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COMMUTATORS OF PARAMETRIC MARCINKIEWICZ INTEGRALS ON GENERALIZED ORLICZ-MORREY SPACES
FATIH DERINGOZ
Abstract. In this paper, we study the boundedness of the commutators of parametric Marcinkiewicz integral operator with smooth kernel on generalized Orlicz-Morrey spaces.
1. Introduction
Suppose that Sn 1is the unit sphere in Rn(n 2) equipped with the normalized Lebesgue measure d = d (x0). Let be a homogeneous function of degree zero on Rn satisfying 2 L1(Sn 1) and the following property
Z
Sn 1
(x0)d (x0) = 0; where x0= x=jxj for any x 6= 0.
The parametric Marcinkiewicz integral is de…ned by Hörmander [12] as follows:
(f )(x) = 0 @Z 1 0 1 t Z jx yj t (x y) jx yjn f (y)dy 2 dt t 1 A 1=2 ;
where 0 < < n. When = 1, we simply denote it by (f ). The operator (f ) is de…ned by Stein in [17].
Let b be a locally integrable function on Rn; the commutator generated by the
parametric Marcinkiewicz integral and b is de…ned by
;b(f )(x) = 0 @Z 1 0 1 t Z jx yj t (x y) jx yjn (b(x) b(y))f (y)dy 2 dt t 1 A 1=2 : In [2], Deringoz et al. introduced generalized Orlicz-Morrey spaces as an exten-sion of generalized Morrey spaces. Other de…nitions of generalized Orlicz-Morrey
Received by the editors: May 05, 2016, Accepted: Aug. 25, 2016. 2010 Mathematics Subject Classi…cation. 42B20, 42B25, 42B35.
Key words and phrases. Parametric Marcinkiewicz integrals, generalized Orlicz-Morrey spaces, commutator, BM O.
c 2 0 1 7 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis t ic s .
spaces can be found in [14] and [15]. In words of [10], our generalized Orlicz-Morrey space is the third kind and the ones in [14] and [15] are the …rst kind and the second kind, respectively. According to the examples in [6], one can say that the gener-alized Orlicz-Morrey spaces of the …rst kind and the second kind are di¤erent and that the second kind and the third kind are di¤erent. However, it is not known that relation between the …rst and the third kind.
Boundedness of commutators of classical operators of harmonic analysis on gen-eralized Orlicz-Morrey spaces were recently studied in various papers, see for ex-ample [3, 8, 9]. In this paper, we consider the boundedness of commutator of parametric Marcinkiewicz integral operator on generalized Orlicz-Morrey space of the third kind.
Everywhere in the sequel B(x; r) is the ball in Rn of radius r centered at x and jB(x; r)j = vnrnis its Lebesgue measure, where vnis the volume of the unit ball in
Rn. By A. B we mean that A CB with some positive constant C independent of appropriate quantities. If A. B and B . A, we write A B and say that A and B are equivalent.
2. Preliminaries We recall the de…nition of Young functions.
De…nition 1. A function : [0; 1) ! [0; 1] is called a Young function if is convex, left-continuous, lim
r!+0 (r) = (0) = 0 and limr!1 (r) = 1.
From the convexity and (0) = 0 it follows that any Young function is increasing. If there exists s 2 (0; 1) such that (s) = 1, then (r) = 1 for r s. The set of Young functions such that
0 < (r) < 1 for 0 < r < 1
will be denoted by Y: If 2 Y, then is absolutely continuous on every closed interval in [0; 1) and bijective from [0; 1) to itself.
For a Young function and 0 s 1, let
1
(s) = inffr 0 : (r) > sg:
If 2 Y, then 1is the usual inverse function of . We note that ( 1(r)) r 1( (r)) for 0 r < 1: It is well known that
r 1(r)e 1(r) 2r for r 0; (2.1)
where e(r) is de…ned by
e(r) = supfrs (s) : s 2 [0; 1)g ; r 2 [0; 1)
A Young function is said to satisfy the 2-condition, denoted also as 2 2,
if
(2r) k (r) for r > 0
for some k > 1. If 2 2, then 2 Y. A Young function is said to satisfy the
r2-condition, denoted also by 2 r2, if
(r) 1
2k (kr); r 0;
for some k > 1.
