The present section consists of two parts which are about the bounded estimates on Morrey and modified spaces for the multilinear fractional integral operator \(T_{\Omega,\alpha}\) and the multilinear fractional maximal operator \(M_{\Omega,\alpha}\), respectively.
Boundedness on Morrey spaces
In this part, we will prove Theorem 1.1. Let us begin with a requisite Hedberg’s type estimates, which plays a key role in proving Theorem 1.1.
Lemma 3.1
Let
\(0<\alpha<n\), \(\Omega\in L^{s}(\mathbb{S}^{n-1})\)
with
\(1< s\leq\infty\), p
be the harmonic mean of
\(p_{1},\ldots,p_{m}>1\), \(0\leq\lambda< n-\alpha p\), \(s'\leq p<{n}/\alpha\)
and satisfy (1.1), then there exists a positive constant
C
such that
$$ \bigl|T_{\Omega,\alpha}{\mathbf{f}}(x)\bigr|\leq C\bigl[M_{\Omega}{\mathbf{f}}(x) \bigr]^{1-p\alpha/(n-\lambda)}\prod_{j=1}^{m} \|f_{j}\| ^{p\alpha/(n-\lambda)}_{L^{p_{j},\lambda_{j}}}. $$
Proof
For any \(\delta>0\), we split the integral into two parts:
$$\begin{aligned} T_{\Omega,\alpha}{\mathbf{f}}(x)&= \biggl( \int_{|y|< \delta}+ \int_{|y|\geq \delta} \biggr) \frac{\Omega(y)}{|y|^{n-\alpha}}\prod _{j=1}^{m}f_{j}(x-\theta_{j}y)\,dy =:A(x,\delta)+S(x,\delta). \end{aligned}$$
For \(A(x,\delta)\), we have
$$\begin{aligned} \bigl|A(x,\delta)\bigr| &\leq \int_{|y|< \delta}\frac{|\Omega(y)|}{|y|^{n-\alpha}}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|\,dy \\ &\leq\sum_{i=0}^{\infty}\int_{2^{-i-1}\delta\leq|y|< 2^{-i}\delta }\frac{|\Omega(y)|}{|y|^{n-\alpha}}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|\,dy \\ &\leq\sum_{i=0}^{\infty}\bigl(2^{-i-1} \delta\bigr)^{\alpha-n} \int _{|y|< 2^{-i}\delta}\bigl|\Omega(y)\bigr|\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|\,dy \\ &\leq\sum_{i=0}^{\infty}\bigl(2^{-i-1} \delta\bigr)^{\alpha-n}\bigl(2^{-i}\delta \bigr)^{n}M_{\Omega}{ \mathbf{f}}(x) \\ &\leq2^{n-\alpha}\delta^{\alpha}M_{\Omega}{\mathbf{f}}(x)\sum _{i=0}^{\infty}2^{-i\alpha} \\ &\leq C\delta^{\alpha}M_{\Omega}{\mathbf{f}}(x). \end{aligned}$$
Recalling the conditions of Lemma 3.1, we can see \(s'\leq p<(n-\lambda)/\alpha\), which implies \(\alpha<(n-\lambda)/p\leq(n-\lambda)/s'\), then we get
$$n-\alpha s'>n-(n-\lambda)s'/p\geq n-(n-\lambda)= \lambda. $$
In order to estimate \(S(x,\delta)\), we choose a real number σ such that
$$n-\alpha s'>\sigma>n-(n-\lambda)s'/p\geq \lambda. $$
One can then see from the choice of σ that
$$ n-\bigl(n-\alpha-\sigma/s'\bigr)s< 0 $$
(3.1)
and
$$ {(n-\sigma)}/{s'}-{(n-\lambda)}/p< 0. $$
(3.2)
Then, using the Hölder inequality, we obtain
$$\begin{aligned} \bigl|S(x,\delta)\bigr| &\leq \int_{|y|\geq\delta}\frac{|\Omega(y)|}{|y|^{n-\alpha-\sigma /s'}}\frac{1}{|y|^{\sigma/s'}}\prod _{j=1}^{m}\bigl|f_{j}(x-\theta_{j}y)\bigr|\,dy \\ &\leq \biggl( \int_{|y|\geq\delta}\frac{|\Omega(y)|^{s}}{|y|^{(n-\alpha -\sigma/s')s}}\,dy \biggr)^{1/s} \Biggl( \int_{|y|\geq\delta}\frac{1}{|y|^{\sigma}}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &=:E_{\sigma}(\delta)\times F_{\sigma}(x,\delta). \end{aligned}$$
For \(E_{\sigma}(\delta)\), by the fact (3.1), we obtain
$$ E_{\sigma}(\delta)= \biggl( \int_{\delta}^{\infty}\int_{\mathbb{S}^{n-1}}\bigl|\Omega (\xi)\bigr|^{s} r^{n-(n-\alpha-\sigma/s')s-1}\,d\xi \,dr \biggr)^{1/s}=C\delta^{\alpha-(n-\sigma)/s'}. $$
For \(F_{\sigma}(x,\delta)\), we have
$$\begin{aligned} F_{\sigma}(x,\delta) &\leq \Biggl(\sum_{i=0}^{\infty}\int_{2^{i}\delta\leq|y|< 2^{i+1}\delta }\frac{1}{|y|^{\sigma}}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &\leq\sum_{i=0}^{\infty}\bigl(2^{i} \delta\bigr)^{-\sigma/s'} \Biggl( \int _{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'}. \end{aligned}$$
If \(p>s'\), applying Hölder’s inequality and the fact (3.2), we have
$$\begin{aligned} F_{\sigma}(x,\delta) &\leq\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'} \biggl( \int _{|y|< 2^{i+1}\delta}\,dy \biggr)^{1/s'-1/p} \Biggl( \int_{|y|< 2^{i+1}\delta }\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{p}\,dy \Biggr)^{1/p} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-n/p} \Biggl( \int_{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{p}\,dy \Biggr)^{1/p} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-(n-\lambda )/p} \Biggl(\frac{1}{(2^{i+1}\delta)^{\lambda}} \int_{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{p}\,dy \Biggr)^{1/p} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-(n-\lambda )/p}\prod _{j=1}^{m} \biggl(\frac{1}{(2^{i+1}\delta)^{\lambda_{j}}} \int _{|y|< 2^{i+1}\delta}\bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-(n-\lambda )/p}\prod _{j=1}^{m}\|f_{j}\|_{L^{p_{j},\lambda_{j}}} \\ &\leq C\delta^{(n-\sigma)/s'-(n-\lambda)/p}\prod_{j=1}^{m} \|f_{j}\| _{L^{p_{j},\lambda_{j}}}. \end{aligned}$$
