Abstract

We introduce Morrey-Campanato spaces of martingales and give their basic properties. Our definition of martingale Morrey-Campanato spaces is different from martingale Lipschitz spaces introduced by Weisz, while Campanato spaces contain Lipschitz spaces as special cases. We also give the relation between these definitions. Moreover, we establish the boundedness of fractional integrals as martingale transforms on these spaces. To do this we show the boundedness of the maximal function on martingale Morrey-Campanato spaces.

1. Introduction

The purpose of this paper is to introduce Morrey-Campanato spaces of martingales. The Lebesgue space 𝐿𝑝 plays an important role in martingale theory as well as in harmonic analysis. Moreover, in martingale theory, Lorentz spaces, Orlicz spaces, Hardy spaces, Lipschitz spaces, and John-Nirenberg space BMO also have been developed by many authors, see [15], and so forth. Recently Kikuchi [6] investigated Banach function spaces of martingales. In this paper we introduce Morrey-Campanato spaces of martingales and give their basic properties. Moreover, we establish the boundedness of fractional integrals as martingale transforms on these spaces. Note that Campanato spaces are not Banach function spaces in general.

We consider a probability space (Ω,,𝑃) such that =𝜎(𝑛𝑛), where {𝑛}𝑛0 is a nondecreasing sequence of sub-𝜎-algebras of . Following Weisz [5], we call {𝑛}𝑛0 a stochastic basis. For the sake of simplicity, let 1=0. We suppose that every 𝜎-algebra 𝑛 is generated by countable atoms, where 𝐵𝑛 is called an atom (more precisely a (𝑛,𝑃)-atom), if any 𝐴𝐵 with 𝐴𝑛 satisfies 𝑃(𝐴)=𝑃(𝐵) or 𝑃(𝐴)=0. Denote by 𝐴(𝑛) the set of all atoms in 𝑛. The expectation operator and the conditional expectation operators relative to 𝑛 are denoted by 𝐸 and 𝐸𝑛, respectively.

We define Morrey-Campanato spaces as the following: let 𝑝[1,) and 𝜆(,). For 𝑓𝐿1, let𝑓𝐿𝑝,𝜆=sup𝑛0sup𝐵𝐴𝑛1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓||𝑝𝑑𝑃1/𝑝,𝑓𝑝,𝜆=sup𝑛0sup𝐵𝐴𝑛1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝,(1.1) and let𝐿𝑝,𝜆=𝑓𝐿0𝑝𝑓𝐿𝑝,𝜆<,𝑝,𝜆=𝑓𝐿0𝑝𝑓𝑝,𝜆<,(1.2) where 𝐿0𝑝 is the set of all 𝑓𝐿𝑝 such that 𝐸0𝑓=0.

We give basic properties of martingale Morrey-Campanato spaces and compare these spaces with martingale Lipschitz spaces introduced by Weisz [7]. It is well known, in harmonic analysis, that Campanato spaces contain Lipschitz spaces as special cases. Recently, martingale Campanato spaces were introduced in [8] as generalization of martingale Lipschitz spaces. While our definition of martingale Morrey-Campanato spaces is different from the one in [8], we can prove that our martingale Morrey-Campanato spaces contain martingale Lipschitz spaces by Weisz as special cases, under the assumption that every 𝜎-algebra 𝑛 is generated by countable atoms.

The fractional integrals are very useful tools to analyse function spaces in harmonic analysis. Actually, Hardy and Littlewood [9, 10] and Sobolev [11] investigated the fractional integrals to establish the theory of Lebesgue spaces and Lipschitz spaces. Stein and Weiss [12], Taibleson and Weiss [13], and Krantz [14] also investigated the fractional integrals to establish the theory of Hardy spaces. See also [15]. The 𝐿𝑝-𝐿𝑞 boundedness of the fractional integrals is well known as the Hardy-Littlewood-Sobolev theorem derived from [911]. This boundedness has been extended to Morrey-Campanato spaces by Peetre [16] and Adams [17], see also [18]. It is known that Morrey-Campanato spaces contain 𝐿𝑝, BMO, and Lip𝛼 as special cases, see for example [16, 19].

On the other hand, in martingale theory, Watari [20] and Chao and Ombe [21] proved the boundedness of the fractional integrals for 𝐿𝑝 (𝐻𝑝), BMO, and Lipschitz spaces of the dyadic martingale. In this paper, we also establish the boundedness of fractional integrals as martingale transforms on Morrey-Campanato spaces. Our result generalizes and improves the results in [20, 21].

For a martingale 𝑓=(𝑓𝑛)𝑛0 relative to {𝑛}𝑛0, denote its martingale difference by 𝑑𝑛𝑓=𝑓𝑛𝑓𝑛1 (𝑛0, with convention 𝑑0𝑓=0). For 𝛼>0, we define the fractional integral 𝐼𝛼𝑓=((𝐼𝛼𝑓)𝑛)𝑛0 of 𝑓 by𝐼𝛼𝑓𝑛=𝑛𝑘=0𝑏𝛼𝑘1𝑑𝑘𝑓,(1.3) where 𝑏𝑘 is an 𝑘-measurable function such that𝑏𝑘(𝜔)=𝑃(𝐵)fora.e.𝜔𝐵with𝐵𝐴𝑘.(1.4) Then ((𝐼𝛼𝑓)𝑛)𝑛0 is a martingale for any martingale 𝑓=(𝑓𝑛)𝑛0, since each 𝑏𝑘 is bounded, that is, 𝐼𝛼 is a martingale transform introduced by Burkholder [22]. This definition of 𝐼𝛼 is an extention of the one in [20, 21] which is for dyadic martingales. We can prove the boundedness of fractional integrals 𝐼𝛼 as martingale transforms from 𝐿𝑝 to 𝐿𝑞, if 1<𝑝<𝑞< and 1/𝑝+𝛼=1/𝑞. That is, if a martingale 𝑓=(𝑓𝑛)𝑛0 is 𝐿𝑝-bounded, then ((𝐼𝛼𝑓)𝑛)𝑛0 is 𝐿𝑞-bounded and the following inequality holds:sup𝑛0𝐼𝛼𝑓𝑛𝐿𝑞𝐶sup𝑛0𝑓𝑛𝐿𝑝,(1.5) where 𝐶 is a positive constant independent of 𝑓. Further, we prove the boundedness of fractional integrals 𝐼𝛼 as martingale transforms on Morrey-Campanato spaces.

To prove the boundedness of fractional integrals we use a different method from [20, 21]. More precisely, under the assumption that every 𝜎-algebra 𝑛 is generated by countable atoms, we first prove the boundedness of the maximal function, and then we use the pointwise estimate by the maximal function and its boundedness, namely, Hedberg’s method in [23]. We also use the method in [24, 25]. By considering sequences of atoms precisely, we can apply these methods to martingale Morrey-Campanato spaces. From this point of view our assumption seems to be natural to define martingale Morrey-Campanato spaces.

We state notation, definitions, and remarks in the next section and give basic properties of Morrey-Campanato spaces in Section 3. We prove the boundedness of the maximal function and fractional integrals in Sections 4 and 5, respectively.

At the end of this section, we make some conventions. Throughout this paper, we always use 𝐶 to denote a positive constant that is independent of the main parameters involved but whose value may differ from line to line. Constants with subscripts, such as 𝐶𝑝, is dependent on the subscripts. If 𝑓𝐶𝑔, we then write 𝑓𝑔 or 𝑔𝑓 and if 𝑓𝑔𝑓, we then write 𝑓𝑔.

2. Notation, Definitions, and Remarks

Recall that (Ω,,𝑃) is a probability space, and {𝑛}𝑛0 a nondecreasing sequence of sub-𝜎-algebras of such that =𝜎(𝑛𝑛). For the sake of simplicity, let 1=0. As in Section 1, we always suppose that every 𝜎-algebra 𝑛 is generated by countable atoms, with denoting by 𝐴(𝑛) the set of all atoms in 𝑛. We define the fractional integral 𝐼𝛼 as a martingale transform by (1.3).

For a martingale 𝑓=(𝑓𝑛)𝑛0 relative to {𝑛}𝑛0, the maximal function 𝑓 of 𝑓 is defined by𝑓𝑛=sup0𝑚𝑛||𝑓𝑚||,𝑓=sup𝑛0||𝑓𝑛||.(2.1)

It is known that if 𝑝(1,), then any 𝐿𝑝-bounded martingale converges in 𝐿𝑝. Moreover, if 𝑓𝐿𝑝, 𝑝[1,), then (𝑓𝑛)𝑛0 with 𝑓𝑛=𝐸𝑛𝑓 is an 𝐿𝑝-bounded martingale and converges to 𝑓 in 𝐿𝑝 (see, e.g., [26]). For this reason a function 𝑓𝐿1 and the corresponding martingale (𝑓𝑛)𝑛0 with 𝑓𝑛=𝐸𝑛𝑓 will be denoted by the same symbol 𝑓. Note also that 𝑓𝐿𝑝=sup𝑛0𝐸𝑛𝑓𝐿𝑝. In this case𝑓𝑛=sup0𝑚𝑛||𝐸𝑚𝑓||,𝑓=sup𝑛0||𝐸𝑛𝑓||for𝑓𝐿1.(2.2)

Let be the set of all martingales such that 𝑓0=0. For 𝑝[1,], let 𝐿0𝑝 be the set of all 𝑓𝐿𝑝 such that 𝐸0𝑓=0. For any 𝑓𝐿0𝑝, let 𝑓𝑛=𝐸𝑛𝑓. Then (𝑓𝑛)𝑛0 is an 𝐿𝑝-bounded martingale in . For this reason we can regard 𝐿0𝑝 as a subset of .

In Section 1 we have introduced Morrey spaces 𝐿𝑝,𝜆 and Campanato spaces 𝑝,𝜆 as the following.

Definition 2.1. Let 𝑝[1,) and 𝜆(,). For 𝑓𝐿1, let 𝑓𝐿𝑝,𝜆=sup𝑛0sup𝐵𝐴(𝑛)1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓||𝑝𝑑𝑃1/𝑝,𝑓𝑝,𝜆=sup𝑛0sup𝐵𝐴𝑛1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝,(2.3) and define 𝐿𝑝,𝜆=𝑓𝐿0𝑝𝑓𝐿𝑝,𝜆<,𝑝,𝜆=𝑓𝐿0𝑝𝑓𝑝,𝜆<.(2.4)
Then functionals 𝑓𝐿𝑝,𝜆 and 𝑓𝑝,𝜆 are norms on 𝐿𝑝,𝜆 and 𝑝,𝜆, respectively. Note that 𝐿𝑝,𝜆 and 𝑝,𝜆 are not always trivial set {0} even if 𝜆>0 and 𝜆>1, respectively. This property is different from classical Morrey-Campanato spaces on 𝑛.
The martingale 𝑓=(𝑓𝑛)𝑛0 is said to be 𝐿𝑝,𝜆-bounded if 𝑓𝑛𝐿𝑝,𝜆 (𝑛0) and sup𝑛0𝑓𝑛𝐿𝑝,𝜆<. Similarly, the martingale 𝑓=(𝑓𝑛)𝑛0 is said to be 𝑝,𝜆-bounded if 𝑓𝑛𝑝,𝜆 (𝑛0) and sup𝑛0𝑓𝑛𝑝,𝜆<.

