IS S N 1 3 0 3 –5 9 9 1
ON THE BOUNDEDNESS OF THE MAXIMAL OPERATOR AND RIESZ POTENTIAL IN THE MODIFIED MORREY SPACES
CANAY AYKOL AND M. ESRA YILDIRIM
Abstract. In this paper we prove the boundedness of the maximal operator M and give the necessary and su¢ cient conditions for the boundedness of Riesz potential operator I in the modi…ed Morrey spaces by using Guliyev method
1. Introduction
Morrey spaces Lp; were introduced by Morrey in 1938 in connection with certain
problems in elliptic partial di¤erential equations and calculus of variations ([19]). Later, Morrey spaces found important applications to Navier Stokes ([18], [25]) and Schrödinger ([21, 22]) equations, elliptic problems with discontinuous coe¢ cients ([5, 8]) and potential theory ([1, 2]). An exposition of the Morrey spaces can be found in the book [17]. Morrey spaces were widely studied during last decades, including the study of classical operators of harmonic analysis such as maximal, singular and potential operators ([1, 3, 4, 6, 7]). Modi…ed Morrey spaces and the boundedness conditions of maximal operators and Riesz potential studied by some authors (see, for example [12, 13, 14, 15, 16]).
In [9] Guliyev considered the generalized Morrey spaces Mp;' with a general
function '(x; r) de…ning the Morrey-type norm. He found the conditions on the pair ('1; '2) without any assumption on monotonicity of '1, '2 which ensures the
boundedness of the maximal operator in generalized Morrey spaces. He also proved the Spanne and Sobolev-Adams type theorems for the Riesz potential operator I . In the present work, we prove the boundedness of the maximal operator M and Riesz potential operator I in modi…ed Morrey spaces by using Guliyev methods given in [9].
Received by the editors July 09, 2015, Accepted: Aug. 04, 2014. 2000 Mathematics Subject Classi…cation. 46E30, 42B20, 42B25.
Key words and phrases. Morrey space, modi…ed Morrey space, maximal operator, fractional maximal operator, Riesz potential
M.E. Yildirim was partially supported by the Scienti…c and Technological Research Coun-cil of Turkey (TUBITAK Programme 2228-B).
c 2 0 1 4 A n ka ra U n ive rsity
2. Definitions and Preliminary Tools Let f 2 L1
loc(Rn). As usual we de…ne the Hardy-Littlewood maximal function
of f , M f , setting M f (x) := sup t>0jB(x; t)j 1 Z B(x;t)jf(y)jdy;
where B(x; t) denotes the open ball centered at x of radius t for x 2 Rn and t > 0.
jB(x; t)j = !ntn and !n denotes the volume of the unit ball in Rn.
For 0 < n, we de…ne the fractional maximal function M f (x) := sup
t>0jB(x; t)j
n 1
Z
B(x;t)jf(y)jdy:
In the case = 0, we get M0f = M f . The fractional maximal function M f is
closely related to the Riesz potential operator I f (x) := Z Rn f (y)dy jx yjn ; 0 < < n; such that M f (x) !n 1 n (I jfj)(x): (2.1)
The operators M and I play important role in real and harmonic analysis (see, for example [1, 20, 23, 24]).