If \(p=s'\), using the Hölder inequality and the fact \(\lambda<\sigma\), we get
$$\begin{aligned} F_{\sigma}(x,\delta) &\leq\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'} \Biggl( \int _{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'+\lambda /s'} \Biggl(\frac{1}{(2^{i+1}\delta)^{\lambda}} \int_{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(\lambda-\sigma)/s'}\prod _{j=1}^{m} \biggl(\frac{1}{(2^{i+1}\delta)^{\lambda_{j}}} \int _{|y|< 2^{i+1}\delta}\bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(\lambda-\sigma)/s'}\prod _{j=1}^{m}\|f_{j}\|_{L^{p_{j},\lambda_{j}}} \\ &\leq C\delta^{(\lambda-\sigma)/s'}\prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}} \\ &= C\delta^{(n-\sigma)/s'-(n-\lambda)/p}\prod_{j=1}^{m} \|f_{j}\| _{L^{p_{j},\lambda_{j}}}. \end{aligned}$$
Hence, for every \(p\geq s'\), we have
$$\bigl|S(x,\delta)\bigr|\leq C\delta^{\alpha-(n-\lambda)/p}\prod_{j=1}^{m} \|f_{j}\| _{L^{p_{j},\lambda_{j}}}. $$
Thus
$$\bigl|T_{\Omega,\alpha}{\mathbf{f}}(x)\bigr|\leq C \Biggl( \delta^{\alpha}M_{\Omega }{ \mathbf{f}}(x)+\delta^{\alpha-(n-\lambda)/p}\prod_{j=1}^{m} \|f_{j}\| _{L^{p_{j},\lambda_{j}}} \Biggr),\quad \delta>0. $$
Now take
$$\delta= \Biggl[\bigl(M_{\Omega}{\mathbf{f}}(x)\bigr)^{-1}\prod _{j=1}^{m}\|f_{j}\| _{L^{p_{j},\lambda_{j}}} \Biggr]^{p/(n-\lambda)}, $$
and then we get the conclusion of Lemma 3.1. □
Now we are ready to prove Theorem 1.1.
Proof of Theorem 1.1
First of all, we will devote our efforts to the proof of (i).
Sufficiency. By Lemma 3.1 and the conclusion (i) of Theorem 2.3, we have
$$\begin{aligned} \biggl(\frac{1}{t^{\lambda}} \int_{B(x,t)}\bigl|T_{\Omega,\alpha}{\mathbf {f}}(y)\bigr|^{q}\,dy \biggr)^{1/q} &\leq C\prod_{j=1}^{m} \|f_{j}\|^{1-p/q}_{L^{p_{j},\lambda_{j}}} \biggl(\frac {1}{t^{\lambda}} \int_{B(x,t)}\bigl(M_{\Omega}{\mathbf{f}}(y) \bigr)^{p}\,dy \biggr)^{1/q} \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|^{1-p/q}_{L^{p_{j},\lambda_{j}}}\prod_{j=1}^{m} \|f_{j}\| ^{p/q}_{L^{p_{j},\lambda_{j}}} \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{L^{p_{j},\lambda_{j}}}. \end{aligned}$$
Taking the supremum for \(x\in \mathbb {R}^{n}\) and \(t>0\), we will get the desired conclusion.
Necessity. Suppose that \(T_{\Omega,\alpha}\) is bounded from \(L^{p_{1},\lambda_{1}}\times\cdots\times L^{p_{m},\lambda_{m}}\) to \(L^{q,\lambda}\). Define \({\mathbf {f}}_{\epsilon}(x)=(f_{1}(\epsilon x),\ldots, f_{m}(\epsilon x))\) for \(\epsilon>0\). Then it is easy to show that
$$ T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}(y)= \epsilon^{-\alpha}T_{\Omega,\alpha }{\mathbf{f}}(\epsilon y). $$
(3.3)
Thus
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}\|_{L^{q,\lambda}} &= \epsilon^{-\alpha}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{t^{\lambda}} \int_{B(x,t)}\bigl|T_{\Omega,\alpha}{\mathbf{f}}(\epsilon y)\bigr|^{q}\,dy \biggr)^{1/q} \\ &=\epsilon^{-\alpha-n/q}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{t^{\lambda}} \int_{B(\epsilon x,\epsilon t)}\bigl|T_{\Omega,\alpha}{\mathbf {f}}(y)\bigr|^{q}\,dy \biggr)^{1/q} \\ &=\epsilon^{-\alpha-n/q+\lambda/q}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{(\epsilon t)^{\lambda}} \int_{B(\epsilon x,\epsilon t)}\bigl|T_{\Omega ,\alpha}{\mathbf{f}}(y)\bigr|^{q}\,dy \biggr)^{1/q} \\ &=\epsilon^{-\alpha-(n-\lambda)/q}\|T_{\Omega,\alpha}{\mathbf{f}}\| _{L^{q,\lambda}}. \end{aligned}$$
Since \(T_{\Omega,\alpha}\) is bounded from \(L^{p_{1},\lambda_{1}}\times\cdots\times L^{p_{m},\lambda_{m}}\) to \(L^{q,\lambda}\), we have
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}}\|_{L^{q,\lambda}}&=\epsilon^{\alpha +(n-\lambda)/q} \|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}\|_{L^{q,\lambda}} \\ &\leq C\epsilon^{\alpha+(n-\lambda)/q}\prod_{j=1}^{m} \bigl\| f_{j}(\epsilon\cdot )\bigr\| _{L^{p_{j},\lambda_{j}}} \\ &=C\epsilon^{\alpha+(n-\lambda)/q} \prod_{j=1}^{m} \sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{t^{\lambda _{j}}} \int_{B(x,t)}\bigl|f_{j}(\epsilon y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &= C\epsilon^{\alpha+(n-\lambda)/q} \prod_{j=1}^{m} \epsilon^{-n/{p_{j}}}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{t^{\lambda_{j}}} \int_{B(\epsilon x,\epsilon t)}\bigl|f_{j}(y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &= C\epsilon^{\alpha+(n-\lambda)/q} \prod_{j=1}^{m} \epsilon^{(\lambda_{j}-n)/{p_{j}}}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{(\epsilon t)^{\lambda_{j}}} \int_{B(\epsilon x,\epsilon t)}\bigl|f_{j}(y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &= C\epsilon^{\alpha+(n-\lambda)/q-(n-\lambda)/{p}} \prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}}, \end{aligned}$$
where C is independent of ϵ.
