4.1 Reduction

We introduce the maximal operator

$$ \begin{align*} {\mathcal{L}}_{\Omega}^{\sharp}\big(f_1,\dots,f_m\big)(x) := \sup_{\tau\in \mathbb Z} \bigg| \sum_{\gamma<\tau} \int_{(\mathbb R^n)^m }{K^{\gamma}(\vec{{y}}\,)\prod_{j=1}^{m}f_j(x-y_j)}~d\,\vec{{y}}\, \bigg| \end{align*} $$

for $x\in \mathbb R^n$ . Then we claim that

(4.2) $$ \begin{align} {\mathcal{L}}_{\Omega}^*\big(f_1,\dots,f_m\big) \le \mathcal{M}_{\Omega}\big(f_1,\dots,f_m\big)(x)+ {\mathcal{L}}_{\Omega}^{\sharp}\big(f_1,\dots,f_m\big). \end{align} $$

To prove (4.2), we introduce the notation

$$ \begin{align*}K^{(\epsilon)} ( \vec y\,) := K ( \vec y\,) \chi_{|\vec y\,| \ge \epsilon}, \qquad \widetilde{K}^{(\epsilon)} ( \vec y\,) := K ( \vec y\,) \big( 1- \widehat{\Theta^{(m)}} (\vec y/\epsilon) \big),\end{align*} $$

setting $\widehat {\Theta ^{(m)}}(\vec {{y}}\,):=1-\sum _{\gamma \in \mathbb {N}}{\widehat {\Phi ^{(m)}}(\vec {{y}}\,/2^{\gamma })}$ so that

$$\begin{align*}\mathrm{Supp}(\widehat{\Theta^{(m)}})\subset \{\vec{{y}}\,\in (\mathbb R^n)^m:|\vec{{y}}\,|\le 2\} \end{align*}$$

and $\widehat {\Theta ^{(m)}}(\vec {{y}}\,)=1$ for $|\vec {{y}}\,|\le 1$ .

Given $\epsilon>0$ , choose $\rho \in \mathbb Z$ such that $2^{\rho }\le \epsilon <2^{\rho +1} $ . Then we write

(4.3) $$ \begin{align} &\bigg| \int_{(\mathbb R^n)^m\setminus B(0,\epsilon)}{K(\vec{{y}}\,)\prod_{j=1}^{m}f_j(x-y_j)}~d\,\vec{{y}}\, \bigg| \notag\\ &\qquad\qquad \le \bigg| \int_{(\mathbb R^n)^m }{\big( K^{(\epsilon)}(\vec{{y}}\,) - \widetilde{K}^{(2^{\rho})}(\vec{{y}}\,) \big) \prod_{j=1}^{m}f_j(x-y_j)}~d\,\vec{{y}}\, \bigg| \end{align} $$

(4.4) $$ \begin{align} &\qquad \qquad\qquad \qquad+ \bigg| \int_{(\mathbb R^n)^m} \widetilde{K}^{(2^{\rho})}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j) ~d\,\vec{{y}}\,\bigg|. \end{align} $$

Term (4.4) is clearly less than

$$ \begin{align*} \bigg|\sum_{\gamma\in\mathbb{Z}: \gamma<-\rho}\int_{(\mathbb R^n)^m}K^{\gamma}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j) ~d\,\vec{{y}}\, \bigg|\le {\mathcal{L}}_{\Omega}^{\sharp}\big(f_1,\dots,f_m\big)(x), \end{align*} $$

while (4.3) is controlled by $ \mathcal {M}_{\Omega }\big (f_1,\dots ,f_m\big )(x) $ as

$$ \begin{align*} \big|K^{(\epsilon)}(\vec{{y}}\,) - \widetilde K^{(2^{\rho})}(\vec{{y}}\,) \big|\lesssim |K(\vec{{y}}\,)| \chi_{|\vec{{y}}\,|\approx 2^{\rho}}\lesssim \frac{|\Omega(\vec{{y}}\,')|}{2^{\rho mn}}\chi_{|\vec{{y}}\,|\lesssim 2^{\rho}}. \end{align*} $$

Thus, (4.2) follows after taking the supremum over all $\epsilon>0$ .

Since the boundedness of $\mathcal {M}_{\Omega }$ follows from Lemma 3.1 with the fact that $q>\frac {2m}{m+1}$ implies $\frac {m}{2} <\frac {1}{q}+\frac {m}{q'}$ , matters reduce to the boundedness of ${\mathcal {L}}_{\Omega }^{\sharp }$ .

For each $\gamma \in \mathbb {Z}$ , let

$$ \begin{align*}K_{\mu}:=\sum_{\gamma\in\mathbb{Z}}K_{\mu}^{\gamma}.\end{align*} $$

In the study of multilinear rough singular integral operators ${\mathcal {L}}_{\Omega }$ in [Reference Grafakos, He, Honzík and Park18] whose kernel is $\sum _{\gamma \in \mathbb {Z}}K^{\gamma }=\sum _{\mu \in \mathbb {Z}}\sum _{\gamma \in \mathbb {Z}}K_{\mu }^{\gamma }=\sum _{\mu \in \mathbb {Z}}K_{\mu }$ , the part where $\mu $ is less than a constant is relatively simple because the Fourier transform of $K_{\mu }$ satisfies the estimate

(4.5) $$ \begin{align} \big|\partial^{\alpha}\widehat{K_{\mu}}(\vec{{\xi}}\,)\big|\lesssim \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}|\vec{{\xi}}\,|^{-|\alpha|}Q(\mu),\qquad 1<q\le \infty, \end{align} $$

for all multi-indices $\alpha $ and $\vec {{\xi }}\,\in \mathbb {R}^{mn}\setminus \{0\}$ , where $Q(\mu )=2^{(mn-\delta ')\mu }$ if $\mu \ge 0$ and $Q(\mu )=2^{\mu (1-\delta ')}$ if $\mu <0$ for some $0<\delta '<1/q'$ , which is the condition of the Coifman–Meyer multiplier theorem ([Reference Coifman and Meyer9], [Reference Grafakos15, Theorem 7.5.3]) with constant $\Vert \Omega \Vert _{L^q(\mathbb S^{mn-1})}Q(\mu )$ . The remaining case when $\mu $ is large enough was handled by using product-type wavelet decompositions. We expect that a similar strategy would work in handling ${\mathcal {L}}_{\Omega }^{\sharp }$ .

To argue strictly, we write

$$ \begin{align*} {\mathcal{L}}_{\Omega}^{\sharp}\big(f_1,\dots,f_m\big) \le \widetilde {\mathcal{L}}_{\Omega}^{\sharp}\big(f_1,\dots,f_m\big) + \sum_{\mu\in\mathbb{Z}:2^{\mu-10}>C_0\sqrt{mn}} {\mathcal{L}}_{\Omega,\mu}^{\sharp}\big(f_1,\dots,f_m\big) , \end{align*} $$

where we set

$$ \begin{align*}\widetilde {\mathcal{L}}_{\Omega}^{\sharp}\big(f_1,\dots,f_m\big) (x) := \sup_{\tau \in \mathbb Z} \bigg| \int_{(\mathbb R^n)^m }{ \sum_{ \gamma<\tau} \sum_{\mu\in\mathbb{Z}:2^{\mu-10}\le C_0\sqrt{mn}} K^{\gamma}_{\mu}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j)}~d\,\vec{{y}}\, \bigg| \end{align*} $$

and

$$ \begin{align*}{\mathcal{L}}_{\Omega,\mu}^{\sharp}\big(f_1,\dots,f_m\big)(x) := \sup_{\tau\in \mathbb Z} \bigg| \sum_{\gamma<\tau} \int_{(\mathbb R^n)^m }{ K^{\gamma}_{\mu}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j)}~d\,\vec{{y}}\, \bigg|. \end{align*} $$

Then Theorem 1.2 follows from the following two propositions.