Proposition 2.2. Let 1𝑝<. Let 𝑓𝐿1 and (𝑓𝑛)𝑛0 be its corresponding martingale with 𝑓𝑛=𝐸𝑛𝑓 (𝑛0).
(i)Assume that 𝜆(,0]. If 𝑓𝐿𝑝,𝜆, then (𝑓𝑛)𝑛0 is 𝐿𝑝,𝜆-bounded and 𝑓𝐿𝑝,𝜆sup𝑛0𝑓𝑛𝐿𝑝,𝜆.(2.5) Conversely, if (𝑓𝑛)𝑛0 is 𝐿𝑝,𝜆-bounded, then 𝑓𝐿𝑝,𝜆 and 𝑓𝐿𝑝,𝜆sup𝑛0𝑓𝑛𝐿𝑝,𝜆.(2.6)(ii)Assume that 𝜆(,). If 𝑓𝑝,𝜆, then (𝑓𝑛)𝑛0 is 𝑝,𝜆-bounded and 𝑓𝑝,𝜆sup𝑛0𝑓𝑛𝑝,𝜆.(2.7) Conversely, if (𝑓𝑛)𝑛0 is 𝑝,𝜆-bounded, then 𝑓𝑝,𝜆 and 𝑓𝑝,𝜆sup𝑛0𝑓𝑛𝑝,𝜆.(2.8)

Proof. (i) Let 𝑓𝐿𝑝,𝜆 and 𝑛0. Fix any 𝐵𝐴(𝑘). If 𝑘𝑛, then 𝐵||𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝𝐵𝐸𝑛||𝑓||𝑝𝑑𝑃1/𝑝=𝐵||𝑓||𝑝𝑑𝑃1/𝑝𝑃(𝐵)𝜆+1/𝑝𝑓𝐿𝑝,𝜆.(2.9) If 𝑛<𝑘, then taking 𝐵𝑛𝐴(𝑛) such that 𝐵𝐵𝑛, we have 𝐸𝑛1𝑓=𝑃𝐵𝑛𝐵𝑛𝑓𝑑𝑃on𝐵𝑛.(2.10) Hence 1𝑃(𝐵)𝐵||𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝=1𝑃𝐵𝑛𝐵𝑛||𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝1𝑃𝐵𝑛𝐵𝑛𝐸𝑛||𝑓||𝑝𝑑𝑃1/𝑝=1𝑃𝐵𝑛𝐵𝑛||𝑓||𝑝𝑑𝑃1/𝑝𝐵𝑃𝑛𝜆𝑓𝐿𝑝,𝜆𝑃(𝐵)𝜆𝑓𝐿𝑝,𝜆.(2.11) Therefore, 𝑓𝑛=𝐸𝑛𝑓𝐿𝑝,𝜆 and 𝑓𝑛𝐿𝑝,𝜆𝑓𝐿𝑝,𝜆 for all 𝑛0. This shows that (𝑓𝑛)𝑛0 is a 𝐿𝑝,𝜆-bounded martingale and sup𝑛0𝑓𝑛𝐿𝑝,𝜆𝑓𝐿𝑝,𝜆.(2.12) Conversely, from the inequality 𝐵||𝑓||𝑝𝑑𝑃liminf𝑛𝐵||𝐸𝑛𝑓||𝑝𝑑𝑃𝐵𝑚𝐴𝑚,(2.13) it follows that 𝑓𝐿𝑝,𝜆sup𝑛0𝐸𝑛𝑓𝐿𝑝,𝜆.(2.14)(ii) Let 𝑓𝑝,𝜆 and 𝑛0. Fix any 𝐵𝐴(𝑘). If 𝑘𝑛, then𝐵||𝐸𝑛𝑓𝐸𝑘𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝=𝐵||𝐸𝑛𝑓𝐸𝑛𝐸𝑘𝑓||𝑝𝑑𝑃1/𝑝𝐵𝐸𝑛||𝑓𝐸𝑘𝑓||𝑝𝑑𝑃1/𝑝=𝐵||𝑓𝐸𝑘𝑓||𝑝𝑑𝑃1/𝑝𝑃(𝐵)𝜆+1/𝑝𝑓𝑝,𝜆.(2.15) If 𝑛<𝑘, then 𝐸𝑘[𝐸𝑛𝑓]=𝐸𝑛𝑓. Hence 1𝑃(𝐵)𝐵||𝐸𝑛𝑓𝐸𝑘𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝=0.(2.16) Therefore, 𝑓𝑛=𝐸𝑛𝑓𝑝,𝜆 and 𝑓𝑛𝑝,𝜆𝑓𝑝,𝜆 for all 𝑛0. This shows that (𝑓𝑛)𝑛0 is a 𝑝,𝜆-bounded martingale and sup𝑛0𝑓𝑛𝑝,𝜆𝑓𝑝,𝜆.(2.17) Conversely, from the inequality 𝐵||𝑓𝑓𝑘||𝑝𝑑𝑃liminf𝑛𝐵||𝐸𝑛𝑓𝑓𝑘||𝑝𝑑𝑃𝐵𝐴𝑘,𝑘=1,2,,(2.18) it follows that 𝑓𝑝,𝜆sup𝑛0𝐸𝑛𝑓𝑝,𝜆.(2.19)

Remark 2.3. In general, for 𝑓𝐿𝑝,𝜆 (resp., 𝑝,𝜆), 𝐸𝑛𝑓 does not always converge to 𝑓 in 𝐿𝑝,𝜆 (resp., 𝑝,𝜆). See Remark 3.7.

Remark 2.4. If 0={,Ω}, then 𝐿𝑝,𝜆𝐿𝑝 and 𝑓𝐿𝑝𝑓𝐿𝑝,𝜆 for 𝑓𝐿𝑝,𝜆. Therefore, if (𝑓𝑛)𝑛0 is an 𝐿𝑝,𝜆-bounded martingale, then it is an 𝐿𝑝-bounded martingale. If 1<𝑝<, from the known result it follows that there exists 𝑓𝐿0𝑝 such that 𝐸𝑛𝑓=𝑓𝑛, 𝑛0, and (𝑓𝑛)𝑛0 converges to 𝑓 in 𝐿𝑝 and a.e. Moreover, we can deduce that 𝑓𝐿𝑝,𝜆, since 𝐵||𝑓𝑛||(𝜔)𝑝𝑑𝑃1/𝑝𝑃(𝐵)𝜆+1/𝑝𝑓𝑛𝐿𝑝,𝜆𝑃(𝐵)𝜆+1/𝑝sup𝑛0𝑓𝑛𝐿𝑝,𝜆,𝐵𝑚𝐴𝑚.(2.20) The stochastic basis {𝑛}𝑛0 is said to be regular, if there exists a constant 𝑅2 such that 𝑓𝑛𝑅𝑓𝑛1(2.21) holds for all nonnegative martingales (𝑓𝑛)𝑛0.

Remark 2.5. In general, 𝐿𝑝,𝜆𝑝,𝜆 with 𝑓𝑝,𝜆2𝑓𝐿𝑝,𝜆. Actually, for any 𝐵𝐴(𝑛), 1𝑃(𝐵)𝐵||𝑓𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝1𝑃(𝐵)𝐵||𝑓||𝑝𝑑𝑃1/𝑝+1𝑃(𝐵)𝐵||𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝12𝑃(𝐵)𝐵||𝑓||𝑝𝑑𝑃1/𝑝.(2.22) Moreover, if {𝑛}𝑛0 is regular and 𝜆<0, then we can prove that 𝐿𝑝,𝜆=𝑝,𝜆 with equivalent norms (Theorem 3.1 and Remark 3.2).

Remark 2.6. By definition, if 0𝜆𝜆, then we have that 𝐿0𝐿𝑝,𝜆𝐿𝑝,𝜆 with 𝑓𝐿𝑝,𝜆𝑓𝐿𝑝,𝜆𝑓𝐿 and 𝐿0𝑝,𝜆𝑝,𝜆 with 𝑓𝑝,𝜆𝑓𝑝,𝜆2𝑓𝐿. If 𝜆1/𝑝, then 𝐿0𝑝𝐿𝑝,𝜆𝑝,𝜆 with 𝑓𝑝,𝜆/2𝑓𝐿𝑝,𝜆𝑓𝐿𝑝.

Remark 2.7. By definition and Remark 2.5, if 0={,Ω}, then 𝐿𝑝,𝜆𝑝,𝜆𝐿0𝑝 with 𝑓𝐿𝑝𝑓𝑝,𝜆2𝑓𝐿𝑝,𝜆.

Definition 2.8. Let BMO=1,0 and Lip𝛼=1,𝛼 if 𝛼>0.
Our definitions of BMO and Lip𝛼 are different from the ones by Weisz [7]. To compare both we give another definition of martingale Morrey and Campanato spaces.

Definition 2.9. Let 𝑝[1,) and 𝜆(,). For 𝑓𝐿1, let 𝑓𝐿𝑝,𝜆,=sup𝑛0sup𝐵𝑛1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓||𝑝𝑑𝑃1/𝑝,𝑓𝑝,𝜆,=sup𝑛0sup𝐵𝑛1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝,(2.23) and define𝐿𝑝,𝜆,=𝑓𝐿0𝑝𝑓𝐿𝑝,𝜆,<,𝑝,𝜆,=𝑓𝐿0𝑝𝑓𝑝,𝜆,<.(2.24)
Note that the spaces 𝐿𝑝,𝜆, and 𝑝,𝜆, can be defined without the assumption that every 𝜎-algebra 𝑛 is generated by countable atoms.

