As is well known, for the real-valued local integrable functions $f$ on $\mathbb{R}^n, $ the classical Hardy-Littlewood maximal operator $M$ is defined by

$ Mf(x)=\sup\limits_{x\in Q}\frac{1}{|Q|}\displaystyle\int_Q|f(y)|dy, $

where $Q$ is a non-degenerate cube with its sides paralleled to the coordinate axes and $|Q|$ is the Lebesgue measure of $Q.$

Suppose that $ {\it\Phi}: [0, \infty)\rightarrow \mathbb{R} $ is nondecreasing and continuous with $ {\it\Phi}(0)=0$ and $\lim\limits_{t\rightarrow+\infty}{\it\Phi}(t)=\infty.$ Let

$ L^{{\it\Phi}}=\Big\{f:~\displaystyle\int_{R^n} {\it\Phi}(\varepsilon|f|)dx <\infty~ \hbox{ for some }\varepsilon>0 \Big\}. $

Then $L^{\it{\Phi}}$ is called an Orlicz space. In Orlicz spaces, Kokilashvili and Krbec [1, Theorem 1.2.1] stated that the following statements are equivalent:

(a) there exists a positive constant $C$ such that

$ \begin{equation} \label{Booka1} \displaystyle\int_{{\mathbb{R}^n}}{\it\Phi}(Mf)dx\leq C\displaystyle\int_{{\mathbb{R}^n}}{\it\Phi}(C|f|)dx, ~~\forall f\in L^1_{\hbox{loc}}; \end{equation} $

(1.1)

(b) the function ${\it\Phi}^\alpha$ is quasi-convex for some $\alpha\in(0, 1).$

When $ {\it\Phi}(t)=\frac{t^p}{p}, ~~p>1, $ $L^{\it{\Phi}}$ is the space $L^p.$ The above result implies that $M$ is bounded on $L^p$ for $1 <p < \infty$ (see Grafakos [2, Theorem 2.1.6]) and unbounded on $L^1$(see Grafakos [2, Example 2.1.2]). In fact, $M$ maps $L^1$ to $L^{1, \infty}$ (see Grafakos [2, Theorem 2.1.6]).

Let $\mathbb{T}$ be the group of real numbers modulo $2\pi$ and $f(x)$ a real-valued integrable function denoted on $\mathbb{T}$ with period $2\pi.$ The Hardy-Littlewood maximal function $M_{\mathbb{T}}f(x)$ is defined by $M_{\mathbb{T}}f(x)=\sup\limits_{x\in I}\frac{1}{|I|}\displaystyle\int_I|f(y)|dy, $ where the supremum taken over all open $I\subseteq\mathbb{T}$ with $x\in I.$

Let $a(s)$ and $b(s)$ be two positive nondecreasing continuous functions defined on $[0, \infty)$ satisfying

$ \displaystyle\int_0^1\frac{a(s)}{s}ds=K <+\infty, $

(1.2)

$ \displaystyle\int_1^\infty\frac{a(s)}{s}ds=+\infty $

(1.3)

and

$ \begin{equation} \label{a2}\lim\limits_{s\rightarrow\infty}b(s)=+\infty. \end{equation} $

(1.4)

Then, we define

$ \begin{equation} \label{a3}{\it\Phi}(t)=\displaystyle\int_0^ta(s)ds, ~~~{\it\Psi}(t)=\displaystyle\int_0^tb(s)ds, ~\forall t\geq0. \end{equation} $

(1.5)

Under hypotheses (1.3)–(1.5), a series of inequalities involving $M_{\mathbb{T}}f$ and $f$ were obtained. Kita [3, Theorem 2.1] gave necessary and sufficient conditions for $a(\cdot)$ and $b(\cdot)$ in order to guarantee that $M_{\mathbb{T}}f$ is in $L^{\it{\Phi}}$ whenever $f$ is in $L^{\it{\Psi}}.$ Conversely, Kita [4, Theorem 2.1] stated necessary and sufficient conditions for $a(\cdot)$ and $b(\cdot)$ in order to guarantee that a function $f$ is in $L^{\it{\Psi}}$ whenever the function $M_{\mathbb{T}}f$ is in $L^{\it{\Phi}}, $ which is a generalization of [5, Theorem 5.4] to functions in an Orlicz space $L^{\it{\Psi}}$ and also a generalization of the result of [6, Theorem 3.1]. These are related to the class of $L\log L.$ The class was introduced by Zygmund to give a sufficient condition on the local integrability of the Hardy-Littlewood maximal operator. The necessity of this condition was observed by Stein [7].

Under the additional hypothesis $(1.2), $ Kita [8, Theorem 2.3, Theorem 2.5] extended the above results [3, Theorem 2.1] and [4, Theorem 2.1] to the cases on $\mathbb{R}^n.$ Moreover, Kita [9, 10] obtained the related weighted theory on $\mathbb{R}^n.$ Specially, Kita [10] also dealt with the iterated maximal operator.

Comparing with the results without weights on $\mathbb{R}^n$ and $\mathbb{T}$, Mei and Liu [11, 12] considered the martingale cases. In our paper, we prove weighted inequalities for maximal operators in Orlicz martingale classes. Let us introduce some preliminaries.

