\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{2})(x)-T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{2})(\tilde{z}_{0})|dx +\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(x)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(\tilde{z}_{0})|dx\\ &&: = \sum\limits_{j = 1}^{4}\tilde{J}_{j}. \end{eqnarray*} $ |
Notice that $ 1 < q < \infty $ and $ 0 < l/q\leq \alpha $, where $ l $ is given as in Definition 1.2. By Lemma 3.1, we have
$ \begin{eqnarray*} \tilde{J}_{1}&\leq& C|B|^{-1}\|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{1})\|_{L^{1}(B)}\\ &\leq& C|B|^{-1/q}\|T(\tilde{f}_{1}^{1}, \tilde{f}_{2}^{1})\|_{L^{q, \infty}(B)}\\ &\leq& C|B|^{-1/q}\|\tilde{f}_{1}^{1}\|_{L^{l_{1}}}\|\tilde{f}_{2}^{1}\|_{L^{l_{2}}}\\ &\leq& C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}r_{B}^{n({\frac{\alpha}{l}}-{\frac{1}{q}})}\\ &\leq& C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. \end{eqnarray*} $ |
For $ x\in B $ and $ y_{1}\in (16\tilde{B})^{c} $, there are $ 2|x-\tilde{z}_{0}|^{\alpha}\leq |y_{1}-\tilde{z}_{0}| $, $ r_{B}\leq |x-\tilde{z}_{0}|\leq 4r_{B} $ and $ |x-\tilde{z}_{0}| < 1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} \tilde{J}_{2} &\leq &\dfrac{1}{|B|}\int_{B} \int_{(16\tilde{B})^{c}}\int_{16\tilde{B}} |K(x, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})|f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dx\\ &\leq &\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{16\tilde{B}}\Big(\int_{2^k|x-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k+1}|x-\tilde{z}_{0}|^\alpha} |K(x, y_{1}, y_{2})-K(x_{0}, y_{1}, y_{2})|^{p_0}dy_{1}\Big)^{\frac{1}{p_0}}\\ &&\times(2^{k+1}|x-\tilde{z}_{0}|^\alpha)^{\frac{n}{p_0'}} dy_{2}dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}|\tilde{B}| 2^{-k(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}|x-\tilde{z}_{0}|^{\varepsilon-{\alpha}(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}(2^{k+1}|x-\tilde{z}_{0}|^\alpha)^{\frac{n}{p_0'}}dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq &C \sum^{\infty}_{k = 1}|\tilde{B}|2^{-k(n+\frac{\varepsilon}{\alpha})}\dfrac{1}{|B|}\int_{B}|x-\tilde{z}_{0}|^{-{\alpha}n}dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq&C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. \end{eqnarray*} $ |
Similarly we can also get that
$ \tilde{J}_{3}\leq C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. $ |
For $ x\in B $ and $ y_{1}, y_{2}\in (16\tilde{B})^{c} $, there are $ |x-\tilde{z}_{0}|^{\alpha}\leq\frac{1}{2}|y_{1}-\tilde{z}_{0}| $, $ |x-\tilde{z}_{0}|^{\alpha}\leq\frac{1}{2}|y_{2}-\tilde{z}_{0}| $, $ r_{B}\leq |x-\tilde{z}_{0}|\leq 4r_{B} $ and $ |x-\tilde{z}_{0}| < 1 $. By Hölder's inequality and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} \tilde{J}_{4} &\leq &\dfrac{1}{|B|}\int_{B} \int_{(16\tilde{B})^{c}}\int_{(16\tilde{B})^{c}} |K(x, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})||f_{1}(y_{1})||f_{2}(y_{2})|dy_{2}dy_{1}dx\\ &\leq &\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\Big(\int_{2^{k_1}|x-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k_1+1}|x-\tilde{z}_{0}|^\alpha}\int_{2^{k_2}|x-\tilde{z}_{0}|^\alpha\leq|y_{2}-\tilde{z}_{0}|\leq2^{k_2+1}|x-\tilde{z}_{0}|^\alpha}\\ &&|K(x, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})|^{p_0}dy_{2}dy_{1}\Big)^{\frac{1}{p_0}} \prod^{2}_{i = 1}(2^{k_i+1}|x-\tilde{z}_{0}|^\alpha)^{\frac{n}{p_0'}}dx\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq&C\dfrac{1}{|B|}\int_{B}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}2^{-k_{i}(n(1-{\frac{1}{p_0}})+\frac{\varepsilon}{2\alpha}))}|x-\tilde{z}_{0}|^{\varepsilon-\alpha(n(2-{\frac{2}{p_0}})+\frac{\varepsilon}{\alpha})}\prod^{2}_{i = 1}(2^{k_i+1}|x-\tilde{z}_{0}|^\alpha)^{\frac{n}{p_0'}}dx \|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq&C\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1}\prod^{2}_{i = 1} 2^\frac{-k_i\varepsilon}{2\alpha}\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}\\ &\leq& C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}. \end{eqnarray*} $ |
Thus, according to the estimates in both cases, there is
$ \|T(f_{1}, f_{2})\|_{BMO}\sim \sup\limits_{B} \inf\limits_{a\in \mathbb{C}}\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(x)-a|dx \leq C\|f_{1}\|_{\infty}\|f_{2}\|_{\infty}, $ |
which completes the proof of the theorem.