Remark 2.10. By the definitions we have the relations 𝐿𝑝,𝜆,𝐿𝑝,𝜆 with 𝑓𝐿𝑝,𝜆𝑓𝐿𝑝,𝜆, and 𝑝,𝜆,𝑝,𝜆 with 𝑓𝑝,𝜆𝑓𝑝,𝜆,. If 𝜆0, then we can prove that 𝐿𝑝,𝜆,=𝐿𝑝,𝜆 and 𝑝,𝜆,=𝑝,𝜆 with the same norms, respectively (see Proposition 3.8). If 1/𝑝<𝜆<0, then 𝐿𝑝,𝜆,𝐿𝑝,𝜆 and 𝑝,𝜆,𝑝,𝜆 in general (see Proposition 3.9).

Remark 2.11. It is known that, if {𝑛}𝑛0 is regular and 𝜆0, then 1,𝜆,=𝑝,𝜆, with 𝑓1,𝜆,𝑓𝑝,𝜆,𝐶𝑝𝑓1,𝜆, for each 𝑝[1,) (see, e.g., [8]).
We also define weak Morrey spaces.

Definition 2.12. For 𝑝[1,) and 𝜆(,), let 𝑓𝑊𝐿𝑝,𝜆=sup𝑛0sup𝐵𝐴(𝑛)1𝑃(𝐵)𝜆sup𝑡>0𝑡𝑃||𝑓||𝐵>𝑡𝑃(𝐵)1/𝑝,(2.25) for measurable functions 𝑓, and define 𝑊𝐿𝑝,𝜆=𝑓𝐿01𝑓𝑊𝐿𝑝,𝜆<.(2.26)

3. Basic Properties of Morrey and Campanato Spaces

In this section we give basic properties of Morrey and Campanato spaces. The following theorem gives the relation between Morrey and Campanato spaces.

Theorem 3.1. Let {𝑛}𝑛0 be regular, 0={,Ω} and (Ω,,𝑃) be nonatomic. Let 𝑝[1,). (i)If 𝜆1/𝑝, then 𝐿𝑝,𝜆=𝑝,𝜆=𝐿0𝑝 and 12𝑓𝑝,𝜆𝑓𝐿𝑝,𝜆=𝑓𝐿𝑝𝑓𝑝,𝜆.(3.1)(ii)If 1/𝑝<𝜆<0, then 𝐿0𝐿𝑝,𝜆=𝑝,𝜆𝐿0𝑝 and 𝑓𝐿𝑝𝑓𝐿𝑝,𝜆𝑓𝐿,12𝑓𝑝,𝜆𝑓𝐿𝑝,𝜆𝐶𝑓𝑝,𝜆.(3.2)(iii)If 𝜆=0, then 𝐿0=𝐿𝑝,0𝑝,0=BMO and 𝑓𝐿𝑝,0=𝑓𝐿,𝑓BMO𝑓𝑝,0𝐶𝑝𝑓𝐵𝑀𝑂.(3.3)(iv)If 𝜆>0, then {0}=𝐿𝑝,𝜆𝑝,𝜆=Lip𝜆 and 𝑓Lip𝜆𝑓𝑝,𝜆𝐶𝑝𝑓Lip𝜆.(3.4)

Remark 3.2. We can prove (i) without the assumption that {𝑛}𝑛0 is regular or that (Ω,,𝑃) is nonatomic. In (ii), we can prove that 𝐿𝑝,𝜆=𝑝,𝜆 and (1/2)𝑓𝑝,𝜆𝑓𝐿𝑝,𝜆𝐶𝑓𝑝,𝜆 without the assumption that 0={,Ω} or that (Ω,,𝑃) is nonatomic. To show 𝐿0𝑝,𝜆 in (ii) and (iii), we can replace the condition that (Ω,,𝑃) is nonatomic by a weaker condition as in Proposition 3.6(ii), which follows from the condition that (Ω,,𝑃) is nonatomic. In (iv), we need the condition that (Ω,,𝑃) is nonatomic to show 𝐿𝑝,𝜆={0}.

To prove the theorem we first prove a lemma and two propositions.

Lemma 3.3. Let {𝑛}𝑛0 be regular. Then every sequence 𝐵0𝐵1𝐵𝑛,𝐵𝑛𝐴𝑛,(3.5) has the following property: for each 𝑛1, 𝐵𝑛=𝐵𝑛11or1+𝑅𝑃𝐵𝑛𝐵𝑃𝑛1𝐵𝑅𝑃𝑛,(3.6) where 𝑅 is the constant in (2.21).

Remark 3.4. Since 𝐵𝑛𝐴(𝑛) is an (𝑛,𝑃)-atom, we always interpret 𝐵𝑛1𝐵𝑛 as the inclusion modulo null sets, that is, 𝐵𝑛1𝐵𝑛 means 𝑃(𝐵𝑛𝐵𝑛1)=0. Therefore, 𝐵𝑛=𝐵𝑛1 means 𝑃(𝐵𝑛𝐵𝑛1)=𝑃(𝐵𝑛1𝐵𝑛)=0.

Remark 3.5. By the lemma we see that there exists 𝑚 such that 𝐵𝑚=𝐵𝑛 for all 𝑛𝑚, if and only if lim𝑛𝑃(𝐵𝑛)>0.

Proof of Lemma 3.3. Let 𝐵𝑛1={𝐸𝑛1[𝜒𝐵𝑛]1/𝑅}. Then 𝐵𝑛1𝑛1. By the regularity we have 𝜒𝐵𝑛𝑅𝐸𝑛1[𝜒𝐵𝑛]. This shows 𝐵𝑛1𝐵𝑛. In this case, 𝐵𝑛1𝐵𝑛1𝐵𝑛, since 𝐵𝑛1𝐴(𝑛1). From the definition of 𝐵𝑛1 it follows that 𝜒𝐵𝑛1𝑅𝐸𝑛1[𝜒𝐵𝑛]. Then we have 𝑃𝐵𝑛1𝐵𝑃𝑛1𝜒𝐵=𝐸𝑛1𝐸𝑅𝐸𝑛1𝜒𝐵𝑛𝜒=𝑅𝐸𝐵𝑛𝐵=𝑅𝑃𝑛.(3.7) Next we show 𝐵𝑛1=𝐵𝑛 or (1+1/𝑅)𝑃(𝐵𝑛)𝑃(𝐵𝑛1). Suppose that 𝑃𝐵𝑛1<11+𝑅𝑃𝐵𝑛.(3.8) Then 𝑃𝐵𝑛1𝐵𝑛𝐵=𝑃𝑛1𝐵𝑃𝑛<𝑃𝐵𝑛𝑅𝑃𝐵𝑛1𝑅.(3.9) Therefore, 𝐸𝑛1𝜒𝐵𝑛1𝐵𝑛=𝑃𝐵𝑛1𝐵𝑛𝑃𝐵𝑛1𝜒𝐵𝑛1<1𝑅𝜒𝐵𝑛1.(3.10) From the regularity and the inequality above it follows that 𝜒𝐵𝑛1𝐵𝑛𝑅𝐸𝑛1𝜒𝐵𝑛1𝐵𝑛<𝜒𝐵𝑛1.(3.11) This means that 𝐵𝑛1=𝐵𝑛.

Proposition 3.6. Let {𝑛}𝑛0 be regular, 1𝑝< and 𝜆>1/𝑝.
(i)For a sequence 𝐵0𝐵1𝐵𝑘, 𝐵𝑘𝐴(𝑘), let 𝑓0=0 and 𝑓𝑛=𝑛𝑘=1𝑃𝐵𝑘𝜆𝑃𝐵𝑘1𝑃𝐵𝑘𝜒𝐵𝑘𝜒𝐵𝑘1,𝑛1.(3.12)Then 𝑓=(𝑓𝑛)𝑛0 is a martingale in and in 𝑝,𝜆.(ii)Let 0𝜆>𝜆>1/𝑝. If there exists a sequence 𝐵0𝐵1𝐵𝑘, 𝐵𝑘𝐴(𝑘) and lim𝑘𝑃(𝐵𝑘)=0, then 𝐿0𝑝,𝜆𝑝,𝜆. If 0={,Ω} also, then 𝐿0𝑝,𝜆𝑝,𝜆𝐿0𝑝.

