Hardy-Littlewood maximal function on noncommutative Lorentz spaces

Abstract

This paper is mainly devoted to the study of the Hardy-Littlewood maximal function on noncommutative Lorentz spaces and to obtaining $(p,q)$-$(p,q)$-type inequality for the Hardy-Littlewood maximal function on noncommutative Lorentz spaces.

MSC: 47A30, 47L05.

1 Introduction

In [1] Nelson defined the measure topology of τ-measurable operators affiliated with a semi-finite von Neumann algebra. Fack and Kosaki [2] investigated generalized s-numbers of τ-measurable operators and proved dominated convergence theorems for a gage and convexity (or concavity) inequality.

As for noncommutative maximal inequalities, a version of ergodic theory was given by Junge [3] and Junge, Xu [4]. In 2007, Mei [5] presented a version of noncommutative Hardy-Littlewood maximal inequality for an operator-valued function. In this paper, we study another version of Hardy-Littlewood maximal inequality introduced by Bekjan [6]. In [6], Bekjan defined the Hardy-Littlewood maximal function for τ-measurable operators and, among other things, obtained weak $(1,1)$-type and $(p,p)$-type inequalities for the Hardy-Littlewood maximal function. In [6], for an operator T affiliated with a semi-finite von Neumann algebra, the Hardy-Littlewood maximal function of T is defined by

$$\mathit{MT}(x)=\underset{r>0}{sup}\frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T|E_{[x-r,x+r]}(|T|)).$$

The classical Hardy-Littlewood maximal function of a Lebesgue measurable function $f:\mathcal{R}\to \mathcal{R}$ denoted by $Mf(x)$ is defined as

$$Mf(x)=\underset{r>0}{sup}\frac{1}{m([x-r,x+r])}{\int }_{[x-r,x+r]}|f(t)|dt,$$

where m is a Lebesgue measure on $(-\mathrm{\infty },+\mathrm{\infty })$ (cf. [7]). Moreover, a natural generalization of this is the case $f:\mathcal{R}\to \mathcal{R}$ and μ, a Borel measure on $(-\mathrm{\infty },+\mathrm{\infty })$, where

$$M_{\mu }f(x)=\underset{r>0}{sup}\frac{1}{\mu ([x-r,x+r])}{\int }_{[x-r,x+r]}|f(t)|d\mu (t).$$

As discussed by Bekjan in [6], let $\mu (A)=\tau (E_A(|T|))$, where A is a Borel subset of $(-\mathrm{\infty },+\mathrm{\infty })$. Then μ is a Borel measure and

$$\mathit{MT}(x)=\underset{r>0}{sup}\frac{1}{\mu ([x-r,x+r])}{\int }_{[x-r,x+r]}td\mu (t),$$

i.e., $\mathit{MT}(x)$ is the Hardy-Littlewood maximal function $M_{\mu }f(x)$ of $f:\mathcal{R}\to \mathcal{R}$ defined by

$$f(t)=\{\begin{array}{cc}t,\hfill & t\in \sigma (|T|),\hfill \\ 0,\hfill & t\notin \sigma (|T|),\hfill \end{array}$$

(1.1)

with respect to μ.

In view of spectral theory, $|T|$ is represented as

$$|T|={\int }_{\sigma (|T|)}td{E}_t,$$

(1.2)

and $\mathit{MT}(|T|)$ is represented as $\mathit{MT}(x)$. Thus, for T, $\mathit{MT}(|T|)$ is considered as the operator analogue of the Hardy-Littlewood maximal function in the classical case. Therefore, roughly speaking, $\mathit{MT}(|T|)$ stands in relation to T as $Mf(x)$ stands in relation to f in classical analysis.

In this paper, we study the Hardy-Littlewood maximal function on noncommutative Lorentz spaces. By primarily adapting the techniques in [8], we obtain the $(p,q)$-$(p,q)$-type inequality for the Hardy-Littlewood maximal function on noncommutative Lorentz spaces.