Proposition 4.1 Let $1<p_1,\dots ,p_m\le \infty $ and $1/p=1/p_1+\cdots +1/p_m$ . Suppose that $1<q<\infty $ and $\Omega \in L^q(\mathbb {S}^{mn-1})$ with $\int _{\mathbb {S}^{mn-1}}\Omega d\sigma =0$ . Then there exists a constant $C>0$ such that

$$ \begin{align*} \big\Vert \widetilde{\mathcal{L}}_{\Omega}^{\sharp}(f_1,\dots,f_m)\big\Vert_{L^p}\le C\Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^{p_j}} \end{align*} $$

for Schwartz functions $f_1,\dots ,f_m$ on $\mathbb R^n$ .

Proposition 4.2 Let $\frac {2m}{m+1}<q\le \infty $ and $\Omega \in L^q(\mathbb {S}^{mn-1})$ with $\int _{\mathbb {S}^{mn-1}}\Omega d\sigma =0$ . Suppose that $\mu \in \mathbb {Z}$ satisfies $2^{\mu -10}>C_0\sqrt {mn}$ . Then there exist $C, \epsilon _0>0$ such that

$$ \begin{align*} \big\| {\mathcal{L}}_{\Omega,\mu}^{\sharp}(f_1,\dots,f_m)\big\|_{L^{2/m} } \lesssim 2^{-\epsilon_0 \mu} \| \Omega\|_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2} \end{align*} $$

for Schwartz functions $f_1,\dots ,f_m$ on $\mathbb R^n$ .

4.2 Proof of Proposition 4.1

We decompose $\widetilde {\mathcal {L}}_{\Omega }^{\sharp }$ further so that the Coifman–Meyer multiplier theorem is involved: setting

$$ \begin{align*}\widetilde{K}(\vec{{y}}\,):=\sum_{\mu\in\mathbb{Z}: 2^{\mu-10}\le C_0\sqrt{mn}}K_{\mu}(\vec{{y}}\,)=\sum_{\mu\in\mathbb{Z}: 2^{\mu-10}\le C_0\sqrt{mn}}\;\sum_{\gamma\in\mathbb{Z}}K_{\mu}^{\gamma}(\vec{{y}}\,),\end{align*} $$

$\widetilde {\mathcal {L}}^{\sharp }_{\Omega }\big (f_1,\dots ,f_m\big )(x)$ is controlled by the sum of

$$ \begin{align*} T_{\widetilde{K}}^*\big(f_1,\dots,f_m\big)(x):= \sup_{\tau\in\mathbb{Z}}\Big|\int_{|y|> 2^{-\tau}}\widetilde{K}(\vec{{y}}\,)\prod_{j=1}^{m}f_j(x-y_j)~d\,\vec{{y}}\, \Big| \end{align*} $$

and

(4.6) $$ \begin{align} \mathfrak{T}_{K}^{**}\big(f_1,\dots,f_m\big)(x):= \sup_{\tau\in\mathbb{Z}}\Big| \int_{(\mathbb R^n)^m} {K}^{**}_{\tau}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j) ~ d\,\vec{{y}}\, \Big|, \end{align} $$

where

(4.7) $$ \begin{align} {K}^{**}_{\tau}(\vec{{y}}\,):=\Big(\sum_{\mu\in\mathbb{Z}: 2^{\mu-10}\le C_0\sqrt{mn}}\; \sum_{\gamma<\tau}K_{\mu}^{\gamma}(\vec{{y}}\,)\Big) -\widetilde{K}(\vec{{y}}\,)\chi_{|\vec{{y}}\,|> 2^{-\tau}}. \end{align} $$

To obtain the boundedness of $T_{\widetilde {K}}^*$ , we claim that $\widetilde {K}$ is an m-linear Calderón–Zygmund kernel with constant $C\Vert \Omega \Vert _{L^q(\mathbb {S}^{mn-1})}$ for $1<q< \infty $ . Indeed, it follows from (4.5) that

$$ \begin{align*} \big| \partial^{\alpha} \widehat{\widetilde{K}}(\vec{{\xi}}\,)\big|\le \sum_{\mu\in\mathbb{Z}: 2^{\mu-10}\le C_0\sqrt{mn}}\big|\partial^{\alpha}\widehat{K_{\mu}}(\vec{{\xi}}\,)\big|\lesssim \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}|\vec{{\xi}}\,|^{-|\alpha|} \end{align*} $$

as the sum of $Q(\mu )$ over $\mu $ satisfying $2^{\mu -10}\le C_0\sqrt {mn}$ converges. Then $\widetilde {K}$ satisfies the size and smoothness conditions for m-linear Calderón–Zygmund kernel with constant $C\Vert \Omega \Vert _{L^q(\mathbb {S}^{mn-1})}$ , as mentioned in the proof of [Reference Grafakos and Torres21, Proposition 6]. Since $\widetilde {K}$ is a Calderón–Zygmund kernel, Cotlar’s inequality in [Reference Grafakos and Torres20, Theorem 1] yields that $T_{\widetilde {K}}^*$ is bounded on the full range of exponents with constant $C\Vert \Omega \Vert _{L^{q}(\mathbb {S}^{mn-1})}$ .

To handle the boundedness of the operator $\mathfrak {T}_K^{**}$ , we observe that the kernel ${K}_{\tau }^{**}$ can be written as

(4.8) $$ \begin{align} {K}_{\tau}^{**}(\vec{{y}}\,)=\sum_{\mu\in\mathbb{Z}:2^{\mu-10}\le C_0\sqrt{mn} }\; \Big( \sum_{\gamma<\tau}K_{\mu}^{\gamma}(\vec{{y}}\,)\chi_{|\vec{{y}}\,|\le 2^{-\tau}}- \sum_{\gamma\ge \tau}K_{\mu}^{\gamma}(\vec{{y}}\,)\chi_{|\vec{{y}}\,|>2^{-\tau}}\Big), \end{align} $$

and thus

$$ \begin{align*} \mathfrak{T}_{K}^{**}\big(f_1,\dots,f_m\big)(x) \le \sup_{\tau\in\mathbb{Z}}\; \sum_{\mu\in\mathbb{Z}:2^{\mu-10}\le C_0\sqrt{mn} } \mathcal{I}_{\mu,\tau}(x)+\mathcal{J}_{\mu,\tau}(x), \end{align*} $$

where

$$ \begin{align*} \mathcal{I}_{\mu,\tau}(x):= \sum_{\gamma<\tau}\Big|\int_{|\vec{{y}}\,|<2^{-\tau}} K_{\mu}^{\gamma}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j) \;d\,\vec{{y}}\, \Big|, \end{align*} $$

$$ \begin{align*} \mathcal{J}_{\mu,\tau}(x):= \sum_{\gamma \ge \tau}\Big|\int_{|\vec{{y}}\,|\ge 2^{-\tau}} K_{\mu}^{\gamma}(\vec{{y}}\,) \prod_{j=1}^{m}f_j(x-y_j) \;d\,\vec{{y}}\, \Big|. \end{align*} $$