De…nition 2. (Orlicz Space). For a Young function , the set L (Rn) = f 2 L1loc(Rn) :
Z
Rn (kjf(x)j)dx < 1 for some k > 0
is called Orlicz space. If (r) = rp; 1 p < 1, then L (Rn) = Lp(Rn). If (r) = 0; (0 r 1) and (r) = 1; (r > 1), then L (Rn) = L1(Rn). The space Lloc(Rn) is de…ned as the set of all functions f such that f
B 2 L (R
n) for all
balls B Rn.
L (Rn) is a Banach space with respect to the norm
kfkL (Rn)= inf > 0 :
Z
Rn
jf(x)j dx 1 : By elementary calculations we have the following.
Lemma 1. Let be a Young function and B be a set in Rn with …nite Lebesgue
measure. Then
k BkL (Rn)=
1
1(jBj 1):
In the next sections where we prove our main estimates, we use the following lemma.
Lemma 2. [2] For a Young function , the following inequality is valid Z
B(x;r)jf(y)jdy 2jB(x; r)j
1
jB(x; r)j 1 kfkL (B(x;r));
where kfkL (B(x;r))= kf BkL (Rn).
Various versions of generalized Orlicz-Morrey spaces were introduced in [14], [15] and [2]. We used the de…nition of [2] which runs as follows.
De…nition 3. Let '(x; r) be a positive measurable function on Rn (0; 1) and
be any Young function. We denote by M ;'(Rn) the generalized Orlicz-Morrey
space, the space of all functions f 2 Lloc(Rn) for which
kfkM ;' = sup x2Rn;r>0
We recall the de…nition of the space of BM O(Rn).
De…nition 4. Suppose that b 2 L1
loc(Rn), let kbk = sup x2Rn;r>0 1 jB(x; r)j Z B(x;r)jb(y) bB(x;r)jdy; where bB(x;r)= 1 jB(x; r)j Z B(x;r) b(y)dy: De…ne BM O(Rn) = fb 2 L1loc(Rn) : kbk < 1g:
To prove our theorems, we need the following lemmas.
Lemma 3. [13] Let b 2 BMO(Rn). Then there is a constant C > 0 such that
bB(x;r) bB(x;t) Ckbk ln
t
r for 0 < 2r < t; where C is independent of b, x, r; and t.
Lemma 4. [8, 11] Let b 2 BMO(Rn) and be a Young function with 2 2, then kbk t sup x2Rn;r>0 1 jB(x; r)j 1 b( ) bB(x;r) L (B(x;r)): (2.2) We will use the following statement on the boundedness of the weighted Hardy operator Hwg(t) := Z 1 t 1 + lns t g(s)w(s)ds; 0 < t < 1; where w is a weight.
Lemma 5. Let v1, v2 and w be weights on (0; 1) and v1(t) be bounded outside a
neighborhood of the origin. The inequality ess sup
t>0
v2(t)Hwg(t) C ess sup t>0
v1(t)g(t) (2.3)
holds for some C > 0 for all non-negative and non-decreasing g on (0; 1) if and only if B := ess sup t>0 v2(t) Z 1 t 1 + lns t w(s)ds ess sup s< <1 v1( )< 1: (2.4) Moreover, the value C = B is the best constant for (2.3).
Note that, Lemma 5 is proved analogously to [7, Theorem 3.1]. Remark 1. In (2.3) and (2.4) it is assumed that 11 = 0 and 0 1 = 0.
3. Main Results
The following result concerning the boundedness of commutator of parametric Marcinkiewicz integral operator ;b on Lp is known.
Theorem A. [16] Suppose that 1 < p; q < 1, 2 Lq(Sn 1), 0 < < n and b 2 BMO(Rn). Then, there is a constant C independent of f such that
k ;b(f )kLp(Rn) CkfkLp(Rn):
The following interpolation result is from [5].