If \(1/p<1/q+\alpha/(n-\lambda)\), then for all \({\mathbf {f}} \in L^{p_{1},\lambda_{1}}\times\cdots\times L^{p_{m},\lambda_{m}}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{L^{q,\lambda}}=0\) as \(\epsilon\rightarrow0\).
If \(1/p>1/q+\alpha/(n-\lambda)\), then for all \({\mathbf {f}} \in L^{p_{1},\lambda_{1}}\times\cdots\times L^{p_{m},\lambda_{m}}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{L^{q,\lambda}}=0\) as \(\epsilon\rightarrow\infty\).
Therefore we get \(1/p=1/q+\alpha/(n-\lambda)\).
We proceed to prove (ii). Sufficiency. For any \(\beta>0\), applying Lemma 3.1 and the conclusion (ii) of Theorem 2.3, we get
$$\begin{aligned} &\|T_{\Omega,\alpha}{\mathbf{f}}\|_{WL^{q,\lambda}} \\ &\quad=\sup_{\beta >0} {\beta}\sup_{x\in \mathbb {R}^{n},t>0} \biggl( \frac{1}{t^{\lambda}}\bigl|\bigl\{ y\in B(x,t):\bigl|T_{\Omega}{\mathbf{f}}(y)\bigr|>\beta \bigr\} \bigr| \biggr)^{1/q} \\ &\quad\leq\sup_{\beta>0} {\beta}\sup_{x\in \mathbb {R}^{n},t>0} \Biggl( \frac{1}{t^{\lambda}}\Biggl|\Biggl\{ y\in B(x,t):\bigl(CM_{\Omega}{\mathbf{f}}(y) \bigr)^{s'/q}\prod_{j=1}^{m} \|f_{j}\|^{1-s'/q}_{L^{p_{j},\lambda_{j}}}>\beta\Biggr\} \Biggr| \Biggr)^{1/q} \\ &\quad\leq\sup_{\beta>0} {\beta}\sup_{x\in \mathbb {R}^{n},t>0} \biggl[ \frac{1}{t^{\lambda}}\biggl\vert \biggl\{ y\in B(x,t):M_{\Omega}{ \mathbf{f}}(y)> \biggl(\frac{\beta}{C\prod_{j=1}^{m}\|f_{j}\|^{1-s'/q}_{L^{p_{j},\lambda_{j}}}} \biggr)^{q/s'} \biggr\} \biggr\vert \biggr]^{1/q} \\ &\quad\leq\sup_{\beta>0} {\beta}\sup_{x\in \mathbb {R}^{n},t>0} \biggl[ \frac{1}{t^{\lambda}}\biggl\vert \biggl\{ y\in B(x,t):M_{\Omega}{ \mathbf{f}}(y)> \biggl(\frac{\beta}{C\prod_{j=1}^{m} \|f_{j}\|^{1-s'/q}_{L^{p_{j},\lambda_{j}}}} \biggr)^{q/s'} \biggr\} \biggr\vert \biggr]^{\frac{1}{s'}\times\frac{s'}{q}} \\ &\quad\leq{C}\sup_{\beta>0}\ {\beta} \Biggl[ \biggl(\frac{\prod_{j=1}^{m} \|f_{j}\|^{1-s'/q}_{L^{p_{j},\lambda _{j}}}}{\beta} \biggr)^{q/s'} \prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}} \Biggr]^{s'/q} \\ &\quad\leq{C}\sup_{\beta>0} {\beta} \Biggl[\frac{\prod_{j=1}^{m}\|f_{j}\| ^{1-s'/q}_{L^{p_{j},\lambda_{j}}}}{\beta}\prod _{j=1}^{m}\|f_{j}\| ^{s'/q}_{L^{p_{j},\lambda_{j}}} \Biggr] \\ &\quad\leq{C} {\prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}}}. \end{aligned}$$
Thus, we complete the sufficiency of (ii).