We claim that there exists $\epsilon>0$ such that

(4.9) $$ \begin{align} \mathcal{I}_{\mu,\tau}+ \mathcal{J}_{\mu,\tau}\lesssim_{C_0,m,n} 2^{\epsilon \mu}\Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\mathcal{M}f_j ~\text{ uniformly in ~} \tau\in\mathbb{Z} \end{align} $$

for $\mu $ satisfying $2^{\mu -10}\le C_0\sqrt {mn}$ , where we recall $\mathcal M$ is the Hardy–Littlewood maximal operator. Then, using Hölder’s inequality and the boundedness of $\mathcal {M}$ , we obtain

(4.10) $$ \begin{align} \big\Vert \mathfrak{T}_{K}^{**}\big(f_1,\dots,f_m\big) \big\Vert_{L^p}\lesssim \Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})} \Big\Vert \prod_{j=1}^{m}\mathcal{M}f_j \Big\Vert_{L^p}\lesssim \Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^{p_j}} \end{align} $$

for $1<p_1,\dots ,p_m\le \infty $ and $0<p\le \infty $ satisfying $1/p=1/p_1+\dots +1/p_m$ as $\sum _{\mu : 2^{\mu -10}\le C_0\sqrt {mn}}2^{\epsilon \mu }$ converges. Therefore, let us prove (4.9).

Using (4.1), we have

$$ \begin{align*} \mathcal{I}_{\mu,\tau}(x)&\lesssim \sum_{\gamma<\tau}\int_{|\vec{{y}}\,|<2^{-\tau}}\int_{|\vec{{z}}\,|\approx 1}2^{\gamma mn}2^{\mu mn}|\Omega(\vec{{z}}\,')|d\vec{{z}}\, \prod_{j=1}^{m}|f_j(x-y_j)|~d\,\vec{{y}}\,\\ &\lesssim 2^{\mu mn}\Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})}\frac{1}{2^{-\tau mn}}\int_{|\vec{{y}}\,|<2^{-\tau}} \prod_{j=1}^{m}|f_j(x-y_j)| ~d\,\vec{{y}}\,\\ &\lesssim 2^{ \mu mn}\Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\mathcal{M}f_j(x), \end{align*} $$

as desired.

In addition,

$$ \begin{align*} \mathcal{J}_{\mu,\tau}(x)\le \sum_{\gamma\ge \tau}\int_{|\vec{{y}}\,|\ge 2^{-\tau}}2^{\gamma mn}\Big|\int_{|\vec{{z}}\,|\approx 1}{\Phi_{\mu}(2^{\gamma}\vec{{y}}\,-\vec{{z}}\,)\Omega(\vec{{z}}\,')}d\,\vec{{z}}\, \Big|\prod_{j=1}^{m}|f_j(x-y_j)| ~ d\,\vec{{y}}\,. \end{align*} $$

Since $\Omega $ has vanishing mean, we have

$$ \begin{align*} &\Big|\int_{|\vec{{z}}\,|\approx 1}{\Phi_{\mu}(2^{\gamma}\vec{{y}}\,-\vec{{z}}\,)\,\Omega(\vec{{z}}\,')}\,d\,\vec{{z}}\, \Big| \\ &\lesssim 2^{\mu (mn+1)}\int_{|\vec{{z}}\,|\approx 1}\int_0^1\big| \nabla \Phi (2^{\mu+\gamma}\vec{{y}}\,-2^{\mu}t\vec{{z}}\,)\big|\;dt\; |\Omega(\vec{{z}}\,')|~ d\, \vec{{z}}\,. \end{align*} $$

Now we choose a constant M such that $mn<M<mn+1$ and see that

$$ \begin{align*} \big| \nabla \Phi (2^{\mu+\gamma}\vec{{y}}\,-2^{\mu}t\vec{{z}}\,)\big|&\lesssim_M \frac{1}{(1+|2^{\mu+\gamma}\vec{{y}}\,-2^{\mu}t\vec{{z}}\,|)^M}\\ &\lesssim_{C_0,m,n,M} \frac{1}{(1+2^{\mu+\gamma}|\vec{{y}}\,|)^M}\le \frac{1}{2^{M(\mu+\gamma)}}\frac{1}{|\vec{{y}}\,|^{M}}, \end{align*} $$

as $|\vec {{z}}\,|\approx 1$ , $0<t<1$ , and $2^{\mu -10}\le C_0\sqrt {mn}$ . This yields that

$$ \begin{align*} \mathcal{J}_{\mu,\tau}(x)&\lesssim 2^{\mu(mn+1-M)}\Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})}\Big(\sum_{\gamma\ge \tau}2^{-\gamma(M- mn)}\Big)\\ &\qquad \qquad \qquad \qquad \times \int_{|\vec{{y}}\,|\ge 2^{-\tau}}{\frac{1}{|\vec{{y}}\,|^M}\prod_{j=1}^{m}|f_j(x-y_j)|}~ d\, \vec{{y}}\,. \end{align*} $$

Since $M>mn$ , the sum over $\gamma \ge \tau $ converges to $2^{-\tau (M-mn)}$ and the integral over $|\vec {{y}}\,|\ge 2^{-\tau }$ is estimated by

$$ \begin{align*} &\sum_{l=0}^{\infty}\int_{2^{-\tau+l}\le |\vec{{y}}\,|<2^{-\tau+l+1}}{\frac{1}{|\vec{{y}}\,|^M}\prod_{j=1}^{m}|f_j(x-y_j)|}~ d\,\vec{{y}}\,\\ &\lesssim 2^{\tau(M-mn)}\sum_{l=0}^{\infty}2^{-l(M-mn)} \bigg(\frac{1}{2^{(-\tau+l+1)mn}}\int_{|\vec{{y}}\,|\le 2^{-\tau+l+1}}\prod_{j=1}^{m}|f_j(x-y_j)| ~ d\,\vec{{y}}\,\bigg)\\ &\lesssim 2^{\tau(M-mn)}\prod_{j=1}^{m}\mathcal{M}f_j(x). \end{align*} $$

Finally, we have

$$ \begin{align*} \mathcal{J}_{\mu,\tau} \lesssim 2^{\mu(mn+1-M)}\Vert \Omega\Vert_{L^1(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\mathcal{M}f_j , \end{align*} $$

which completes the proof of (4.9).