Let $(\Omega, \mathcal {F}, \mu)$ be a complete probability space, and $(\mathcal {F}_n)_{n\geq0}$ be an increasing sequence of sub-$\mathcal {F}$-fields of $\mathcal{F}$ with $\mathcal{F}=\bigvee\mathcal{F}_n.$ If $f=(f_n)_{n\geq0}$ is a martingale adapted to $(\mathcal {F}_n)_{n\geq0}, $ the maximal function (operator) $Mf$ for martingale $f=(f_n)$ is defined by $Mf=\sup\limits_{n\geq 0}|f_n|.$ In this paper, the limit of $f_n$ is also denoted by $f.$ Specifically, $\big({\Omega}, (\mathcal {F}_n)_{n\geq0}, \mathcal{F}, \nu\big)$ is said to satisfy the $R\hbox{-condition}$ (or simply $\nu\in R$), if for all $n\geq1, ~F_n\in\mathcal {F}_n, $ there exists a $G_n\in \mathcal {F}_{n-1}$ such that $F_n\subset G_n$ and $\nu(G_n)\leq d\cdot\nu(F_n).$ In our paper, we assume that $\mathcal {F}_{0}=\{\emptyset, \Omega\}.$

A weight $\omega$ is a positive random variable with $E \omega <\infty.$ For convenience, we assume $E \omega=1$ in this paper. In addition, we say $\omega\in A_1$ (or $B_1$), if $\omega^{-1}\in L^{\infty}(d\mu)$ and there exists a constant $C\geq1$ such that $\omega_n\leq C \omega$ (or $C \omega_n\geq\omega$). Moreover, for the weights as above, a function $f\in L^1(\omega d\mu)$ implies $f\in L^1(d\mu).$

Let $a(s), $ $b(s), $ $\it{\Phi(t)}$ and $\it{\Psi(t)}$ satisfy (1.2)-(1.5). It is clear that $\it{\Phi}$ and $\it{\Psi}$ are convex. Recall that $L^{\it{\Phi}}(d\mu)\cup L^{\it{\Psi}}(d\mu)\subset L^1(d\mu)$ when $\mu({\Omega}) <\infty.$

Then we have the following Theorems 1.1 and 1.2, which are martingale versions of [8, Theorem 2.3]. Kita [8, Theorem 2.3] gave an assumption that the function $\frac{a(s)}{s}$ is bounded in a neighborhood of zero. However, we do not need the assumption in our Theorems 1.1 and 1.2.

Theorem 1.1 Let $\omega\in A_1.$ Suppose that there exists a positive constant $C_1$ such that

$ \begin{equation} \label{2b1} \displaystyle\int_1^s\frac{a(t)}{t}dt\leq C_1b(C_1s), ~s\geq1. \end{equation} $

(1.6)

Then there exist positive constants $C_2$ and $C_3$ such that

$ \begin{equation} \label{2bb1}\displaystyle\int_{{\Omega}}{\it\Phi}(Mf)\omega d\mu\leq C_2\|f\|_{L^1(\omega d\mu)}+C_2\displaystyle\int_{{\Omega}}{\it\Psi}(C_3|f|)\omega d\mu, \end{equation} $

(1.7)

where $C_2$ and $C_3$ are constants independent of $f=(f_n).$

Theorem 1.2 Let $(\Omega, \mathcal {F}, \omega)$ be non-atomic and $\omega\in R\cap B_1.$ Suppose that $f\in L^{\it{\Psi}}(\omega d\mu)$ implies $Mf\in L^{{\it\Phi}}(\omega d\mu).$ Then $(1.6)$ holds.

Following from Theorems 1.1 and 1.2, we easily have Corollaries 1.3 and 1.4, respectively.

Corollary 1.3 Let $\omega\in A_1.$ Suppose that $(1.6)$ is valid, then $Mf\in L^{\it{\Phi}}(\omega d\mu)$ for all $f\in L^{\it{\Psi}}(\omega d\mu).$

Corollary 1.4 Let $(\Omega, \mathcal {F}, \omega)$ be non-atomic and $\omega\in R\cap B_1.$ Suppose that $(1.7)$ is valid, then we have $(1.6).$

We consider the converse inequality of maximal functions. Theorems 1.5 and 1.6 are martingale versions of [8, Theorem 2.5] and involve weights.

Theorem 1.5 Let $\omega\in B_1\cap R.$ Suppose that there exist positive constants $C_6$ and $s_0>1$ such that

$ \begin{equation} \label{add2} \displaystyle\int_1^s\frac{a(t)}{t}dt\geq C_6b(C_6s), ~s\geq s_0. \end{equation} $

(1.8)

Then there exist positive constants $C_7$ and $C_8$ such that

$ \begin{equation} \label{add3}\displaystyle\int_{{\Omega}}{\it\Psi}(C_7|f|)\omega d\mu\leq C_8\|f\|_{L^1(\omega d\mu)}+C_8 \displaystyle\int_{{\Omega}}{\it\Phi}(M(|f|))\omega d\mu, \end{equation} $

(1.9)

where $C_7$ and $C_8$ are constants independent of $f=(f_n)$ and $\|f\|_{L^1(\omega d\mu)}\leq1.$

Theorem 1.6 Let $(\Omega, \mathcal {F}, \omega)$ be non-atomic and $\omega\in A_1.$ Suppose that $(1.9)$ is valid. Then $(1.8)$ holds.