Proof of Theorem $ 2.4 $. We only give the proof when $ m = 2 $.
Let $ f_1, f_2\in BMO $, then for any ball $ B = B(x_{0}, r_{B}) $ with $ r_{B} > 0 $, we will consider two cases, respectively.
Case 1: $ r_{B}\geq \frac{1}{4} $.
Write
$ f_{1} = (f_{1})_{32B}+(f_{1}-(f_{1})_{32B})\chi_{32B}+(f_{1}-(f_{1})_{32B})\chi_{(32B)^{c}}: = f_{1}^{1}+f_{1}^{2}+f_{1}^{3}, $ |
$ f_{2} = (f_{2})_{32B}+(f_{2}-(f_{2})_{32B})\chi_{32B}+(f_{2}-(f_{2})_{32B})\chi_{(32B)^{c}}: = f_{2}^{1}+f_{2}^{2}+f_{2}^{3}. $ |
It follows from the hypothesis of the theorem that
$ T(f_{1}, f_{2}) = T(f_{1}^{2}, f_{2}^{2})+T(f_{1}^{2}, f_{2}^{3})+T(f_{1}^{3}, f_{2}^{2})+T(f_{1}^{3}, f_{2}^{3}). $ |
Choose a $ z_{0} \in 6B\setminus5B $ and take a $ d_{0} = T(f_{1}^{2}, f_{2}^{3})(z_{0})+T(f_{1}^{3}, f_{2}^{2})(z_{0})+T(f_{1}^{3}, f_{2}^{3})(z_{0}) $, then
$ \begin{eqnarray*} &&\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(x)-d_{0}|dx\\ &&\leq \dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{2})(x)|dx+\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{3})(x)-T(f_{1}^{2}, f_{2}^{3})(z_{0})|dx\\ &&\quad+\dfrac{1}{|B|}\int_{B}|T(f_{1}^{3}, f_{2}^{2})(x)-T(f_{1}^{3}, f_{2}^{2})(z_{0})|dx +\dfrac{1}{|B|}\int_{B}|T(f_{1}^{3}, f_{2}^{3})(x)-T(f_{1}^{3}, f_{2}^{3})(z_{0})|dx\\ &&: = \sum\limits_{j = 1}^{4}L_{j}. \end{eqnarray*} $ |
Take $ p_{1}, \dots, p_{m} $ such that $ \max\{p_0', m\} < p_{1}, \dots, p_{m} < \infty $. Let $ 1/p = 1/p_{1}+\cdots+1/p_{m} $, then $ 1 < p < \infty $. By Theorem 2.1, we obtain the boundedness of $ T $ from $ L^{p_{1}}\times \cdots \times L^{p_{m}} $ into $ L^{p} $. By Hölder's inequality, we have
$ \begin{eqnarray*} L_{1} &\leq& \biggl(\dfrac{1}{|B|}\int_{B}|T(f_{1}^{2}, f_{2}^{2})(x)|^{p}dx\biggr)^{\frac{1}{p}}\\ &\leq& C\biggl(\dfrac{1}{|32B|}\int_{32B}|f_{1}(y_{1})-(f_{1})_{32B}|^{p_{1}}dy_{1}\biggr)^{\frac{1}{p_{1}}} \biggl(\dfrac{1}{|32B|}\int_{32B}|f_{2}(y_{2})-(f_{2})_{32B}|^{p_{2}}dy_{2}\biggr)^{\frac{1}{p_{2}}}\\ &\leq& C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. \end{eqnarray*} $ |
For $ x\in B $ and $ y_{2}\in (32B)^{c} $, there are $ 2|x-z_{0}|\leq |y_{2}-z_{0}| $, $ 4r_{B}\leq |x-z_{0}|\leq 7r_{B} $ and $ |x-z_{0}|\geq1 $. By Hölder's inequality, Lemma 3.10 and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} L_{2} &\leq& \dfrac{1}{|B|}\int_{B} \int_{(32B)^{c}}\int_{32B} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})||f_{1}(y_{1})-(f_{1})_{32B}||f_{2}(y_{2})-(f_{2})_{32B}|dy_{1}dy_{2}dx\\ &\leq & \dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{32B}\Big(\int_{2^k|x-z_{0}|\leq|y_{2}-z_{0}|\leq2^{k+1}|x-z_{0}|} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|^{p_0}dy_{2}\Big)^{\frac{1}{p_0}}\\ &&\times\Big(\int_{2^k|x-z_{0}|\leq|y_{2}-z_{0}|\leq2^{k+1}|x-z_{0}|}|f_{2}(y_{2})-(f_{2})_{32B}|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}} |f_{1}(y_{1})-(f_{1})_{32B}|dy_{1}dx\\ &\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}|B| 2^{-k(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}|x-z_{0}|^{\varepsilon-(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}dx \Big(\int_{2^{k+5}B}|f_{2}(y_{2})-(f_{2})_{32B}|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}}\|f_{1}\|_{BMO} \\ &\leq &C r_B^{\frac{n}{p_0'}}\sum^{\infty}_{k = 1}2^{-k(n+\frac{\varepsilon}{\alpha})}\Big(1+\ln\dfrac{2^{k+5}r_B}{32r_B}\Big)\int_{B}|x-z_{0}|^{-2n-\frac{\varepsilon}{\alpha}+\varepsilon+{\frac{n}{p_0}}}dx \|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq &C\sum^{\infty}_{k = 1}k2^{-k(n+\frac{\varepsilon}{\alpha})} r_B^{-\frac{\varepsilon}{\alpha}+\varepsilon}\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq& C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. \end{eqnarray*} $ |
Similarly we also can get that
$ L_{3}\leq C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. $ |
For $ x\in B $ and $ y_{1}, y_{2}\in (32B)^{c} $, there are $ 2|x-z_{0}|\leq |y_{1}-z_{0}| $, $ 2|x-z_{0}|\leq |y_{2}-z_{0}| $, $ 4r_{B}\leq |x-z_{0}|\leq 7r_{B} $ and $ |x-z_{0}|\geq1 $. By Hölder's inequality, Lemma 3.10 and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} L_{4} &\leq& \dfrac{1}{|B|}\int_{B} \int_{(32B)^{c}}\int_{(32B)^{c}} |K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})||f_{1}(y_{1})-(f_{1})_{32B}|\\ &&\times|f_{2}(y_{2})-(f_{2})_{32B}|dy_{2}dy_{1}dx\\ &\leq &\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\Big(\int_{2^{k_1}|x-z_{0}|\leq|y_{1}-z_{0}|\leq2^{{k_1}+1}|x-z_{0}|}\int_{2^{k_2}|x-z_{0}|\leq|y_{2}-z_{0}|\leq2^{{k_2}+1}|x-z_{0}|}\\ &&|K(x, y_{1}, y_{2})-K(z_{0}, y_{1}, y_{2})|^{p_0}dy_{2}dy_{1}\Big)^{\frac{1}{p_0}}\Big(\int_{2^{k_1+5}B}|f_{1}(y_{1})-(f_{1})_{32B}|^{p_0'}dy_{1}\Big)^{\frac{1}{p_0'}}\\ &&\times\Big(\int_{2^{k_2+5}B}|f_{2}(y_{2})-(f_{2})_{32B}|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}}dx\\ &\leq&C\dfrac{1}{|B|}\int_{B}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}2^{-k_{i}(n(1-{\frac{1}{p_0}})+\frac{\varepsilon}{2\alpha})}|x-z_{0}|^{\varepsilon-(n(2-{\frac{2}{p_0}})+\frac{\varepsilon}{\alpha})}\prod^{2}_{i = 1}\Big((2^{k_i+5}r_B)^{\frac{n}{p_0'}}\\ &&\times\Big(1+\ln\dfrac{2^{k_i+5}r_B}{32r_B}\Big)\Big)dx\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq&C\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}k_{i}2^{\frac{-k_{i}\varepsilon}{2\alpha}}r_B^{-\frac{\varepsilon}{\alpha}+\varepsilon}\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq& C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. \end{eqnarray*} $ |
Case 2: $ 0 < r_{B} < \frac{1}{4} $.