Lemma 6. Let T be a sublinear operator of weak type (p; p) for any p 2 (1; 1). Then T is bounded on L (Rn), where is a Young function satisfying that 2
2\ r2.
As a consequence of Lemma 6 and Theorem A, we get the following result. Corollary 1. Let be a Young function, b 2 BMO(Rn), 2 Lq(Sn 1) (q > 1) and 0 < < n. If 2 2\ r2, then ;b is bounded on L (Rn).
The following lemma is a generalization of the [1, Lemma 4.3] for Orlicz spaces. Lemma 7. Let be a Young function with 2 2\ r2, b 2 BMO(Rn), B =
B(x0; r) and 2 L1(Sn 1). Then k ;bf kL (B). 1kbk jBj 1 Z 1 r 1 + lnt r kfkL (B(x0;t)) 1 jB(x0; t)j 1 dt t (3.1) holds for any ball B, 0 < < n, and for all f 2 Lloc(Rn).
Proof. For B = B(x0; r) and 2B = B(x0; 2r) write f = f1+ f2with f1= f 2B and
f2= f { (2B)
. Hence
;bf
L (B) ;bf1 L (B)+ ;bf2 L (B):
Since L1(Sn 1) $ Lq(Sn 1) for 1 q < 1, from the boundedness of
;b in
L (Rn) provided by Corollary 1, it follows that
It’s clear that x 2 B, y 2 {(2B) implies 12jx0 yj jx yj 32jx0 yj. Then by
the Minkowski inequality and conditions on , we get j ;b(f2) (x)j Z Rn j (x y)j jx yjn jf2(y)jjb(x) b(y)j Z 1 jx yj dt t1+2 !1=2 dy .Z{ (2B) j (jx yjjx yjx y )jjf(y)j jx yjn jb(x) b(y)jdy . Z{ (2B) jf(y)j jx0 yjnjb(x) b(y)jdy: Then k ;bf2kL (B). Z {(2B) jb(y) b( )j jx0 yjn jf(y)jdy L (B) . Z{ (2B) jb(y) bBj jx0 yjn jf(y)jdy L (B)+ Z {(2B) jb( ) bBj jx0 yjn jf(y)jdy L (B) = I1+ I2:
For the term I1 we have
I1t 1 1 jBj 1 Z { (2B) jb(y) bBj jx0 yjn jf(y)jdy t 1 1 jBj 1 Z {(2B)jb(y) bBjjf(y)j Z 1 jx0 yj dt tn+1dy = 1 1 jBj 1 Z 1 2r Z 2r jx0 yj t jb(y) bBjjf(y)jdy dt tn+1 . 1 1 jBj 1 Z 1 2r Z B(x0;t) jb(y) bBjjf(y)jdy dt tn+1: Hence I1. 1 1 jBj 1 Z 1 2r Z B(x0;t) jb(y) bB(x0;t)jjf(y)jdy dt tn+1 + 1 1 jBj 1 Z 1 2r jb B(x0;r) bB(x0;t)j Z B(x0;t) jf(y)jdy dt tn+1:
Applying Hölder’s inequality, by (2.1) and Lemmas 2, 3 and 4 we get I1. 1 1 jBj 1 Z 1 2r b( ) bB(x0;t) Le(B(x0;t))kfkL (B(x0;t)) dt tn+1 + 1 1 jBj 1 Z 1 2r jb B(x0;r) bB(x0;t)jkfkL (B(x0;t)) 1 jB(x0; t)j 1 dt t
. 1kbk jBj 1 Z 1 2r 1 + lnt r kfkL (B(x0;t)) 1 jB(x0; t)j 1 dt t : For I2 we obtain I2t kb( ) bBkL (B) Z {(2B) jf(y)j jx0 yjn dy: By Lemma 4, we get I2. 1kbk jBj 1 Z { (2B) jf(y)j jx0 yjn dy: (3.3)
To the remaining integral we use the same trick as above in the estimation of I1:
Z {(2B) jf(y)j jx0 yjn dy = n Z {(2B)jf(y)j Z 1 jx0 yj dt tn+1dy = n Z 1 2r Z 2r jx0 yj<t jf(y)jdytn+1dt n Z 1 2r Z B(x0;t) jf(y)jdytn+1dt