De…nition 2.1. Let 1 p < 1, 0 n, [t]1 = minf1; tg. We de…ne the
Morrey space Lp; (Rn), and the modi…ed Morrey space eLp; (Rn) as the set of
locally integrable functions f with the …nite norms kfkLp; := sup x2Rn;t>0 t pkfk Lp(B(x;t)); (2.2) kfkLep; := sup x2Rn;t>0 [t] p 1 kfkLp(B(x;t)); (2.3) respectively. Note that e Lp;0(Rn) = Lp;0(Rn) = Lp(Rn); e Lp; (Rn) ,! Lp; (Rn) \ Lp(Rn) and maxfkfkLp; ; kfkLpg kfkLep;
and if < 0 or > n, then Lp; (Rn) = eLp; = , where is the set of all functions
De…nition 2.2. [3, 4, 10, 11] Let 1 p < 1 , 0 n. We de…ne the weak Morrey space W Lp; (Rn), and the modi…ed weak Morrey space W eLp; (Rn) as the
set of all locally integrable functions f with …nite norms kfkW Lp; := sup x2Rn;t>0 t pkfk W Lp(B(x;t)); kfkW eLp; := sup x2Rn;t>0 [t] p 1 kfkW Lp(B(x;t)); respectively. Note that W Lp(Rn) = W Lp;0(Rn) = W eLp;0(Rn); Lp; (Rn) W Lp; (Rn) and kfkW Lp; kfkLp; e Lp; (Rn) W eLp; (Rn) and kfkW eLp; kfkLep; :
The following lemmas give some embeddings between Morrey spaces which were proved in [14, 15].
Lemma 2.3. Let 0 < n and 0 < n . Then for p = n ; Lp; (Rn) ,! L1;n (Rn); kfkL1;n !
1=p0
n kfkLp; :
Lemma 2.4. Let 0 < n and 0 < n . Then for n p < n e
Lp; (Rn) ,! L1;n (Rn); kfkL1;n ! 1=p0
n kfkLep; :
The following theorems proved by Guliyev in [9] will be our main tools to obtain the boundedness of maximal operator M and Riesz potential I in modi…ed Morrey spaces, respectively.
Theorem A. Let 1 p < 1 and f 2 Lloc
p (Rn). Then for p > 1 kMfkLp(B(x;t)) Ct n p Z 1 t r np 1kfk Lp(B(x;r))dr; (2.4) and for p = 1 kMfkW L1(B(x;t)) Ct n p Z 1 t r np 1kfk L1(B(x;r))dr; (2.5)
where C is a constant independent of f , x 2 Rn and t > 0.
Theorem B. Let 1 p < 1, 0 < < n p and f 2 L loc p (Rn). Then jI f(x)j Ct M f (x) + C Z 1 t r np 1kfk Lp(B(x;r))dr; (2.6)
where C is a constant independent of f , x and t.
3. Boundedness of the maximal operator and the Riesz potential in the modified Morrey spaces
In this section we prove the boundedness of the maximal operator and the Riesz potential in modi…ed Morrey spaces eLp; . We prove Theorems 3.1 and 2 with the
help of Theorems A and B, respectively.
Theorem 3.1. Let 1 p < 1, 0 < n and f 2 eLp; (Rn).
( i) If p > 1, then the maximal operator M is bounded in eLp; (Rn).
( ii)If p = 1, then M is bounded from eL1; (Rn) to W eL1; (Rn).
Proof. (i ) Let 1 < p < 1. From the inequality (2.4) we get
kMfkLep; = sup x2Rn;t>0 [t] p 1 kMfkLp(B(x;t)) C sup x2Rn;t>0 [t] p 1 t n p Z 1 t r np 1kfk Lp(B(x;r))dr C sup x2Rn;t>0 [t] p 1 t n pkfk e Lp; minf Z 1 t r np 1dr; Z 1 t r pn 1drg = C sup x2Rn;t>0 [t] p 1 t n pkfk e Lp; minft n p; t n p g = C sup x2Rn;t>0 [t] p 1 t n pkfk e Lp; [t] p 1t n p = CkfkLep; ;
(ii ) Let p = 1. From the inequality (2.5) we get kMfkW eL1; = sup x2Rn;t>0[t]1 kMfkW L1(B(x;t)) C sup x2Rn;t>0 [t]1 tn Z 1 t r n 1kfkL1(B(x;r))dr C sup x2Rn;t>0 [t]1 tnkfkLe 1; minf Z 1 t r n 1dr; Z 1 t r n 1drg = C sup x2Rn;t>0 [t]1 tnkfkLe 1; minft n; t n g = C sup x2Rn;t>0[t]1 t n kfkLe1; [t]1 t n = CkfkLe1; ;
which implies that M is bounded from eL1; (Rn) to W eL1; (Rn).