The remainder of this paper is organized as follows. Section 2 consists of some notations and preliminaries, including the noncommutative Lorentz spaces and their properties. In Section 3, we present the main result of this paper.

2 Preliminaries

Throughout the paper, let ℳ be a finite von Neumann algebra acting on the Hilbert space ℋ with a normal faithful tracial state τ, and C will be a numerical constant not necessarily the same in each instance. The identity in ℳ is denoted by 1, and we denote by ${\mathcal{M}}_{\mathrm{proj}}$ the lattice of (orthogonal) projections in ℳ. A linear operator $T:dom(T)\to \mathcal{H}$, with domain $dom(T)\subseteq \mathcal{H}$, is said to be affiliated with ℳ if $uT=Tu$ for all unitary u in the commutant ${\mathcal{M}}^{\mathrm{\prime }}$ of ℳ. The closed densely defined linear operator T affiliated with ℳ is called τ-measurable if for every $ϵ>0$ there exists an orthogonal projection $P\in {\mathcal{M}}_{\mathrm{proj}}$ such that $P(H)\subseteq dom(T)$ and $\tau (1-P)<ϵ$. The collection of all τ-measurable operators is denoted by $\tilde{\mathcal{M}}$. With the sum and product defined as the respective closures of the algebraic sum and product, $\tilde{\mathcal{M}}$ is an ∗-algebra. For a positive self-adjoint operator T affiliated with ℳ, we set

$$\tau (T)=\underset{n}{sup}\tau \left({\int }_0^n\lambda d{E}_{\lambda }\right)={\int }_0^{\mathrm{\infty }}\lambda d\tau (E_{\lambda }),$$

where $T={\int }_0^{\mathrm{\infty }}\lambda d{E}_{\lambda }$ is the spectral decomposition of T.

Let T be a τ-measurable operator and $t>0$. The ‘t th singular number (or generalized s-number) of T’ is defined by

$${\mu }_t(T)=inf\{\parallel \mathit{TE}\parallel :E\in {\mathcal{M}}_{\mathrm{proj}},\tau (1-E)\le t\}.$$

By Proposition 2.2 of [2], we have

$${\mu }_t(T)=inf\{s⩾0:{\lambda }_s(T)⩽t\}(t>0),$$

where ${\lambda }_s(T)=\tau (E_{(s,\mathrm{\infty })}(|T|))$ ($s⩾0$) and $E_{(s,\mathrm{\infty })}(|T|)$ is the spectral projection of $|T|$ corresponding to the interval $(s,\mathrm{\infty })$. The reader is referred to [2] for basic properties and detailed information on generalized s-numbers and the distribution function of τ-measurable operators.

Definition 2.1 (See, e.g., [9])

Let T be a τ-measurable operator affiliated with a finite von Neumann algebra ℳ, and let $0<p,q\le \mathrm{\infty }$. Define

$${\parallel T\parallel }_{L^{p,q}(\mathcal{M})}=\{\begin{array}{cc}{({\int }_0^{\mathrm{\infty }}{(t^{\frac{1}{p}}{\mu }_t(T))}^q\frac{dt}{t})}^{\frac{1}{q}}\hfill & \text{if }q<\mathrm{\infty },\hfill \\ {sup}_{t>0}{t}^{\frac{1}{p}}{\mu }_t(T)\hfill & \text{if }q=\mathrm{\infty }.\hfill \end{array}$$

(2.1)

The set of all $T\in \tilde{\mathcal{M}}$ with ${\parallel T\parallel }_{L^{p,q}(\mathcal{M})}<\mathrm{\infty }$ is called the noncommutative Lorentz space, denoted by $L^{p,q}(\mathcal{M})$ with indices p and q.

For convenience, we need the following Hardy inequalities in [10].

