Abstract

Let {𝜑𝛼𝑛}𝑛 be the Laguerre functions of Hermite type with index 𝛼. These are eigenfunctions of the Laguerre differential operator 𝐿𝛼=1/2(𝑑2/𝑑𝑦2+𝑦2+1/𝑦2(𝛼21/4)). In this paper, we investigate the boundedness of the Hardy-Littlewood maximal function, the heat maximal function, and the Littlewood-Paley 𝑔-function associated with 𝐿𝛼 in the localized BMO space BMO𝐿𝛼, which is the dual space of the Hardy space 𝐻1𝐿𝛼.

1. Introduction

Let 𝑛, 𝛼>1. The Laguerre function of Hermite type 𝜑𝛼 on (0,) is defined as 𝜑𝛼𝑛(𝑦)=Γ(𝑛+1)Γ(𝑛+1+𝛼)1/2𝑒𝑦2/2𝑦𝛼𝐿𝛼𝑛𝑦2(2𝑦)1/2,𝑦(0,),(1.1) where 𝐿𝛼𝑛(𝑥) denotes the Laguerre polynomial of degree 𝑛 and order 𝛼, see [1]. It is well known that for every 𝛼>1 the system {𝜑𝛼𝑛}𝑛=0 forms an orthonormal basis of 𝐿2(0,). Moreover, these functions are eigenfunctions of the Laguerre differential operator 𝐿𝛼=12𝑑2𝑑𝑦2+𝑦2+1𝑦2𝛼214(1.2) satisfying 𝐿𝛼𝜑𝛼𝑛=(2𝑛+𝛼+1)𝜑𝛼𝑛. The operator 𝐿𝛼 can be extended to a positive self-adjoint operator on 𝐿2(0,) by giving a suitable domain of definition, see [2]; we also denote the extension by 𝐿𝛼. Let {𝑇𝛼𝑡}𝑡0 be the heat-diffusion semigroup generated by 𝐿𝛼. More precisely, for 𝑓𝐿2(0,), we define 𝑇𝛼𝑡𝑓(𝑥)=0𝑊𝛼𝑡(𝑥,𝑦)𝑓(𝑦)𝑑𝑦,(1.3) where 𝑊𝛼𝑡(𝑥,𝑦)=2𝑒𝑡1𝑒2𝑡1/22𝑥𝑦𝑒𝑡1𝑒2𝑡1/2𝐼𝛼2𝑥𝑦𝑒𝑡1𝑒2𝑡1exp21+𝑒2𝑡1𝑒2𝑡𝑥2+𝑦2.(1.4)𝐼𝛼 is the modified Bessel function of the first kind and order 𝛼.

In [3], we introduced and developed a localized BMO space BMO𝐿𝛼 associated with the operator 𝐿𝛼, which is the dual space of the Hardy space 𝐻1𝐿𝛼 introduced by Dziubański [4]. More precisely, let 𝜌𝐿𝛼1(𝑥)=81min𝑥,𝑥,𝑥>0.(1.5)

Definition 1.1. Let 𝛼>1/2, 𝐵𝑠(𝑦) be any ball in (0,) with the center 𝑦 and the radius 𝑠 and 𝑓 a locally integrable function on (0,). We say 𝑓BMO𝐿𝛼 if there exists a constant 𝐶0 independent of 𝑠 and 𝑦 such that 1||𝐵𝑠||(𝑦)𝐵𝑠(𝑦)||𝑓𝑓𝐵𝑠(𝑦)||𝐶,if𝑠<𝜌𝐿𝛼(1𝑦),||𝐵𝑠||(𝑦)𝐵𝑠(𝑦)||𝑓||𝐶,if𝑠𝜌𝐿𝛼(𝑦).(1.6) Here, 𝑓𝐵𝑠(𝑦)=(1/|𝐵𝑠(𝑦)|)𝐵𝑠(𝑦)𝑓𝑑𝑥. We let 𝑓BMO𝐿𝛼 denote the smallest 𝐶 in the two inequalities above.

It is readily seen that BMO𝐿𝛼 is a Banach space with norm BMO𝐿𝛼.

In this paper, we obtain the boundedness on BMO𝐿𝛼 of several operators including the Hardy-Littlewood maximal operator defined on (0,), the heat maximal function, and the Littlewood-Paley 𝑔-function associated with 𝑇𝛼𝑡.

These results were investigated by Dziubański et al. in [5] for Schrödinger operators on 𝑑 with 𝑑3 and with potentials satisfying a reverse Hölder's inequality. Recently, a theory of localized BMO spaces on RD-spaces associated with an admissible function 𝜌 was investigated in [6]; the authors also established the similar results above for their BMO spaces. The admissible function 𝜌 in [6] is required to satisfy 1𝜌(𝑥)𝐶01𝜌(𝑦)1+𝑑(𝑥,𝑦)𝜌(𝑦)𝑘0.(1.7) Obviously, our 𝜌𝐿𝛼 in (1.5) does not satisfy this condition. Indeed, let 𝑥 tend to zero and 𝑦=1; then the left side becomes greater than the right.

It is notable the generalized square functions associated to Schrödinger operators are studied in [7]. The authors of [7] gave several of equivalent conditions for BMO-boundedness of square functions.

In this paper, in order to obtain some key estimates, we will employ the differences in integral kernels (the heat kernel, the 𝑔-function kernel) associated with the Hermite operator and the Laguerre operator, respectively (see [8, 9]).

The paper is organized as follows. In the next section we present some preliminary lemmas and collect some useful estimates of the kernels associated with the heat semigroups and the 𝑔-functions. In Section 3, we establish the boundedness of two maximal operators (the Hardy-Littlewood maximal operator and the heat maximal function) from BMO𝐿𝛼 to BMO𝐿𝛼. In Section 4, we obtain the boundedness on BMO𝐿𝛼 of the Littlewood-Paley 𝑔-function associated with the heat semigroup for 𝐿𝛼. We make some conventions. Throughout this paper by 𝐶 we always denote a positive constant that may vary at each occurrence; 𝐵𝑟(𝑦0) stands for {𝑦>0,|𝑦𝑦0|𝑟}; 𝐴𝐵 means (1/𝐶)𝐴𝐵𝐶𝐴, and the notation 𝑋𝑌 is used to indicate that 𝑋𝐶𝑌 with an independent positive constant 𝐶.

2. Preliminaries

Now we give the following covering lemma for (0,) which will be used frequently below. The proof is trivial and left to the reader.

Lemma 2.1. Let 𝑥0=1,   𝑥𝑗=𝑥𝑗1+𝜌𝐿𝛼(𝑥𝑗1) for 𝑗1, and 𝑥𝑗=𝑥𝑗+1𝜌𝐿𝛼(𝑥𝑗+1) for 𝑗<0. One defines the family of “critical balls” of ={𝐵𝑘}𝑘=, where 𝐵𝑘={𝑥(0,)|𝑥𝑥𝑘|<𝜌𝐿𝛼(𝑥𝑘)}. Then one has(a)𝑘=𝐵𝑘=(0,), (b)for every 𝑘, 𝐵𝑘𝐵𝑗= provided that 𝑗{𝑘1,𝑘,𝑘+1}, (c)for any 𝑦0(0,), at most three balls in have nonempty intersection with 𝐵(𝑦0,𝜌𝐿𝛼(𝑦0)).