In the following we give the necessary and su¢ cient conditions for the bounded-ness of the Riesz potential in modi…ed Morrey spaces.
Theorem 3.2. Let 0 < < n, 0 < n and 1 p < n .
( i) If 1 < p < n , then condition n 1p 1q n is necessary and su¢ cient for the boundedness of the operator I from eLp; (Rn) to eLq; (Rn).
( ii) If p = 1 < n , then condition n 1 1q n is necessary and su¢ cient for the boundedness of the operator I from eL1; (Rn) to W eLq; (Rn).
Proof. (i ) Su¢ ciency. Let 1 < p < n , n 1p 1q n and f 2 eLp; (Rn).
kI fkLeq; = sup x2Rn;t>0 [t] q 1 kI fkLq(B(x;t)) = sup x2Rn;t>0 [t]1 Z B(x;t)jI f(y)j qdy !1 q C sup x2Rn;t>0 [t] q 1 Z B(x;t) r M f (y) + Z 1 r n p 1kfk Lp(B(x; ))d q dy !1=q C sup x2Rn;t>0 [t] q 1 Z B(x;t) r M f (y) + kfkLep; minf Z 1 r n p 1d ; Z 1 r + n p 1d g q dy !1=q = C sup x2Rn;t>0 [t] q 1 Z B(x;t) r M f (y) + kfkLep; minfr n p; r + n p g q dy !1=q : Minimizing with respect to r, with
r = " kfkLep; M f (y) # p n and r = " kfkLep; M f (y) #p n we have kI fkLeq; C sup x2Rn;t>0 [t] q 1 0 @ Z B(x;t) 0 @minf M f (y) kfkLep; !1 p n ; M f (y) kfkLep; !1 p n gkfkLep; 1 A q dy 1 A 1=q Ckfk1 p q e Lp; sup x2Rn;t>0 [t] q 1 kMfk p q Lp(B(x;t)):
Hence by Theorem 3.1(i ) we have
kI fkLeq; Ckfk 1 p q e Lp; kfk p q e Lp; = CkfkLep; ;
which implies that I is bounded from eLp; (Rn) to eLq; (Rn).
Necessity. Let 1 < p < n , f 2 eLp; (Rn). Suppose that I is bounded from
e
kfskLep; = sup x2Rn;t>0 [t] p 1 kfskLp(B(x;t)) = sup x2Rn;t>0 [t]1 Z B(x;t)jf s(y)jpdy !1=p = s np sup x2Rn;t>0 [t]1 Z B(x;st)jf(y)j pdy !1=p = s npsup t>0 [st]1 [t]1 p sup x2Rn;t>0 [st]1 Z B(x;st)jf(y)j pdy !1=p = s np[s]p 1;+kfkLep; ; (3.1) and I fs(x) = s I f (sx); kI fskLeq; = s sup x2Rn;t>0 [t]1 Z B(x;t)jI f(sy)j qdy !1=q = s nq sup t>0 [st]1 [t]1 q sup x2Rn;t>0 [st]1 Z B(sx;st)jI f(y)j qdy !1=q = s nq[s]q 1;+kI fkLeq; :
By the boundedness of I from eLp; (Rn) to eLq; (Rn) we get
kI fkLeq; = s +n q[s] q 1;+kI fskLeq; s +nq[s] q 1;+kfskLep; Cs +nq n p[s]p q 1;+ kfkLep; :
If 1p < 1q+n, then in the case t ! 0 we have kI fkLeq; = 0 for all f 2 eLp; (R n).
If 1 p >
1
q + n , then in the case t ! 1 we have kI fkLeq; = 0 for all f 2
e
Lp; (Rn).