Lemma 2.2 If $q\ge 1$, $r>0$ and $f\ge 0$, then

$${\left({\int }_0^{\mathrm{\infty }}{[{\int }_0^{t}f(y)dy]}^q{t}^{-r-1}dt\right)}^{\frac{1}{q}}\le \frac{q}{r}{\left({\int }_0^{\mathrm{\infty }}{[yf(y)]}^q{y}^{-r-1}\right)}^{\frac{1}{q}}$$

and

$${\left({\int }_0^{\mathrm{\infty }}{[{\int }_t^{\mathrm{\infty }}f(y)dy]}^q{t}^{r-1}dt\right)}^{\frac{1}{q}}\le \frac{q}{r}{\left({\int }_0^{\mathrm{\infty }}{[yf(y)]}^q{y}^{r-1}\right)}^{\frac{1}{q}}.$$

Lemma 2.3 Let $0<r_2<p<\mathrm{\infty }$ and $0<q,s<\mathrm{\infty }$, then

$$L^{p,q}(\mathcal{M})\subset L^{r_2,s}(\mathcal{M}).$$

Let $L_{\mathrm{loc}}(\mathcal{M};\tau )$ be the set of all τ-measurable operators such that

$$\tau (|T|E_I(|T|))<+\mathrm{\infty }$$

for all bounded intervals $I\subset [0,+\mathrm{\infty })$.

Definition 2.4 (See, e.g., [6])

Let $T\in L_{\mathrm{loc}}(\mathcal{M};\tau )$, the maximal function of T is defined by

$$\mathit{MT}(x)=\underset{r>0}{sup}\frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T|E_{[x-r,x+r]}(|T|))$$

(let $\frac{0}{0}=0$). M is called the Hardy-Littlewood maximal operator.

Remark 2.5 By the introduction of [6], we know that $\mathit{MT}(|T|)$ is represented as $\mathit{MT}(x)$. Hence, for $T\in L_{\mathrm{loc}}(\mathcal{M};\tau )$, $\mathit{MT}(|T|)$ is considered as the operator analogue of the Hardy-Littlewood maximal function in the classical case. Therefore, roughly speaking, $\mathit{MT}(|T|)$ stands in relation to T as $Mf(x)$ stands in relation to f in classical analysis. Also, in [6], it was proved that $\mathit{MT}(|T|)$ defined in Definition 2.4 was weak $(1,1)$-type and $(p,p)$-type. We refer the readers to [6] for more details and basic properties of $\mathit{MT}(|T|)$.

3 Main result

Lemma 3.1 Let $0<q<\mathrm{\infty }$, $1\le p,p_0,p_1<\mathrm{\infty }$ and $p_0\ne p_1$ such that

$$\frac{1}{p}=\frac{1-\theta }{p_0}+\frac{\theta }{p_1}\mathit{\text{for some }}0<\theta <1.$$

Assume that ℳ has no minimal projection, then there exists a constant C such that $\mathrm{\forall }T\in L^{p,q}(\mathcal{M})$ we have

$${\parallel \mathit{MT}\parallel }_{p,q}\le C{\parallel T\parallel }_{p,q}.$$

(3.1)

Proof We assume that $p_0<p_1$. Theorem 2 of [6] and Lemma 2.3 imply that

$${\parallel \mathit{MT}\parallel }_{p_0,\mathrm{\infty }}\le {\parallel \mathit{MT}\parallel }_{p_0}\le C{\parallel T\parallel }_{p_0}\le C{\parallel T\parallel }_{p_0,m}$$

(3.2)

and

$${\parallel \mathit{MT}\parallel }_{p_1,\mathrm{\infty }}\le {\parallel \mathit{MT}\parallel }_{p_1}\le C{\parallel T\parallel }_{p_1}\le C{\parallel T\parallel }_{p_1,m},$$

(3.3)

where $m=\frac{1}{2}min(1,q)$.