Corollary 2.2. There exists a constant 𝐶>0 such that for every 𝐵𝑅(𝑥)(0,) with 𝑅>𝜌𝐿𝛼(𝑥), one has ||𝐵𝑅(||𝑥)𝐵𝑘𝐵𝑘𝐵𝑅(𝑥)||𝐵𝑘||||𝐵𝐶𝑅(||.𝑥)(2.1)

Corollary 2.3. There exists a constant 𝐶 such that, for 𝑓BMO𝐿𝛼, one has 𝑓BMO𝐿𝛼𝐶sup𝑘||𝑓||𝐵𝑘+𝑓BMO(𝐵𝑘),(2.2) where, for any ball 𝐵, the norm BMO(𝐵) is given by 𝑓BMO(𝐵)=sup𝐵𝑟(𝑥)𝐵1𝐵𝑟(𝑥)𝐵𝑟(𝑥)||𝑓𝑓𝐵𝑟(𝑥)||𝑑𝑦sup𝐵𝑟(𝑥)𝐵inf𝑐1𝐵𝑟(𝑥)𝐵𝑟(𝑥)||||𝑓𝑐𝑑𝑦.(2.3)

Corollary 2.4 (see [3, Corollary  3]). Let 𝐵=𝐵𝑟(𝑦0)(0,). There exists a constant 𝐶>0 such that, for all 𝑓BMO𝐿𝛼, one has (1)if 𝑟𝜌𝐿𝛼(𝑦0)/2, then ((1/|𝐵|)𝐵|𝑓(𝑥)|2𝑑𝑥)1/2𝐶𝑓BMO𝐿𝛼, (2)if 𝑟<𝜌𝐿𝛼(𝑦0)/2, then ((1/|𝐵|)𝐵|𝑓(𝑥)𝑓𝐵|2𝑑𝑥)1/2𝐶𝑓BMO𝐿𝛼. We give two elementary lemmas, which will be used frequently in next section. The proofs are trivial, and the reader also refer to Lemmas 9 and 2 in [5].

Lemma 2.5. Let BMO(𝐵𝑘) and 𝑔1 and 𝑔2 be functions in 𝐿(0,). If 𝑓 is any measurable function satisfying +𝑔1𝑓+𝑔2,a.e.,(2.4) then 𝑓BMO(𝐵𝑘) and 𝑓BMO(𝐵𝑘)BMO(𝐵𝑘)+max(𝑔1,𝑔2).

Lemma 2.6. For all 𝑓BMO𝐿𝛼 and 𝐵=𝐵𝑟(𝑦0) with 𝑟<𝜌𝐿𝛼(𝑦0). There exists a constant 𝐶>0 such that ||𝑓𝐵||𝜌𝐶1+log𝐿𝛼𝑦0𝑟𝑓BMO𝐿𝛼.(2.5)

Let 𝐻 be the Hermite operator 1𝐻=2𝑑2𝑑𝑥2+𝑥2.(2.6) One considers the heat diffusion semigroup {𝑊𝑡}𝑡>0 associated with 𝐻 and defined by, for every 𝑓𝐿2(), 𝑊𝑡𝑓(𝑥)=𝑊𝑡(𝑥,𝑦)𝑓(𝑦)𝑑𝑦,𝑥,(2.7) where for each 𝑥,𝑦 and 𝑡>0, 𝑊𝑡𝑒(𝑥,𝑦)=𝑡𝜋1𝑒2𝑡1/21exp21+𝑒2𝑡1𝑒2𝑡𝑥2+𝑦2𝑒+2𝑥𝑦𝑡1𝑒2𝑡(2.8) (see [10]).

Proposition 2.7. Let 𝛼>1/2, 𝑊𝑡(𝑥,𝑦) be in (2.8). There exists 𝐶>0 such that, for 𝑡>0, (a)𝑊𝛼𝑡(𝑥,𝑦)𝐶𝑦𝛼+1/2𝑥𝛼3/2, 0<𝑦<𝑥/2, (b)𝑊𝛼𝑡(𝑥,𝑦)𝐶𝑥𝛼+1/2𝑦𝛼3/2, 0<2𝑥<𝑦, (c)|𝑊𝛼𝑡(𝑥,𝑦)𝑊𝑡(𝑥,𝑦)|𝐶(1/𝑦), 𝑥/2<𝑦<2𝑥, (d)|𝑊𝛼𝑡(𝑥,𝑦)|𝐶(1/𝑡)𝑒|𝑥𝑦|2/10𝑡.

Parts (a), (b), and (c) are the contents of Lemma 2.11 in [8]. Part (d) is from (2.6) in [4].

Remark 2.8. The ranges 0<𝑦<𝑥/2 and 0<2𝑥<𝑦 are not critical; Proposition 2.7 also holds when 0<𝑦<𝑥/𝑐 and 0<𝑐𝑥<𝑦, where 𝑐>1.

Now we consider the estimates of the integral kernel for the 𝑔-function, which will be defined in Section 4: 𝑄𝑡(𝑥,𝑦)=𝑡2𝜕𝑊𝛼𝑠(𝑥,𝑦)||||𝜕𝑠𝑠=𝑡2,(2.9)𝑃𝑡(𝑥,𝑦)=𝑡2𝜕𝑊𝑠(𝑥,𝑦)||||𝜕𝑠𝑠=𝑡2.(2.10)

Proposition 2.9. One has,(a)for every 𝑡,𝑥,𝑦(0,) such that 𝑒𝑡2𝑥𝑦/(1𝑒2𝑡2)1, ||𝑄𝑡(||𝑥,𝑦)𝐶𝑡2(𝑥𝑦)(𝛼+1/2)𝑒(𝑥2+𝑦2)/8𝑡2𝑒(𝛼+1)𝑡21𝑒2𝑡2𝛼+2,(2.11)(b)for every 𝑡,𝑥,𝑦(0,) such that 𝑒𝑡2𝑥𝑦/(1𝑒2𝑡2)>1, ||𝑄𝑡(𝑥,𝑦)𝑃𝑡(||𝑥,𝑦)𝐶𝑡2𝑒(𝑥𝑦)2/2𝑡2𝑒𝑡2/2𝑥𝑦1𝑒𝑡21/2.(2.12)

Parts (a) and (b) are contained in [9, (3.4) and (3.6)].

Proposition 2.10. For every 𝑁1, there is a constant 𝐶𝑁 such that (a)if 𝑡>0, |𝑃𝑡(𝑥,𝑦)|𝐶𝑁𝑒𝑡2/8(1/𝑡)𝑒|𝑥𝑦|2/10𝑡2(1+𝑡|𝑥|)𝑁; (b)for ||𝑡, |𝑃𝑡(𝑥+,𝑦)𝑃𝑡(𝑥,𝑦)|𝐶(||/𝑡)(1/𝑡)𝑒|𝑥𝑦|2/20𝑡2, 𝐶 is independent of 𝑥,𝑦,𝑡; (c)|𝑃𝑡(𝑥,𝑦)𝑑𝑦|𝐶(𝑡/𝜌𝐿𝛼(𝑥))2, 𝐶 is independent of 𝑥 and 𝑡.