We easily obtain Corollary 1.7 by Theorems 1.5. In addition, following the proof of Theorem 1.6, we have Corollary 1.8.

Corollary 1.7 Let $\omega\in B_1\cap R.$ Suppose that $(1.8)$ is valid. Then $f\in L^{\it{\Psi}}(\omega d\mu)$ for all $f$ satisfying $Mf\in L^{\it{\Phi}}(\omega d\mu).$

Corollary 1.8 Let $(\Omega, \mathcal {F}, \omega)$ be non-atomic and $\omega\in A_1.$ Suppose that $Mf\in L^{\it{\Phi}}(\omega d\mu)$ implies $f\in L^{\it{\Psi}}(\omega d\mu).$ Then $(1.8)$ holds.

First, we give some lemmas which will be used in our proofs several times.

Lemma 2.1 Suppose that $\nu$ is an finite measure on $({\Omega}, \mathcal {F}), $ then for each function $f\in L^1(d\nu), $ we have

$ \begin{equation} \label{3a1}\displaystyle\int_{\{|f|>t\}}|f|d\nu=t\cdot\nu\{|f|>t\}+\displaystyle\int^\infty_t\nu\{|f|>s\}ds, ~t>0. \end{equation} $

(2.1)

Proof The proof follows in the same way as the proof of [9, Lemma 3.1].

Lemma 2.2 Let $\omega\in A_1$. Then there exists a positive constant $C_0$ such that

$ \begin{equation}\omega\{Mf>\lambda\} \leq\frac{C_0}{\lambda}\displaystyle\int_{\{|f|>\frac{\lambda}{2}\}}|f|\omega d\mu \end{equation} $

(2.2)

for all $\lambda>0$ and $f\in L^1(\omega d\mu).$

Proof By the assumption that $\omega\in A_1, $ there exists a positive constant $C$ such that

$ \omega\{Mg>\lambda\} \leq\frac{C}{\lambda}\|g\|_{L^1(\omega)}, ~~g\in L^1(\omega) $

in view of [13, Theorem 6.6.2]. For $\lambda>0, $ we set $f_\lambda=f\cdot\chi_{\{|f|\leq\frac{\lambda}{2}\}}$ and $f^\lambda=f-f_\lambda.$ Then $f^\lambda\in L^1(\omega)$ and $ Mf_\lambda=\sup\limits_{n\geq0}|E_n(f_\lambda)| \leq\sup\limits_{n\geq0}E_n(|f_\lambda|) \leq\frac{\lambda}{2}.$ It follows that

$ \omega\{Mf>\lambda\} \leq\omega\{Mf_\lambda>\frac{\lambda}{2}\}+ \omega\{Mf^\lambda>\frac{\lambda}{2}\} \leq\frac{C_0}{\lambda}\displaystyle\int_{\{|f|>\frac{\lambda}{2}\}}|f|\omega d\mu, $

where $C_0=2C.$

Lemma 2.3 Let $\omega\in A_1$. Then there exists a positive constant $C$ such that

$ \begin{equation}\label{add3d1}\omega\{Mf>t\} \leq\frac{C}{t}\displaystyle\int_{\frac{t}{2}}^\infty\omega\{|f|>s\} ds \end{equation} $

(2.3)

for all $t>0$ and $f\in L^{\it{\Psi}}(\omega).$

Proof Fix $f\in L^{\it{\Psi}}(\omega d\mu).$ For $t>0, $ set $ f_t=(|f|\wedge\frac{t}{2})\cdot\mathop{\hbox{sign}}f$ and $f^t=f-f_t.$ Trivially, we have $f^t\in L^1(\omega).$ By Lemma $2.2, $ we also have

$ \begin{equation}\label{3e1}\omega\{Mf^t>\lambda\} \leq\frac{C_0}{\lambda}\displaystyle\int_{\{|f^t|>\frac{\lambda}{2}\}}|f^t|\omega d\mu. \end{equation} $

(2.4)

Let $\lambda=\frac{t}{2}$ in $(2.4).$ It follows that

$ \begin{equation}\label{3f1}\omega\{Mf^t>\frac{t}{2}\} \leq\frac{2C_0}{t}\displaystyle\int_{{\Omega}}|f^t|\omega d\mu. \end{equation} $

(2.5)