and by Lemma 2 we then get Z {(2B) jf(y)j jx0 yjn dy. Z 1 2r kfk L (B(x0;t)) 1 jB(x0; t)j 1 dt t : (3.4)
Therefore, by (3.4) and (3.3) we have I2. 1kbk jBj 1 Z 1 2r kfk L (B(x0;t)) 1 jB(x0; t)j 1 dt t gathering estimates for I1and I2; we obtain
k ;bf2kL (B). 1kbk jBj 1 Z 1 2r 1 + lnt r kfkL (B(x0;t)) 1 jB(x0; t)j 1 dt t : (3.5) Now collect the estimates (3.2) and (3.5):
k ;bf kL (B). kbk kfkL (2B) + kbk 1 jBj 1 Z 1 2r 1 + lnt r kfkL (B(x0;t)) 1 jB(x0; t)j 1 dt t : To …nalize the proof, it remains to note that the …rst term here may be estimated in the form similar to the second one:
kfkL (2B) C 1 jBj 1 Z 1 2r kfkL (B(x0;t)) 1 jB(x0; t)j 1 dt t : (3.6)
To prove (3.6), observe that since 1 is concave and nonnegative it follows that
1(u) u v 1(v) for u v; whence 1 (jBj 1) t n rn 1 (jB(x0; t)j 1); r t:
Then 1 jBj 1 = n 1 jBj 1 (2r)n Z 1 2r dt tn+1 . Z 1 2r 1 jB(x0; t)j 1 dt t ; from which (3.6) follows by the monotonicity of the norm kfkL (B(x0;t))with respect
to t, and this completes the proof.
Theorem 1. Let 0 < < n, b 2 BMO(Rn), be any Young function, '
1; '2 and
satisfy the condition Z 1 r 1 + lnt r ess inft<s<1 '1(x; s) 1 jB(x; s)j 1 1 jB(x; t)j 1 dt t C '2(x; r); (3.7) where C does not depend on x and r. Let also 2 L1(Sn 1). If 2 2\ r2,
then the operator ;b is bounded from M ;'1(Rn) to M ;'2(Rn).
Proof. The proof follows from the Lemmas 5 and 7. We can also give the following alternative proof for Theorem 1 by inspiring the ideas in [4].
Since f 2 M ;'1(Rn) and '
1; '2 and satisfy the condition (3.7), we have
Z 1 r 1 + lnt r kfkL (B(x;t)) 1 jB(x; t)j 1 dtt t Z 1 r 1 + lnt r kfkL (B(x;t)) ess inf t<s<1 '1(x;s) 1 jB(x;s)j 1 ess inf t<s<1 '1(x; s) 1 jB(x; s)j 1 1 jB(x; t)j 1 dtt . kfkM ;'1 Z 1 r 1 + lnt r ess inft<s<1 '1(x; s) 1 jB(x; s)j 1 1 jB(x; t)j 1 dtt . kfkM ;'1'2(x; r):
Then from (3.1) we get k ;bkM ;'2 = sup x2Rn;r>0 '2(x; r) 1 1(jB(x; r)j 1)k ;bkL (B(x;r)) . kbk sup x2Rn;r>0 '2(x; r) 1 Z 1 r 1 + lnt r kfkL (B(x;t)) 1 jB(x; t)j 1 dt t . kbk kfkM ;'1
Acknowledgements. The research of F. Deringoz was partially supported by the grant of Ahi Evran University Scienti…c Research Project (FEF.A3.16.011).
References
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Current address : Department of Mathematics, Ahi Evran University, Kirsehir, Turkey E-mail address : deringoz@hotmail.com
