Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 117-125 , 2010 Applied Mathematics
The Boundedness of Maximal Operators Associated with the Hankel Transform in the Lorentz Spaces
Cuma Bolat1, Ayhan Serbetci2
1Sütçü ·Imam University, Faculty of Arts and Sciences, Department of Mathematics,
Kahramanmara¸s, Türkiye e-mail: b olatcum a@ ksu.edu.tr
2Ankara University, Faculty of Sciences, Department of Mathematics, Ankara, Türkiye
e-mail:serb etci@ science.ankara.edu.tr
Received Date: July 7, 2009 Accepted Date: August 3, 2009
Abstract. In this study the boundedness conditions for the maximal operator M associated with the Hankel transform h in the Lorentz spaces Lpq; (0; 1) are found.
Key words: Bessel di¤erential operator; Hankel transform; -rearrangement; Lorentz spaces; fractional maximal function.
2000 Mathematics Subject Classi…cation: 42B25; 42B35; 47G10. 1. Introduction
Assume > 1=2 and de…ne as the class of all measurable functions f de…ned on (0; 1) Lp; Lp; (0; 1) Lp (0; 1) ; x2x+1dx ; (1 p 1); which kfkLp; 0 @ 1 Z 0 jf (x)jpx2 +1dx 1 A 1=p < 1;
and by L1; (0; 1) = L1(0; 1) the space of the essentially bounded measur-able functions with respect to the Lebesgue measure on (0; 1).
We recall some basic results in harmonic analysis related to the Hankel (Fourier-Bessel) transform. The Hankel transform appears taking di¤erent forms in
the literature (see, for instance, [8, 12, 14, 15]). Here we de…ne the Hankel transformation h through [14, 15] h (f ) (x) = 1 Z 0
j (xy) f (y) y2 +1dy;x 2 (0; 1) ;
where j (s) = 2 ( + 1) s J (s), with J the Bessel function of the …rst kind and index .
Remark 1. As is well known ([14]), the function j is just the solution of the di¤erential equation
L u = u; u (0) = 1; u0(0) = 0; where L is the Bessel di¤erential operator given by
(L )x= d 2 dx+ 2 + 1 x d dx; > 1=2:
De…nition 1. 1) The generalized translation operator Ty, y 0 is de…ned for smooth functions on (0; 1) by
Tyf (x) = C Z
0
f px2+ y2+ 2xy cos sin2 d ;
where C = ( + 1) [p ( ( + 1=2))] 1 The operator Ty is the solution of the
(L )xu = (L )yu; u (x; 0) = f (x) ; uy(x; 0) = 0 di¤ erential equation.
2) The generalized convolution operator of two smooth functions f; g on (0; 1) is de…ned by (f # g) (x) = 1 Z 0 Tyf (x) g (y) y2 +1dy; x 2 (0; 1):
For 0 < 2 + 2, the fractional maximal function M ; associated with Hankel transform h is de…ned at f 2 Lloc
1; (0; 1) by M ; f (x) = sup r>0 1 r2 +2 r Z 0 Tyjf (x)j y2 +1dy; x 2 (0; 1) :
For = 0 we get the maximal function M0; f = M f associated with the Hankel transform ([5, 6, 13]) such that
M f (x) = sup r>0 1 r2 +2 r Z 0 Tyjf (x)j y2 +1dy; x 2 (0; 1) :
It is well known that for the classical Hardy-Littlewood maximal operator the rearrangement inequality
cf (t) (M f ) (t) Cf (t) ; t 2 (0; 1)
holds, where f (t) is the nonincreasing rearrangement of f and f (t) =1tR1 0
f (t) dt. Similar sharp rearrangement estimates are obtained for the fractional maximal function of f 2 Lloc
1 (Rn) in [2, 10, 11], and for the fractional maximal func-tion of f 2 Lp; (Rn) ; > 0, associated with the Laplace-Bessel di¤erential operator Bn= n X j=1 @2 @x2 j + xn @ @x
in [7]. These estimates are of great importance in the study of operators on rearrangement-invariant spaces as well as in interpolation theory.
In [1] the sharp rearrangement estimates for the fractional maximal function M ; f associated with the Hankel transform h are found. In this paper, as consequence of these results the boundedness conditions for the maximal opera-tor M associated with the Hankel transform in the Lorentz spaces Lpq; (0; 1) are found.
Throughout the paper, we denote M+(0; 1) the set of all non-negative measur-able functions on (0; 1) with respect to the measure x2 +1dx, and M+(0; 1; #) the set of all non-increasing functions from M+(0; 1). We use the letters c; C for positive constants, independent of appropriate parameters and not necessary the same at each occurrence.