4.3 Proof of Proposition 4.2

The proof is based on the wavelet decomposition and the recent developments in [Reference Grafakos, He, Honzík and Park18]. Recalling that $\widehat {K_{\mu }^0}\in L^{q'}$ , we apply the wavelet decomposition (2.1) to write

$$ \begin{align*} \widehat{K_{\mu}^0}(\vec{{\xi}}\,)=\sum_{\lambda\in\mathbb{N}_0}\sum_{\vec{{G}}\in\mathcal{I}^{\lambda}}\sum_{\vec{{k}}\,\in (\mathbb Z^n)^m}b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}\Psi_{G_1,k_1}^{\lambda}(\xi_1)\cdots \Psi_{G_m,k_m}^{\lambda}(\xi_m), \end{align*} $$

where

$$ \begin{align*} b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}:=\int_{(\mathbb R^n)^m}{\widehat{K_{\mu}^0}(\vec{{\xi}}\,)\Psi_{\vec{{G}},\vec{{k}}\,}^{\lambda}(\vec{{\xi}}\,)}~ d\,\vec{{\xi}}\,. \end{align*} $$

It is known in [Reference Grafakos, He, Honzík and Park18] that for any $0<\delta <1/q'$ ,

(4.11) $$ \begin{align} \big\Vert \{b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}\}_{\vec{{k}}\,}\big\Vert_{\ell^{\infty}}\lesssim 2^{-\delta {\mu}}2^{-\lambda (M+1+mn)} \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}, \end{align} $$

where M is the number of vanishing moments of $\Psi _{\vec {{G}}}$ . Moreover, it follows from the inequality (2.2), the Hausdorff–Young inequality, and Young’s inequality that

(4.12) $$ \begin{align} \big\Vert \{b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}\}_{\vec{{k}}\,}\big\Vert_{\ell^{q'}}&\lesssim 2^{-\lambda mn (1/2-1/q')}\Vert \widehat{K_{\mu}^0}\Vert_{L^{q'}}\lesssim 2^{-\lambda mn(1/q-1/2)}\Vert \Omega \Vert_{L^q(\mathbb{S}^{mn-1})}. \end{align} $$

Now we may assume that $2^{\lambda +\mu -2}\le |\vec k|\le 2^{\lambda +\mu +2}$ due to the compact supports of $\widehat {K_{\mu }^0}$ and $\Psi _{\vec {{G}},\vec {{k}}\,}^{\lambda }$ . In addition, by symmetry, it suffices to focus on the case $|k_1|\ge \cdots \ge |k_m|$ . Since $\widehat {K_{\mu }^{\gamma }}(\vec {{\xi }}\,)=\widehat {K_{\mu }^0}(\vec {{\xi }}\,/2^{\gamma })$ , the boundedness of $ {\mathcal {L}}_{\Omega ,\mu }^{\sharp }$ is reduced to the inequality

(4.13) $$ \begin{align} &\bigg\Vert \sup_{\tau\in\mathbb{Z}}\Big| \sum_{\lambda\in\mathbb{N}_0}\sum_{\vec{{G}}\in\mathcal{I}^{\lambda}} \sum_{\gamma\in\mathbb{Z}: \gamma<\tau}\sum_{\vec{{k}}\,\in\mathcal{U}^{\lambda+\mu}} b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}\prod_{j=1}^{m} L_{G_j,k_j}^{\lambda,\gamma}f_j\Big|\bigg\Vert_{L^{2/m}}\nonumber\\ &\lesssim 2^{-\epsilon_0 \mu}\Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2}, \end{align} $$

where the operators $L_{G_j,k_j}^{\lambda ,\gamma }$ and the set $\mathcal {U}^{\lambda +\mu }$ are defined as in (2.5) and (2.4). We split $\mathcal {U}^{\lambda +{\mu }}$ into m disjoint subsets $\mathcal {U}_l^{\lambda +{\mu }}$ ( $1\le l\le m$ ) as before such that for $k\in \mathcal {U}^{\lambda +\mu }_l$ , we have

$$ \begin{align*}|k_1|\ge\cdots \ge |k_l|\ge 2C_0\sqrt n\ge |k_{l+1}|\ge\cdots\ge |k_m|.\end{align*} $$

Then the left-hand side of (4.13) is estimated by

$$ \begin{align*} \bigg( \sum_{l=1}^{m}\sum_{\lambda\in\mathbb{N}_0}\sum_{\vec{{G}}\in\mathcal{I}^{\lambda}} \bigg\Vert \sup_{\tau\in\mathbb{Z}}\Big| \sum_{\gamma\in\mathbb{Z}:\gamma<\tau} \mathcal{T}_{\vec{{G}},l}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) \Big| \bigg\Vert_{L^{2/m}}^{2/m}\bigg)^{m/2}, \end{align*} $$

where $\mathcal {T}_{\vec {{G}},l}^{\lambda ,\gamma ,\mu }$ is defined by

$$ \begin{align*} \mathcal{T}_{\vec{{G}},l}^{\lambda,\gamma,\mu}\big(f_1,\dots,f_m\big):=\sum_{\vec{{k}}\,\in\mathcal{U}_l^{\lambda+{\mu}}}b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu} \bigg(\prod_{j=1}^{m}L_{G_j,k_j}^{\lambda,\gamma}f_j \bigg). \end{align*} $$

We claim that for each $1\le l\le m$ , there exists $\epsilon _0, M_0>0$ such that

(4.14) $$ \begin{align} \begin{aligned} \bigg\Vert \sup_{\tau\in\mathbb{Z}}\Big| &\sum_{\gamma\in\mathbb{Z}:\gamma<\tau} \mathcal{T}_{\vec{{G}},l}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) \Big| \bigg\Vert_{L^{2/m}} \\ &\lesssim 2^{-\epsilon_0 \mu_0}2^{-\lambda M_0}\Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2}, \end{aligned} \end{align} $$

which concludes (4.13). Therefore, it remains to prove (4.14).

4.3.1 Proof of (4.14)

When $2\le l\le m$ , we apply (2.8) with $2<q'<\frac {2m}{m-1}$ , along with (4.11), and (4.12) to obtain

$$ \begin{align*} &\bigg\Vert \sup_{\tau\in\mathbb{Z}}\Big| \sum_{\gamma\in\mathbb{Z}:\gamma<\tau}\mathcal{T}_{\vec{{G}},l}^{\lambda,\gamma,\mu}\big(f_1,\dots,f_m\big) \bigg\Vert_{L^{2/m}}\le \bigg\Vert \sum_{\gamma\in\mathbb{Z}} \big| \mathcal{T}_{\vec{{G}},l}^{\lambda,\gamma,\mu}\big(f_1,\dots,f_m\big) \big| \bigg\Vert_{L^{2/m}} \\ &\lesssim \big\Vert \big\{ b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}\big\}_{\vec{{k}}\,}\big\Vert_{\ell^{\infty}}^{1-\frac{(m-1)q'}{2m}} \big\Vert \big\{ b_{\vec{{G}},\vec{{k}}\,}^{\lambda,\mu}\big\}_{\vec{{k}}\,}\big\Vert_{\ell^{q'}}^{\frac{(m-1)q'}{2m}} 2^{\lambda mn/2}(\lambda+1)^{l/2}{\mu}^{l/2}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2}\\ &\lesssim \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}2^{-\delta {\mu}(1-\frac{(m-1)q'}{2m})}{\mu}^{m/2}2^{-\lambda C_{M,m,n,q}}(\lambda+1)^{m/2}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2} , \end{align*} $$

where

$$ \begin{align*} C_{M,m,n,q}:=(M+1+mn)\bigg(1-\frac{(m-1)q'}{2m}\bigg)+mn(1/q-1/2)\frac{(m-1)q'}{2m}-\frac{mn}{2}. \end{align*} $$

Here, we used the fact that $\frac {l-1}{2l}\le \frac {m-1}{2m}$ for $l\le m$ . Then (4.14) follows from choosing M sufficiently large so that $C_{M,m,n,q}>0$ since $1-\frac {(m-1)q'}{2m}>0$ .