Necessity. Let \(T_{\Omega,\alpha}\) be bounded from \(L^{p_{1},\lambda_{1}}\times\cdots\times L^{p_{m},\lambda_{m}}\) to \(WL^{q,\lambda}\). Because we have (3.3) for \({\mathbf {f}}_{\epsilon}(x)=(f_{1}(\epsilon x),\ldots, f_{m}(\epsilon x))\) with \(\epsilon>0\), then we obtain
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}\|_{WL^{q,\lambda}} &=\sup _{r>0} r\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{t^{\lambda}} \int_{\{y\in B(x,t):|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}(y)|>r\}}\,dy \biggr)^{\frac{1}{q}} \\ &=\sup_{r>0} r\sup_{x\in \mathbb {R}^{n},t>0} \biggl( \frac {1}{t^{\lambda}} \int_{\{y\in B( x,t):|T_{\Omega,\alpha}{\mathbf{f}}(\epsilon y)|>r\epsilon^{\alpha}\}}\,dy \biggr)^{\frac{1}{q}} \\ &=\epsilon^{-n/q}\sup_{r>0}\ r\sup _{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{t^{\lambda}} \int_{\{y\in B(\epsilon x,\epsilon t):|T_{\Omega,\alpha}{\mathbf{f}}(y)|>r\epsilon^{\alpha}\}}\,dy \biggr)^{\frac{1}{q}} \\ &=\epsilon^{-\alpha-n/q+\lambda/q}\sup_{r>0}\ r\epsilon^{\alpha}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{(\epsilon t)^{\lambda}} \int_{\{y\in B(\epsilon x,\epsilon t):|T_{\Omega,\alpha}{\mathbf{f}}(y)|>r\epsilon^{\alpha}\}}\,dy \biggr)^{\frac{1}{q}} \\ &=\epsilon^{-\alpha-(n-\lambda)/q}\|T_{\Omega,\alpha}{\mathbf{f}}\| _{WL^{q,\lambda}}. \end{aligned}$$
Since \(T_{\Omega,\alpha}\) is bounded from \(L^{p_{1},\lambda_{1}}\times\cdots\times L^{p_{m},\lambda_{m}}\) to \(WL^{q,\lambda}\), we have
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}}\|_{WL^{q,\lambda}}&=\epsilon^{\alpha +(n-\lambda)/q} \|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}\|_{WL^{q,\lambda}} \leq C\epsilon^{\alpha+(n-\lambda)/q}\prod_{j=1}^{m} \bigl\| f_{j}(\epsilon\cdot )\bigr\| _{L^{p_{j},\lambda_{j}}} \\ &\leq C\epsilon^{\alpha+(n-\lambda)/q-(n-\lambda)/p} \prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}}, \end{aligned}$$
where C is independent of ϵ.
If \(1/p<1/q+\alpha/(n-\lambda)\), then for all \({\mathbf {f}} \in L^{p_{1},\lambda}\times\cdots\times L^{p_{m},\lambda}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{WL^{q,\lambda}}=0\) as \(\epsilon\rightarrow0\).
If \(1/p>1/q+\alpha/(n-\lambda)\), then for all \({\mathbf {f}} \in L^{p_{1},\lambda}\times\cdots\times L^{p_{m},\lambda}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{WL^{q,\lambda}}=0\) as \(\epsilon\rightarrow\infty\).
Consequently, we get \(1/p=1/q+\alpha/(n-\lambda)\).
Next, we prove conclusions (i) and (ii) hold for \(M_{\Omega,\alpha}\). By the same arguments as above we get the necessity part and the sufficiency part follows from the conclusion of \(T_{\Omega,\alpha}\) and the following lemma.
Lemma 3.2
[9]
Suppose that
\(0<\alpha<n\), \(\Omega\in L^{s}(\mathbb{S}^{n-1})\)
with
\(1< s\leq\infty\). Then
$$ M_{\Omega,\alpha}({\mathbf {f}}) (x)\leq C_{\alpha,n}T_{|\Omega|,\alpha}\bigl({ |\mathbf {f}|}\bigr) (x), $$
where
\({|\mathbf {f}|}=(|f_{1}|,\ldots,|f_{m}|)\).
Then the proof of Theorem 1.1 is completed. □
As an application of Theorem 1.1, we get Spanne type estimates, which can be seen a multi-version of Proposition 1.1.
Corollary 3.1
Let
α, Ω, s, \(p_{j}\), \(\lambda_{j}\), p, and
λ
be as in Theorem
1.1, \(1/q=1/p-\alpha/{n}\), \(\mu/q=\lambda/p\).
-
(i)
If
\(p>s'\), then
\(T_{\Omega,\alpha}\)
is bounded from
\(L^{p_{1},\lambda_{1}}(\mathbb{R}^{n})\times\cdots\times L^{p_{m},\lambda_{m}}(\mathbb{R}^{n})\)
to
\(L^{q,\mu}(\mathbb{R}^{n})\).
-
(ii)
If
\(p=s'\), then
\(T_{\Omega,\alpha}\)
is bounded from
\(L^{p_{1},\lambda_{1}}(\mathbb{R}^{n})\times\cdots\times L^{p_{m},\lambda_{m}}(\mathbb{R}^{n})\)
to
\(WL^{q,\mu}(\mathbb{R}^{n})\).
Moreover, similar estimates hold for
\(M_{\Omega,\alpha}\).
Proof
From Lemma 3.2, we only need to show the boundedness of \(T_{\Omega,\alpha}\).
First, we choose t to satisfy \((n-\mu)/q=(n-\lambda)/t\), then we get
$${1}/{t}=(n-\mu)/q(n-\lambda)={1}/{p}-{\alpha}/(n-\lambda )< {1}/{p}-{ \alpha}/n=1/q. $$
Then Hölder’s inequality implies \(L^{t,\lambda}(\mathbb{R}^{n})\subset L^{q,\mu}(\mathbb{R}^{n})\) and \(WL^{t,\lambda}(\mathbb{R}^{n})\subset WL^{q,\mu}(\mathbb{R}^{n})\). In fact, there exists a constant \(C>0\) such that
$$ \|T_{\Omega,\alpha}{\mathbf{f}}\|_{L^{q,u}}\leq C\|T_{\Omega,\alpha}{ \mathbf{f}}\|_{L^{t,\lambda}} $$
and
$$ \|T_{\Omega,\alpha}{\mathbf{f}}\|_{WL^{q,u}}\leq C\|T_{\Omega,\alpha}{ \mathbf{f}}\|_{WL^{t,\lambda}}. $$
Then, by Theorem 1.1, we have
$$ \|T_{\Omega,\alpha}{\mathbf{f}}\|_{L^{q,u}}\leq C\|T_{\Omega,\alpha}{ \mathbf{f}}\|_{L^{t,\mu}}\leq C\prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}} \quad\mbox{for } p>s' $$
and
$$ \|T_{\Omega,\alpha}{\mathbf{f}}\|_{WL^{q,u}}\leq C\|T_{\Omega,\alpha}{ \mathbf{f}}\|_{WL^{t,\mu}}\leq C\prod_{j=1}^{m} \|f_{j}\|_{L^{p_{j},\lambda_{j}}} \quad\mbox{for } p=s'. $$
Thus, the proof of Corollary 3.1 is completed. □
As an another application, by Hölder’s inequality, we obtain an Olsen’s inequality as in the following corollary, which is a multi-version of the results in considered by Olsen in [12] in the study of the Schrödinger equation with perturbed potentials W.