Hence,

$ \omega\{Mf>t\} \leq \omega\{Mf^t>\frac{t}{2}\}+\omega\{Mf_t>\frac{t}{2}\} =\omega\{Mf^t>\frac{t}{2}\}\\ \leq \frac{2C_0}{t}\displaystyle\int_{{\Omega}}|f^t|\omega d\mu =\frac{2C_0}{t}\displaystyle\int_{\{|f|>\frac{t}{2}\}} |f-\frac{t}{2}\cdot\mathop{\hbox{sign}}f|\omega d\mu\\ = \frac{2C_0}{t}\displaystyle\int_{\{|f|>\frac{t}{2}\}} \Big||f|-\frac{t}{2}\Big|\omega d\mu =\frac{2C_0}{t}\displaystyle\int_{\{|f|>\frac{t}{2}\}} (|f|-\frac{t}{2})\omega d\mu. $

Combining with (2.1), we have (2.3) is valid with $C=2C_0.$

Proof of Theorem 1.1 For $C_3>0$ (which will be determined later), fix $f\in L^{\it{\Psi}}(\omega d\mu)$ such that $\displaystyle\int_{{\Omega}}{\it\Psi}(C_3|f|)\omega d\mu <\infty.$ With the constant $C$ in Lemma $2.3, $ it follows from (1.2), (1.3) and (1.6) that

$ \displaystyle\int_{{\Omega}}\it{\Phi}(Mf)\omega d\mu = \displaystyle\int_0^\infty\omega\{Mf>t\}a(t)dt\\ \leq C\displaystyle\int_0^\infty\frac{a(t)}{t} \big(\displaystyle\int_{\frac{t}{2}}^\infty\omega\{|f|>s\} ds\big) dt\\ = C\displaystyle\int_0^\infty\omega\{|f|>s\} \big(\displaystyle\int_0^{2s}\frac{a(t)}{t}dt\big)ds\\ = C\displaystyle\int_0^\infty\omega\{|f|>s\} \big(\displaystyle\int_0^{1}\frac{a(t)}{t}dt+\displaystyle\int_1^{2s}\frac{a(t)}{t}dt\big)ds\\ \leq KC\|f\|_{L^1(\omega d\mu)}+CC_1\displaystyle\int_0^\infty\omega\{|f|>s\} b(2C_1s)ds\\ = KC\|f\|_{L^1(\omega d\mu)}+ \frac{C}{2}\displaystyle\int_{{\Omega}}{\it\Psi}(2C_1|f|)\omega d\mu. $

Let $C_2=(KC)\vee\frac{C}{2}$ and $C_3=2C_1.$ The we have (1.7).

In order to give the proof of Theorem $1.2, $ we need the following lemmas.

Lemma 2.4 Let $\hat{E}_n(\cdot)$ be the conditional expectation with respect to $(\Omega, \mathcal ({\mathcal {F}}_n)_{n\geq0}, $ $\mathcal {F}, \omega d\mu)$ and $\hat{M}(\cdot)=\sup\limits_{n\geq0}\hat{E}_n(\cdot).$ Suppose $\omega\in R.$ Then

$ \begin{equation}\label{3j1}\displaystyle\int_{\{|f|>\lambda\}}|f|\omega d\mu\leq d\lambda\cdot\omega\{\hat{M}(|f|)>\lambda\}, ~f\in L^1(\omega d\mu), ~ \lambda\geq\|f\|_{L^1(\omega d\mu)}. \end{equation} $

(2.6)

Proof For $f\in L^1(\omega d\mu), $ the sequence $\big(\hat{E}_n(|f|)\big)_{n\geq0}$ is a uniformly integral martingale with respect to $(\Omega, \mathcal ({\mathcal {F}}_n)_{n\geq0}, \mathcal {F}, \omega d\mu)$. In view of [13, Theorem 1.3.2.9], we have $\lim\limits_{n\rightarrow\infty}\hat{E}_n(|f|)=|f|~~{\rm a.e.}.$ Then it follows that $|f|\leq \hat{M}(|f|).$

Because of $\omega\in R, $ we have

$ \begin{equation*}\displaystyle\int_{\{\hat{M}f >\lambda\}}|f|\omega d\mu\leq d\lambda\cdot\omega\{\hat{M}f>\lambda\}, ~f\in L^1(\omega d\mu), ~ \lambda\geq\|f\|_{L^1(\omega d\mu)}, \end{equation*} $

where we use [13, Theorem 7.1.2]. Combining this with $|f|\leq \hat{M}(|f|), $ we have

$ \begin{equation*}\displaystyle\int_{\{|f|>\lambda\}}|f|\omega d\mu \leq\displaystyle\int_{\{\hat{M}f >\lambda\}}|f|\omega d\mu\leq d\lambda\cdot\omega\{\hat{M}(|f|)>\lambda\}, ~f\in L^1(\omega d\mu), ~ \lambda\geq\|f\|_{L^1(\omega d\mu)}. \end{equation*} $

Lemma 2.5 Let $\omega$ be a weight. Then

$ \hat{E}(f\cdot\omega^{-1}|\mathcal{F}_n)\cdot E(\omega|\mathcal{F}_n)=E(f|\mathcal{F}_n), ~f\in L^1(d\mu). $

Proof For all $ A\in \mathcal{F}_n$, we have

$ \displaystyle\int_Afd\mu = \displaystyle\int_Af\cdot\omega^{-1}\omega d\mu =\displaystyle\int_A\hat{E}(f\cdot\omega^{-1}| \mathcal{F}_n)\cdot\omega d\mu\\ = \displaystyle\int_A\hat{E}(f\cdot\omega^{-1}| \mathcal{F}_n)\cdot{E}(\omega| \mathcal{F}_n)d\mu. $

It follows that $\hat{E}(f\cdot\omega^{-1}|\mathcal{F}_n)\cdot E(\omega|\mathcal{F}_n)=E(f|\mathcal{F}_n)$.