Denote by $ \tilde{B} = B(x_{0}, r_{B}^{\alpha}) $. Write
$ f_{1} = (f_{1})_{16\tilde{B}}+(f_{1}-(f_{1})_{16\tilde{B}})\chi_{16\tilde{B}}+(f_{1}-(f_{1})_{16\tilde{B}})\chi_{(16\tilde{B})^{c}} : = \tilde{f}_{1}^{1}+\tilde{f}_{1}^{2}+\tilde{f}_{1}^{3}, $ |
$ f_{2} = (f_{2})_{16\tilde{B}}+(f_{2}-(f_{2})_{16\tilde{B}})\chi_{16\tilde{B}}+(f_{2}-(f_{2})_{16\tilde{B}})\chi_{(16\tilde{B})^{c}} : = \tilde{f}_{2}^{1}+\tilde{f}_{2}^{2}+\tilde{f}_{2}^{3}. $ |
It follows from the hypothesis of the theorem that
$ T(f_{1}, f_{2}) = T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})+T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{3})+T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{2}) +T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{3}). $ |
Choose a $ \tilde{z}_{0} \in 3B\setminus2B $ and take a $ \tilde{d}_{0} = T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{3})(\tilde{z}_{0})+T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{2})(\tilde{z}_{0}) +T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{3})(\tilde{z}_{0}) $, then
$ \begin{eqnarray*} &&\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(x)-\tilde{d}_{0}|dx\\ &&\leq \dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})(x)|dx +\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{3})(x)-T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{3})(\tilde{z}_{0} )|dx\\ &&\quad+\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{2})(x)-T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{2})(\tilde{z}_{0} )|dx +\dfrac{1}{|B|}\int_{B}|T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{3})(x)-T(\tilde{f}_{1}^{3}, \tilde{f}_{2}^{3})(\tilde{z}_{0} )|dx\\ &&: = \sum\limits_{j = 1}^{4}\tilde{L}_{j}. \end{eqnarray*} $ |
Notice that $ 1 < q < \infty $ and $ 0 < l/q\leq \alpha $, where $ l $ is given as in Definition 1.2. By Lemma 3.1, we have
$ \begin{eqnarray*} \tilde{L}_{1}&\leq& |B|^{-1}\|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})\|_{L^{1}(B)}\\ &\leq& C|B|^{-1/q}\|T(\tilde{f}_{1}^{2}, \tilde{f}_{2}^{2})\|_{L^{q, \infty}(B)}\\ &\leq& C|B|^{-1/q}|\tilde{B}|^{1/l}\biggl(\dfrac{1}{|16\tilde{B}|}\int_{16\tilde{B}}|f_{1}(y_{1})-(f_{1})_{16\tilde{B}}|^{l_{1}}dy_{1}\biggr)^{\frac{1}{l_{1}}}\biggl(\dfrac{1}{|16\tilde{B}|}\int_{16\tilde{B}}|f_{2}(y_{2})-(f_{2})_{16\tilde{B}}|^{l_{2}}dy_{2}\biggr)^{\frac{1}{l_{2}}}\\ &\leq& C r_{B}^{n({\frac{\alpha}{l}}-{\frac{1}{q}})}\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq& C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. \end{eqnarray*} $ |
For $ x\in B $ and $ y_{2}\in (16\tilde{B})^{c} $, there is $ 2|x-\tilde{z}_{0}|^{\alpha}\leq|y_{2}-\tilde{z}_{0}| $, $ r_{B}\leq |x-\tilde{z}_{0}|\leq 4r_{B} $ and $ |x-\tilde{z}_{0}| < 1 $. By Hölder's inequality, Lemma 3.10 and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} \tilde{L}_{2} &\leq &\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}\int_{16\tilde{B}}\Big(\int_{2^k|x-\tilde{z}_{0}|^\alpha\leq|y_{2}-\tilde{z}_{0}|\leq2^{k+1}|x-\tilde{z}_{0}|^\alpha} |K(x, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})|^{p_0}dy_{2}\Big)^{\frac{1}{p_0}}\\ &&\times\Big(\int_{2^{k+4}\tilde{B}}|f_{2}(y_{2})-(f_{2})_{16\tilde{B}}|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}}|f_{1}(y_{1})-(f_{1})_{16\tilde{B}}|dy_{1}dx\\&\leq &C\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k = 1}|\tilde{B}| 2^{-k(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}|x-\tilde{z}_{0}|^{\varepsilon-{\alpha}(n(2-{\frac{1}{p_0}})+\frac{\varepsilon}{\alpha})}dx(2^{k+4}r_B^\alpha)^{\frac{n}{p_0'}}\Big(1+\ln\dfrac{2^{k+4}r_B^\alpha}{16r_B^\alpha}\Big)\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq &C\sum^{\infty}_{k = 1}k2^{-k(n+\frac{\varepsilon}{\alpha})}\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq& C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. \end{eqnarray*} $ |
Similarly we can also get that
$ \tilde{L}_{3}\leq C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. $ |
For $ x\in B $ and $ y_{1}, y_{2}\in (16\tilde{B})^{c} $, there are $ 2|x-\tilde{z}_{0}|^{\alpha}\leq|y_{1}-\tilde{z}_{0}| $, $ 2|x-\tilde{z}_{0}|^{\alpha}\leq|y_{2}-\tilde{z}_{0}| $, $ r_{B}\leq |x-\tilde{z}_{0}|\leq 4r_{B} $ and $ |x-\tilde{z}_{0}| < 1 $. By Hölder's inequality, Lemma 3.10 and the condition $ (1.3) $ of the kernel, we have
$ \begin{eqnarray*} \tilde{L}_{4} &\leq &\dfrac{1}{|B|}\int_{B} \sum^{\infty}_{k_1 = 1} \sum^{\infty}_{k_2 = 1}\Big(\int_{2^{k_1}|x-\tilde{z}_{0}|^\alpha\leq|y_{1}-\tilde{z}_{0}|\leq2^{k_1+1}|x-\tilde{z}_{0}|^\alpha}\int_{2^{k_2}|x-\tilde{z}_{0}|^\alpha\leq|y_{2}-\tilde{z}_{0}|\leq2^{k_2+1}|x-\tilde{z}_{0}|^\alpha}\\ &&|K(x, y_{1}, y_{2})-K(\tilde{z}_{0}, y_{1}, y_{2})|^{p_0}dy_{2}dy_{1}\Big)^{\frac{1}{p_0}}\Big(\int_{2^{k+4}\tilde{B}}|f_{1}(y_{1})-(f_{1})_{16\tilde{B}}|^{p_0'}dy_{1}\Big)^{\frac{1}{p_0'}}\\ &&\times\Big(\int_{2^{k+4}\tilde{B}}|f_{2}(y_{2})-(f_{2})_{16\tilde{B}}|^{p_0'}dy_{2}\Big)^{\frac{1}{p_0'}}dx\\ &\leq&C\dfrac{1}{|B|}\int_{B}\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1} \prod^{2}_{i = 1}2^{-k_{i}(n(1-{\frac{1}{p_0}})+\frac{\varepsilon}{2\alpha})}|x-\tilde{z}_{0}|^{\varepsilon-\alpha(n(2-{\frac{2}{p_0}})+\frac{\varepsilon}{\alpha})}dx\prod^{2}_{i = 1}\Big((2^{k_i+4}r_B^\alpha)^{\frac{n}{p_0}}(1+\ln\dfrac{2^{k_i+4}r_B^\alpha}{16r_B^\alpha})\Big)\\ &&\times\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq&C\sum^{\infty}_{k_1 = 1}\sum^{\infty}_{k_2 = 1}\prod^{2}_{i = 1} k_i2^\frac{-k_i\varepsilon}{2\alpha}\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}\\ &\leq& C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}. \end{eqnarray*} $ |
Thus, according to the estimates in both cases, there is
$ \|T(f_{1}, f_{2})\|_{BMO}\sim \sup\limits_{B} \inf\limits_{a\in \mathbb{C}}\dfrac{1}{|B|}\int_{B}|T(f_{1}, f_{2})(x)-a|dx\leq C\|f_{1}\|_{BMO}\|f_{2}\|_{BMO}, $ |
which completes the proof of the theorem.
Compared with the multilinear strongly singular Calderón-Zygmund operator in [15], the kernel function of the multilinear strongly singular integral operator with generalized kernel in the present article satisfies the weaker smoothness condition. Our results further generalize the corresponding results in [15].
We thank the referees for their time and comments. This work was supported by the National Natural Science Foundation of China (No. 12071052).
The authors declare that there is no conflict of interests regarding the publication of this article.
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