Proof. (i) By the definition of the sequence (𝑓𝑛)𝑛0, we have 𝐸𝑛1𝑑𝑛𝑓𝐵=𝑃𝑛𝜆𝑃𝐵𝑛1𝑃𝐵𝑛𝐸𝑛1𝜒𝐵𝑛𝜒𝐵𝑛1=0,(3.13) for every 𝑛1. Hence, we obtain that (𝑓𝑛)𝑛0 is a martingale.
We next show that the sequence (𝑓𝑛)𝑛0 converges in 𝐿𝑝. If lim𝑘𝑃(𝐵𝑘)>0 then the convergence is clear by Remark 3.5. We assume that lim𝑘𝑃(𝐵𝑘)=0. Then, by Lemma 3.3, we can take a sequence of integers 0=𝑘0<𝑘1<<𝑘𝑗< that satisfies 11+𝑅𝑃𝐵𝑘𝑗𝐵𝑃𝑘𝑗1𝐵𝑅𝑃𝑘𝑗,(3.14) and 𝐵𝑘𝑗1=𝐵𝑘 if 𝑘𝑗1𝑘<𝑘𝑗. In this case we can write 𝑓𝑛=𝑘𝑗𝑛𝑃𝐵𝑘𝑗𝜆𝑃𝐵𝑘𝑗1𝑃𝐵𝑘𝑗𝜒𝐵𝑘𝑗𝜒𝐵𝑘𝑗1.(3.15) Using (3.14) and the assumption 𝜆>1/𝑝, we have 𝑘𝑗>𝑛𝑃𝐵𝑘𝑗𝜆𝑃𝐵𝑘𝑗1𝑃𝐵𝑘𝑗𝜒𝐵𝑘𝑗𝜒𝐵𝑘𝑗1𝐿𝑝𝑘𝑗>𝑛𝑃𝐵𝑘𝑗𝜆𝑅𝜒𝐵𝑘𝑗𝐿𝑝+𝜒𝐵𝑘𝑗1𝐿𝑝2𝑅𝑘𝑗>𝑛𝑃𝐵𝑘𝑗𝜆+1/𝑝2𝑅𝑗=011+𝑅(𝜆+1/𝑝)𝑗𝑃𝐵𝑛𝜆+1/𝑝𝐵𝑃𝑛𝜆+1/𝑝.(3.16) Therefore, (𝑓𝑛)𝑛0 converges in 𝐿𝑝. We denote by 𝑓 the limit of (𝑓𝑛)𝑛0.
We can also deduce from (3.16) that 1𝑃𝐵𝑛𝐵𝑛||𝑓𝐸𝑛𝑓||𝑝𝑑𝑃1/𝑝𝐵𝑃𝑛𝜆.(3.17) On the other hand, for 𝐵𝐴(𝑛), we have 𝑓𝐸𝑛𝑓𝜒𝐵=𝑓𝐸𝑛𝑓,𝐵=𝐵𝑛,0,𝐵𝐵𝑛.(3.18)
Combining (3.17) and (3.18), we have 𝑓𝜆,𝑝1, that is, we get 𝑓𝑝,𝜆.
(ii) First we show 𝐿0𝑝,𝜆𝑝,𝜆. By Remark 2.6 we have 𝐿0𝑝,𝜆𝑝,𝜆. Then we need to show 𝑝,0𝐿0 and 𝑝,𝜆𝑝,𝜆.
We consider 𝑓𝑛 in (3.12) for the sequence 𝐵𝑘, 𝑘=0,1,. Then we can write 𝑓𝑛=𝑘𝑗𝑛𝑃𝐵𝑘𝑗1𝑃𝐵𝑘𝑗𝜒𝐵𝑘𝑗𝜒𝐵𝑘𝑗1(3.19) for 𝜆=0 and we have 𝑓=(𝑓𝑛)𝑛0𝑝,0. On the other hand, for a.e. 𝜔𝐵𝑘𝑗𝐵𝑘𝑗+1, 𝑓=𝑓𝑘𝑗+1=𝑓𝑘𝑗𝜒𝐵𝑘𝑗=𝑗𝑖=0𝑃𝐵𝑘𝑖1𝑃𝐵𝑘𝑖111𝑅(𝑗+1)1.(3.20)
Then 𝑓𝐿0 and 𝑝,0𝐿0.
Next, let 𝑔𝑛=𝑘𝑗𝑛𝑃𝐵𝑘𝑗𝜆𝑃𝐵𝑘𝑗1𝑃𝐵𝑘𝑗𝜒𝐵𝑘𝑗𝜒𝐵𝑘𝑗1.(3.21) Then 𝑔=(𝑔𝑛)𝑛0𝑝,𝜆. On the other hand, since |||𝑔𝐸𝑘𝑗1𝑔|||𝐵=𝑃𝑘𝑗𝜆on𝐵𝑘𝑗1𝐵𝑘𝑗,(3.22) we have 1𝑃𝐵𝑘𝑗1𝜆1𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1|||𝑔𝐸𝑘𝑗1𝑔|||𝑝𝑑𝑃1/𝑝1𝑃𝐵𝑘𝑗1𝜆1𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1𝐵𝑘𝑗|||𝑔𝐸𝑘𝑗1𝑔|||𝑝𝑑𝑃1/𝑝𝑃𝐵𝑘𝑗𝜆𝑃𝐵𝑘𝑗1𝜆𝑃𝐵𝑘𝑗1𝐵𝑘𝑗𝑃𝐵𝑘𝑗11/𝑝𝑅𝜆(1+𝑅)𝜆1/𝑝𝑃𝐵𝑘𝑗1𝜆𝜆as𝑗,(3.23) since 𝜆<𝜆. This shows 𝑝,𝜆𝑝,𝜆.
Finally, if 0={,Ω} also, then by Remark 2.7 and 𝑝,𝜆𝑝,𝜆 we have 𝑝,𝜆𝐿0𝑝. This shows the conclusion.

Remark 3.7. In Proposition 3.6, 𝑓=(𝑓𝑛)𝑛0 in (3.12) converges in 𝐿𝑝 as in the above proof. Moreover, the limit belongs to both 𝑝,𝜆 and 𝐿𝑝,𝜆 when 1/𝑝<𝜆<0, since we will show that 𝐿𝑝,𝜆=𝑝,𝜆 in the proof of Theorem 3.1. However, it converges in neither 𝑝,𝜆 nor 𝐿𝑝,𝜆. Actually, by a similar calculation to 𝑔=(𝑔𝑛)𝑛0 in the above proof, we have |||𝑓𝑓𝑛𝐸𝑘𝑗1𝑓𝑓𝑛|||=|||𝑓𝐸𝑘𝑗1𝑓|||𝐵=𝑃𝑘𝑗𝜆on𝐵𝑘𝑗1𝐵𝑘𝑗,(3.24) for 𝑛𝑘𝑗1, and then 1𝑃𝐵𝑘𝑗1𝜆1𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1|||𝑓𝐸𝑘𝑗1𝑓|||𝑝𝑑𝑃1/𝑝𝑅𝜆(1+𝑅)𝜆1/𝑝.(3.25) By Remark 2.5, we have 2𝑓𝑓𝑛𝐿𝑝,𝜆𝑓𝑓𝑛𝑝,𝜆𝑅𝜆(1+𝑅)𝜆1/𝑝.(3.26) This shows that (𝑓𝑛)𝑛0 converges in neither 𝑝,𝜆 nor 𝐿𝑝,𝜆.

Proposition 3.8. Let 1𝑝<.(i)If 0={,Ω} and 𝜆1/𝑝, then 𝐿0𝑝=𝐿𝑝,𝜆=𝐿𝑝,𝜆,=𝑝,𝜆=𝑝,𝜆, with𝑓𝐿𝑝,𝜆,=𝑓𝐿𝑝,𝜆=𝑓𝐿𝑝𝑓𝑝,𝜆𝑓𝑝,𝜆,2𝑓𝐿𝑝,𝜆,.(3.27)(ii)If 𝜆0, then 𝐿𝑝,𝜆=𝐿𝑝,𝜆,𝑝,𝜆=𝑝,𝜆,𝐿0𝑝 with𝑓𝐿𝑝𝑓𝐿𝑝,𝜆,=𝑓𝑝,𝜆𝑓𝐿𝑝,𝜆,=𝑓𝐿𝑝,𝜆.(3.28)

Proof. (i) Let 𝜆1/𝑝. For any 𝑓𝐿0𝑝 and any 𝐵𝑛, 1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓||𝑝𝑑𝑃1/𝑝=𝑃(𝐵)𝜆1/𝑝𝐵||𝑓||𝑝𝑑𝑃1/𝑝𝑃(Ω)𝜆1/𝑝Ω||𝑓||𝑝𝑑𝑃1/𝑝=𝑓𝐿𝑝=𝑃(Ω)𝜆1/𝑝Ω|𝑓𝐸0𝑓|𝑝𝑑𝑃1/𝑝.(3.29) Then by the definition of the norms and the assumption 0={,Ω} we see that 𝑓𝐿𝑝,𝜆,=𝑓𝐿𝑝,𝜆=𝑓𝐿𝑝𝑓𝑝,𝜆𝑓𝑝,𝜆,. By the same observation as Remark 2.5 we have 𝑓𝑝,𝜆,2𝑓𝐿𝑝,𝜆,.
(ii) Let 𝜆0. By Remark 2.10 we need to show only 𝐿𝑝,𝜆𝐿𝑝,𝜆, with 𝑓𝐿𝑝,𝜆,𝑓𝐿𝑝,𝜆 and 𝑝,𝜆𝑝,𝜆,𝐿0𝑝 with 𝑓𝐿𝑝𝑓𝑝,𝜆,𝑓𝑝,𝜆.
Let 𝑓𝐿𝑝,𝜆. For any 𝐵𝑛, there exists a sequence of atoms 𝐵𝐴(𝑛), =1,2,, such that 𝐵=𝐵 and 𝑃(𝐵)=𝑃(𝐵). Then 𝐵||𝑓||𝑝𝑑𝑃=𝐵||𝑓||𝑝𝑑𝑃𝑃𝐵𝜆𝑝+1𝑓𝑝𝐿𝑝,𝜆𝐵𝑃𝜆𝑝+1𝑓𝑝𝐿𝑝,𝜆,(3.30) since 𝜆𝑝+11. Therefore 𝑓𝐿𝑝,𝜆, and 𝑓𝐿𝑝,𝜆,𝑓𝐿𝑝,𝜆. Similarly, we have 𝑓𝑝,𝜆,𝑓𝑝,𝜆 for 𝑓𝑝,𝜆. By the definition of 𝑝,𝜆, norm we have 𝑓𝐿𝑝=𝑓𝐸0𝑓𝐿𝑝𝑃(Ω)𝜆+1/𝑝𝑓𝑝,𝜆,=𝑓𝑝,𝜆, for 𝑓𝑝,𝜆,.