Reforming Lemma $2.5, $ we have

$ \begin{equation}\label{lemma1}\hat{E}(g|\mathcal{F}_n)\cdot E(\omega|\mathcal{F}_n)=E(g\omega|\mathcal{F}_n), ~g\in L^1(\omega d\mu). \end{equation} $

(2.7)

Lemma 2.6 Suppose $\omega \in B_1\cap R.$ Then there exist constants $C_4$ and $C_5$ such that

$ \begin{equation}\label{3k1}\displaystyle\int_{\{|f|>C_4\lambda\}}|f|\omega d\mu\leq C_5\lambda\cdot\omega\{M(|f|)>\lambda\}, ~~f\in L^1(\omega d\mu), ~\lambda\geq\frac{1}{C}\|f\|_{L^1(\omega d\mu)}, \end{equation} $

(2.8)

where the constant $C$ is the one in the definition of $B_1.$

Proof Fix $f\in L^1(\omega d\mu)$ and $\lambda\geq \frac{1}{C}\|f\|_{L^1(\omega d\mu)}.$ Since $\omega\in B_1, $ there exists a constant $C\geq1$ such that $C\cdot\omega_n\geq\omega.$ Combining this with $(2.7), $ we have

$ \hat{E}_n(|f|)=\omega_n^{-1}{E}_n(|f|\omega)={E}_n(|f|\omega_n^{-1}\omega) \leq{E}_n(|f|\omega_n^{-1}\omega)\leq C{E}_n(|f|). $

Thus

$ \begin{equation}\label{add4}\hat{M}(|f|)\leq CM(|f|).\end{equation} $

(2.9)

Since $C\lambda\geq\|f\|_{L^1(\omega d\mu)}, $ we have

$ \begin{equation}\label{3k2}\displaystyle\int_{\{|f|>C\lambda\}}|f|\omega d\mu\leq C d\cdot\lambda\cdot\omega\{CM(|f|)>C\lambda\} \end{equation} $

(2.10)

in view of Lemma $2.4.$ Thus, $(2.8)$ is valid with $C_4=C$ and $C_5=Cd.$

Proof of Theorem 1.2 Assume for contradiction that $(1.6)$ does not hold. Then there exists a sequence of positive numbers $s_k\geq1$ such that

$ \begin{equation}\label{3l2}\displaystyle\int_1^{s_k}\frac{a(t)}{t}dt> 2^kb(2^ks_k), ~k\geq1. \end{equation} $

(2.11)

Set $\alpha_k=\frac{1}{\alpha2^k{\it\Psi}(2^ks_k)}.$ It is clear that we can choose $\alpha$ big enough to make $\sum\limits_{k=1}\limits^{\infty}\alpha_k\leq1$ and $\frac{1}{\alpha}\leq{\it\Psi}(1).$ By the assumption that $(\Omega, \mathcal {F}, \omega)$ is non-atomic, we have a family of measurable sets $Q_k\in\mathcal {F}$ such that

$ \begin{equation}\label{3m1} \omega(Q_k)=\alpha_k~\hbox{and}~ Q_{k_1}\cap Q_{k_2}=\emptyset, ~k_1\neq k_2. \end{equation} $

(2.12)

Set $f=\sum\limits_{k=1}\limits^{\infty}2^ks_k\cdot\chi_{Q_k}, $ then

$ \displaystyle\int_{{\Omega}}{\it\Psi}(|f|)\omega d\mu = \sum\limits_{k=1}\limits^{\infty} \displaystyle\int_{Q_k}{\it\Psi}(|f|)\omega d\mu\\ = \sum\limits_{k=1}\limits^{\infty} {\it\Psi}(2^ks_k)\omega(Q_k)\\ = \sum\limits_{k=1}\limits^{\infty} {\it\Psi}(2^ks_k)\frac{1}{\alpha2^k{\it\Psi}(2^ks_k)}\\ = \frac{1}{\alpha}\leq{\it\Psi}(1). $

Thus $f\in L^{\it{\Psi}}(\omega d\mu).$ By Jensen's inequality, we have $f\in L^1(\omega d\mu) \hbox{~and~} \|f\|_{L^1(\omega d\mu)}\leq1.$

On the other side, we claim that the function $Mf$ is not in $L^{\it{\Phi}}(\omega d\mu).$ Then we get a contradiction. To show the claim in the following way: fix a constant $\varepsilon\in(0, 1]$ for $\varepsilon f, $ with $C, ~C_4 \hbox{ and }C_5$ in Lemma 2.6, we have