Corollary 3.2
Let
α, Ω, s, \(p_{j}\), \(\lambda_{j}\), p, and
λ
be as in Theorem
1.1
and let
\(W\in L^{(n-\lambda)/\alpha,\lambda}\). If
\(p>s'\)
and
\(1/p-1/q=\alpha/(n-\lambda)\), then there exists a positive constant
C
such that
$$ \|W\cdot T_{\Omega,\alpha}{\mathbf{f}}\|_{L^{p,\lambda}(\mathbb{R}^{n})} \leq C \|W\|_{L^{(n-\lambda)/\alpha,\lambda}(\mathbb{R}^{n})} \prod_{j=1}^{m}\|f_{j}\|_{L^{p_{j},\lambda_{j}}(\mathbb{R}^{n})}. $$
Moreover, similar estimates hold for
\(M_{\Omega,\alpha}\).
Boundedness on modified Morrey spaces
This part we will devote to the boundedness on modified Morrey spaces and show the proof Theorem 1.2 and 1.3. With the same arguments on Morrey spaces, we also begin with a requisite Hedberg’s type estimates on modified Morrey spaces.
Lemma 3.3
Let
\(0<\alpha<n\), \(\Omega\in L^{s}(\mathbb{S}^{n-1})\)
with
\(1< s\leq\infty\), p
be the harmonic mean of
\(p_{1},\ldots,p_{m}>1\), \(0\leq\lambda< n-\alpha p\), \(s'\leq p<{n}/\alpha\)
and satisfy (1.1), then there exists a positive constant
C
such that
$$ \bigl|T_{\Omega,\alpha}{\mathbf{f}}(x)\bigr|\leq C\bigl(M_{\Omega}{\mathbf{f}}(x) \bigr)^{p/q}\prod_{j=1}^{m} \|f_{j}\|^{1-p/q}_{\widetilde {L}^{p_{j},\lambda_{j}}}. $$
Proof
For any \(\delta>0\), we do the same decomoposition of \(T_{\Omega,\alpha}\) as in the proof of Lemma 3.1, then we only need to estimate \(F_{\sigma}(x,\delta)\). We also choose the same σ during the proof of Lemma 3.1, then we get
$$ (n-\sigma)/{s'}-{n}/p\leq(n-\sigma)/{s'}-(n- \lambda)/p< 0. $$
(3.4)
If \(p>s'\), by the Hölder inequality and the fact (3.4), we obtain
$$\begin{aligned} F_{\sigma}(x,\delta) &\leq\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'} \Biggl( \int _{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-n/p} \Biggl( \int_{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{p}\,dy \Biggr)^{1/p} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma )/s'-n/p}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda/p} \Biggl(\frac{1}{[2^{i+1}\delta ]_{1}^{\lambda}} \int_{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta _{j}y)\bigr|^{p}\,dy \Biggr)^{1/p} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma )/s'-n/p}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda/p}\prod_{j=1}^{m} \biggl(\frac {1}{[2^{i+1}\delta]_{1}^{\lambda_{j}}} \int_{|y|< 2^{i+1}\delta}\bigl|f_{j}(x-\theta _{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-n/p}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda/p}\prod_{j=1}^{m} \|f_{j}\|_{\widetilde{L}^{p_{j},\lambda_{j}}} \\ &= C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}}\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-n/p}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda /p} \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sum_{i=0}^{\infty}\bigl(2^{i}\delta \bigr)^{(n-\sigma)/s'-n/p}, &\mbox{if } \delta\geq1/2,\\ \sum_{i=0}^{[\log_{2}{\frac{1}{2\delta}}]}\bigl(2^{i} \delta\bigr)^{(n-\sigma )/s'-(n-\lambda)/p}\\ \quad{}+\sum^{\infty}_{i=[\log_{2}{\frac{1}{2\delta }}]+1} \bigl(2^{i}\delta\bigr)^{(n-\sigma)/s'-n/p}, & \mbox{if } 0< \delta< 1/2 \end{array}\displaystyle \right . \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \delta^{(n-\sigma)/s'-n/p}, &\mbox{if } \delta\geq1/2,\\ \delta^{(n-\sigma)/s'-(n-\lambda)/p}+\delta^{(n-\sigma)/s'-n/p}, &\mbox{if } 0< \delta< 1/2 \end{array}\displaystyle \right . \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \delta^{(n-\sigma)/s'-n/p}, &\mbox{if }\delta\geq1/2,\\ \delta^{(n-\sigma)/s'-(n-\lambda)/p}, &\mbox{if } 0< \delta< 1/2 \end{array}\displaystyle \right . \\ &\leq C\delta^{(n-\sigma)/s'-n/p}[2\delta]_{1}^{\lambda/p}\prod _{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}}. \end{aligned}$$
If \(p=s'\), using the Hölder inequality and the fact \(0\leq\lambda<\sigma\), we get