By Lemma 1.8 of [11], for all $t\in (0,1)$, we can take $P\in {\mathcal{M}}_{\mathrm{proj}}$ such that $P|T|=|T|P$ and $\tau (P)=t$. Set $T_1=|T|P$, $T_2=|T|-T_1$, it is easy to check that $T_1\in L^{p_0,m}(\mathcal{M})$ and $T_2\in L^{p_1,m}(\mathcal{M})$. Indeed, we see that ${\mu }_v(T_1)={\mu }_v(|T|P)={\mu }_v(T){\chi }_{[0,t]}$ and ${\mu }_v(T_2)={\mu }_v(|T|P^{\perp })={\mu }_{v+t}(T)$. Thus we obtain

$$\begin{array}{rl}{\parallel T_1\parallel }_{p_0,m}^m& ={\int }_0^{\mathrm{\infty }}{v}^{\frac{m}{p_0}-1}{\mu }_v{(|T|P)}^{m}dv={\int }_0^t{v}^{\frac{m}{p_0}-1}{\mu }_v{(T)}^{m}dv\\ ={\int }_0^t{(v^{\frac{1}{p}}{\mu }_v(T))}^m{v}^{\frac{m}{p_0}-\frac{m}{p}-1}dv\\ \le {\parallel T\parallel }_{p,\mathrm{\infty }}^m{\int }_0^t{v}^{\frac{m}{p_0}-\frac{m}{p}-1}dv\\ \le {\left(\frac{q}{p}\right)}^{\frac{m}{q}}{\parallel T\parallel }_{p,q}^m\frac{1}{\frac{m}{p_0}-\frac{m}{p}}{t}^{\frac{m}{p_0}-\frac{m}{p}}<\mathrm{\infty }\end{array}$$

and

$$\begin{array}{rl}{\parallel T_2\parallel }_{p_1,m}^m& ={\int }_0^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-1}{\mu }_v{(|T|P^{\perp })}^{m}dv={\int }_0^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-1}{\mu }_{v+t}{(T)}^{m}dv\\ \le {\int }_0^t{v}^{\frac{m}{p_1}-1}{\mu }_t{(T)}^{m}dv+{\int }_t^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-1}{\mu }_v{(T)}^{m}dv\\ \le \frac{p_1}{m}{t}^{\frac{m}{p_1}}{\mu }_t{(T)}^m+\underset{v>t}{sup}{(v^{\frac{1}{p}}{\mu }_v(T))}^m{\int }_t^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-\frac{m}{p}-1}dv\\ =\frac{p_1}{m}{t}^{\frac{m}{p_1}}{\mu }_t{(T)}^m+{\parallel T\parallel }_{p,\mathrm{\infty }}^m\frac{1}{\frac{m}{p}-\frac{m}{p_1}}{t}^{\frac{m}{p_1}-\frac{m}{p}}\\ \le \frac{p_1}{m}{t}^{\frac{m}{p_1}-\frac{m}{p}}{(\underset{t>0}{sup}{t}^{\frac{1}{p}}{\mu }_t(T))}^m+{\left(\frac{q}{p}\right)}^{\frac{m}{q}}\frac{1}{\frac{m}{p}-\frac{m}{p_1}}{t}^{\frac{m}{p_1}-\frac{m}{p}}{\parallel T\parallel }_{p,q}^m\\ =\left[{\left(\frac{q}{p}\right)}^{\frac{m}{q}}{t}^{\frac{m}{p_1}-\frac{m}{p}}(\frac{p_1}{m}+\frac{1}{\frac{m}{p}-\frac{m}{p_1}})\right]{\parallel T\parallel }_{p,q}^m<\mathrm{\infty }.\end{array}$$