$ \begin{equation}\label{3n1}\frac{\varepsilon}{\lambda} \displaystyle\int_{\{\varepsilon f>C_4\lambda\}}f\omega d\mu\leq C_5\cdot\omega\{\varepsilon Mf>\lambda\}, ~\lambda\geq\frac{\varepsilon}{C}\|f\|_{L^1(\omega d\mu)}. \end{equation} $

(2.13)

Note that $1\geq\frac{\varepsilon}{C}\|f\|_{L^1(\omega d\mu)}, $ we have

$ \displaystyle\int_{\Omega}\it{\Phi}(\varepsilon\cdot Mf)\omega d\mu = \displaystyle\int_0^\infty\omega\{\varepsilon\cdot Mf>\lambda\} a(\lambda)d\lambda\\ \geq \displaystyle\int_1^\infty\omega\{\varepsilon\cdot Mf>\lambda\} a(\lambda)d\lambda \\ \geq \displaystyle\int_1^\infty\Big(\frac{\varepsilon}{C_5\lambda} \displaystyle\int_{\{\varepsilon\cdot |f|>C_4\lambda\}}|f|\omega d\mu \Big)a(\lambda)d\lambda\\ = \frac{\varepsilon}{C_5}\displaystyle\int_\Omega|f| \Big(\displaystyle\int_1^{\frac{\varepsilon\cdot |f|}{C_4}}\frac{a(\lambda)}{\lambda}d\lambda\Big)\omega d\mu\\ \geq \frac{\varepsilon}{C_5}\sum\limits_{k=1}\limits^{\infty}\displaystyle\int_{Q_k}2^ks_k \Big(\displaystyle\int_1^{\frac{\varepsilon\cdot 2^ks_k}{C_4}} \frac{a(\lambda)}{\lambda}d\lambda\Big)\omega d\mu $

For the above $\varepsilon, $ we can choose $k(\varepsilon)$ such that $\frac{\varepsilon\cdot 2^k}{C_4}>1, ~k\geq k(\varepsilon).$ Put

$ L=\frac{\varepsilon}{C_5}\sum\limits_{k=1}\limits^{k(\varepsilon)-1}\displaystyle\int_{Q_k}2^ks_k \Big(\displaystyle\int_1^{\frac{\varepsilon\cdot 2^ks_k}{C_4}} \frac{a(\lambda)}{\lambda}d\lambda\Big)\omega d\mu , $

then $L$ is a real number. Therefore, it follows from $(2.11)\hbox{ and }(2.12)$ that

$ \displaystyle\int_{\Omega}{\it\Phi}(\varepsilon\cdot Mf)\omega d\mu \geq L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)}\limits^{\infty} \displaystyle\int_{Q_k}2^ks_k \Big(\displaystyle\int_1^{\frac{\varepsilon\cdot 2^ks_k}{C_4}}\frac{a(\lambda)}{\lambda}d\lambda\Big)\omega d\mu\\ \geq L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)}\limits^{\infty} \displaystyle\int_{Q_k}2^ks_k \Big(\displaystyle\int_1^{s_k}\frac{a(\lambda)}{\lambda}d\lambda\Big)\omega d\mu \geq L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)}\limits^{\infty} \displaystyle\int_{Q_k}2^ks_k \big(2^kb(2^ks_k)\big)\omega d\mu\\ = L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)} \limits^{\infty}2^k2^ks_kb(2^ks_k)\omega(Q_k) =L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)} \limits^{\infty}2^k 2^ks_kb(2^ks_k)\frac{1}{\alpha2^k{\it\Psi}(2^ks_k)}\\ \geq L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)} \limits^{\infty}2^k2^ks_k b(2^ks_k)\frac{1}{\alpha2^k(2^ks_k)b(2^ks_k)} =L+\frac{\varepsilon}{C_5}\sum\limits_{k=k(\varepsilon)} \limits^{\infty}\frac{1}{\alpha}=\infty. $

The claim is proved and the proof is finished.

Proof of Theorem 1.5 For $C_7=\frac{C_6}{C}, $ fix $f\in L^{\it{\Phi}}(\omega d\mu)$ such that $\displaystyle\int_{{\Omega}}{\it\Phi}(M(|f|))\omega d\mu <\infty$ and $\|f\|_{L^1(\omega d\mu)}\leq1, $ where the constant $C$ is the one in the definition of $B_1.$ Then

$ \displaystyle\int_{{\Omega}}{\it\Psi}(\frac{C_6}{C}|f|)\omega d\mu = \frac{C_6}{C}\displaystyle\int_0^\infty\omega\{|f|>t\}b(\frac{C_6}{C}t)dt\\ = C_6\displaystyle\int_0^\infty\omega\{|\frac{f}{C}|>t\}b(C_6t)dt\\ = C_6\displaystyle\int_0^{s_0}\omega\{|\frac{f}{C}|>t\}b(C_6t)dt +C_6\displaystyle\int_{s_0}^\infty\omega\{|\frac{f}{C}|>t\}b(C_6t)dt. $

Denote

$ I_0=C_6\displaystyle\int_0^{s_0}\omega\{|\frac{f}{C}|>t\}b(C_6t)dt ~\hbox{and}~I_1=C_6\displaystyle\int_{s_0}^\infty\omega\{\frac{f}{C}|>t\}b(C_6t)dt. $