Proof. By using (2.8) we can write, for every 𝑥,𝑦 and 𝑠>0, 𝜕𝑊𝜕𝑠𝑠1(𝑥,𝑦)=2𝜋𝑒((𝑥𝑒𝑠𝑦)2+(𝑦𝑒𝑠𝑥)2)/2(1𝑒2𝑠)𝑒𝑠/21𝑒2𝑠3/2×1+𝑒2𝑠+2𝑒𝑠(𝑦(𝑥𝑒𝑠𝑦)+𝑥(𝑦𝑒𝑠𝑥))2𝑒2𝑠(𝑥𝑒𝑠𝑦)2+(𝑦𝑒𝑠𝑥)21𝑒2𝑠.(2.13)
By the simple fact (𝑥𝑒𝑠𝑦)2+(𝑦𝑒𝑠𝑥)2=2(𝑥𝑦)2𝑒𝑠+𝑥2+𝑦2(1𝑒𝑠)2,(2.14) a straightforward manipulation leads to |||𝜕𝑊𝜕𝑠𝑠|||(𝑥,𝑦)𝐶𝑒((𝑥𝑦)2+(𝑥2+𝑦2)(1𝑒𝑠)2)/8(1𝑒2𝑠)𝑒𝑠/21𝑒2𝑠3/2𝐶𝑁𝑒𝑠/81𝑒2𝑠3/2𝑒𝑐(|𝑥𝑦|2/10𝑠)1+𝑠|𝑥|𝑁,(2.15) which implies (a).
To prove (b), we also directly compute the 𝑥 partial derivative: 𝜕𝑊𝜕𝑥𝜕𝑠𝑠1(𝑥,𝑦)=2𝜋𝑒((𝑥𝑒𝑠𝑦)2+(𝑦𝑒𝑠𝑥)2)/2(1𝑒2𝑠)𝑒𝑠/21𝑒2𝑠3/2×4𝑒𝑠(𝑦𝑒𝑠𝑥)4𝑒2𝑠(𝑥𝑦)1+𝑒𝑠+12𝜋𝑒((𝑥𝑒𝑠𝑦)2+(𝑦𝑒𝑠𝑥)2)/2(1𝑒2𝑠)𝑒𝑠/21𝑒2𝑠5/2((𝑥𝑒𝑠𝑦)𝑒𝑠(𝑦𝑒𝑠𝑥×))1+𝑒2𝑠+2𝑒𝑠(𝑦(𝑥𝑒𝑠𝑦)+𝑥(𝑦𝑒𝑠𝑥))2𝑒2𝑠(𝑥𝑒𝑠𝑦)2+(𝑦𝑒𝑠𝑥)21𝑒2𝑠.(2.16) By an elementary manipulation and (2.14), we have |||𝜕𝑊𝜕𝑥𝜕𝑠𝑠|||1(𝑥,𝑦)𝐶𝑠2𝑒(𝑥𝑦)2/16𝑠.(2.17) This together with the mean value theorem and the condition ||𝑡 leads to (b).
Let 𝜙𝑛(𝑦)=𝜙(𝑦/𝑛); 𝜙(𝑦) is a smooth function satisfying 𝜙(𝑦)=1 for |𝑦|1, 𝜙(𝑦)=0 for |𝑦|2 and Δ𝜙(𝑦)1 for 𝑦. From the above, for fixed 𝑠 and 𝑥, a straightforward manipulation shows that +||||𝜕𝑊𝑠(𝑥,𝑦)||||𝜕𝑠𝑑𝑦<.(2.18) Hence, we have ||||𝜕𝑊𝑠(𝑥,𝑦)||||=||||𝜕𝑠𝑑𝑦lim𝑛𝜕𝑊𝑠(𝑥,𝑦)𝜙𝜕𝑠𝑛||||=||||(𝑦)𝑑𝑦lim𝑛𝑊𝑠(𝑥,𝑦)𝐻𝜙𝑛(||||𝑦)𝑑𝑦𝐶𝑊𝑠(𝑥,𝑦)𝑦2𝑑𝑦.(2.19) Using (2.8) again, ||||𝐼=𝜕𝑊𝑠(𝑥,𝑦)||||𝜕𝑠𝑑𝑦𝐶𝑒𝑠/4𝑠exp(𝑥𝑦)2𝑒𝑠+𝑥2+𝑦2(1𝑒𝑠)221𝑒2𝑠(𝑦𝑥)2+𝑥2𝑑𝑦,(2.20) which implies (c).

Lemma 2.11 (see [3, Theorem 2]). For all 𝑓BMO𝐿𝛼 and 𝐵=𝐵𝑟(𝑦0)(0,), there exists a constant 𝐶>0 such that 1||𝐵||𝑟0𝐵𝑄2𝑡𝑓(𝑥)𝑑𝑥𝑑𝑡𝑡𝐶𝑓2BMO𝐿𝛼.(2.21)

3. Maximal Operators

First of all, we define the following notions: 𝑀+𝑓(𝑥)=sup𝑥𝐵(0,)1||𝐵||𝐵||||𝑓(𝑦)𝑑𝑦,(3.1)𝛼𝑓(𝑥)=sup𝑡>0||𝑇𝛼𝑡||.𝑓(𝑥)(3.2)

In this section, we will show 𝛼 and 𝑀+ are bounded on BMO𝐿𝛼.

Theorem 3.1. There exists a constant 𝐶>0 such that, for all 𝑓BMO𝐿𝛼, 𝑀+𝑓<, for a.e. 𝑥(0,), and 𝑀+𝑓BMO𝐿𝛼𝐶𝑓BMO𝐿𝛼.(3.3)

Proof. First of all, we show that for a.e. 𝑥(0,), 𝑀+𝑓<. To do this, we only need to show that for, at almost 𝑥𝐵𝑘 in Lemma 2.1, 𝑀+𝑓(𝑥)<. Let us split 𝑓=𝑓1+𝑓2 with 𝑓1=𝑓𝜒𝐵𝑘. Obviously, since 𝑓 is locally integrable, we have 𝑀+𝑓1< for a.e. 𝑥(0,). For 𝑓2, if 𝑥𝐵 and 𝐵𝐵𝑘=, since supp 𝑓2 is in the complement of 𝐵𝑘, we have (1/|𝐵|)𝐵|𝑓(𝑦)|𝑑𝑦=0. Otherwise, by the definition of BMO𝐿𝛼, (1/|𝐵|)𝐵|𝑓(𝑦)|𝑑𝑦(4/|𝐵4𝑟(𝑥𝑘)|)𝐵4𝑟(𝑥𝑘)|𝑓(𝑦)|𝑑𝑦𝑐𝑓BMO𝐿𝛼.
We turn to the boundedness in BMO𝐿𝛼. Let 𝑀 denote the Hardy-Littlewood function on ; it is well known in [11] that 𝑀 is bounded on BMO(). Let 𝑓0 be a function defined on which is 𝑓 on (0,) and 0 on (,0]. Notice that 𝑀+𝑓=𝑀𝑓0, for 𝑥(0,), so 𝑀+𝑓BMO(𝐵𝑘)=𝑀𝑓0BMO(𝐵𝑘)𝑓𝐶0BMO.(3.4) Now, we need to show that 𝑓0BMO𝐶𝑓BMO𝐿𝛼. Indeed, if 𝐵(0,), it is obvious that (1/|𝐵|)𝐵|𝑓0(𝑓0)𝐵|𝑑𝑦𝑓BMO𝐿𝛼. If 𝐵(0,)=, then (1/|𝐵|)𝐵|𝑓0(𝑓0)𝐵|𝑑𝑦=0. If 𝐵(0,) and 𝐵(,0), let 𝐵=𝐵1𝐵2, here 𝐵1=𝐵(,0) and 𝐵2=𝐵(0,), then 1||𝐵||𝐵||𝑓0𝑓0𝐵||1𝑑𝑦2||𝐵||𝐵2||𝑓0||𝑑𝑦2𝑓BMO𝐿𝛼.(3.5) On the other hand, we again split 𝑓=𝑓1+𝑓2 with 𝑓1=𝑓𝜒𝐵𝑘, from the argument above, 𝑀+𝑓2(𝑥)𝑐𝑓BMO𝐿𝛼, for a.e 𝑥𝐵𝑘. So 1||𝐵𝑘||𝐵𝑘||𝑀+𝑓||1𝑑𝑦||𝐵𝑘||𝐵𝑘||𝑀+𝑓1||1𝑑𝑦+||𝐵𝑘||𝐵𝑘||𝑀+𝑓2||1𝑑𝑦||𝐵𝑘||𝐵𝑘||𝑀+𝑓1||2𝑑𝑦1/2+𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼,(3.6) where in the last inequality we have used Corollary 2.4.