Since

$$\begin{array}{r}\frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T|E_{[x-r,x+r]}(|T|))\\ \le \frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T|P{E}_{[x-r,x+r]}(|T|))\\ +\frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T|P^{\perp }{E}_{[x-r,x+r]}(|T|))\\ =\frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T_1|E_{[x-r,x+r]}(|T_1|))\\ +\frac{1}{\tau (E_{[x-r,x+r]}(|T|))}\tau (|T_2|E_{[x-r,x+r]}(|T_2|))\\ \le \frac{1}{\tau (E_{[x-r,x+r]}(|T_1|))}\tau (|T_1|E_{[x-r,x+r]}(|T_1|))\\ +\frac{1}{\tau (E_{[x-r,x+r]}(|T_2|))}\tau (|T_2|E_{[x-r,x+r]}(|T_2|)),\end{array}$$

taking supremum, we get

$$\mathit{MT}(x)\le {\mathit{MT}}_1(x)+{\mathit{MT}}_2(x),$$

which implies that

$${\parallel \mathit{MT}\parallel }_{p,q}\le C({\parallel {\mathit{MT}}_1\parallel }_{p,q}+{\parallel {\mathit{MT}}_2\parallel }_{p,q}).$$

We estimate each term separately. For the first term, using (3.2) we get

$$\begin{array}{rl}{\parallel {\mathit{MT}}_1\parallel }_{p,q}& ={\left\{{\int }_0^{\mathrm{\infty }}{t}^{\frac{q}{p}}{({\mu }_t({\mathit{MT}}_1))}^q\frac{dt}{t}\right\}}^{\frac{1}{q}}\\ ={\left\{{\int }_0^{\mathrm{\infty }}{t}^{\frac{q}{p}-\frac{q}{p_0}}{(t^{\frac{1}{p_0}}{\mu }_t({\mathit{MT}}_1))}^q\frac{dt}{t}\right\}}^{\frac{1}{q}}\\ \le C{\left\{{\int }_0^{\mathrm{\infty }}{t}^{q(\frac{1}{p}-\frac{1}{p_0})}{\parallel T_1\parallel }_{p_0,m}^q\frac{dt}{t}\right\}}^{\frac{1}{q}}\\ =C{\left\{{\left[{\int }_0^{\mathrm{\infty }}{t}^{-q(\frac{1}{p_0}-\frac{1}{p})-1}{\left({\int }_0^t{v}^{\frac{m}{p_0}-1}{\mu }_v{(T)}^{m}dv\right)}^{\frac{q}{m}}dt\right]}^{\frac{m}{q}}\right\}}^{\frac{1}{m}}.\end{array}$$

After replacing r and q respectively with $q(\frac{1}{p_0}-\frac{1}{p})$ and $\frac{q}{m}$ in the first inequality in Lemma 2.2, we see that the last expression is estimated as follows:

$$\begin{array}{r}\le \frac{C}{{(\frac{m}{p_0}-\frac{m}{p_1})}^{\frac{1}{m}}}{\{{\int }_0^{\mathrm{\infty }}{[v\cdot v^{\frac{m}{p_0}-1}\cdot {\mu }_v{(T)}^m]}^{\frac{q}{m}}\cdot v^{-q(\frac{1}{p_0}-\frac{1}{p})-1}dv\}}^{\frac{1}{q}}\\ =C{\left({\int }_0^{\mathrm{\infty }}{v}^{\frac{q}{p}-1}{\mu }_v{(T)}^{q}dv\right)}^{\frac{1}{q}}\\ =C{\parallel T\parallel }_{p,q},\end{array}$$

i.e., ${\parallel {\mathit{MT}}_1\parallel }_{p,q}\le C{\parallel T\parallel }_{p,q}$. To estimate the second term, by applying (3.3) we obtain