(ii ) Su¢ ciency. Let p = 1 and n 1 1
q n . From the inequality (2.6) we
have jI f(x)j Ct M f (x) + C Z 1 t r n 1kfkL1(B(x;r))dr Ct M f (x) + CkfkLe1; minft n; t + n g: Minimizing with respect to t, with
t = " kfkLe1; M f (x) # 1 n and t = " kfkLe1; M f (x) #1 n we have jI f(x)j C min 8 < : M f (x) kfkLe1; !1 n ; M f (x) kfkLe1; !1 n 9 = ;kfkLe1; : Therefore we get jI f(x)j C(M f (x))1=qkfk1 1=qe L1; : (3.2)
Using the inequality (3.2) and from Theorem 3.1(ii ) we get
kI fkqW eLq; = sup x2Rn;t>0 [t]1 kI fkqW L q(B(x;t)) = sup r>0 rq sup x2Rn;t>0[t]1 jfy 2 B(x; t) : jI f(y)j > rgj sup r>0 rq sup x2Rn;t>0 [t]1 jfy 2 B(x; t) : C(Mf(y))1=qkfk1 1=qe L1; > rgj = sup r>0 rq sup x2Rn;t>0 [t]1 8 < :y 2 B(x; t) : Mf(y) > 0 @ r Ckfk1 1=qLe1; 1 A q9 = ; C sup r>0 rq 0 B @kfk 1 1 q e L1; r 1 C A q kfkLe1; = CkfkqLe1; ;
which implies that I is bounded from eL1; (Rn) to W eLq; (Rn).
kI fskW eLq; = sup x2Rn;t>0 [t] q 1 kI fskW eLq; (B(x;t)) = sup r>0 r sup x2Rn;t>0 [t]1 Z fy2B(x;t):jI fs(y)j>rg dy !1=q = sup r>0 r sup x2Rn;t>0 [t]1 Z fy2B(xs;t):jI f (sy)j>rs g dy !1=q = s nq sup t>0 [ts]1 [t]1 q sup r>0 rs sup x2Rn;t>0 [ts]1 Z fy2B(x;ts):jI f (y)j>rs g !1=q = s nq[s]q 1;+kI fkW eLq; :
By using the boundedness of I from eL1; (Rn) to W eLq; (Rn) we get
kI fkW eLq; = s +n q[s] q 1;+kI fskW eLq; Cs +nq[s] q 1;+kfskLe1; = Cs +nq n[s] q 1;+ kfkLe1; : If 1 < 1
q+n, then in the case t ! 0 we have kI fkW eLq; = 0 for all f 2 eL1; (R n).
If 1 > 1q + n , then in the case t ! 1 we have kI fkW eL
q; = 0 for all
f 2 eL1; (Rn).
Therefore n 1 1q n .
Corollary 1. Let 0 < < n, 0 < n and 1 p < n .
( i) If 1 < p < n , then condition n 1p 1q n is necessary and su¢ cient for the boundedness of the operator M from eLp; (Rn) to eLq; (Rn).
( ii) If p = 1 < n , then condition n 1 1q n is necessary and su¢ cient for the boundedness of the operator M from eL1; (Rn) to W eLq; (Rn).
Proof. Su¢ ciency of Corollary 1 is obtained from Theorem 3.2 and inequality (2.1). Necessity. For the fractional maximal operator M the following equality
M fs(x) = s M f (sx)
(i ) Let 1 < p < n and M be bounded from eL p; (Rn) to eLq; (Rn). Then we have kM fskLeq; = s n q[s]q 1;+kM fkLeq; :
By similar methods in Theorem 3.2 we obtain n 1p 1q n . (i ) Let M be bounded from eL1; (Rn) to W eLq; (Rn).
Then we have kM fskW eLq; = s n q[s]q 1;+kM fkW eLq; : Therefore we get n 1 1q n . References
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Current address : Ankara University, Faculty of Sciences, Dept. of Mathematics, Ankara, TURKEY
E-mail address : [email protected], [email protected] URL: http://communications.science.ankara.edu.tr/index.php?series=A1