Proof of Theorem 3.1. (i) We have the conclusion by Proposition 3.8 without the assumption that {𝑛}𝑛0 is regular or that (Ω,,𝑃) is nonatomic.
(ii) By Proposition 3.6 we only need to prove 𝐿𝑝,𝜆=𝑝,𝜆 with (1/2)𝑓𝑝,𝜆𝑓𝐿𝑝,𝜆𝐶𝑓𝑝,𝜆. The first norm inequality follows from Remark 2.5. We show the second one. Note that we do not need the assumption that 0={,Ω} or that (Ω,,𝑃) is nonatomic.
Let 𝑓𝑝,𝜆. Then, for any 𝐵𝐴(𝑛), 1𝑃(𝐵)𝐵|𝑓|𝑝𝑑𝑃1/𝑝1𝑃(𝐵)𝐵|𝑓𝐸𝑛𝑓|𝑝𝑑𝑃1/𝑝+||||1𝑃(𝐵)𝐵||||𝑓(𝜔)𝑑𝑃𝑃(𝐵)𝜆𝑓𝑝,𝜆+||||1𝑃(𝐵)𝐵||||,𝑓(𝜔)𝑑𝑃(3.31) since 𝐸𝑛1𝑓=𝑃(𝐵)𝐵𝑓(𝜔)𝑑𝑃on𝐵.(3.32) If 𝐵0, then 𝐸𝑛𝑓=0 on 𝐵. Assume that 𝐵0. By Lemma 3.3 we can choose 𝐵𝑘𝑗𝐴(𝑘𝑗), 0=𝑘0<𝑘1<<𝑘𝑚𝑛, such that 𝐵𝑘0𝐵𝑘1𝐵𝑘2𝐵𝑘𝑚=𝐵 and that (1+1/𝑅)𝑃(𝐵𝑘𝑗)𝑃(𝐵𝑘𝑗1)𝑅𝑃(𝐵𝑘𝑗). Then, since 1𝑃𝐵𝑘0𝐵𝑘0𝑓(𝜔)𝑑𝑃=0,(3.33) we have 1𝑃(𝐵)𝐵1𝑓(𝜔)𝑑𝑃=𝑃(𝐵)𝐵1𝑓(𝜔)𝑑𝑃𝑃𝐵𝑘0𝐵𝑘0=𝑓(𝜔)𝑑𝑃𝑚𝑗=11𝑃𝐵𝑘𝑗𝐵𝑘𝑗1𝑓(𝜔)𝑑𝑃𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1=𝑓(𝜔)𝑑𝑃𝑚𝑗=11𝑃𝐵𝑘𝑗𝐵𝑘𝑗𝑓𝐸𝑘𝑗1𝑓(𝜔)𝑑𝑃.(3.34) By Hölder’s inequality and the assumption 𝜆<0 we have ||||1𝑃(𝐵)𝐵𝑓||||(𝜔)𝑑𝑃𝑚𝑗=11𝑃𝐵𝑘𝑗𝐵𝑘𝑗|||𝑓𝐸𝑘𝑗1𝑓|||𝑝𝑑𝑃1/𝑝𝑚𝑗=11𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1|||𝑓𝐸𝑘𝑗1𝑓|||𝑝𝑑𝑃1/𝑝𝑚𝑗=1𝑃𝐵𝑘𝑗1𝜆𝑓𝑝,𝜆𝑚𝑗=111+R𝑚𝑗+1𝑃𝐵𝑘𝑚𝜆𝑓𝑝,𝜆𝑃(𝐵)𝜆𝑓𝑝,𝜆.(3.35) Therefore we have 𝑓𝐿𝑝,𝜆𝑓𝑝,𝜆.
(iii) Let 𝜆=0. By Remark 2.6 and Proposition 3.6 we have 𝐿0𝐿𝑝,0 with 𝑓𝐿𝑝,0𝑓𝐿 and 𝐿0𝑝,0.
Next we show 𝐿𝑝,0𝐿0 and 𝑓𝐿𝑓𝐿𝑝,0.
Let 𝑓𝐿𝑝,0 and 𝑓0 a.e. Take a positive number 𝑟 such that 𝑃(|𝑓|>𝑟)>0. For any 𝜖>0, there exists 𝑛 and 𝐵𝑛 such that 𝑃||𝑓||>𝐵>𝑟(1𝜖)𝑃(𝐵),(3.36) because is generated by 𝑛𝑛. For the above 𝐵, we can take a sequence of atoms 𝐵𝐴(𝑛), =1,2,, such that 𝐵=𝐵 and 𝑃(𝐵)=𝑃(𝐵). Hence, by the pigeonhole principle, there exists 𝐵𝐴(𝑛) such that 𝑃𝐵||𝑓||>𝐵>𝑟(1𝜖)𝑃.(3.37) Therefore, we have 𝑓𝑝𝐿𝑝,01𝑃𝐵𝐵||𝑓||𝑝1𝑑𝑃𝑃𝐵𝐵{|𝑓|>𝑟}||𝑓||𝑝𝑃𝐵𝑑𝑃||𝑓||>𝑟𝑃𝐵𝑟𝑝(1𝜖)𝑟𝑝.(3.38) This shows that 𝑃(|𝑓|>𝑟)>0 implies 𝑓𝐿𝑝,0𝑟. Then we have the conclusion.
By Proposition 3.8 and Remark 2.11 we have that BMO(=1,𝜆)=𝑝,𝜆 with equivalent norms.
(iv) Let 𝜆>0. For 𝑓𝐿𝑝,𝜆, we suppose that there exists 𝑟>0 such that 𝑃(|𝑓|>𝑟)>0. Then, for any 𝜖>0, we have the same estimate as (3.37). Moreover, we can decompose 𝐵 in (3.37) to atoms in 𝐴(𝑛+1) and we have the same estimate as (3.37) for some atom in 𝐴(𝑛+1). Therefore, we can take an atom 𝐵 in 𝐴(𝑛+𝑚) for large enough 𝑚 such that 𝐵 satisfies (3.37) and 𝑃(𝐵)<𝜖. Hence we have 𝑓𝑝𝐿𝑝,𝜆1𝑃𝐵1+𝜆𝑝𝐵||𝑓||𝑝𝑑𝑃(1𝜖)𝑟𝑝𝑃𝐵𝜆𝑝(1𝜖)𝑟𝑝𝜀𝜆𝑝,(3.39) that is, 𝑟𝜖𝜆/(1𝜖)1/𝑝𝑓𝐿𝑝,𝜆. This contradicts that 𝑟>0. Then 𝑓=0 a.e. and 𝐿𝑝,𝜆={0}. By Proposition 3.6 we have that {0}𝑝,𝜆.
Finally, by Proposition 3.8 and Remark 2.11 we have that 1,𝜆=𝑝,𝜆 with equivalent norms.

Next we prove that 𝐿𝑝,𝜆,𝐿𝑝,𝜆 and 𝑝,𝜆,𝑝,𝜆 in general by an example.

Proposition 3.9. Let (Ω,,𝑃) be as follows: [Ω=0,1),𝐴𝑛=𝐼𝑛,𝑗=𝑗2𝑛,(𝑗+1)2𝑛)𝑗=0,1,,2𝑛1,(3.40)𝑛𝐴=𝜎𝑛,=𝜎𝑛𝑛,𝑃=theLebesguemeasure.(3.41) If 1/𝑝<𝜆<0, then 𝐿𝑝,𝜆,𝐿𝑝,𝜆 and 𝑝,𝜆,𝑝,𝜆.

Proof. We construct 𝑓 such that 𝑓𝐿𝑝,𝜆𝐿𝑝,𝜆,,𝑓𝑝,𝜆𝑝,𝜆,.(3.42)
Step 1. Denote the characteristic function of 𝐼𝑛,𝑗 by 𝜒𝑛,𝑗 and let 𝑓𝑛=2𝑛1𝑗=0𝑓𝑛+𝑚,2𝑚𝑗,𝑓𝑛,𝑗𝐼=𝑃𝑛,𝑗𝜆𝜒𝑛+1,2𝑗𝜒𝑛+1,2𝑗+1,(3.43) where we choose 𝑚 such that 𝑃𝐼𝑛+𝑚,0𝑝𝜆+1𝐼𝑃𝑛,0.(3.44)
Note that 𝐼𝑛,𝑗=𝐼𝑛+1,2𝑗𝐼𝑛+1,2𝑗+1 and |𝑓𝑛,𝑗|=𝑃(𝐼𝑛,0)𝜆𝜒𝑛,𝑗.
If 𝑘𝑛+𝑚, then 𝐸𝑘𝑓𝑛=0, and |𝑓𝑛𝐸𝑘𝑓𝑛|=|𝑓𝑛|=𝑃(𝐼𝑛+𝑚,0)𝜆2𝑛1𝑗=1𝜒𝑛+𝑚,2𝑚𝑗. Hence1𝑃𝐼𝑘,𝐼𝑘,||𝑓𝑛𝐸𝑘𝑓𝑛||𝑝𝑑𝑃1/𝑝=1𝑃(𝐼𝑘,)𝐼𝑘,||𝑓𝑛||𝑝𝑑𝑃1/𝑝𝐼=𝑃𝑛+𝑚,0𝜆𝑗𝐼𝑚𝑗𝑛+𝑚,2𝐼𝑘,𝑃𝐼𝑛+𝑚,2𝑚𝑗𝑃𝐼𝑘,1/𝑝.(3.45)
If 𝑘𝑛, then the number of the elements of {𝑗𝐼𝑛+𝑚,2𝑚𝑗𝐼𝑘,} is the same as of {𝑗𝐼𝑛,𝑗𝐼𝑘,}. Hence 1𝑃(𝐼𝑘,)𝐼𝑘,||𝑓𝑛𝐸𝑘𝑓𝑛||𝑝𝑑𝑃1/𝑝=1𝑃𝐼𝑘,𝐼𝑘,||𝑓𝑛||𝑝𝑑𝑃1/𝑝𝐼𝑃𝑛+𝑚,0𝜆𝑃𝐼𝑛+𝑚,0𝑃𝐼𝑛,01/𝑝𝐼1𝑃𝑘,𝜆,(3.46) where we use (3.44) and 𝜆<0 for the last two inequalities. If 𝑛<𝑘𝑛+𝑚, then the number of the elements of {𝑗𝐼𝑛+𝑚,2𝑚𝑗𝐼𝑘,} is one at most. Hence 1𝑃𝐼𝑘,𝐼𝑘,||𝑓𝑛𝐸𝑘𝑓𝑛||𝑝𝑑𝑃1/𝑝=1𝑃𝐼𝑘,𝐼𝑘,||𝑓𝑛||𝑝𝑑𝑃1/𝑝𝐼𝑃𝑛+𝑚,0𝜆𝑃𝐼𝑛+𝑚,0𝑃𝐼𝑘,1/𝑝𝐼𝑃𝑘,𝜆,(3.47) where we use 1/𝑞<𝜆<0 for the last inequality. In the above, if 𝐼𝑘,=𝐼𝑛+𝑚,0, then the equality holds.
If 𝑘>𝑛+𝑚, then 𝑓𝑛𝐸𝑘𝑓𝑛=0 and 1𝑃𝐼𝑘,𝐼𝑘,||𝑓𝑛||𝑝𝑑𝑃1/𝑝𝐼𝑃𝑛+𝑚,0𝜆𝐼𝑃𝑘,0𝜆.(3.48) Therefore, 𝑓𝑛𝐿𝑝,𝜆𝑝,𝜆,𝑓𝑛𝐿𝑝,𝜆=𝑓𝑛𝑝,𝜆=1.(3.49) On the other hand, for the set 𝐵=2𝑛1𝑗=0𝐼𝑛+𝑚,2𝑚𝑗𝑛+𝑚, 1𝑃(𝐵)𝐵||𝑓𝑛𝐸𝑛+𝑚𝑓𝑛||𝑝𝑑𝑃1/𝑝=1𝑃(𝐵)𝐵||𝑓𝑛||𝑝𝑑𝑃1/𝑝𝐼=𝑃𝑛+𝑚,0𝜆=(2𝑛𝑃(𝐵))𝜆.(3.50)
Therefore, 𝑓𝑛𝐿𝑝,𝜆,,𝑓𝑛𝑝,𝜆,2𝑛𝜆 as 𝑛.
Step 2. Let 𝑓𝑛 be as in Step 1. If 𝑘<𝑛, then, by the same observation as Step 1 we also have that 𝑓𝑛𝜒𝑘,𝐿𝑝,𝜆𝑝,𝜆,𝑓𝑛𝜒𝑘,𝐿𝑝,𝜆=𝑓𝑛𝜒𝑘,𝑝,𝜆=1.(3.51) Moreover, if 𝑘<𝑛, then, for the set 𝐵=(2𝑛1𝑗=0𝐼𝑛+𝑚,2𝑚𝑗)𝐼𝑘,𝑛+𝑚, 1𝑃(𝐵)𝐵||𝑓𝑛𝜒𝑘,𝐸𝑛+𝑚𝑓𝑛𝜒𝑘,||𝑝𝑑𝑃1/𝑝=1𝑃(𝐵)𝐵||𝑓𝑛𝜒𝑘,||𝑝𝑑𝑃1/𝑝𝐼=𝑃𝑛+𝑚,0𝜆=2𝑛+𝑘𝑃(𝐵)𝜆.(3.52) Therefore, 𝑓𝑛𝜒𝑘,𝐿𝑝,𝜆,,𝑓𝑛𝜒𝑘,𝑝,𝜆,2(𝑛+𝑘)𝜆.Step 3. Let 𝑓𝑛 be as in Step 1 and let 𝑓=𝑛=12𝑛𝜆/2𝑓2𝑛𝜒𝑛,1.(3.53) Note that 𝑛=1𝜒𝑛,11. Then 𝑓𝐿𝑝,𝜆,𝑓𝑝,𝜆𝑛=12𝑛𝜆/2=2𝜆/212𝜆/2.(3.54) On the other hand, by the same observation as Step 2, we have that 𝑓𝐿𝑝,𝜆,,𝑓𝑝,𝜆,2𝑛𝜆/2×2(2𝑛+𝑛)𝜆=2𝑛𝜆/2,(3.55) for all 𝑛. This shows (3.42).