$$\begin{array}{rl}{\parallel {\mathit{MT}}_2\parallel }_{p,q}=& {\left\{{\int }_0^{\mathrm{\infty }}{t}^{\frac{q}{p}}{({\mu }_t({\mathit{MT}}_2))}^q\frac{dt}{t}\right\}}^{\frac{1}{q}}\\ =& {\left\{{\int }_0^{\mathrm{\infty }}{t}^{\frac{q}{p}-\frac{q}{p_1}}{(t^{\frac{1}{p_1}}{\mu }_t({\mathit{MT}}_2))}^q\frac{dt}{t}\right\}}^{\frac{1}{q}}\\ \le & C{\left\{{\int }_0^{\mathrm{\infty }}{t}^{q(\frac{1}{p}-\frac{1}{p_1})}{\parallel T_2\parallel }_{p_1,m}^q\frac{dt}{t}\right\}}^{\frac{1}{q}}\\ =& C{\left\{{\left[{\int }_0^{\mathrm{\infty }}{t}^{q(\frac{1}{p}-\frac{1}{p_1})-1}{\left({\int }_0^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-1}{\mu }_v{(T_2)}^{m}dv\right)}^{\frac{q}{m}}dt\right]}^{\frac{m}{q}}\right\}}^{\frac{1}{m}}\\ \le & C{\left\{{\int }_0^{\mathrm{\infty }}{t}^{q(\frac{1}{p}-\frac{1}{p_1})-1}{\left({\int }_0^t{v}^{\frac{m}{p_1}-1}{\mu }_t{(T)}^{m}dv\right)}^{\frac{q}{m}}dt\right\}}^{\frac{1}{q}}\\ +C{\left\{{\left[{\int }_0^{\mathrm{\infty }}{t}^{q(\frac{1}{p}-\frac{1}{p_1})-1}{\left({\int }_t^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-1}{\mu }_v{(T)}^{m}dv\right)}^{\frac{q}{m}}dt\right]}^{\frac{m}{q}}\right\}}^{\frac{1}{m}}.\end{array}$$

For the second term ${\{{[{\int }_0^{\mathrm{\infty }}{t}^{q(\frac{1}{p}-\frac{1}{p_1})-1}{({\int }_t^{\mathrm{\infty }}{v}^{\frac{m}{p_1}-1}{\mu }_v{(T)}^{m}dv)}^{\frac{q}{m}}dt]}^{\frac{m}{q}}\}}^{\frac{1}{m}}$, replace r and q respectively with $q(\frac{1}{p}-\frac{1}{p_1})$ and $\frac{q}{m}$ in the second inequality in Lemma 2.2, and we estimate the last expression as follows:

$$\begin{array}{r}\le C{\left\{\frac{p_1}{m}{\mu }_t{(T)}^q{\int }_0^{\mathrm{\infty }}{t}^{\frac{q}{p}-1}dt\right\}}^{\frac{1}{q}}+C{\left\{{\int }_0^{\mathrm{\infty }}{[v\cdot v^{\frac{m}{p_1}-1}\cdot {\mu }_v{(T)}^m]}^{\frac{q}{m}}{v}^{q(\frac{1}{p}-\frac{1}{p_1})-1}dv\right\}}^{\frac{1}{q}}\\ =C{\left\{\frac{p_1}{m}{\mu }_t{(T)}^q{\int }_0^{\mathrm{\infty }}{t}^{\frac{q}{p}-1}dt\right\}}^{\frac{1}{q}}+C{\parallel T\parallel }_{p,q}\\ \le C{\parallel T\parallel }_{p,q},\end{array}$$

i.e., ${\parallel {\mathit{MT}}_2\parallel }_{p,q}\le C{\parallel T\parallel }_{p,q}$.

For the case of $p_0>p_1$, we may simply reverse the roles of $p_0$ and $p_1$ in the above proof.

We have now shown that

$${\parallel \mathit{MT}\parallel }_{p,q}\le C{\parallel T\parallel }_{p,q}.$$

 □

Theorem 3.2 Let $0<q<\mathrm{\infty }$, $1\le p,p_0,p_1<\mathrm{\infty }$ and $p_0\ne p_1$ be such that

$$\frac{1}{p}=\frac{1-\theta }{p_0}+\frac{\theta }{p_1}\mathit{\text{for some }}0<\theta <1.$$

Assume that ℳ has minimal projections, then there exists a constant C such that for all $T\in L^{p,q}(\mathcal{M})$ we have

$${\parallel \mathit{MT}\parallel }_{p,q}\le C{\parallel T\parallel }_{p,q}.$$

Proof Since ℳ has minimal projections, we consider the von Neumann algebra tensor product $\mathcal{M}\overline{\otimes }{L}^{\mathrm{\infty }}([0,1];dm)$ denoted by $\overline{\mathcal{M}}$, equipped with the tensor product trace $\tau \otimes dm$, where dm is the Lebesgue measure on $[0,1]$, then $\overline{\mathcal{M}}$ has no minimal projection.