Then, for $I_0, $ we have

$ I_0 \leq C_6b(C_6s_0)\displaystyle\int_0^{s_0}\omega\{|\frac{f}{C}|>t\}dt \leq C_6b(C_6s_0)\displaystyle\int_0^\infty\omega\{|\frac{f}{C}|>t\}dt\\ = \frac{1}{C}\cdot C_6b(C_6s_0)\|f\|_{L^1(\omega d\mu)}. $

It follows from $(1.8)$ that

$ I_1 \leq \displaystyle\int_{s_0}^\infty\omega\{|\frac{f}{C}|>t\} \Big(\displaystyle\int_1^t\frac{a(s)}{s} ds\Big) dt \leq\displaystyle\int_1^\infty\omega\{|\frac{f}{C}|>t\}\Big(\displaystyle\int_1^t\frac{a(s)}{s} ds\Big) dt \\ = \displaystyle\int_1^\infty\frac{a(s)}{s}\Big(\displaystyle\int_s^\infty\omega\{|\frac{f}{C}|>t\}dt\Big)ds. $

It follows from (2.1), (2.6) and (2.9) that

$ I_1 \leq \displaystyle\int_1^\infty\frac{a(s)}{s} \Big(\displaystyle\int_{\{|\frac{f}{C}|>s\}}|\frac{f}{C}|\omega d\mu\Big)ds \leq\displaystyle\int_1^\infty\frac{a(s)}{s} \Big(d\cdot s\cdot\omega\{\hat{M}(|\frac{f}{C}|)>s\}\Big)ds\\ = d\cdot\displaystyle\int_1^\infty a(s) \cdot\omega\{\hat{M}(|\frac{f}{C}|)>s\}ds \leq d\cdot\displaystyle\int_1^\infty a(s) \cdot\omega\{C\cdot M(\frac{|f|}{C})>s\}ds\\ \leq d\cdot\displaystyle\int_\Omega{\it\Phi}(M(|f|)) \omega d\mu, $

where we have used $\|\frac{f}{C}\|_{L^1(\omega d\mu)}\leq\|f\|_{L^1(\omega d\mu)}\leq1.$ Therefore, $(1.9)$ is valid with $C_7=\frac{C_6}{C}$ and $C_8=\frac{C_6b(C_6s_0)}{C}\vee d.$

Proof of Theorem 1.6 We proceed by contradiction and assume that $(1.8)$ dose not hold. Then we get a sequence $s_k>1$ satisfying

$ \frac{4}{3} <s_1 <s_2 <\cdot\cdot\cdot <s_k <s_{k+1} <\cdot\cdot\cdot ~\hbox{and}~\lim\limits_{k\rightarrow\infty}s_k=\infty; $

(2.14)

$ \displaystyle\int_1^{s_k}\frac{a(t)}{t}dt < \frac{1}{2^k}b(\frac{1}{2^k}s_k), ~k\geq1; $

(2.15)

$ s_{k+1}>4s_k, ~k\geq1; $

(2.16)

$ b(\frac{s_k}{2^k})\geq k2^k, ~k\geq1. $

(2.17)

Let $\alpha_k=\frac{1}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})}, $ then

$ \sum\limits_{k=1}\limits^{\infty}\alpha_k =\sum\limits_{k=1}\limits^{\infty} \frac{1}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})} \leq\sum\limits_{k=1}\limits^{\infty} \frac{1}{\frac{s_k}{2^k}k2^k} =\sum\limits_{k=1}\limits^{\infty} \frac{1}{ks_k}\leq\sum\limits_{k=1}\limits^{\infty} \frac{1}{s_k} \leq \frac{1}{s_1}\sum\limits_{k=1}\limits^{\infty} \frac{1}{4^{k-1}}\leq1. $

By the assumption that $(\Omega, \mathcal {F}, \omega)$ is non-atomic, we have a family of measurable sets $Q_k\in\mathcal {F}$ such that

$ \begin{equation}\label{extra7} \omega(Q_k)=\alpha_k~\hbox{and}~ Q_{k_1}\cap Q_{k_2}=\emptyset, ~k_1\neq k_2. \end{equation} $

(2.18)

Set $f=\sum\limits_{k=1}\limits^{\infty}\frac{ks_k}{2^k}\cdot\chi_{Q_k}, $ then

$ \displaystyle\int_{{\Omega}}|f|\omega d\mu =\sum\limits_{k=1}\limits^{\infty} \displaystyle\int_{Q_k}|f|\omega d\mu =\sum\limits_{k=1}\limits^{\infty} \frac{ks_k}{2^k}\omega(Q_k) =\sum\limits_{k=1}\limits^{\infty} \frac{ks_k}{2^k}\frac{1}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})} \leq\sum\limits_{k=1}\limits^{\infty}\frac{1}{2^k}=1. $

Thus $f\in L^1(\omega d\mu)$ and $\|f\|_{L^1(\omega d\mu)}\leq1.$ Moreover, we also have $\displaystyle\int_{\Omega}\it{\Phi}(Mf)\omega d\mu <\infty.$ In fact, for the sequence $s_k, $ we have