Theorem 3.2. Let 𝛼>1/2. There exists a constant 𝐶>0 such that 𝛼𝑓BMO𝐿𝛼𝐶𝑓BMO𝐿𝛼.(3.7)

Proof. By the definition of BMO𝐿𝛼 and Corollary 2.3, it suffices to prove the following: for every fixed “critical ball” 𝐵𝑘 (see Lemma 2.1) we have (1)(1/|𝐵_𝑘|)𝐵𝑘|𝛼𝑓|𝑑𝑥𝐶𝑓BMO𝐿𝛼, (2)𝛼𝑓BMO(𝐵𝑘)𝐶𝑓BMOL𝛼.
Let us start to prove (1). It is immediate from Theorem 3.1 and (d) of Proposition 2.7; since 𝛼𝑓(𝑥)𝑀+𝑓(𝑥), for 𝑥>0, therefore, 1||𝐵𝑘||𝐵𝑘||𝑀+𝑓||𝑑𝑦𝐶𝑓BMO𝐿𝛼.(3.8)
It remains to show (2). By Lemma 2.5, we split 𝛼𝑓(𝑥) into several parts. First, we shall show sup𝑡>𝜌2𝐿𝛼𝑥𝑘||𝑇𝛼𝑡||𝑓(𝑥)𝐿(𝐵𝑘)𝐶𝑓BMO𝐿𝛼.(3.9)
From (d) of Proposition 2.7, we have ||𝑇𝛼𝑡||𝑓(𝑥)0||||1𝑓(𝑦)𝑡1/2||||1+𝑥𝑦𝑡𝑁𝑑𝑦𝑗=012𝑗𝑁1𝑡1/2{𝑦>0,|𝑦𝑥|<2𝑗𝑡}||||𝑓(𝑦)𝑑𝑦.(3.10)
Notice that, for 𝑗0 and 𝑡>𝜌2𝐿𝛼(𝑥𝑘), we have 2𝑗𝑡𝜌𝐿𝛼(𝑥)𝜌𝐿𝛼(𝑥𝑘), for 𝑥𝐵𝑘. Thus 1𝑡{𝑦>0,|𝑦𝑥|<2𝑗𝑡}||||𝑓(𝑦)𝑑𝑦𝐶2𝑗𝑓BMO𝐿𝛼.(3.11)
Therefore, sup𝑡>𝜌2𝐿𝛼𝑥𝑘||𝑇𝛼𝑡||𝑓(𝑥)𝑗=012𝑁1𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(3.12)
By Lemma 2.5, it suffices to show that sup0<𝑡𝜌2𝐿𝛼𝑘)(𝑥|𝑇𝛼𝑡𝑓(𝑥)| satisfies (2). Write 𝑓=𝑓𝜒{𝑥𝑘/2𝑦2𝑥𝑘}+𝑓𝜒{𝑦<𝑥𝑘/2}+𝑓𝜒{𝑦>2𝑥𝑘}=𝑓1+𝑓2+𝑓3.(3.13)
By Proposition 2.7, it easily follows that sup0<𝑡𝜌2𝐿𝛼𝑥𝑘||𝑇𝛼𝑡𝑓2||(𝑥)𝐿(𝐵𝑘)𝐶𝑓BMO𝐿𝛼,sup0<𝑡𝜌2𝐿𝛼𝑥𝑘||𝑇𝛼𝑡𝑓3||(𝑥)𝐿(𝐵𝑘)𝐶𝑓BMO𝐿𝛼.(3.14)
Indeed, since 𝑥𝑥𝑘, for 𝑥𝐵𝑘, by (a) of Proposition 2.7 and Remark 2.8, we have ||𝑇𝛼𝑡𝑓2||(𝑥)𝑥𝑘0/2𝑦𝛼+1/2𝑥𝑘𝛼3/2||||1𝑓(𝑦)𝑑𝑦𝑥𝑘𝑥𝑘0/2||𝑓||(𝑦)𝑑𝑦𝑓BMO𝐿𝛼.(3.15)
Similarly, ||𝑇𝛼𝑡𝑓3(||𝑥)𝑛=12𝑛𝑥𝑘𝛼3/2𝑥𝑘𝛼+1/22𝑛+1𝑥𝑘2𝑛𝑥𝑘||||𝑓(𝑦)𝑑𝑦𝑛=1(2𝑛)𝛼1/2𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(3.16)
Now, we come to treat 𝑓1. We make further decompositions. Split 𝑇𝛼𝑡𝑓1=𝑇𝛼𝑡𝑓1𝑊𝑡𝑓1+𝑊𝑡𝑓1𝐻𝑡𝑓1+𝐻𝑡𝑓1,(3.17)
where 𝐻𝑡𝑔(𝑥)=0𝑒𝑡𝜋1𝑒2𝑡1/21exp21+𝑒2𝑡1𝑒2𝑡(𝑥𝑦)2𝑔(𝑦)𝑑𝑦.(3.18)
For the first term, by (c) of Proposition 2.7, we have ||𝑇𝛼𝑡𝑓1𝑊𝑡𝑓1||1𝐶𝑥𝑘2𝑥𝑘𝑥2/2||||𝑓(𝑦)𝑑𝑦𝐶𝑓BMO𝐿𝛼.(3.19)