Now let us prove (4.14) for $l=1$ . In this case, we first see the estimate

(4.15) $$ \begin{align} \Big\Vert \Big( \sum_{\gamma\in\mathbb{Z}}\big| \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) \big|^2\Big)^{1/2}\Big\Vert_{L^{2/m}} \lesssim 2^{-\epsilon_0\mu} 2^{-M_0\lambda} \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2} \end{align} $$

for some $\epsilon _0, M_0>0$ , which can be proved, as in [Reference Grafakos, He, Honzík and Park18, Section 6], by using (2.7) and (4.11).

Choose a Schwartz function $\Gamma $ on $\mathbb R^n$ whose Fourier transform is supported in the ball $ \{\xi \in \mathbb R^n: |\xi |\le 2\}$ and is equal to $1$ for $|\xi |\le 1$ , and define $\Gamma _k:=2^{kn}\Gamma (2^k\cdot )$ so that $\mathrm {Supp}(\widehat {\Gamma _k})\subset \{\xi \in \mathbb R^n: |\xi |\le 2^{k+1}\}$ and $\widehat {\Gamma _k}(\xi )=1$ for $|\xi |\le 2^k$ .

Since the Fourier transform of $\mathcal {T}_{\vec {{G}},1}^{\lambda ,\gamma ,\mu }(f_1,\dots ,f_m)$ is supported in the set $\big \{ \xi \in \mathbb R^n: 2^{\gamma +\mu -5}\le |\xi |\le 2^{\gamma +\mu +4} \big \}$ , we can write

$$ \begin{align*} \sum_{\gamma\in\mathbb{Z}: \gamma<\tau}\mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m)=\Gamma_{\tau+\mu+3}\ast \Big( \sum_{\gamma\in\mathbb{Z}: \gamma<\tau}\mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m)\Big) \end{align*} $$

and then split the right-hand side into

$$ \begin{align*} & \Gamma_{\tau+\mu+3}\ast \Big( \sum_{\gamma\in\mathbb{Z}}\mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m)\Big)-\Gamma_{\tau+\mu+3}\ast \Big( \sum_{ \gamma\in\mathbb{Z} :\gamma\ge \tau}\mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m)\Big). \end{align*} $$

Due to the Fourier support conditions of $\Gamma _{\tau +\mu +3}$ and $\mathcal {T}_{\vec {{G}},1}^{\lambda ,\gamma ,\mu }(f_1,\dots ,f_m)$ , the sum in the second term can be actually taken over $\tau \le \gamma \le \tau +9$ . Therefore, the left-hand side of (4.14) is controlled by the sum of

(4.16) $$ \begin{align} I:=\bigg\Vert \sup_{\nu\in\mathbb{Z}}\Big| \Gamma_{\nu}\ast \Big(\sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } \Big)\Big| \bigg\Vert_{L^{2/m}} \end{align} $$

and

(4.17) $$ \begin{align} II:=\sum_{\gamma=0}^{9}\Big\Vert \sup_{\tau\in\mathbb{Z}}\big| \Gamma_{\tau+\mu+3}\ast T_{\vec{{G}},1}^{\lambda,\tau+\gamma,\mu}(f_1,\dots,f_m)\big|\Big\Vert_{L^{2/m}}. \end{align} $$

First of all, when $0\le \gamma \le 9$ , the Fourier supports of both $\Gamma _{\tau +\mu +3}$ and $T_{\vec {{G}},1}^{\lambda ,\tau +\gamma ,\mu }(f_1,\dots ,f_m)$ are $\{\xi \in \mathbb R^n: |\xi |\sim 2^{\tau +\mu }\}$ . This implies that for any $0<r<1$ ,

$$ \begin{align*} &\big| \Gamma_{\tau+\mu+3}\ast T_{\vec{{G}},1}^{\lambda,\tau+\gamma,\mu}(f_1,\dots,f_m)(x)\big|\\ &\lesssim 2^{(\tau+\mu)(n/r-n)}\Big( \int_{\mathbb R^n} \big| \Gamma_{\tau+\mu+3}(x-y)\big|^r \big| T_{\vec{{G}},1}^{\lambda,\tau+\gamma,\mu}(f_1,\dots,f_m)(y)\big|^r ~dy\Big)^{1/r}\\ &\lesssim \Big(\mathcal{M}\big(|T_{\vec{{G}},1}^{\lambda,\tau+\gamma,\mu}(f_1,\dots,f_m)|^r\big)(x)\Big)^{1/r}, \end{align*} $$

where the Nikolskii inequality (see [Reference Triebel28, Proposition 1.3.2]) is applied in the first inequality. Setting $0<r<2/m$ , and using the maximal inequality for $\mathcal {M}$ and the embedding $\ell ^2\hookrightarrow \ell ^{\infty }$ we obtain

(4.18) $$ \begin{align} II&\lesssim \big\Vert \sup_{\tau\in\mathbb{Z}}\big| T_{\vec{{G}},1}^{\lambda,\tau,\mu}(f_1,\dots,f_m) \big| \big\Vert_{L^{2/m}} \end{align} $$

$$\begin{align*} &\le \Big\Vert \Big( \sum_{\gamma\in\mathbb{Z}}\big| \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) \big|^2\Big)^{1/2}\Big\Vert_{L^{2/m}}. \end{align*}$$

Then the $L^{2/m}$ norm is bounded by the right-hand side of (4.14), thanks to (4.15). This completes the estimate for $II$ defined in (4.17), and we turn our attention to I defined in (4.16).

In the sequel, we will make use of the following inequality: if $\widehat {g_{\gamma }}$ is supported on $\{\xi \in \mathbb {R}^n : C^{-1} 2^{\gamma +\mu }\leq |\xi |\leq C2^{\gamma +\mu }\}$ for some $C>1$ and $\mu \in \mathbb {Z}$ , then

(4.19) $$ \begin{align} \Big\Vert \Big\{ \Phi^{(1)}_j\!\ast\! \Big(\sum_{\gamma\in\mathbb{Z}}{g_{\gamma}}\!\Big)\!\Big\}_{\! j\in\mathbb{Z}}\Big\Vert_{L^p(\ell^q)}\lesssim_{C} \big\Vert \big\{ g_j\big\}_{j\in\mathbb{Z}}\big\Vert_{L^p(\ell^q)} \quad \text{uniformly in }~\mu \end{align} $$

for $0<p<\infty $ . The proof of (4.19) is elementary and standard, so it is omitted here (see [Reference Grafakos, He and Honzík16, equation (13)] and [Reference Yamazaki31, Theorem 3.6] for related arguments).