$ \frac{(k+1)s_{k+1}}{2^{k+1}} \geq\frac{(k+1)4s_k}{2^{k+1}} >\frac{4ks_k}{2^{k+1}}\geq2\frac{ks_k}{2^k} >\frac{ks_k}{2^k}, ~k\geq1. $

It is clear that $\frac{s_{k+1}}{2^{k+1}}>\frac{s_k}{2^k}, $ thus

$ \omega(Q_{k+1}) =\frac{1}{\frac{s_{k+1}}{2^{k+1}}b(\frac{s_{k+1}}{2^{k+1}})} \leq\frac{1}{\frac{4s_k}{2^{k+1}}b(\frac{s_k}{2^k})} \leq\frac{1}{\frac{2s_k}{2^k}b(\frac{s_k}{2^k})}=\frac{1}{2}\omega(Q_k), ~k\geq1. $

Let $\beta_k=\frac{ks_k}{2^k}, ~k\geq1$ and $\beta_0=0.$ Obviously, $\beta_k\uparrow\infty.$ For $k\geq1, $ when $s\in[\beta_{k-1}, \beta_k), $ we have

$ \omega\{|f|>s\}=\sum\limits_{i=0}\limits^{\infty}\omega(Q_{k+i}) \leq\omega(Q_k)\sum\limits_{i=0} \limits^{\infty}\frac{1}{2^i}=2\omega(Q_k) $

and

$ \displaystyle\int_1^{2s}\frac{a(t)}{t}dt\leq0\leq\frac{1}{2^k}b(\frac{s_k}{2^k}), ~\hbox{if }2s\leq1;\\ \displaystyle\int_1^{2s}\frac{a(t)}{t}dt \leq\displaystyle\int_1^{2\beta_k}\frac{a(t)}{t}dt \leq\displaystyle\int_1^{s_k}\frac{a(t)}{t}dt <\frac{1}{2^k}b(\frac{s_k}{2^k}), ~\hbox{if }2s>1. $

Combining (2.3), (1.2) and (1.3), we have

$ \displaystyle\int_{\Omega}\it{\Phi}(Mf)\omega d\mu =\displaystyle\int_0^\infty\omega\{Mf>t\} a(t)dt\\ \leq \displaystyle\int_0^\infty\Big(\frac{C}{t} \displaystyle\int_{\frac{t}{2}}^\infty \omega\{|f|>s\} ds \Big)a(t)dt \\ = C\displaystyle\int_0^\infty \omega\{|f|>s\}\Big( \displaystyle\int_{0}^{2s}\frac{a(t)}{t} dt \Big) ds\\ = C\displaystyle\int_0^\infty \omega\{|f|>s\}\Big( \displaystyle\int_{0}^{1}\frac{a(t)}{t} dt +\displaystyle\int_{1}^{2s}\frac{a(t)}{t} dt \Big) ds\\ = KC+C\sum\limits_{k=1}^\infty\displaystyle\int_{\beta_{k-1}}^{\beta_k} \omega\{|f|>s\}\Big( \displaystyle\int_{1}^{2s}\frac{a(t)}{t} dt \Big) ds\\ \leq KC+2C\sum\limits_{k=1}^\infty \omega(Q_k)\Big(\frac{1}{2^k}b(\frac{s_k}{2^k}) \Big)\beta_k \\ = KC+2C\sum\limits_{k=1}^\infty \frac{1}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})} \Big(\frac{1}{2^k}b(\frac{s_k}{2^k}) \Big)\frac{ks_k}{2^k}\\ = KC+2C\sum\limits_{k=1}^\infty\frac{k}{2^k} <\infty. $

On the other hand, we have

$ \displaystyle\int_{\Omega}{\it\Psi}(C_7\cdot |f|)\omega d\mu =\sum\limits_{k=1}\limits^{\infty}{\it\Psi} (C_7\cdot \frac{ks_k}{2^k})\omega({Q_k}) =\sum\limits_{k=1}\limits^{\infty}{\it\Psi} (C_7\cdot \frac{ks_k}{2^k})\frac{1}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})}. $

For the above $C_7, $ we can choose a constant $k(C_7)>0$ such that $k\cdot C_7\geq2, ~k>k(C_7).$ Therefore,

$ \displaystyle\int_{\Omega}{\it\Psi}(C_7\cdot |f|)\omega d\mu \geq \sum\limits_{k=k(C_7)}\limits^{\infty}{\it\Psi} (C_7\cdot \frac{ks_k}{2^k})\frac{1}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})}\\ \geq \sum\limits_{k=k(C_7)} \limits^{\infty}\displaystyle\int_{\frac{s_k}{2^k}}^{2\frac{s_k}{2^k}} \frac{b(t)}{\frac{s_k}{2^k}b(\frac{s_k}{2^k})}dt \geq\sum\limits_{k=k(C_7)}\limits^{\infty}1=\infty. $

The result contradicts assumption $(1.9).$ This completes the proof.