$$\begin{aligned} F_{\sigma}(x,\delta) &\leq\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'} \Biggl( \int _{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda/s'} \Biggl(\frac{1}{[2^{i+1}\delta]_{1}^{\lambda}} \int _{|y|< 2^{i+1}\delta}\prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{s'}\,dy \Biggr)^{1/s'} \\ &\leq C\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda/s'}\prod_{j=1}^{m} \biggl(\frac{1}{[2^{i+1}\delta]_{1}^{\lambda _{j}}} \int_{|y|< 2^{i+1}\delta}\bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}}\sum_{i=0}^{\infty}\bigl(2^{i}\delta\bigr)^{-\sigma/s'}\bigl[2^{i+1}\delta \bigr]_{1}^{\lambda/s'} \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sum_{i=0}^{\infty}\bigl(2^{i}\delta \bigr)^{-\sigma/s'}, &\mbox{if } \delta\geq1/2,\\ \sum_{i=0}^{[\log_{2}{\frac{1}{2\delta}}]}\bigl(2^{i} \delta\bigr)^{(\lambda -\sigma)/s'}\\ \quad{}+\sum^{\infty}_{i=[\log_{2}{\frac{1}{2\delta }}]+1} \bigl(2^{i}\delta\bigr)^{-\sigma/s'}, & \mbox{if } 0< \delta< 1/2 \end{array}\displaystyle \right . \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \delta^{-\sigma/s'}, &\mbox{if }\delta\geq1/2,\\ \delta^{(\lambda-\sigma)/s'}+\delta^{-\sigma/s'}, &\mbox{if } 0< \delta< 1/2 \end{array}\displaystyle \right . \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}} \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \delta^{-\sigma/s'}, &\mbox{if } \delta\geq1/2,\\ \delta^{(\lambda-\sigma)/s'}, &\mbox{if } 0< \delta< 1/2 \end{array}\displaystyle \right . \\ &\leq C\delta^{-\sigma/s'}[2\delta]_{1}^{\lambda/p}\prod _{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}} \\ &\leq C\delta^{(n-\sigma)/s'-n/p}[2\delta]_{1}^{\lambda/p}\prod _{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}}. \end{aligned}$$
Then, combining with the estimates \(E_{\sigma}(\delta)\leq C\delta^{\alpha-(n-\sigma)/s'}\), we have
$$ \bigl|S(x,\delta)\bigr|\leq C\delta^{\alpha-n/p}[2\delta]_{1}^{\lambda/p} \prod_{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}}, \quad\mbox{for every } p\geq s'. $$
Thus
$$\begin{aligned} \bigl|T_{\Omega,\alpha}{\mathbf{f}}(x)\bigr| \leq{}& C \Biggl( \delta^{\alpha}M_{\Omega}{ \mathbf{f}}(x)+\delta^{\alpha-n/p}[2\delta ]_{1}^{\lambda/p} \prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda _{j}}} \Biggr) \\ \leq {}&C \min \Biggl\{ \delta^{\alpha}M_{\Omega}{\mathbf{f}}(x)+ \delta^{\alpha-n/p}\prod_{j=1}^{m}\| f_{j}\|_{\widetilde{L}^{p_{j},\lambda_{j}}},\\ &{}\delta^{\alpha}M_{\Omega}{\mathbf{f}}(x)+\delta ^{\alpha-(n-\lambda)/p}\prod_{j=1}^{m} \|f_{j}\|_{\widetilde{L}^{p_{j},\lambda _{j}}} \Biggr\} . \end{aligned}$$
Minimizing with respect to δ, at
$$\delta= \Biggl[\bigl(M_{\Omega}{\mathbf{f}}(x)\bigr)^{-1}\prod _{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}} \Biggr]^{p/n} $$
and
$$\delta= \Biggl[\bigl(M_{\Omega}{\mathbf{f}}(x)\bigr)^{-1}\prod _{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}} \Biggr]^{p/(n-\lambda)} $$
we have
$$\begin{aligned} \bigl|T_{\Omega,\alpha}{\mathbf{f}}(x)\bigr|&\leq C \min \biggl\{ \biggl( \frac{M_{\Omega}{\mathbf{f}}(x)}{\prod_{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}}} \biggr)^{1-\frac {p\alpha}{n}}, \biggl(\frac{M_{\Omega}{\mathbf{f}}(x)}{\prod_{j=1}^{m}\|f_{j}\|_{\widetilde {L}^{p_{j},\lambda_{j}}}} \biggr)^{1-\frac{p\alpha}{n-\lambda}} \biggr\} \prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda _{j}}} \\ &\leq C\bigl(M_{\Omega}{\mathbf{f}}(x)\bigr)^{p/q}\prod _{j=1}^{m}\|f_{j}\|^{1-p/q}_{\widetilde {L}^{p_{j},\lambda_{j}}}. \end{aligned}$$
This is the conclusion of Lemma 3.3. □
Now we give the proof of Theorem 1.2.
Proof of Theorem 1.2
Similarly to the proof of sufficiency in Theorem 1.1, by the boundedness of \(M_{\Omega}\) in Theorem 2.4, we will get the sufficiency. Now, we give only the proof of necessity.