By (2.8), 𝑊𝑡𝑓1(𝑥)𝐻𝑡𝑓1=(𝑥)0𝑒𝑡𝜋1𝑒2𝑡1/21exp21+𝑒2𝑡1𝑒2𝑡(𝑥𝑦)2𝑒2𝑥𝑦(1𝑒𝑡)2/1𝑒2𝑡𝑓11(𝑦)𝑑𝑦.(3.20)
Notice that |𝑒2𝑥𝑦(1𝑒𝑡)2/(1𝑒2𝑡)1|𝑐𝑡𝑥2𝑘, when 𝑡𝜌𝐿𝛼(𝑥𝑘)2, 𝑥𝑘/2𝑦2𝑥𝑘 and 𝑥𝐵𝑘. Therefore, for 𝑥𝐵𝑘 and 𝑡𝜌𝐿𝛼(𝑥𝑘)2, we obtain ||𝑊𝑡𝑓1(𝑥)𝐻𝑡𝑓1||(𝑥)𝑡𝑥2𝑘01𝑡𝑒𝑐0|𝑥𝑦|2/𝑡||𝑓1||(𝑦)𝑑𝑦𝑡𝑥2𝑘12𝑗𝜌𝐿𝛼𝑥𝑘/𝑡2𝑗(𝑁1)12𝑗𝑡{𝑦>0,|𝑦𝑥|<2𝑗𝑡}||||𝑓(𝑦)𝑑𝑦+𝑡𝑥2𝑘𝜌𝐿𝛼𝑥𝑘/𝑡<2𝑗2𝑗(𝑁1)12𝑗𝑡{𝑦>0,|𝑦𝑥|<2𝑗𝑡}||||𝑓(𝑦)𝑑𝑦12𝑗𝜌𝐿𝛼𝑥𝑘/𝑡2𝑗(𝑁1)𝑡𝑥2𝑘𝜌1+log𝐿𝛼𝑥𝑘2𝑗𝑡𝑓BMO𝐿𝛼+𝑗=02𝑗(𝑁1)𝑓BMO𝐿𝛼𝑗=02𝑗(𝑁1)𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(3.21)
Finally, by Lemma 2.5 again, we need to show that sup0<𝑡𝜌2𝐿𝛼𝑘)(𝑥|𝐻𝑡𝑓1(𝑥)| satisfies (2). Consider 𝐵=𝐵𝑟(𝑥0)𝐵𝑘 and write 𝑓1=𝑓1𝑓𝐵𝜒𝐵+𝑓1𝑓𝐵𝜒(𝐵)𝑐(0,)+𝑓𝐵𝜒(0,)=𝑓11+𝑓12.(3.22) By Corollary 2.3, we choose a constant 𝐶𝐵=sup0<𝑡𝜌𝐿𝛼(𝑥𝑘)2|𝐻𝑡𝑓12(𝑥0)|, 1||𝐵||𝐵|||||sup0<𝑡𝜌𝐿𝛼𝑥𝑘2||𝐻𝑡𝑓1||(𝑥)𝐶𝐵|||||1𝑑𝑥||𝐵||𝐵sup0<𝑡𝜌𝐿𝛼𝑥𝑘2||𝐻𝑡𝑓1(𝑥)𝐻𝑡𝑓12𝑥0||1𝑑𝑥||𝐵||𝐵sup0<𝑡𝜌𝐿𝛼𝑥𝑘2||𝐻𝑡𝑓11||1(𝑥)𝑑𝑥+||𝐵||𝐵sup0<𝑡𝜌𝐿𝛼𝑥𝑘2||𝐻𝑡𝑓12(𝑥)𝐻𝑡𝑓12𝑥0||𝑑𝑥.(3.23) For the first integral, by Corollary 2.4 it easily follows that 1||𝐵||𝐵sup0<𝑡𝜌𝐿𝛼𝑥𝑘2||𝐻𝑡𝑓11||1(𝑥)𝑑𝑥||𝐵||𝐵||𝑓11||2𝑑𝑥1/2𝐶𝑓BMO𝐿𝛼.(3.24) For the second integral, ||𝐻𝑡𝑓1(𝑥)𝐻𝑡𝑓12𝑥0||||||(𝐵)𝑐(0,)𝐻𝑡(𝑥,𝑦)𝐻𝑡𝑥0𝑓,𝑦1(𝑦)𝑓𝐵||||+||||𝑑𝑦0𝐻𝑡(𝑥,𝑦)𝐻𝑡𝑥0𝑓,𝑦𝐵||||𝑑𝑦=𝐼𝑡1(𝑥)+𝐼𝑡2(𝑥).(3.25) By the mean value theorem and the elementary inequality 12𝑛𝑟{𝑦>0,|𝑦𝑥0|<2𝑛𝑟}||𝑓(𝑦)𝑓𝐵||𝑑𝑦𝑐𝑛𝑓BMO𝐿𝛼,(3.26) we have 𝐼𝑡1(𝑥)(𝐵)𝑐(0,)1𝑡𝑒|𝑦𝑥0|2/10𝑡||𝑥𝑥0||𝑡||𝑓1(𝑦)𝑓𝐵||𝑑𝑦(𝐵)𝑐(0,)||𝑥𝑥0||||𝑦𝑥0||2||𝑓1(𝑦)𝑓𝐵||𝑑𝑦𝑟𝑛=01(2𝑛𝑟)2{𝑦>0,|𝑦𝑥0|<2𝑛𝑟}||𝑓(𝑦)𝑓𝐵||+||𝑓(𝑦)𝑓1||(𝑦)𝑑𝑦𝑛=02𝑛(𝑛+1)𝑓BMO𝐿𝛼.(3.27) On the other hand, by the fact |𝑓𝐵|𝐶(1+log(𝜌𝐿𝛼(𝑥0)/𝑟))𝑓BMO𝐿𝛼 in Lemma 2.6, we obtain 𝐼𝑡2||||(𝑥)0𝐻𝑡(𝑥,𝑦)𝐻𝑡𝑥0𝑓,𝑦𝐵||||𝑑𝑦{𝑦<0}||𝑥𝑥0||||𝑦𝑥0||2||𝑓𝐵||𝑟𝑑𝑦𝜌𝐿𝛼𝑥0𝜌1+log𝐿𝛼𝑥0𝑟𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(3.28) Therefore, we obtain 1||𝐵||𝐵sup0<𝑡𝜌𝐿𝛼𝑥𝑘2||𝐻𝑡𝑓12(𝑥)𝐻𝑡𝑓12𝑥0||𝑑𝑥𝑓BMO𝐿𝛼,(3.29) which establishes the proof.