Let $|T|={\int }_{\sigma (|T|)}\lambda d{E}_{\lambda }(|T|)$ be the spectral decomposition of T. Since

$$\sigma (|T|)=\sigma (|T|\otimes 1),$$

we have

$$|T\otimes 1|=|T|\otimes 1={\int }_{\sigma (|T|)}\lambda d(E_{\lambda }(|T|)\otimes 1)={\int }_{\sigma (|T|\otimes 1)}\lambda d(E_{\lambda }(|T|)\otimes 1).$$

It is easy to check that $E_{\lambda }(|T|)\otimes 1$ is a spectral series for each $\lambda \ge 0$. Hence, for any interval

$$I\subset \sigma (|T|)=\sigma (|T\otimes 1|)=\sigma (|T|\otimes 1),$$

by the uniqueness of the spectral decomposition, we see that

$$E_I(|T\otimes 1|)=E_I(|T|)\otimes 1.$$

For $\mathrm{\forall }r>0$, since

$$\tau \left(E_{[x-r,x+r]}(|T|)\right)={\int }_0^1\tau \left(E_{[x-r,x+r]}(|T|)\right)dm=\tau \otimes dm(E_{[x-r,x+r]}(|T|)\otimes 1)$$

and

$$\begin{array}{r}\tau \otimes dm(|T\otimes 1|E_{[x-r,x+r]}(|T|)\otimes 1)\\ =\tau \otimes dm{\left\{{(|T\otimes 1|E_{[x-r,x+r]}(|T|)\otimes 1)}^{\ast }(|T\otimes 1|E_{[x-r,x+r]}(|T|)\otimes 1)\right\}}^{\frac{1}{2}}\\ =\tau \otimes dm{\{{(E_{[x-r,x+r]}(|T|)\otimes 1)}^{\ast }|T\otimes 1|(|T\otimes 1|E_{[x-r,x+r]}(|T|)\otimes 1)\}}^{\frac{1}{2}}\\ =\tau \otimes dm{\{(E_{[x-r,x+r]}(|T|)\otimes 1){|T\otimes 1|}^2{E}_{[x-r,x+r]}(|T|)\otimes 1\}}^{\frac{1}{2}}\\ =\tau \otimes dm{\{E_{[x-r,x+r]}(|T|)\otimes 1({|T|}^2\otimes 1)E_{[x-r,x+r]}(|T|)\otimes 1\}}^{\frac{1}{2}}\\ =\tau \otimes dm{\left\{(E_{[x-r,x+r]}(|T|)\otimes 1)(|T|\otimes 1)(|T|\otimes 1)(E_{[x-r,x+r]}(|T|)\otimes 1)\right\}}^{\frac{1}{2}}\\ =\tau \otimes dm(|T|E_{[x-r,x+r]}(|T|)\otimes 1)\\ =\tau (|T|E_{[x-r,x+r]}(|T|)),\end{array}$$

which implies that

$$M(T\otimes 1)(x)=\mathit{MT}(x).$$

By an adaptation of the proof of Lemma 3.1, we deduce that

$${\parallel M(T\otimes 1)(T\otimes 1)\parallel }_{p,q}\le C{\parallel T\otimes 1\parallel }_{p,q}.$$

With the trivial fact ${\mu }_t(T)={\mu }_t(T\otimes 1)$, we know

$${\parallel T\otimes 1\parallel }_{p,q}={\parallel T\parallel }_{p,q}.$$

Combing this result with $M(T\otimes 1)(|T\otimes 1|)=\mathit{MT}(|T|)$, we infer that

$${\parallel \mathit{MT}\parallel }_{p,q}\le C{\parallel T\parallel }_{p,q}.$$

 □