At the end of this section we prove the relation of 𝑓𝐿1,𝜆 and 𝑓𝑊𝐿𝑝. Recall that𝑓𝑊𝐿𝑝=sup𝑡>0||𝑓||𝑡𝑃>𝑡1/𝑝.(3.56)

The following is well known for classical Morrey spaces on 𝑛. We give a proof for convenience, though it is the same as the proof for classical Morrey spaces on 𝑛.

Proposition 3.10. If 1<𝑞<𝑝< and 1/𝑝=𝜆, then 𝑓𝐿1,𝜆𝑓𝐿𝑞,𝜆𝐶𝑓𝑊𝐿𝑝.(3.57)

Proof. The first inequality followed from Hölder’s inequality. To show the second inequality, we assume that 𝑓𝑊𝐿𝑝=1 and prove that 𝑓𝐿𝑞,𝜆𝐶. From 𝑓𝑊𝐿𝑝=1, we have 𝑃(|𝑓|>𝑡)𝑡𝑝. For any atom 𝐵𝐴(𝑛), let 𝜂=𝑃(𝐵)𝜆=𝑃(𝐵)1/𝑝 and 𝑓=𝑓𝜂+𝑓𝜂,𝑓𝜂||||||||(𝜔)=𝑓(𝜔),𝑓(𝜔)>𝜂,0,𝑓(𝜔)𝜂.(3.58) Then 1𝑃(𝐵)𝑞𝜆+1𝐵||𝑓𝜂||(𝜔)𝑞1𝑑𝑃𝑃(𝐵)𝑞𝜆+10𝑃||𝑓𝜂||>𝑡𝑞𝑡𝑞11𝑑𝑡𝑃(𝐵)𝑞𝜆+1𝜂0𝑃||𝑓𝜂||>𝜂𝑞𝑡𝑞1𝑑𝑡+𝜂𝑡𝑝𝑞𝑡𝑞11𝑑𝑡𝑃(𝐵)𝑞𝜆+1𝑞1+𝜂𝑝𝑞𝑞𝑝=𝑝.𝑝𝑞(3.59) Since |𝑓𝜂(𝜔)|𝜂 we have 1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵||𝑓𝜂||(𝜔)𝑞𝑑𝑃1/𝑞𝜂𝑃(𝐵)𝜆=1.(3.60) We get the conclusion.

4. Maximal Function

It is known as Doob’s inequality that (see for example [5, Pages 20-21])𝑓𝐿𝑝𝑝𝑝1𝑓𝐿𝑝,𝑓𝐿𝑝(𝑝>1),(4.1)𝑓𝑊𝐿1𝑓𝐿1,𝑓𝐿1.(4.2)

In this section we extend (4.1) and (4.2) to Morrey norms. Note that we do not need the regularity of the stochastic basis {𝑛}𝑛0.

Theorem 4.1. Let 1𝑝< and 1/𝑝𝜆<0. Then, for 𝑓𝐿1, 𝑓𝐿𝑝,𝜆𝐶𝑝𝑓𝐿𝑝,𝜆,if𝑝>1,𝑓𝑊𝐿1,𝜆𝐶1𝑓𝐿1,𝜆,if𝑝=1.(4.3)

Proof. Case 1 (𝑝>1). For any 𝐵𝐴(𝑚) and 𝑚0, let 𝑓=𝑔+ and 𝑔=𝑓𝜒𝐵. Then, using (4.1), we have 𝐵𝑔𝑝𝑑𝑃Ω𝑔𝑝𝑑𝑃Ω||𝑔||𝑝𝑑𝑃=𝐵||𝑓||𝑝𝑑𝑃.(4.4) Hence 1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵𝑔𝑝𝑑𝑃1/𝑝𝑓𝐿𝑝,𝜆.(4.5)
Next, take 𝐵𝑛𝐴(𝑛), 𝑛=0,1,,𝑚, such that 𝐵=𝐵𝑚𝐵𝑚1𝐵0. Then, for a.e. 𝜔𝐵, 𝐸𝑛1(𝜔)=0,(𝑛𝑚),𝑃𝐵𝑛𝐵𝑛𝑑𝑃,(𝑛<𝑚).(4.6) If 𝑛<𝑚, then ||𝐸𝑛||1(𝜔)𝑃𝐵𝑛𝐵𝑛||||𝑝𝑑𝑃1/𝑝𝐵𝑃𝑛𝜆𝑓𝐿𝑝,𝜆𝑃(𝐵)𝜆𝑓𝐿𝑝,𝜆,(4.7) since 𝜆<0. Hence 𝑃(𝐵)𝜆𝑓𝐿𝑝,𝜆on𝐵.(4.8)
Then 1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵𝑝𝑑𝑃1/𝑝𝑓𝐿𝑝,𝜆.(4.9)
By (4.5), (4.9), and the inequality 𝑓𝑔+, we have 1𝑃(𝐵)𝜆1𝑃(𝐵)𝐵𝑓𝑝𝑑𝑃1/𝑝𝑓𝐿𝑝,𝜆,(4.10) which shows the conclusion.
Case 2 (𝑝=1). Let 𝑔 and be as in Case 1. Using (4.2), we have, for all 𝑡>0, 𝑔𝑡𝑃𝐵𝑔>𝑡𝑡𝑃>𝑡Ω||𝑔||𝑑𝑃=𝐵||𝑓||𝑑𝑃(4.11) and then 1𝑃(𝐵)𝜆𝑔𝑡𝑃𝐵>𝑡𝑃(𝐵)𝑓𝐿1,𝜆.(4.12) We also have (4.8) for the case 𝑝=1. Then 1𝑃(𝐵)𝜆𝑡𝑃𝐵>𝑡𝑃(𝐵)𝑓𝐿1,𝜆.(4.13) Therefore we have the conclusion.

5. Fractional Integrals

In this section we establish the boundedness of the fractional integrals. To do this we first prove norm inequalities for functions, and then we get the boundedness of 𝐼𝛼 as a martingale transform.

For normed spaces 𝑀1 and 𝑀2 of functions, we denote by 𝐵(𝑀1,𝑀2) the set of all bounded martingale transforms from 𝑀1 to 𝑀2, that is, 𝐼𝛼𝐵(𝑀1,𝑀2) means thatsup𝑛0𝐼𝛼𝑓𝑛𝑀2𝐶sup𝑛0𝑓𝑛𝑀1,(5.1) for all 𝑀1-bounded martingales 𝑓=(𝑓𝑛)𝑛0.

We state our results in Section 5.1. To prove Theorems 5.1 and 5.5 we show the pointwise estimate for 𝐼𝛼𝑓 with the assumptions 𝑓𝐿𝑝< and 𝑓𝐿𝑝,𝜆<, respectively. To avoid repetition we prove first Theorem 5.5 in Section 5.2 and then Theorem 5.1 in Section 5.3. The proof of Theorem 5.8 is in Section 5.4.

5.1. Boundedness of Fractional Integrals

The following is for 𝐿𝑝-𝐿𝑞 boundedness.

Theorem 5.1. Assume that {𝑛}𝑛0 is regular. Let 1𝑝<𝑞<, 1/𝑝+𝛼=1/𝑞. Then, for 𝑓𝐿1, 𝐼𝛼𝑓𝐿𝑞𝐶𝑓𝐿𝑝𝐼,if𝑝>1,(5.2)𝛼𝑓𝑊𝐿𝑞𝐶𝑓𝐿1,if𝑝=1.(5.3)

Remark 5.2. Let 𝑓=(𝑓𝑛)𝑛0 be an 𝐿1-bounded martingale. For each 𝐿1-function 𝑓𝑚, 𝑚0, consider the corresponding martingale 𝑓𝑚=(𝐸𝑛[𝑓𝑚])𝑛0=(𝑓min(𝑛,𝑚))𝑛0. Then (𝐼𝛼𝑓𝑚)𝑛=(𝐼𝛼𝑓)𝑛 for 𝑚𝑛. Therefore, from (5.3) it follows that, for 𝑚𝑛, 𝐼𝛼𝑓𝑛𝑊𝐿𝑞=𝐼𝛼𝑓𝑚𝑛𝑊𝐿𝑞𝑓𝑚𝐿1sup𝑛0𝑓𝑛𝐿1.(5.4) This shows that sup𝑛0𝐼𝛼𝑓𝑛𝑊𝐿𝑞𝐶sup𝑛0𝑓𝑛𝐿1.(5.5)

Remark 5.3. Let a martingale 𝑓=(𝑓𝑛)𝑛0 be 𝐿1-bounded. Since 𝑊𝐿𝑞𝐿𝑞1 and (𝐼𝛼𝑓)𝑛𝐿𝑞1(𝐼𝛼𝑓)𝑛𝑊𝐿𝑞 for 1<𝑞1<𝑞, from Remark 5.2 it follows that the martingale ((𝐼𝛼𝑓)𝑛)𝑛0 is 𝐿𝑞1-bounded and that it converges in 𝐿𝑞1. Denote this limit by 𝐼𝛼𝑓. Then, 𝐼𝛼𝑓=𝑘=0𝑏𝛼𝑘1𝑑𝑘𝑓 and 𝐸𝑛[𝐼𝛼𝑓]=(𝐼𝛼𝑓)𝑛.
If 𝑝>1, then the same observation as in Remark 5.2 with (5.2) shows that sup𝑛0𝐼𝛼𝑓𝑛𝐿𝑞𝐶sup𝑛0𝑓𝑛𝐿𝑝.(5.6) Hence we have the following corollary.