To obtain the bound of I, we note that

$$ \begin{align*} I \approx \Big\Vert \sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) }\Big\Vert_{H^{2/m}}, \end{align*} $$

where $H^{2/m}$ is the Hardy space. We refer to [Reference Grafakos15, Corollary 2.1.8] for the above estimate. Then, using the Littlewood–Paley theory for Hardy space (see, for instance, [Reference Grafakos15, Theorem 2.2.9]) and (4.19), there exists a unique polynomial $Q^{\lambda ,\mu ,\vec {{G}}}(x)$ such that

(4.20) $$ \begin{align} \Big\Vert \sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } -Q^{\lambda,\mu,\vec{{G}}} \Big\Vert_{H^{2/m}}\notag &\lesssim \Big\Vert \Big( \sum_{\gamma\in\mathbb{Z}}\big| \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) \big|^2\Big)^{1/2}\Big\Vert_{L^{2/m}}\nonumber\\ & \lesssim 2^{-\epsilon_0\mu} 2^{-M_0\lambda} \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2}, \end{align} $$

where (4.15) is applied. Furthermore,

$$ \begin{align*} &\Big\Vert \sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } \Big\Vert_{H^{2/m}}\\ &\approx \bigg\Vert \sup_{\nu\in\mathbb{Z}}\Big|\Gamma_{\nu}\ast\Big( \sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } \Big)\Big|\bigg\Vert_{L^{2/m}}\\ &=\bigg\Vert \sup_{\nu\in\mathbb{Z}}\Big|\Gamma_{\nu}\ast\Big( \sum_{\gamma\in\mathbb{Z}: \gamma\le \nu-\mu+5}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } \Big)\Big|\bigg\Vert_{L^{2/m}}\\ &\lesssim \bigg\Vert \sup_{\nu\in\mathbb{Z}} \Big| \sum_{\gamma\in\mathbb{Z}: \gamma\le \nu-\mu+5}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) }\Big|\bigg\Vert_{L^{2/m}}\\ &\le \Big\Vert \sum_{\gamma\in\mathbb{Z}}{ \big| \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) \big| }\Big\Vert_{L^{2/m}}, \end{align*} $$

where the argument that led to (4.18) is applied in the first inequality. As we have discussed in [Reference Grafakos, He, Honzík and Park18, Section 6.1], this quantity is finite for all Schwartz functions $f_1,\dots ,f_m$ . Accordingly, we have

$$ \begin{align*}\sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } -Q^{\lambda,\mu,\vec{{G}}}\in H^{2/m}\end{align*} $$

and

$$ \begin{align*}\sum_{\gamma\in\mathbb{Z}}{ \mathcal{T}_{\vec{{G}},1}^{\lambda,\gamma,\mu}(f_1,\dots,f_m) } \in H^{2/m},\end{align*} $$

and thus $Q^{\lambda ,\mu ,\vec {{G}}}=0$ . Now it follows from (4.20) that

$$ \begin{align*} I \lesssim 2^{-\epsilon_0\mu} 2^{-M_0\lambda} \Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})}\prod_{j=1}^{m}\Vert f_j\Vert_{L^2}, \end{align*} $$

as expected. This completes the proof of (4.14).

5.1 A variant of Theorem 1.2 for general $L^2$ functions

We note that when $f_j\in L^{2}(\mathbb {R}^n)$ , by Lemma 3.1, there exists a set $E^{\Omega }_{f_1,\dots , f_m}$ of measure $0$ such that

$$ \begin{align*}\mathcal M_{\Omega} \big(f_1,\dots , f_m\big)(x)<\infty \end{align*} $$

when $x\notin E^{\Omega }_{f_1,\dots , f_m}$ . Therefore, for $x\notin E^{\Omega }_{f_1,\dots , f_m}$ , we have

(5.1) $$ \begin{align} \int_{\epsilon_0 \le |\vec{{y}}\,| \le \epsilon_0^{-1}} \frac{|\Omega (\vec{{y}}\,')|}{|\vec{{y}}\,|^{mn}}\, \prod_{j=1}^{m}\big|f_j(x-y_j)\big|\; d\vec{{y}}\, \le C \frac{1 }{\epsilon_0^{2mn}} \mathcal M_{\Omega} \big(f_1,\dots , f_m\big)(x), \end{align} $$

and thus ${\mathcal {L}}_{\Omega }^{ ( \epsilon , \epsilon ^{-1}) }(f_1,\dots , f_m)(x)$ and ${\mathcal {L}}_{\Omega }^{*, \epsilon _0 }(f_1,\dots , f_m)(x)$ are well defined, for $f_j$ in $L^2(\mathbb {R}^n)$ and $\Omega $ in $L^q(\mathbb {S}^{mn-1})$ . Moreover,

$$\begin{align*}{\mathcal{L}}_{\Omega}^{(\epsilon , \epsilon^{-1})}(f_1, \dots , f_m)(x)\le {\mathcal{L}}_{\Omega}^{*,\epsilon_0}(f_1, \dots , f_m)(x) \end{align*}$$

pointwise for $x\in \mathbb R^n\setminus E_{f_1,\dots , f_m}^{\Omega }$ .

For given $f_j\in L^2(\mathbb {R}^n)$ , we pick sequences of Schwartz functions $f_j^k$ converging to $f_j$ in $L^2$ as $k\to \infty $ , by density. Using the identity

(5.2) $$ \begin{align} a_1a_2\cdots a_m - b_1b_2\cdots b_m= \sum_{j=1}^m b_1 \cdots b_{j-1} (a_j-b_j) a_{j+1}\cdots a_m \end{align} $$

(with the obvious modification when $j=1$ or $j=m$ ), the inequality

$$\begin{align*}{\mathcal{L}}_{\Omega}^{*,\epsilon_0 }\big(f_1,\dots , f_m\big)\le 2{\mathcal{L}}_{\Omega}^{*}(f_1^k,\dots , f_m^k) +\sum_{j=1}^m {\mathcal{L}}_{\Omega}^{*,\epsilon_0 }(f_1^k,\dots f_{j-1}^k, f_j-f_j^k, f_{j+1}\dots , f_m ) \end{align*}$$

is valid pointwise on the complement of the set

(5.3) $$ \begin{align} E^{\Omega}:=E_{f_1,\dots, f_m}^{\Omega} \cup \left( \bigcup_{k=1}^{\infty} E_{f_1^k,\dots, f_m^k}^{\Omega}\right) \cup \bigg( \bigcup_{j=1}^m\bigcup_{k=1}^{\infty} E_{f_1^k,\dots f_{j-1}^k, f_j-f_j^k, f_{j+1}\dots , f_m }^{\Omega} \bigg), \end{align} $$

which is a set of measure zero. Now we take the $L^{2/m} $ quasi-norm on both sides and apply Theorem 1.3 for Schwartz functions and the estimate (5.1), combined with Lemma 3.1, for the terms in the sum. Taking $k\to \infty $ , we obtain

$$ \begin{align*} \big\|{\mathcal{L}}_{\Omega}^{*,\epsilon_0 }(f_1,\dots , f_m)\big\|_{L^{2/m}(\mathbb{R}^n) } \le C\Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})} \prod_{j=1}^{\infty}\| f_j\|_{L^{2}(\mathbb{R}^n)}. \end{align*} $$

Finally, letting $\epsilon _0 \to 0$ and using the monotone convergence theorem, we conclude

(5.4) $$ \begin{align} \big\|{\mathcal{L}}_{\Omega}^{** }(f_1,\dots , f_m)\big\|_{L^{2/m}(\mathbb{R}^n) } \le C\Vert \Omega\Vert_{L^q(\mathbb{S}^{mn-1})} \prod_{j=1}^{\infty}\| f_j\|_{L^{2}(\mathbb{R}^n)} \end{align} $$

for $f_j\in L^2(\mathbb {R}^n)$ .