2. Preliminaries
It is well known that the following properties for the operator Ty are satis…ed ([9,15]):
1 For 2 C and x; y 2 (0; 1)
Ty(j ( )) (x) = j ( x) j ( y)
2. If f belongs to Lp; , 1 p 1, then for all y 0, the function Tyf belongs Lp; to and
kTyf kLp; kfkLp; :
Lemma 1. For any measurable set A (0; 1) and y 0, the following equality holds (1) Z A Tyf (x) y2 +1dy = C Z x+ ~A f q z2+ z2 1 d (z; z1) ; where ~A = A [0; m) ; m = sup A; d (z; z1) = z12 dzdz1:
The proof of Lemma 1 is straightforward after applying the following substitu-tions
(2) z = x + y cos ; z1= y sin
Let f : (0; 1) ! R be a measurable function and for any measurable set E; jEj = x2 +1dx. We de…ne rearrangement of f in decreasing order by
f (t) = inf fs > 0 : f ; (s) tg ; 8t 2 (0; 1) where f; (s) denotes the distribution function of f given by
f ; (s) = jfx 2 (0; 1) : jf (x)j > sgj :
Some properties of rearrangement of functions is given as follows ([3, 7]): i) if 0 < p < 1;then (3) 1 Z 0 jf (x)jpx2 +1dx = 1 Z 0 (f (t))pdt ;
ii) for any t > 0
(4) sup jEj =t Z E jf (x)j x2 +1dx = t Z 0 f (s) ds; iii) (5) 1 Z 0 jf (x) g (x)j x2 +1dx 1 Z 0 f (t) g (t) dt ;
The function f : (0; 1) ! [0; 1] is de…ned as f (t) = 1t t R 0
f (s) ds:
De…nition 2. The Lorentz space Lpq; (0; 1) is the collection of all measurable functions f on (0; 1) such the quantity
kfkpq; = 8 > > < > > : 1 R 0 tp1f (t) q dt t 1 q ; 0 < p < 1; 0 < q < 1 sup t>0 1 pf (t) ; 0 < p 1; q = 1 is …nite.
Also we give a functional k : kpq; by
kfkpq; = 8 > > < > > : 1 R 0 t1pf (t) q dt t 1 q ; 0 < p < 1; 0 < q < 1 sup t>0 t1pf (t) ; 0 < p 1; q = 1
which is equivalent with k : kpq; such that
(6) kfkpq; kfkpq; p
p 1kfkpq; ; where 1 < p < 1; 1 q 1 or p = q = 1:
We denote by W Lp; (0; 1) the weak Lp; space of all measurable functions f with …nite norm
kfkW Lp; = sup
t>0
tp1f (t) ; 1 p < 1:
Lemma 2. ([7]) For any measurable set A (0; 1) and for any x > 0 the following inequality holds
(7) sup jAj =t Z A Tyjf (x)j y2 +1dy = C t Z 0 f (s) ds:
Proof: By Lemma 1 we have
(8) Z A Tyjf (x)j y2 +1dy = C Z x+ ~A ~ f (z; z1) d (z; z1) ;
where ~f (z; z1) = f p
z2+ z2
1 : For the function ~f (z; z1) the analogue of the equality (4) is valid (see [3])
(9) sup (A)=t Z ~ A ~ f (z; z1) d (z; z1) = t Z 0 ~ f (s) ds; where f~ (s) = infnt > 0 : n(z; z1) : ~f (z; z1) > t o so:
Note that x + ~A = jAj and f~ (s) = f (s). From the equalities (8) and (9) we have sup jAj =t R A Tyjf (x)j y2 +1dy = C sup (A~)=t R x+ ~A ~ f (z; z1) d (z; z1) = C t R 0 ~ f (s) ds = C t R 0 f (s) ds: Thus Lemma 2 is proved.
3. Main Results
The following the sharp pointwise rearrangement estimates of the fractional maximal function M ; f associated with Hankel transform are proved in [1]. Lemma 3. [1] Let 0 < 2 + 2: Then there exists a positive constant C, depending on and , such that
(10) (M ; f ) (t) C sup
t< <1
2 +2f ( ) ; t > 0;
for every f 2 Lloc
1; (0; 1). Inequality (10) is sharp in the sense that for every ' 2 M+(0; 1; #) there exists a function f on (0; 1) such that f = ' a:e: on (0; 1) and
(11) (M ; f ) (t) c sup
t< <1
2 +2f ( ) ; t > 0;
where, again, c is a positive constant which depends on and : In Lemma 3 if we take the limit as ! 0;then we get
lim
and
lim !0t< <sup1
2 +2f ( ) = f (t) ;
and therefore we have the following:
Corollary 1. There exist positive constants c and C, depending on and , such that
(12) cf (t) (M f ) (t) Cf (t)
for every f 2 Lloc
1 (0; 1) and t > 0: Here left hand side of the inequality (12) holds in the sense that for every ' 2 M+(0; 1; #) there exists a function f on (0; 1) such that f = ' a:e: on (0; 1) :
We can now give the boundedness conditions of M f maximal function in Lpq; Lorentz spaces.
Theorem 1. For f 2 L1; and C is a positive constant independent of f , kM fk11; C kfk1;
holds.
Proof. For f 2 L1; we have
kM fk11; = supt>0t (M f ) (t) C sup t>0 t f (t) = C sup t>0 t R 0 f (s) ds = C kfkL1; :
Theorem 2. If 1 < p < 1; 1 q 1 or p = q = 1, then there is a positive constant C independent of f and for all f 2 Lpq; such that
kM fkpq; C kfkpq; :
kM fkpq; = 1 R 0 t1p(M f ) (t) q dt t 1 q C R1 0 t1pf (t) q dt t 1 q = C kfkpq; Cp 1p kfkpq; = C1kfkpq; ;
where C1= Cp 1p . Then M f 2 Lpq; for all f 2 Lpq; If p = q = 1, then the following inequalities hold
sup t>0 (M f ) (t) C sup t>0 f (t) = C kfkpq; Cp 1p kfkpq; = C1kfkpq; : This completes the proof.
4. Acknowledgements.
The authors would like to express their thanks to Professor V.S. Guliyev for his attention to this work and he is also thankful to the referees for valuable comments and suggestions.
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