Corollary 5.4. Assume that {𝑛}𝑛0 is regular. Let 1𝑝<𝑞<, 1/𝑝+𝛼=1/𝑞. Then 𝐼𝛼𝐿𝐵𝑝,𝐿𝑞𝐼,if𝑝>1,𝛼𝐿𝐵1,𝑊𝐿𝑞,if𝑝=1.(5.7)

For Morrey norms, one has the following.

Theorem 5.5. Assume that {𝑛}𝑛0 is regular. Let 1𝑝<, 1/𝑝𝜆<0, and 𝜇=𝜆+𝛼<0. Then, for 𝑓𝐿1, 𝐼𝛼𝑓𝐿𝑞,𝜇𝐶𝑓𝐿𝑝,𝜆𝐼,if𝑝>1,1𝑞(𝜆/𝜇)𝑝,(5.8)𝛼𝑓𝐿𝑞,𝜇𝐶𝑓𝐿1,𝜆𝐼,if𝑝=1,1𝑞<(𝜆/𝜇)𝑝,(5.9)𝛼𝑓𝑊𝐿𝑞,𝜇𝐶𝑓𝐿1,𝜆,if𝑝=1,𝑞=(𝜆/𝜇)𝑝.(5.10)

Note that Theorem 5.1 is not a corollary of Theorem 5.5, since 𝑓𝐿𝑝𝑓𝐿𝑝,1/𝑝𝑓𝐿𝑝 in general.

Remark 5.6. In order to prove (5.8) it suffices to prove it in the case where 𝑞=(𝜆/𝜇)𝑝, since 𝑓𝐿𝑞1,𝜆𝑓𝐿𝑞,𝜆 for 𝑞1𝑞 by Hölder’s inequality. The inequality (5.9) follows from (5.10), since 𝑓𝐿𝑞1,𝜆𝑓𝑊𝐿𝑞,𝜆 for 𝑞1<𝑞.

By the same observation as in Remark 5.2 we have the following.

Corollary 5.7. Assume that {𝑛}𝑛0 is regular. Let 1𝑝<, 1/𝑝𝜆<0, and 𝜇=𝜆+𝛼<0. Then 𝐼𝛼𝐿𝐵𝑝,𝜆,𝐿𝑞,𝜇𝐼,if𝑝>1,1𝑞(𝜆/𝜇)𝑝,𝛼𝐿𝐵1,𝜆,𝐿𝑞,𝜇𝐼,if𝑝=1,1𝑞<(𝜆/𝜇)𝑝,𝛼𝐿𝐵1,𝜆,𝑊𝐿𝑞,𝜇,if𝑝=1,𝑞=(𝜆/𝜇)𝑝.(5.11)

For Campanato spaces, one has the following.

Theorem 5.8. Assume that {𝑛}𝑛0 is regular. Let 1𝑝<, 𝜆1/𝑝, 1𝑞<, and 𝜇=𝜆+𝛼0. Then 𝐼𝛼𝑓𝑞,𝜇𝐶𝑓𝑝,𝜆,𝑓𝐿1,(5.12) and 𝐼𝛼𝐵𝑝,𝜆,𝑞,𝜇,(5.13) where 𝐼𝛼𝑓 in (5.12) denotes the limit function described in Remark 5.3, for 𝑓=(𝐸𝑛𝑓)𝑛0, 𝑓𝐿1.

Remark 5.9. If 𝜇<0, then, by Theorem 3.1 (ii), the inequalities (5.8) and (5.9) in Theorem 5.5 hold with 𝐿𝑝,𝜆 and 𝐿𝑞,𝜇 replaced by 𝑝,𝜆 and 𝑞,𝜇, respectively.
If 1<𝑝< and 1/𝑝=𝜆, then 𝑓1,𝜆𝑓𝐿1,𝜆𝑓𝑊𝐿𝑝𝑓𝐿𝑝 by Remark 2.5 and Proposition 3.10. Since 𝐿𝑝𝐿1,𝜆 in general, the following is an improvement of the results in [21].

Corollary 5.10. Assume that {𝑛}𝑛0 is regular. Then, for 𝑓𝐿1, 𝐼𝛼𝑓BMO𝐶𝑓𝐿1,1/𝑝1,if𝑝𝐼+𝛼=0,1<𝑝<,𝛼𝑓Lip𝛽𝐶𝑓𝐿1,1/𝑝1,if𝑝𝐼+𝛼=𝛽>0,1<𝑝<,𝛼𝑓Lip𝛼𝐶𝑓BMO,𝐼𝛼𝑓Lip𝛾𝐶𝑓Lip𝛽𝐼,if𝛽+𝛼=𝛾,𝛽>0,(5.14)𝛼𝐿𝐵1,1/𝑝1,BMO,if𝑝𝐼+𝛼=0,1<𝑝<,𝛼𝐿𝐵1,1/𝑝,Lip𝛽1,if𝑝𝐼+𝛼=𝛽>0,1<𝑝<,𝛼𝐵BMO,Lip𝛼,𝐼𝛼𝐵Lip𝛽,Lip𝛾,if𝛽+𝛼=𝛾,𝛽>0.(5.15)

5.2. Proof of Theorem 5.5

By Remark 5.6 we only need to prove (5.8) and (5.10) in the case 𝑞=(𝜆/𝜇)𝑝. So we always assume that 𝑞=(𝜆/𝜇)𝑝 in this proof.

First we show the pointwise estimate𝐼𝛼𝑓(𝜔)𝐶𝑓(𝜔)𝜇/𝜆𝑓𝐿𝛼/𝜆𝑝,𝜆,(5.16)

where 𝐶 is a positive constant independent of 𝜔. To do this we prove that, for any 𝑚1 and any 𝐵𝑚𝐴(𝑚),||𝐼𝛼𝑓𝑚||(𝜔)𝐶𝑓(𝜔)𝜇/𝜆𝑓𝐿𝛼/𝜆𝑝,𝜆,𝜔𝐵𝑚.(5.17)

Take 𝐵𝑘𝐴(𝑘), 0𝑘<𝑚, such that 𝐵𝑚𝐵𝑚1𝐵0 and let𝐾=𝑘0<𝑘𝑚,𝐵𝑘𝐵𝑘1=𝑘1,𝑘2,,𝑘,(5.18)

where 0=𝑘0<𝑘1<𝑘2<<𝑘. Then, by Lemma 3.3,11+𝑅𝑏𝑘𝑗𝑏𝑘𝑗1𝑅𝑏𝑘𝑗on𝐵𝑚.(5.19)

Hence, we can write𝐼𝛼𝑓𝑚=0<𝑘𝑗𝑚𝑏𝛼𝑘𝑗1𝑑𝑘𝑗𝑓=𝑗=1𝑏𝛼𝑘𝑗1𝑑𝑘𝑗𝑓on𝐵𝑚,(5.20)

since 𝑏𝑘=𝑏𝑘1 and 𝑑𝑘𝑓=0 for 𝑘𝐾. Note that for 𝜔𝐵𝑚,|||𝑑𝑘𝑗|||=|||𝑓𝑓(𝜔)𝑘𝑗(𝜔)𝑓𝑘𝑗1||||||𝑓(𝜔)𝑘𝑗|||+|||𝑓(𝜔)𝑘𝑗1|||=||||||1(𝜔)𝑃𝐵𝑘𝑗𝐵𝑘𝑗||||||+||||||1𝑓𝑑𝑃𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1||||||1𝑓𝑑𝑃𝑃𝐵𝑘𝑗𝐵𝑘𝑗||𝑓||𝑝𝑑𝑃1/𝑝+1𝑃𝐵𝑘𝑗1𝐵𝑘𝑗1||𝑓||𝑝𝑑𝑃1/𝑝𝑃𝐵𝑘𝑗𝜆𝐵+𝑃𝑘𝑗1𝜆𝑓𝐿𝑝,𝜆11+𝑅𝜆𝑏𝑘𝑗1(𝜔)𝜆𝑓𝐿𝑝,𝜆.(5.21)

Then, for 𝜔𝐵𝑚 and for 0<𝑛𝑚,|||||𝑛𝑘=0𝑏𝑘1(𝜔)𝛼𝑑𝑘|||||1𝑓(𝜔)1+𝑅𝜆0<𝑘𝑗𝑛𝑏𝑘𝑗1(𝜔)𝜇𝑓𝐿𝑝,𝜆1+(1/𝑅)𝜆1(1+1/𝑅)𝜇𝑏𝑛(𝜔)𝜇𝑓𝐿𝑝,𝜆.(5.22)

On the other hand, for 0𝑛<𝑚, letting 𝑗(𝑛)=min{𝑗𝑛<𝑘𝑗}, we have, for 𝜔𝐵𝑚,𝑚𝑘=𝑛+1𝑏𝑘1(𝜔)𝛼𝑑𝑘𝑓(𝜔)=𝑗=𝑗(𝑛)𝑏𝑘𝑗1(𝜔)𝛼𝑑𝑘𝑗=𝑓(𝜔)𝑗=𝑗(𝑛)𝑏𝑘𝑗1(𝜔)𝛼𝑓𝑘𝑗(𝜔)𝑗=𝑗(𝑛)𝑏𝑘𝑗1(𝜔)𝛼𝑓𝑘𝑗1(𝜔)=𝑏𝑘1(𝜔)𝛼𝑓𝑘(𝜔)+1𝑗=𝑗(𝑛)𝑏𝑘𝑗1(𝜔)𝛼𝑏𝑘𝑗(𝜔)𝛼𝑓𝑘𝑗(𝜔)𝑏𝑘𝑗(𝑛)1(𝜔)𝛼𝑓𝑘𝑗(𝑛)1|||||(𝜔),(5.23)𝑚𝑘=𝑛+1𝑏𝑘1(𝜔)𝛼𝑑𝑘|||||𝑓(𝜔)𝑏𝑘1(𝜔)𝛼𝑓+(𝜔)1𝑗=𝑗(𝑛)|||𝑏𝑘𝑗1(𝜔)𝛼𝑏𝑘𝑗(𝜔)𝛼|||𝑓(𝜔)+𝑏𝑘𝑗(𝑛)1(𝜔)𝛼𝑓(𝜔)2𝑏𝑘𝑗(𝑛)1(𝜔)𝛼𝑓(𝜔)=2𝑏𝑛(𝜔)𝛼𝑓(𝜔).(5.24)

Here, letΩ1=𝑓𝜔Ω(𝜔)𝑓𝐿𝑝,𝜆1/𝜆𝑏0(𝜔),Ω2=ΩΩ1.(5.25)