Let \([\epsilon]_{1,+}=\max\{1,\epsilon\}\), by (3.3), for \({\mathbf {f}}_{\epsilon}(x)\) with \(\epsilon>0\), we get
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}\|_{\widetilde{L}^{q,\lambda}} ={}& \epsilon^{-\alpha}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{[t]_{1}^{\lambda}} \int_{B(x,t)}\bigl|T_{\Omega,\alpha}{\mathbf{f}}(\epsilon y)\bigr|^{q}\,dy \biggr)^{1/q} \\ ={}&\epsilon^{-\alpha-n/q}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{[t]_{1}^{\lambda}} \int_{B(\epsilon x,\epsilon t)}\bigl|T_{\Omega,\alpha}{\mathbf {f}}(y)\bigr|^{q}\,dy \biggr)^{1/q} \\ ={}&\epsilon^{-\alpha-n/q}\sup_{t>0} \biggl(\frac{[\epsilon t]_{1}}{[t]_{1}} \biggr)^{\lambda/q}\\ &{}\times\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{[\epsilon t]_{1}^{\lambda}} \int_{B(\epsilon x,\epsilon t)}\bigl|T_{\Omega ,\alpha}{\mathbf{f}}(y)\bigr|^{q}\,dy \biggr)^{1/q} \\ ={}&\epsilon^{-\alpha-n/q}[\epsilon]_{1,+}^{\frac{\lambda}{q}} \|T_{\Omega ,\alpha}{\mathbf{f}}\|_{\widetilde{L}^{q,\lambda}} \end{aligned}$$
and
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}\|_{W\widetilde{L}^{q,\lambda}} ={}&\sup _{r>0}\ r\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{[t]_{1}^{\lambda}} \int_{\{y\in B(x,t):|T_{\Omega,\alpha}{\mathbf{f}_{\epsilon}}(y)|>r\}}\,dy \biggr)^{\frac{1}{q}} \\ ={}&\sup_{r>0}\ r\sup_{x\in \mathbb {R}^{n},t>0} \biggl( \frac{1}{[t]_{1}^{\lambda}} \int_{\{y\in B( x,t):|T_{\Omega,\alpha}{\mathbf{f}}(\epsilon y)|>r\epsilon ^{\alpha}\}}\,dy \biggr)^{\frac{1}{q}} \\ ={}&\epsilon^{-n/q}\sup_{r>0}\ r\sup _{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{[t]_{1}^{\lambda}} \int_{\{y\in B(\epsilon x,\epsilon t):|T_{\Omega,\alpha}{\mathbf{f}}(y)|>r\epsilon^{\alpha}\}}\,dy \biggr)^{\frac{1}{q}} \\ ={}&\epsilon^{-\alpha-n/q}\sup_{t>0} \biggl(\frac{[\epsilon t]_{1}}{[t]_{1}} \biggr)^{\lambda/q} \\ &{} \times\sup_{r>0} r\epsilon^{\alpha}\sup _{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{[\epsilon t]_{1}^{\lambda}}\bigl\vert \bigl\{ y\in B( \epsilon x,\epsilon t):\bigl|T_{\Omega,\alpha}{\mathbf{f}}(y)\bigr|>r\epsilon^{\alpha}\bigr\} \bigr\vert \biggr)^{\frac{1}{q}} \\ ={}&\epsilon^{-\alpha-n/q}[\epsilon]_{1,+}^{\frac{\lambda}{q}} \|T_{\Omega ,\alpha}{\mathbf{f}}\|_{W\widetilde{L}^{q,\lambda}}. \end{aligned}$$
(i) Assume that \(T_{\Omega,\alpha}\) is bounded from \(\widetilde{L}^{p_{1},\lambda_{1}}\times\cdots\times \widetilde{L}^{p_{m},\lambda_{m}}\) to \(\widetilde{L}^{q,\lambda}\), we get
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}}\|_{\widetilde{L}^{q,\lambda}} ={}&\epsilon^{\alpha+n/q}[ \epsilon]_{1,+}^{-\frac{\lambda}{q}}\|T_{\Omega ,\alpha}{\mathbf{f}_{\epsilon}} \|_{\widetilde{L}^{q,\lambda}} \\ \leq{}& C\epsilon^{\alpha+n/q}[\epsilon]_{1,+}^{-\frac{\lambda}{q}}\prod _{j=1}^{m}\bigl\| f_{j}(\epsilon\cdot) \bigr\| _{\widetilde{L}^{p_{j},\lambda_{j}}} \\ ={}&C\epsilon^{\alpha+n/q}[\epsilon]_{1,+}^{-\frac{\lambda}{q}} \prod _{j=1}^{m}\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{[t]_{1}^{\lambda _{j}}} \int_{B(x,t)}\bigl|f_{j}(\epsilon y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ ={}& C\epsilon^{\alpha+n/q}[\epsilon]_{1,+}^{-\frac{\lambda}{q}} \prod _{j=1}^{m}\epsilon^{-n/{p_{j}}}\sup _{x\in \mathbb {R}^{n},t>0} \biggl(\frac{1}{[t]_{1}^{\lambda_{j}}} \int_{B(\epsilon x,\epsilon t)}\bigl|f_{j}(y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ \leq{}& C\epsilon^{\alpha+n/q}[\epsilon]_{1,+}^{-\frac{\lambda}{q}} \prod _{j=1}^{m}\epsilon^{-n/{p_{j}}}\sup _{t>0} \biggl(\frac{[\epsilon t]_{1}}{[t]_{1}} \biggr)^{\lambda_{j}/p_{j}} \\ &{}\times\sup_{x\in \mathbb {R}^{n},t>0} \biggl(\frac {1}{[\epsilon t]^{\lambda_{j}}} \int_{B(\epsilon x,\epsilon t)}\bigl|f_{j}(y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ \leq{}& C\epsilon^{\alpha+n/q-n/p}[\epsilon]_{1,+}^{-\frac{\lambda}{q}} [ \epsilon]_{1,+}^{\frac{\lambda}{p}} \prod_{j=1}^{m} \|f_{j}\|_{\widetilde{L}^{p_{j},\lambda_{j}}} \\ \leq{}& C\epsilon^{\alpha+n/q-n/p}[\epsilon]_{1,+}^{\frac{\lambda}{p}-\frac {\lambda}{q}} \prod _{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}}, \end{aligned}$$
where C is independent of ϵ.
When \(1/p<1/q+\alpha/n\), then for all \({\mathbf {f}} \in \widetilde{L}^{p_{1},\lambda_{1}}\times\cdots\times \widetilde{L}^{p_{m},\lambda_{m}}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{\widetilde{L}^{q,\lambda}}=0\) as \(\epsilon\rightarrow0\).
When \(1/p>1/q+\alpha/(n-\lambda)\), then for all \({\mathbf {f}} \in \widetilde{L}^{p_{1},\lambda_{1}}\times\cdots\times \widetilde{L}^{p_{m},\lambda_{m}}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{\widetilde{L}^{q,\lambda}}=0\) as \(\epsilon\rightarrow\infty\).