4. 𝑔-Function

For all 𝑓𝐿1loc(0,) and 𝑥(0,), define the Littlewood-Paley 𝑔-function by 𝑔(𝑓)(𝑥)0||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡1/2,(4.1) where, {𝑄𝑡}𝑡>0 is a family of operators with the integral kernels 𝑄𝑡(𝑥,𝑦)=𝑡2𝜕𝑊𝛼𝑠(𝑥,𝑦)||||𝜕𝑠𝑠=𝑡2.(4.2)

Theorem 4.1. Let 𝛼>1/2. There exists a constant 𝐶>0 such that, for all 𝑓BMO𝐿𝛼, 𝑔(𝑓)BMO𝐿𝛼 and 𝑔(𝑓)BMO𝐿𝛼𝐶𝑓BMO𝐿𝛼.

Proof. By Proposition 2.9 and (a) of Proposition 2.10, we have 𝑄𝑡1(𝑥,𝑦)𝑐𝑡𝑒𝑐1|𝑥𝑦|2/𝑡2.(4.3) For 𝑓BMO𝐿𝛼, because of this and the integrability of (1+|𝑥|)2𝑓(𝑥) (see [12, page 141]), 𝑄𝑡𝑓(𝑥)=0𝑄𝑡(𝑥,𝑦)𝑓(𝑦)𝑑𝑦(4.4) is well defined absolutely convergent integral for all (𝑥,𝑡)(0,)×(0,). Similar to the proof of Theorem 3.2, we will try to show that, for 𝐵𝑘 in Lemma 2.1, (1)(1/|𝐵𝑘|)𝐵𝑘|𝑔(𝑓)(𝑥)|𝑑𝑥𝐶𝑓BMO𝐿𝛼, (2)𝑔(𝑓)(𝑥)BMO(𝐵𝑘)𝐶𝑓BMO𝐿𝛼. We split []𝑔(𝑓)(𝑥)2=𝑔1(𝑓)(𝑥)2+𝑔2(𝑓)(𝑥)2=20𝜌𝐿𝛼(𝑥𝑘)0||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡+20𝜌𝐿𝛼𝑥𝑘||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡.(4.5) By Lemma 2.11 and Hölder inequality, assertion (1) holds for 𝑔1(𝑓)(𝑥). To finish the proof of (1), it suffices to show that 𝑔2(𝑓)𝐿(𝐵𝑘)𝑐𝑓BMO𝐿𝛼.(4.6)
In the next proof, for the sake of brevity we introduce the additional notations: 𝑋𝑡1𝑒(𝑥)=𝑦(0,)𝑡2𝑥𝑦1𝑒2𝑡2𝑋1,(4.7)𝑡2𝑒(𝑥)=&𝑦(0,)𝑡2𝑥𝑦1𝑒2𝑡2>1.(4.8) By 𝑋𝑡1(𝑥) and 𝑋𝑡1(𝑥), we split 𝑄𝑡𝑓(𝑥) as ||𝑄𝑡||𝑓(𝑥)𝑋𝑡1(𝑥)||||||𝑄𝑓(𝑦)𝑡(||𝑥,𝑦)𝑑𝑦+𝑋𝑡2(𝑥)||||||𝑄𝑓(𝑦)𝑡(𝑥,𝑦)𝑃𝑡(||+𝑥,𝑦)𝑑𝑦𝑋𝑡2(𝑥)||𝑓||||𝑃(𝑦)𝑡||(𝑥,𝑦)𝑑𝑦=𝐼𝑡1(𝑥)+𝐼𝑡2(𝑥)+𝐼𝑡3(𝑥).(4.9)
For 𝐼𝑡1(𝑥) and 𝑥𝐵𝑘, we shall first show the inequality 𝐽1(𝑥)=20𝜌𝐿𝛼𝑥𝑘||𝐼𝑡1||𝑓(𝑥)2𝑑𝑡𝑡𝑐𝑓2BMO𝐿𝛼.(4.10) Using (a) of Proposition 2.9, if 𝑥𝑘1, 𝜌𝐿𝛼(𝑥𝑘)𝑥𝑘, we get ||𝑄𝑡(||𝑥,𝑦)𝐶𝑡2𝛼+3𝑥𝑦𝑡2𝛼+1/2𝑒(𝑥2+𝑦2)/8𝑡2𝑒(𝛼+1)𝑡21𝑒2𝑡2𝛼+2𝑥𝐶𝑘𝑡𝛼+1/21𝑡𝑒|𝑥𝑦|2/16𝑡2.(4.11)
If 𝑥𝑘1, 𝜌𝐿𝛼(𝑥𝑘)1/𝑥𝑘, we have ||𝑄𝑡||1(𝑥,𝑦)𝐶𝑡𝑒(𝑥2+𝑦2)/8𝑡2𝑒((𝛼+1)/2)𝑡21𝐶𝑡𝑥𝑘1𝑡𝑒|𝑥𝑦|2/16𝑡2.(4.12) The previous two inequalities above imply 𝐽1(𝑥)20𝜌𝐿𝛼𝑥𝑘|||||0𝜌𝐿𝛼𝑥𝑘𝑡𝜎1𝑡𝑒|𝑥𝑦|2/16𝑡2|||||||||𝑓(𝑦)𝑑𝑦2𝑑𝑡𝑡20𝜌𝐿𝛼𝑥𝑘𝜌𝐿𝛼𝑥𝑘𝑡2𝜎|||||𝑗=02𝑗(𝑁1)12𝑗𝑡{𝑦>0,|𝑦𝑥|<2𝑗𝑡}|||||||||𝑓(𝑦)𝑑𝑦2𝑑𝑡𝑡𝑓2BMO𝐿𝛼.(4.13)
For 𝐼𝑡2(𝑥) and 𝑥𝐵𝑘, we shall also prove the inequality 𝐽2(𝑥)=20𝜌𝐿𝛼𝑥𝑘||𝐼𝑡2||𝑓(𝑥)2𝑑𝑡𝑡𝑐𝑓2BMO𝐿𝛼.(4.14) We split this integral as 𝐽2(𝑥)=120𝜌𝐿𝛼𝑥𝑘||𝐼𝑡2||𝑓(𝑥)2𝑑𝑡𝑡+1||𝐼𝑡2||𝑓(𝑥)2𝑑𝑡𝑡=𝐽3(𝑥)+𝐽4(𝑥).(4.15) To deal with 𝐽3(𝑥), we discuss two cases. In the first case of 𝑥𝑘1, notice that 𝑦>𝑥, when 𝑥𝐵𝑘, 𝑦𝑋𝑡2(𝑥𝑘) and 𝑡20𝜌𝐿𝛼(𝑥𝑘). According to (b) of Proposition 2.9, 𝐽3(𝑥)120𝜌𝐿𝛼𝑥𝑘|||||1𝑒22𝑡𝑒𝑡2/𝑥1𝑡𝑒|𝑥𝑦|2/4𝑡2𝑥𝑡1𝑒2𝑡2𝑒𝑡2𝑥2|||||||||𝑓(𝑦)𝑑𝑦2𝑑𝑡𝑡20𝜌𝐿𝛼𝑥𝑘𝑥2𝑘𝑡2||||01𝑡𝑒|𝑥𝑦|2/4𝑡2||||||||𝑓(𝑦)𝑑𝑦2𝑑𝑡𝑡𝑓2BMO𝐿𝛼.(4.16) The last inequality is from the same proof of 𝐽1(𝑥). In the second case of 𝑥𝑘>1, using (b) of Proposition 2.9 again, for 𝑡>20𝜌𝐿𝛼(𝑥𝑘) we obtain ||𝐼𝑡2||1(𝑥)𝑥𝑘2𝑥𝑘𝑥𝑘/2𝑡𝑥𝑘||||𝑓(𝑦)𝑑𝑦+(0,)(𝑥𝑘/2,2𝑥𝑘)𝑐1𝑡𝑒|𝑥𝑦|2/2𝑡2||||𝑡𝑓(𝑦)𝑑𝑦𝑥𝑘𝑓BMO𝐿𝛼+𝑡𝑥𝑘01𝑡𝑒|𝑥𝑦|2/4𝑡2||||𝑡𝑓(𝑦)𝑑𝑦𝑥𝑘𝑓BMO𝐿𝛼.