If 𝜔Ω1𝐵𝑚 and𝑓(𝜔)𝑓𝐿𝑝,𝜆1/𝜆𝑏𝑚(𝜔),(5.26)

then, by (5.22) we have||𝐼𝛼𝑓𝑚||(𝜔)𝑏𝑚(𝜔)𝜇𝑓𝐿𝑝,𝜆𝑓(𝜔)𝑓𝐿𝑝,𝜆𝜇/𝜆𝑓𝐿𝑝,𝜆=𝑓(𝜔)𝜇/𝜆𝑓𝐿𝛼/𝜆𝑝,𝜆,(5.27)

since 𝜇<0 and 1𝜇/𝜆=𝛼/𝜆. If 𝜔Ω1𝐵𝑚 and𝑏𝑚𝑓(𝜔)<(𝜔)𝑓𝐿𝑝,𝜆1/𝜆,(5.28)

choosing 𝑛 such that𝑅1𝑏𝑛𝑓(𝜔)<(𝜔)𝑓𝐿𝑝,𝜆1/𝜆𝑏𝑛(𝜔),(5.29)

we have by (5.22) and (5.24)||𝐼𝛼𝑓𝑚||(𝜔)𝑏𝑛(𝜔)𝜇𝑓𝐿𝑝,𝜆+𝑏𝑛(𝜔)𝛼𝑓𝑓(𝜔)(𝜔)𝑓𝐿𝑝,𝜆𝜇/𝜆𝑓𝐿𝑝,𝜆+𝑓(𝜔)𝑓𝐿𝑝,𝜆𝛼/𝜆𝑓(𝜔)𝑓(𝜔)𝜇/𝜆𝑓𝐿𝛼/𝜆𝑝,𝜆.(5.30)

If 𝜔Ω2𝐵𝑚, then by (5.24) we have||𝐼𝛼𝑓𝑚||=|||||(𝜔)𝑚𝑘=1𝑏𝑘1(𝜔)𝛼𝑑𝑘|||||𝑓(𝜔)𝑏0(𝜔)𝛼𝑓𝑓(𝜔)(𝜔)𝑓𝐿𝑝,𝜆𝛼/𝜆𝑓(𝜔)=𝑓(𝜔)𝜇/𝜆𝑓𝐿𝛼/𝜆𝑝,𝜆.(5.31)

Therefore, we have (5.17).

Next, applying the boundedness of the maximal function (Theorem 4.1), we show1𝑃(𝐵)𝐵𝐼𝛼𝑓(𝜔)𝑞𝑑𝑃1/𝑞𝑃(𝐵)𝜇𝑓𝐿𝑝,𝜆,if𝑝>1,(5.32)sup𝑡>0𝑡𝑃𝐼𝐵𝛼𝑓>𝑡𝑃(𝐵)1/𝑞𝑃(𝐵)𝜇𝑓𝐿1,𝜆,if𝑝=1,(5.33)

for any 𝑛 and any 𝐵𝐴(𝑛). To show (5.32), using (5.16) and the relations 𝑞𝜇/𝜆=𝑝 and 𝛼/𝜆=1𝑝/𝑞, we have1𝑃(𝐵)𝐵𝐼𝛼𝑓(𝜔)𝑞𝑑𝑃1/𝑞1𝑃(𝐵)𝐵𝑓(𝜔)𝑞𝜇/𝜆𝑑𝑃1/𝑞𝑓𝐿𝛼/𝜆𝑝,𝜆=1𝑃(𝐵)𝐵𝑓(𝜔)𝑝𝑑𝑃(1/𝑝)(𝑝/𝑞)𝑓𝐿1𝑝/𝑞𝑝,𝜆.(5.34)

Applying the boundedness of the maximal function, we have1𝑃(𝐵)𝐵𝑓(𝜔)𝑝𝑑𝑃(1/𝑝)(𝑝/𝑞)𝑃(𝐵)𝜆𝑓𝐿𝑝,𝜆𝑝/𝑞𝑃(𝐵)𝜇𝑓𝐿𝑝/𝑞𝑝,𝜆.(5.35)

Then we have (5.32). To show (5.33), using (5.16) and the relations 𝑝=1, 𝜇/𝜆=1/𝑞 and 𝛼/𝜆=11/𝑞, we havesup𝑡>0𝑡𝑃𝐼𝐵𝛼𝑓>𝑡𝑃(𝐵)1/𝑞sup𝑡>0𝑡𝑃𝑓𝐵𝜇/𝜆𝑓𝐿𝛼/𝜆1,𝜆>𝑡𝑃(𝐵)1/𝑞=sup𝑡>0𝑡𝑃𝑓𝐵1/𝑞𝑓𝐿11/𝑞1,𝜆>𝑡𝑃(𝐵)1/𝑞=sup𝑡>0𝑡𝑃𝑓𝐵1/𝑞>𝑡𝑃(𝐵)1/𝑞𝑓𝐿11/𝑞1,𝜆=sup𝑡>0𝑡𝑃𝑓𝐵>𝑡𝑃(𝐵)1/𝑞𝑓𝐿11/𝑞1,𝜆.(5.36)

Applying the boundedness of the maximal function, we havesup𝑡>0𝑡𝑃𝑓𝐵>𝑡𝑃(𝐵)1/𝑞𝑃(𝐵)𝜆𝑓𝑊𝐿1,𝜆1/𝑞𝑃(𝐵)𝜇𝑓𝐿1/𝑞𝑝,𝜆.(5.37)

Then we have (5.33). The proof is complete.

5.3. Proof of Theorem 5.1

Let 𝜆=1/𝑝 and 𝜇=1/𝑞. Then 𝜇/𝜆=𝑝/𝑞 and 𝛼/𝜆=(1/𝑝1/𝑞)𝑝=1𝑝/𝑞. In this case the pointwise estimate (5.16) implies𝐼𝛼𝑓(𝜔)𝐶𝑓(𝜔)𝑝/𝑞𝑓𝐿1𝑝/𝑞𝑝,1/𝑝𝐶𝑓(𝜔)𝑝/𝑞𝑓𝐿1𝑝/𝑞𝑝.(5.38)

Applying the boundedness of the maximal function on 𝐿𝑝, we get the conclusion.

5.4. Proof of Theorem 5.8

Note that, for 𝑓𝐿1, 𝑓𝑝,𝜆=𝑓𝐸0𝑓𝑝,𝜆 and 𝐼𝛼𝑓=𝐼𝛼(𝑓𝐸0𝑓). Then it is sufficient to prove (5.12) only for 𝑓𝐿01. Therefore, (5.12) follows from (5.13) and Proposition 2.2(ii).

By Hölder’s inequality we have 𝑓1,𝜆𝑓𝑝,𝜆 for 𝑝1. Furthermore, by Theorem 3.1(iii) and (iv) we have 𝐼𝛼𝑓𝑞,𝜇𝐼𝛼𝑓1,𝜇 for 𝑞1. So we only need to prove thatsup𝑚0𝐼𝛼𝑓𝑚1,𝜇𝐶sup𝑚0𝑓𝑚1,𝜆,(5.39)

for all 1,𝜆-bounded martingales 𝑓=(𝑓𝑚)𝑚0.

To prove (5.39) we show that, for any 𝑛, any 𝑚>𝑛 and any 𝐵𝑛𝐴(𝑛),𝐵𝑛||𝐼𝛼𝑓𝑚𝐼𝛼𝑓𝑛||𝐵𝑑𝑃𝑃𝑛1+𝜇𝑓𝑚1,𝜆.(5.40)

Note that𝐼𝛼𝑓𝑚𝐼𝛼𝑓𝑛=𝑚𝑘=𝑛+1𝑏𝛼𝑘1𝑑𝑘𝑓=𝑚1𝑘=𝑛+1𝑏𝛼𝑘1𝑏𝛼𝑘𝑓𝑘𝑓𝑛+𝑏𝛼𝑚1𝑓𝑚𝑓𝑛.(5.41)

Since 𝑏𝑚 is nonincreasing with respect to 𝑚, we have𝐵𝑛𝑏𝛼𝑚1||𝑓𝑚𝑓𝑛||𝐵𝑑𝑃𝑃𝑛𝛼𝐵𝑛||𝑓𝑚𝑓𝑛||𝐵𝑑𝑃𝑃𝑛1+𝜇𝑓𝑚1,𝜆.(5.42)

Therefore, to prove (5.40), we only need to show that𝑚𝑘=𝑛+1𝐵𝑛𝑏𝛼𝑘1𝑏𝛼𝑘||𝑓𝑘𝑓𝑛||𝐵𝑑𝑃𝑃𝑛1+𝜇𝑓𝑚1,𝜆𝑚>𝑛.(5.43)

Using the inequality |𝑓𝑘𝑓𝑛|𝐸𝑘|𝑓𝑚𝑓𝑛| for 𝑛<𝑘𝑚, we have𝐵𝑛𝑏𝛼𝑘1𝑏𝛼𝑘||𝑓𝑘𝑓𝑛||𝑑𝑃𝐵𝑛𝑏𝛼𝑘1𝑏𝛼𝑘𝐸𝑘||𝑓𝑚𝑓𝑛||𝑑𝑃=𝐵𝑛𝑏𝛼𝑘1𝑏𝛼𝑘||𝑓𝑚𝑓𝑛||𝑑𝑃.(5.44)

Therefore, we have𝑚𝑘=𝑛+1𝐵𝑛𝑏𝛼𝑘1𝑏𝛼𝑘||𝑓𝑘𝑓𝑛||𝑑𝑃𝑚𝑘=𝑛+1𝐵𝑛𝑏𝛼𝑘1𝑏𝛼𝑘||𝑓𝑚𝑓𝑛||=𝑑𝑃𝐵𝑛𝑏𝛼𝑛𝑏𝛼𝑚||𝑓𝑚𝑓𝑛||𝐵𝑑𝑃𝑃𝑛𝛼𝐵𝑛||𝑓𝑚𝑓𝑛||𝐵𝑑𝑃𝑃𝑛1+𝜇𝑓𝑚1,𝜆.(5.45)

We have obtained (5.43). The proof of Theorem 5.8 is completed.

Acknowledgments

The authors wish to express their deep thanks to the referees for their very careful reading and also their many valuable and suggested remarks, which led them to simplify the proofs of the main results and improve the presentation of this article. The first author was supported by Grant-in-Aid for Scientific Research (C), no. 20540167 and no. 24540159, Japan Society for the Promotion of Science. The second author was supported by Grant-in-Aid for Scientific Research (C), no. 24540171, Japan Society for the Promotion of Science.