5.2 Proof of Theorem 1.1

Let $f_1,\dots ,f_m$ be given $L^2$ functions and pick sequences $\{f_j^k\}$ of Schwartz functions such that $f_j^k$ converges to $f_j$ in $L^2(\mathbb {R}^n)$ as $k\to \infty $ . Recall that ${\mathcal {L}}_{\Omega }(f_1 ,\dots , f_m )$ is defined as the $L^{2/m}$ limit of ${\mathcal {L}}_{\Omega }(f_1^k ,\dots , f_m^k )$ as $k\to \infty $ . Then there exists a subsequence $\{k_l\}$ of $\{k\}$ such that ${\mathcal {L}}_{\Omega }(f_1^{k_l} ,\dots , f_m^{k_l} ) \to {\mathcal {L}}_{\Omega }(f_1 ,\dots , f_m )$ pointwise on $\mathbb R^n\setminus E$ , for some set E of measure zero. Let us denote the subsequence $\{k_l\}$ still by $\{k\}$ for notational convenience. Then, on $\mathbb R^n\setminus (E\cup E^{\Omega })$ , where $E^{\Omega }$ is as in (5.3), we have

$$ \begin{align*} \big|{\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}&(f_1,\dots, f_m) -{\mathcal{L}}_{\Omega}(f_1 ,\dots, f_m )\big|\\ &\le \big|{\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}(f_1,\dots, f_m) -{\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}(f_1^k ,\dots, f_m^k )\big|\\ & \quad + \big| {\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}(f_1^k ,\dots, f_m^k ) - {\mathcal{L}}_{\Omega} (f_1^k ,\dots, f_m^k ) \big|\\ & \quad \quad + \big| {\mathcal{L}}_{\Omega} (f_1^k ,\dots, f_m^k ) -{\mathcal{L}}_{\Omega}(f_1 ,\dots, f_m ) \big|. \end{align*} $$

We first take the $\limsup _{\epsilon \to 0}$ on both sides and then the middle term on the right vanishes. Then we apply $\liminf _{k\to \infty }$ so that the last term vanishes. Consequently, we have

$$ \begin{align*} \limsup_{\epsilon \to 0}|&{\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}(f_1,\dots, f_m) -{\mathcal{L}}_{\Omega}(f_1 ,\dots, f_m )|\\ &\le \liminf_{k\to \infty} \sum_{j=1}^m {\mathcal{L}}_{\Omega}^{**} (f_1,\dots, f_{j-1} , f_j-f_j^k, f_{j+1}^k, \dots , f_m^k) \end{align*} $$

on $\mathbb R^n\setminus \big (E\cup E^{\Omega }\big )$ , where the identity (5.2) is applied. It follows that

$$ \begin{align*} & \Big|\big\{x:\,\,\limsup_{\epsilon \to 0}|{\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}(f_1,\dots, f_m) -{\mathcal{L}}_{\Omega}(f_1 ,\dots, f_m )|>\lambda\big\}\Big| \\ &\le \Big|\big\{x:\,\, \liminf_{k\to \infty} \sum_{j=1}^m {\mathcal{L}}_{\Omega}^{**} (f_1,\dots, f_{j-1} , f_j-f_j^k, f_{j+1}^k, \dots , f_m^k)>\lambda\big\}\Big| \\ &\le \lambda^{-\frac 2m} \sum_{j=1}^m \Big\| \liminf_{k\to \infty} \sum_{j=1}^m {\mathcal{L}}_{\Omega}^{**} (f_1,\dots, f_{j-1} , f_j-f_j^k, f_{j+1}^k, \dots , f_m^k)\big\|_{L^{\frac 2m}}^{\frac 2m} \end{align*} $$

by Chebyshev’s inequality. But this last expression tends to zero as $k\to \infty $ in view of Fatou’s lemma and (5.4). We conclude that

$$ \begin{align*}\limsup_{\epsilon \to 0}|{\mathcal{L}}_{\Omega}^{(\epsilon,\epsilon^{-1})}(f_1,\dots, f_m) -{\mathcal{L}}_{\Omega}(f_1 ,\dots, f_m )| \end{align*} $$

equals zero a.e. and this finishes the proof.

7.1 Extension of Theorem 1.4 to general $f_j\in L^2(\mathbb {R}^n)$

Let $f_1,\dots ,f_m$ be given $L^2$ functions on $\mathbb {R}^n$ . As $S_{\sigma }^{\nu }(f_1 ,\dots , f_m ) $ is finite a.e. for each $\nu \in \mathbb {Z}$ , there exists a set $E_{\nu }$ of measure zero so that $\big |S_{\sigma }^{\nu }(f_1 ,\dots , f_m ) \big |<\infty $ on the complement of $E_{\nu }$ . Since $\nu $ ranges over a countable set $\mathbb {Z}$ , the measure of $E:=\bigcup _{\nu \in \mathbb {Z}}E_{\nu }$ is clearly zero and thus we can define

$$\begin{align*}\mathscr{M}_{\sigma}(f_1,\dots , f_m)= \sup_{\nu\in\mathbb{Z}} | S_{\sigma}^{\nu}(f_1 ,\dots, f_m )| \end{align*}$$

on $E^c$ . That is, $\mathscr {M}_{\sigma }(f_1,\dots , f_m)$ is well defined pointwise a.e. Moreover, it controls every $\big | S_{\sigma }^{\nu }(f_1 ,\dots , f_m )\big |$ pointwise, whenever the latter is finite.

We now extend Theorem 1.5 to $f_j\in L^2(\mathbb R^n)$ using the above definition. Without loss of generality, we only consider the case when $\nu \to -\infty $ as the case $\nu \to \infty $ follows similarly. As every sequence that converges in $L^{2/m}$ has a subsequence that converges a.e., there are a set of measure zero $E_{f_1,\dots , f_m}^1 $ and a subsequence

$$\begin{align*}k_1^1<k_2^1<k_3^1<\cdots <k_{\ell}^1<\cdots \end{align*}$$

of the sequence of k’s such that

$$\begin{align*}S_{\sigma}^{-1}(f_1^{k_{\ell}^1},\dots, f_m^{k_{\ell}^1}) (x) \to S_{\sigma}^{-1}(f_1,\dots, f_m)(x) \end{align*}$$

for all $x\in \mathbb R^n\setminus E_{f_1,\dots , f_m}^1$ . Next, there are a set of measure zero $E_{f_1,\dots , f_m}^2$ and a subsequence

$$\begin{align*}k_1^2<k_2^2<k_3^2<\cdots <k_{\ell}^2<\cdots \end{align*}$$

of

$$\begin{align*}k_1^1<k_2^1<k_3^1<\cdots <k_{\ell}^1<\cdots \end{align*}$$

such that

$$\begin{align*}S_{\sigma}^{-2}(f_1^{k_{\ell}^2},\dots, f_m^{k_{\ell}^2}) (x) \to S_{\sigma}^{-2}(f_1,\dots, f_m)(x) \end{align*}$$