(4.17) The last inequality is also from the same proof of 𝐽1(𝑥). Inserting this into 𝐽3(𝑥) leads to 𝐽3(𝑥)𝐶𝑓2BMO𝐿𝛼. Now, it remains to show 𝐽4(𝑥)𝑐𝑓2BMO𝐿𝛼. Using (b) of Proposition 2.9, by the standard argument it easily follows that 𝐽4(𝑥)1𝑒𝑡/10𝑡||||01𝑡𝑒|𝑥𝑦|2/2𝑡2||||||||𝑓(𝑦)𝑑𝑦2𝑑𝑡𝑓2BMO𝐿𝛼.(4.18) To complete the proof of (4.6), we need to show that 𝐽5(𝑥)=20𝜌𝐿𝛼𝑥𝑘||𝐼𝑡3||𝑓(𝑥)2𝑑𝑡𝑡𝑐𝑓2BMO𝐿𝛼.(4.19) We also consider two cases of 𝑥𝑘1 and 𝑥𝑘>1. When 𝑥𝑘1, repeating the above argument for 𝐽3(𝑥) and using (a) of Proposition 2.10, we have 𝐽5𝑐𝑓2BMO𝐿𝛼. When 𝑥𝑘>1, using (a) of Proposition 2.10 again, for 𝑡20𝜌𝐿𝛼(𝑥𝑘), we obtain ||𝐼𝑡3||1(𝑥)𝑥𝑘𝑡01𝑡𝑒𝑐|𝑦𝑥|2/𝑡2||||1𝑓(𝑦)𝑑𝑦𝑡𝑥𝑘𝑗=02𝑗(𝑁1)12𝑗𝑡{𝑦>0,|𝑦𝑥|<2𝑗𝑡}||||𝜌𝑓(𝑦)𝑑𝑦𝐿𝛼𝑥𝑘𝑡𝑓BMO𝐿𝛼,(4.20) which shows that (4.19) holds.
Next, we come to prove assertion (2). By (4.6) and Lemma 2.5, we only need to show 20𝜌𝐿𝛼(𝑥𝑘)0||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡1/2BMO(𝐵𝑘)𝐶𝑓BMO𝐿𝛼.(4.21)
Consider any ball 𝐵=𝐵𝑟(𝑥0)𝐵𝑘. By Lemma 2.11, we have 1||𝐵||𝐵𝑟0||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡1/21𝑑𝑥||𝐵||𝐵𝑟0||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡𝑑𝑥1/2𝐶𝑓BMO𝐿𝛼.(4.22) Therefore, by Lemma 2.5 and Corollary 2.3, it suffices to prove 1||𝐵||𝐵|||||20𝜌𝐿𝛼(𝑥𝑘)𝑟||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡1/220𝜌𝐿𝛼(𝑥𝑘)𝑟||𝑄𝑡𝑓𝑥0||2𝑑𝑡𝑡1/2|||||𝑑𝑥𝐶𝑓BMO𝐿𝛼.(4.23) To prove (4.23), we first claim that, for all 𝑓BMO𝐿𝛼, 𝑥𝐵𝑘, and 𝑡20𝜌𝐿𝛼(𝑥𝑘), ||𝑄𝑡||𝑓(𝑥)𝐶𝑓BMO𝐿𝛼.(4.24)
We shall split into three different estimates: 𝑋𝑡1(𝑥)||||||𝑄𝑓(𝑦)𝑡(||𝑥,𝑦)𝑑𝑦𝐶𝑓BMO𝐿𝛼,(4.25)𝑋𝑡2(𝑥)||𝑓||||𝑄(𝑦)𝑡(𝑥,𝑦)𝑃𝑡||(𝑥,𝑦)𝑑𝑦𝐶𝑓BMO𝐿𝛼,||||(4.26)𝑋𝑡2(𝑥)𝑓(𝑦)𝑃𝑡||||(𝑥,𝑦)𝑑𝑦𝐶𝑓BMO𝐿𝛼.(4.27) Let us first treat (4.25). Since 𝑦𝑐(𝜌𝐿𝛼(𝑥𝑘)2/𝑥𝑘), when 𝑦𝑋𝑡1(𝑥), notice that 𝑥𝑥𝑘 when 𝑥𝐵𝑘, using (a) of Proposition 2.9, and recalling the definition of 𝜌𝐿𝛼(𝑥), we have 𝑋𝑡1(𝑥)||||||𝑄𝑓(𝑦)𝑡||(𝑥,𝑦)𝑑𝑦𝐶𝑐𝜌𝐿𝛼(𝑥𝑘)2/𝑥𝑘0𝑡𝑥2𝑘||||𝑡𝑓(𝑦)𝑑𝑦𝐶𝑥𝑘𝑓BMO𝐿𝛼𝐶𝑓BMO𝐿𝛼.(4.28) For (4.26), using (b) of Proposition 2.9, for 𝑥𝐵𝑘, the left side of (4.26) is controlled by 1𝑥𝑘2𝑥𝑘𝑥𝑘/2𝑡𝑥𝑘||||𝑓(𝑦)𝑑𝑦+(0,)(𝑥𝑘/2,2𝑥𝑘)𝑐1𝑡𝑒|𝑥𝑦|2/2𝑡2||||𝑡𝑓(𝑦)𝑑𝑦𝑥𝑘𝑓BMO𝐿𝛼+𝑡𝑥𝑘01𝑡𝑒|𝑥𝑦|2/4𝑡2||||𝑡𝑓(𝑦)𝑑𝑦𝑥𝑘𝑓BMO𝐿𝛼+𝑡𝑥𝑘12𝑓BMO𝐿𝛼.(4.29) The third inequality is from the same argument for dealing with (3.21) in the proof of Theorem 3.2.
For (4.27), we write ||||𝑋𝑡2(𝑥)𝑓(𝑦)𝑃𝑡||||||||(𝑥,𝑦)𝑑𝑦𝑋𝑡2(𝑥)||𝑓(𝑦)𝑓𝐵𝑡(𝑥)||||𝑃𝑡||||||+||||(𝑥,𝑦)𝑑𝑦𝑋𝑡2(𝑥)𝑓𝐵𝑡(𝑥)𝑃𝑡||||.(𝑥,𝑦)𝑑𝑦(4.30)
By (a), (c) of Proposition 2.10 and the fact that |𝑓𝐵𝑡(𝑥)|(1+log20𝜌𝐿𝛼(𝑥𝑘)/𝑡)𝑓BMO𝐿𝛼, we have ||||𝑋𝑡2(𝑥)𝑓𝐵𝑡(𝑥)𝑃𝑡||||||||(𝑥,𝑦)𝑑𝑦𝑓𝐵𝑡(𝑥)𝑃𝑡||||+|||||(𝑥,𝑦)𝑑𝑦𝑒𝑡2(1𝑒22𝑡)/𝑥𝑓𝐵𝑡(𝑥)𝑃𝑡|||||𝑡(𝑥,𝑦)𝑑𝑦2𝜌𝐿𝛼𝑥𝑘21+log20𝜌𝐿𝛼𝑥𝑘𝑡𝑓BMO𝐿𝛼+(1𝑒22𝑡)𝑒𝑡2/𝑥1𝑡𝑒(𝑥𝑦)2/16𝑡2𝑑𝑦1+log20𝜌𝐿𝛼𝑥𝑘𝑡𝑓BMO𝐿𝛼(1𝑒22𝑡)𝑒𝑡2/𝑡𝑥𝑥/𝑡𝑒(1/16)𝑧2𝑑𝑧1+log20𝜌𝐿𝛼𝑥𝑘𝑡𝑓BMO𝐿𝛼+𝑓BMO𝐿𝛼𝑒(1/20)((1𝑒22𝑡)𝑒𝑡2/𝑡𝑥𝑥𝑡)1+log20𝜌𝐿𝛼𝑥𝑘𝑡𝑓BMO𝐿𝛼+𝑓BMO𝐿𝛼𝑡2𝜌𝐿𝛼𝑥𝑘21+log20𝜌𝐿𝛼𝑥𝑘𝑡𝑓BMO𝐿𝛼+𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(4.31) The third and fourth inequalities come from the fact (1𝑒2𝑡2)𝑒𝑡2/𝑥𝑡𝑥/𝑡 and 𝑥𝑥𝑘, when 𝑡20𝜌𝐿𝛼(𝑥𝑘) and 𝑥𝐵𝑘. It is notable that, for 𝑡20𝜌𝐿𝛼(𝑥𝑘) and 𝑥𝐵𝑘, the inequality above also implies ||||𝑋𝑡2(𝑥)𝑃𝑡||||𝑡(𝑥,𝑦)𝑑𝑦𝐶𝜌𝐿𝛼𝑥𝑘.