Therefore we get \(\alpha/n\leq1/p-1/q\leq\alpha/(n-\lambda)\).
(ii) Assume that \(T_{\Omega,\alpha}\) is bounded from \(\widetilde{L}^{p_{1},\lambda_{1}}\times\cdots\times \widetilde{L}^{p_{m},\lambda_{m}}\) to \(W\widetilde{L}^{q,\lambda}\), we have
$$\begin{aligned} \|T_{\Omega,\alpha}{\mathbf{f}}\|_{W\widetilde{L}^{q,\lambda}} &=\epsilon^{\alpha+n/q}[ \epsilon]_{1,+}^{\frac{-\lambda}{q}}\|T_{\Omega ,\alpha}{\mathbf{f}_{\epsilon}} \|_{W\widetilde{L}^{q,\lambda}} \leq C\epsilon^{\alpha+n/q}[\epsilon]_{1,+}^{-\frac{\lambda}{q}}\prod _{j=1}^{m}\bigl\| f_{j}(\epsilon\cdot) \bigr\| _{\widetilde{L}^{p_{j},\lambda_{j}}} \\ &\leq C\epsilon^{\alpha+n/q-n/p}[\epsilon]_{1,+}^{\frac{\lambda}{p}-\frac {\lambda}{q}} \prod _{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}}, \end{aligned}$$
where C is independent of ϵ.
When \(1/p<1/q+\alpha/n\), then for all \({\mathbf {f}} \in \widetilde{L}^{p_{1},\lambda}\times\cdots\times \widetilde{L}^{p_{m},\lambda}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{W\widetilde{L}^{q,\lambda}}=0\) as \(\epsilon\rightarrow0\).
When \(1/p>1/q+\alpha/(n-\lambda)\), then for all \({\mathbf {f}} \in \widetilde{L}^{p_{1},\lambda}\times\cdots\times \widetilde{L}^{p_{m},\lambda}\), we have \(\|T_{\Omega,\alpha}{\mathbf{f}}\|_{W\widetilde{L}^{q,\lambda}}=0\) as \(\epsilon\rightarrow\infty\).
Consequently, we get \(\alpha/n\leq1/p-1/q\leq\alpha/(n-\lambda)\).
Next, we prove conclusions (i) and (ii) hold for \(M_{\Omega,\alpha}\). By the same arguments as above we get the necessity part and the sufficiency part follows from the conclusion of \(T_{\Omega,\alpha}\) and Lemma [9].
This completes the proof of Theorem 1.2. □
Finally we show the proof of Theorem 1.3.
Proof of Theorem 1.3
By the Hölder inequality, we have
$$\begin{aligned} M_{\Omega,\alpha}{\mathbf{f}}(x) &=\sup_{r>0} \frac{1}{r^{n-\alpha}} \int_{|y|< r} \bigl|\Omega(y)\bigr| \prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|\,dy \\ &\leq C\sup_{r>0}\frac{1}{r^{n-\alpha}} \biggl( \int_{|y|< r} \bigl|\Omega(y)\bigr|^{p'}\,dy \biggr)^{1/p'} \Biggl( \int_{|y|< r} \prod_{j=1}^{m}\bigl|f_{j}(x- \theta_{j}y)\bigr|^{p} \,dy \Biggr)^{1/p} \\ &\leq C\sup_{r>0}\frac{1}{r^{n-\alpha}} \biggl( \int_{|y|< r} \bigl|\Omega(y)\bigr|^{p'}\,dy \biggr)^{1/p'} \prod_{j=1}^{m} \biggl( \int_{|y|< r} \bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\sup_{r>0}{r^{\alpha-n/p}} \biggl( \frac{1}{r^{n}} \int_{|y|< r} \bigl|\Omega(y)\bigr|^{s}\,dy \biggr)^{1/s} \prod_{j=1}^{m} \biggl( \int_{|y|< r} \bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\sup_{r>0}{r^{\alpha-n/p}}\prod _{j=1}^{m} \biggl( \int_{|y|< r} \bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}}. \end{aligned}$$
(i) If \(p=(n-\lambda)/\alpha\geq s'\), by the fact (1.1), we obtain
$$\begin{aligned} M_{\Omega,\alpha}{\mathbf{f}}(x)&\leq C\sup_{r>0}{r^{\alpha-(n-\lambda)/p}} \prod_{j=1}^{m} \biggl(\frac {1}{r^{\lambda_{j}}} \int_{|y|< r} \bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &=C\prod_{j=1}^{m}\|f_{j} \|_{L^{p_{j},\lambda_{j}}}. \end{aligned}$$
(ii) If \((n-\lambda)/\alpha\leq p\leq n/\alpha\), using the fact (1.1), we get
$$\begin{aligned} M_{\Omega,\alpha}{\mathbf{f}}(x)&\leq C\sup_{r>0}{r^{\alpha-n/p}[r]_{1}^{\lambda/p}} \prod_{j=1}^{m} \biggl(\frac {1}{[r]_{1}^{\lambda_{j}}} \int_{|y|< r} \bigl|f_{j}(x-\theta_{j}y)\bigr|^{p_{j}}\,dy \biggr)^{1/{p_{j}}} \\ &\leq C\sup_{r>0}{r^{\alpha-n/p}[r]_{1}^{\lambda/p}} \prod_{j=1}^{m}\|f_{j}\| _{\widetilde{L}^{p_{j},\lambda_{j}}} \\ &\leq C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}}\max \Bigl\{ \sup_{0< r< 1}r^{\alpha-\frac{n-\lambda}{p}},\sup _{r\geq1}r^{\alpha-n/p} \Bigr\} \\ &=C\prod_{j=1}^{m}\|f_{j} \|_{\widetilde{L}^{p_{j},\lambda_{j}}}. \end{aligned}$$
Therefore, we complete the proof of Theorem 1.3. □
Finally, we would like to remark that our theorems generalize the relevant results in [13–15].