for all $x\in \mathbb R^n \setminus (E_{f_1,\dots , f_m}^2\cup E_{f_1,\dots , f_m}^2)$ . There are a set of measure zero $E_{f_1,\dots , f_m}^3$ and a subsequence

$$\begin{align*}k_1^3<k_2^3<k_3^3<\cdots <k_{\ell}^3<\cdots \end{align*}$$

of

$$\begin{align*}k_1^2<k_2^2<k_3^2<\cdots <k_{\ell}^2<\cdots \end{align*}$$

such that

$$\begin{align*}S_{\sigma}^{-3}(f_1^{k_{\ell}^3},\dots, f_m^{k_{\ell}^3}) (x) \to S_{\sigma}^{-3}(f_1,\dots, f_m)(x) \end{align*}$$

for all $x\in \mathbb R^n\setminus (E_{f_1,\dots , f_m}^1 \cup E_{f_1,\dots , f_m}^2 \cup E_{f_1,\dots , f_m}^3)$ . Iterating this process, we can take a diagonal sequence

$$\begin{align*}k_1^1<k_2^2<k_3^3<\cdots <k_{\ell}^{\ell}<\cdots, \end{align*}$$

which is a subsequence of all subsequences, for which $f_j^{k_{\ell }^{\ell }}(x)\to f_j(x)$ for all $1\le j\le m$ and

$$\begin{align*}S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) (x) \to S_{\sigma}^{\nu}(f_1,\dots, f_m)(x), \end{align*}$$

as $\ell \to \infty $ for $x\in \mathbb R^n\setminus \bigcup _{\rho =1}^{\infty } E_{f_1,\dots , f_m}^{\rho }$ and all $\nu =-1,-2,\dots $ .

Now, on the set $ \mathbb R^n\setminus \bigcup _{\rho =1}^{\infty } E_{f_1,\dots , f_m}^{\rho }$ , we have

$$\begin{align*}\big|S_{\sigma}^{\nu}(f_1,\dots, f_m) \big|= \lim_{\ell\to \infty} \big|S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) \big|= \liminf_{\ell\to \infty} \big|S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) \big|, \end{align*}$$

for every $\nu =-1,-2,\dots $ , and thus

$$\begin{align*}\big|S_{\sigma}^{\nu}(f_1,\dots, f_m) \big|\le \liminf_{\ell\to \infty} \mathscr{M}_{\sigma}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}). \end{align*}$$

This deduces

$$\begin{align*}\mathscr{M}_{\sigma}(f_1,\dots, f_m) |\le \liminf_{\ell\to \infty} \mathscr{M}_{\sigma}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) \end{align*}$$

on $ \mathbb R^n\setminus \bigcup _{\rho =1}^{\infty } E_{f_1,\dots , f_m}^{\rho }$ . Taking the $L^{2/m} $ quasi-norm on the both sides and using Fatou’s lemma and Theorem 1.4 for Schwartz functions, we finally obtain

(7.2) $$ \begin{align} \big\Vert \mathscr{M}_{\sigma}(f_1,\dots, f_m) \big\Vert_{L^{2/m}(\mathbb{R}^n)}\lesssim \prod_{j=1}^{m}\Vert f_j\Vert_{L^2(\mathbb R^n)} \end{align} $$

for $f_j\in L^2(\mathbb R^n)$ .

7.2 Proof of Theorem 1.3

Let $f_j$ , $j=1,\dots ,m$ , be functions in $L^2(\mathbb R^n)$ and $\{f_j^{k_{\ell }^{\ell }}\}_{\ell }$ be sequences that appeared above so that on $(\bigcup _{\rho =1}^{\infty }E_{f_1,\dots ,f_m}^{\rho })^c$ ,

(7.3) $$ \begin{align} \lim_{\ell\to \infty} S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) = S_{\sigma}^{\nu}(f_1,\dots, f_m) \end{align} $$

for each $\nu =-1,-2,\dots $ .

On $(\bigcup _{\rho =1}^{\infty }E_{f_1,\dots ,f_m}^{\rho })^c$ , we write

$$ \begin{align*} &\big| S_{\sigma}^{\nu}(f_1,\dots, f_m)(x) - \sigma(0)f_1(x)\cdots f_m(x) \big|\\ &\le \big| S_{\sigma}^{\nu}(f_1,\dots, f_m) - S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) \big|\\ &\quad + \big| S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}})- \sigma(0) f_1^{k_{\ell}^{\ell}} \cdots f_m^{k_{\ell}^{\ell}} \big|\\ &\quad \quad + \big| \sigma(0) f_1^{k_{\ell}^{\ell}} \cdots f_m^{k_{\ell}^{\ell}}-\sigma(0) f_1\cdots f_m \big|. \end{align*} $$

We first take $\limsup _{\nu \to -\infty }$ and use (7.1) to make the middle term on the right vanish, and then apply $\liminf _{\ell \to \infty }$ to handle the last term on the right which will vanish as well. As a result, we obtain

$$ \begin{align*} &\limsup_{\nu \to -\infty}\big| S_{\sigma}^{\nu}(f_1,\dots, f_m) - \sigma(0) f_1\cdots f_m \big|\\ & \le \liminf_{\ell \to \infty} \sup_{\nu<0}\big| S_{\sigma}^{\nu}(f_1,\dots, f_m) - S_{\sigma}^{\nu}(f_1^{k_{\ell}^{\ell}},\dots, f_m^{k_{\ell}^{\ell}}) \big|. \end{align*} $$

Using the identity (5.2), we control the preceding expression pointwise by

$$\begin{align*}\liminf_{\ell \to \infty} \sum_{i=1}^m \mathscr M_{\sigma} (f_1^{k_{\ell}^{\ell}},\dots, f_{i-1}^{k_{\ell}^{\ell}} ,f_i^{k_{\ell}^{\ell}} -f_i, f_{i+1},\dots, f_m ) \end{align*}$$

on the complement of the set

$$\begin{align*}\bigcup_{\rho=1}^{\infty} \bigg[E_{f_1,\dots, f_m}^{\rho} \cup \bigg( \bigcup_{i=1}^m \bigcup_{\ell=1}^{\infty} E_{f_1^{k_{\ell}^{\ell}},\dots, f_{i-1}^{k_{\ell}^{\ell}} ,f_i^{k_{\ell}^{\ell}} -f_i, f_{i+1},\dots, f_m}^{\rho} \bigg)\bigg], \end{align*}$$

which has full measure. Since

$$\begin{align*}\Big\| \liminf_{\ell \to \infty}\sum_{i=1}^m \mathscr M_{\sigma} (f_1^{k_{\ell}^{\ell}},\dots, f_{i-1}^{k_{\ell}^{\ell}} ,f_i^{k_{\ell}^{\ell}} -f_i, f_{i+1},\dots, f_m ) \Big\|_{L^{2/m}(\mathbb R^n)} =0 \end{align*}$$

in view of Fatou’s lemma and (7.2), we finally obtain

$$\begin{align*}\limsup_{\nu \to -\infty}\big| S_{\sigma}^{\nu}(f_1,\dots, f_m) - \sigma(0) f_1\cdots f_m \big|=0 \end{align*}$$

for almost all points in $ \mathbb R^n $ , which proves one part of the claimed a.e. convergences.