(4.32) On the other hand, by (a) of Proposition 2.10 and the simple fact that |𝑓2𝑗𝐵𝑡(𝑥)𝑓𝐵𝑡(𝑥)|𝐶𝑗𝑓BMO𝐿𝛼, we obtain ||||𝑋𝑡2(𝑥)||𝑓(𝑦)𝑓𝐵𝑡(𝑥)||||𝑃𝑡||||||(𝑥,𝑦)𝑑𝑦01𝑡𝑒|𝑦𝑥|2/10𝑡2||𝑓(𝑦)𝑓𝐵𝑡(𝑥)||𝑑𝑦𝑗=02𝑗(𝑁1)12𝑗𝑡{𝑦>0,|𝑦𝑥|2𝑗𝑡}||𝑓(𝑦)𝑓2𝑗𝐵𝑡(𝑥)||+𝑑𝑦𝑗=02𝑗(𝑁1)||𝑓2𝑗𝐵𝑡(𝑥)𝑓𝐵𝑡(𝑥)||𝑗=0𝑗2𝑗(𝑁1)𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(4.33) From (4.24), we deduce that |||||20𝜌𝐿𝛼(𝑥𝑘)𝑟||𝑄𝑡||𝑓(𝑥)2𝑑𝑡𝑡1/220𝜌𝐿𝛼(𝑥𝑘)𝑟||𝑄𝑡𝑓𝑥0||2𝑑𝑡𝑡1/2|||||20𝜌𝐿𝛼(𝑥𝑘)𝑟||𝑄𝑡𝑓(𝑥)𝑄𝑡𝑓𝑥0||2𝑑𝑡𝑡1/2𝑓1/2BMO𝐿𝛼20𝜌𝐿𝛼(𝑥𝑘)𝑟||𝑄𝑡𝑓(𝑥)𝑄𝑡𝑓𝑥0||𝑑𝑡𝑡1/2.(4.34) By (4.28) and (4.29), for 𝑧𝐵𝑘, we have 20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑋𝑡1(𝑧)||||||𝑄𝑓(𝑦)𝑡||(𝑧,𝑦)𝑑𝑦𝑑𝑡𝑡𝐶𝑓BMO𝐿𝛼,20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑋𝑡2(𝑧)||||||𝑄𝑓(𝑦)𝑡(𝑧,𝑦)𝑃𝑡||(𝑧,𝑦)𝑑𝑦𝑑𝑡𝑡𝐶𝑓BMO𝐿𝛼.(4.35) To finish the proof of (4.23), it suffices to show 20𝜌𝐿𝛼(𝑥𝑘)𝑟||||𝑋𝑡2(𝑥)𝑃𝑡(𝑥,𝑦)𝑃𝑡𝑥0𝑓||||,𝑦(𝑦)𝑑𝑦𝑑𝑡𝑡𝐶𝑓BMO𝐿𝛼,(4.36)20𝜌𝐿𝛼(𝑥𝑘)𝑟||||𝑋𝑡2(𝑥)𝑃𝑡𝑥0𝑓,𝑦(𝑦)𝑑𝑦𝑋𝑡2(𝑥0)𝑃𝑡𝑥0𝑓||||,𝑦(𝑦)𝑑𝑦𝑑𝑡𝑡𝐶𝑓BMO𝐿𝛼.(4.37) It is easy to check (4.37). Indeed, without loss of generality, we assume 𝑥<𝑥0. According to (a) of Proposition 2.10, ||||𝑋𝑡2(𝑥)𝑃𝑡𝑥0𝑓,𝑦(𝑦)𝑑𝑦𝑋𝑡2(𝑥0)𝑃𝑡𝑥0||||,𝑦𝑓(𝑦)𝑑𝑦(1𝑒22𝑡)/𝑒𝑡2𝑥1𝑒22𝑡/𝑒𝑡2𝑥01𝑡||𝑓||(𝑦)𝑑𝑦𝑡𝑟𝑥20𝑓BMO𝐿𝛼.(4.38) Inserting this into (4.37) gives the desired result. Moreover, from the above we also have proved ||||𝑋𝑡2(𝑥)𝑃𝑡𝑥0𝑓,𝑦(𝑦)𝑑𝑦𝑋𝑡2(𝑥0)𝑃𝑡𝑥0||||𝑡,𝑦𝑑𝑦𝐶𝜌𝐿𝛼𝑥0.(4.39)
For 𝑟𝑡20𝜌𝐿𝛼(𝑥𝑘), we write ||||𝑋𝑡2(𝑥)𝑃𝑡(𝑥,𝑦)𝑃𝑡𝑥0𝑓||||||||,𝑦(𝑦)𝑑𝑦𝑋𝑡2(𝑥)𝑃𝑡(𝑥,𝑦)𝑃𝑡𝑥0𝑓,𝑦(𝑦)𝑓𝐵||||+||𝑓𝑑𝑦𝐵||||||𝑋𝑡2(𝑥)𝑃𝑡(𝑥,𝑦)𝑃𝑡𝑥0||||,𝑦𝑑𝑦=𝑆𝑡1(𝑥)+𝑆𝑡2(𝑥).(4.40) By (b) of Proposition 2.10, we obtain 𝑆𝑡1(𝑥){𝑦>0,|𝑦𝑥0|>𝑟}𝑟𝑡1𝑡𝑒𝑐|𝑦𝑥0|2/𝑡2||𝑓(𝑦)𝑓𝐵||1𝑑𝑦+𝑡{𝑦>0,|𝑦𝑥0|𝑟}||𝑓(𝑦)𝑓𝐵||𝑑𝑦𝑗=02𝑗+1𝐵𝑟𝑡1/212𝑗𝑟3/2||𝑓(𝑦)𝑓2𝑗+1𝐵+𝑓2𝑗+1𝐵𝑓𝐵||𝑟𝑑𝑦+𝑡𝑓BMO𝐿𝛼𝑟𝑡1/2𝑗=02𝑗/2𝑗𝑓BMO𝐿𝛼+𝑟𝑡𝑓BMO𝐿𝛼𝑟𝑡+𝑟𝑡1/2𝑓BMO𝐿𝛼.(4.41)
From the inequality above, we deduce that 20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑆𝑡1(𝑥)𝑑𝑡𝑡20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑟𝑡+𝑟𝑡1/2𝑑𝑡𝑡𝑓BMO𝐿𝛼𝐶𝑓BMO𝐿𝛼.(4.42)
On the other hand, by using (b) of Proposition 2.10 again, we also have ||||𝑋𝑡2(𝑥)𝑃𝑡(𝑥,𝑦)𝑃𝑡𝑥0||||𝑟,𝑦𝑑𝑦𝐶𝑡.(4.43) This together with (4.32), (4.39), and |𝑓𝐵|(1+log(𝜌𝐿𝛼(𝑥0)/𝑟))𝑓BMO𝐿𝛼 gives that 20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑆𝑡2(𝑥)𝑑𝑡𝑡20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑟𝑡2/3𝑡𝜌𝐿𝛼𝑥𝑘1/3𝜌1+log𝐿𝛼𝑥0𝑟𝑑𝑡𝑡𝑓BMO𝐿𝛼20𝜌𝐿𝛼(𝑥𝑘)𝑟𝑟𝑡1/3𝑑𝑡𝑡𝑓BMO𝐿𝛼𝑓BMO𝐿𝛼.(4.44) Thus we finish the proof of (4.36) and, hence, finish the proof of Theorem 4.1.

Acknowledgments

The authors would like to express their gratitude to the referee for many valuable suggestions and revisions which improve the presentation of this paper. Supported by National Natural Science Foundation of China under Grant no. 10990012.