Jacobi-Dunkl Dini Lipschitz Functions In The Space L p ( R ; A ; (x)dx)
Salah El Ouadih y , Radouan Daher z
Received 06 July 2015
Abstract
In this paper, we obtain an analog of Younis Theorem 5.2 in [7] for the Jacobi- Dunkl transform on the real line for functions satisfying the ( ; )-Jacobi-Dunkl Lipschitz condition in the space L
p( R ; A
;(x)dx) where
21and 6 =
21.
1 Introduction
Younis Theorem 5.2 [7] characterized the set of functions in L 2 ( R ) satisfying the Dini Lipschitz condition by means of an asymptotic estimate growth of the norm of their Fourier transforms, namely we have the following Theorem 1.1.
THEOREM 1.1 ([7, Theorem 5.2]). Let f 2 L 2 ( R ). Then the following statements are equivalents
(a) k f (x + h) f (x) k = O (log h
1h
) as h ! 0; where 0 < < 1 and 0.
(b) R
j j r j f b ( ) j 2 d = O (log r r)
22as r ! 1 , where f b stands for the Fourier trans- form of f .
In this paper, we obtain an analog of Theorem 1.1 for the Jacobi-Dunkl transform on the real line. For this purpose, we use a generalized translation operator. In this section, we recapitulate from [1, 2, 3, 5] some results related to the harmonic analysis associated with Jacobi-Dunkl operator ; . The Jacobi-Dunkl function with parameters ( ; ); 2 1 , and 6 = 2 1 , de…ned by the formula
8 x 2 R ; ; (x) =
( ' ; (x) i d dx ' ; (x) , if 2 Cnf 0 g ;
1 , if = 0:
Mathematics Sub ject Classi…cations: 42B37, 42B10, 43A90, 42C15.
y
Departement of Mathematics, Faculty of Sciences Aïn Chock, University Hassan II, Casablanca, Morocco
z
Departement of Mathematics, Faculty of Sciences Aïn Chock, University Hassan II, Casablanca, Morocco
88
where 2 = 2 + 2 , = + + 1, F is the Gausse hypergeometric function (see [1,6]), and ' ; is the Jacobi function given by
' ; (x) = F + i
2 ; i
2 ; + 1; (sinh(x)) 2 :
Then ; is the unique C 1 -solution on R of the di¤erentiel-di¤erence equation (
; U = i U 2 C ; U (0) = 1;
where ; is the Jacobi-Dunkl operator given by
; U (x) = d U (x)
dx + [(2 + 1) coth x + (2 + 1) tanh x] U (x) U ( x)
2 :
The operator ; is a particular case of the operator D given by D U (x) = d U (x)
dx + A 0 (x) A(x)
U (x) U ( x)
2 ;
where A(x) = j x j 2 +1 B(x), and B a function of class C 1 on R , even and positive. The operator ; corresponds to the function
A(x) = A ; (x) = 2 (sinh j x j ) 2 +1 (cosh j x j ) 2 +1 : Using the relation
d
dx ' ; (x) =
2 + 2
4( + 1) sinh(2x)' +1; +1 (x);
the function ; can be written in the form above (see [2])
; (x) = ' ; (x) + i
4( + 1) sinh(2x)' +1; +1 (x); x 2 R :
Denote L p ; ( R ) = L p ; ( R ; A ; (x)dx), 1 < p 2, the space of measurable functions f on R such that
k f k p; ; = Z
R j f (x) j p A ; (x)dx
1=p
< + 1 :
Using the eigenfunctions ; of the operator ; called the Jacobi-Dunkl kernels, we de…ne the Jacobi-Dunkl transform of a function f 2 L p ; ( R ) by
F ; f ( ) = Z
R
f (x) ; (x)A ; (x)dx; 2 R ; and the inversion formula
f (x) = Z
R F ; f ( ) ; (x)d ( );
where
d ( ) = j j
8 p 2
2 j C ; ( p 2
2 ) j 1 Rn ] ; [ ( )d : Here,
C ; ( ) = 2 i ( + 1) (i )
( 1 2 ( + i )) ( 1 2 ( + 1 + i )) ; 2 Cn (i N );
and 1 Rn ] ; [ is the characteristic function of Rn ] ; [. The Jacobi-Dunkl transform is a unitary isomorphism from L 2 ; ( R ) onto L 2 ( R ; d ( )), i.e.,
k f k 2; ; = kF ; (f ) k L
2( R ;d ( )) : (1) Plancherel’s theorem (1) and the Marcinkiewics interpolation theorem (see [8]) we get for f 2 L p ; ( R ) with 1 < p 2 and q such that 1 p + 1 q = 1,
kF ; (f ) k L
q( R ;d ( )) K k f k p; ; ; (2) where K is a positive constant (see [9]).
The operator of Jacobi-Dunkl translation is de…ned by T x f (y) =
Z
R
f(z)d x;y ; (z); 8 x; y 2 R ;
where x;y ; (z); x; y 2 R are the signed measures given by
d x;y ; (z) = 8 >
> <
> >
:
K ; (x; y; z)A ; (z)dz , if x; y 2 R ;
x , if y = 0;
y , if x = 0:
Here, x is the Dirac measure at x. And,
K ; (x; y; z) = M ; (sinh j x j sinh j y j sinh j z j ) 2 1 I
x;yZ
0
(x; y; z) (g (x; y; z)) + 1 sin 2 d ;
I x;y = [ j x j j y j ; jj x j j y jj ] [ [ jj x j j y jj ; j x j + j y j ];
(x; y; z) = 1 x;y;z + z;x;y + z;y;x ;
8 z 2 R ; 2 [0; ]; x;y;z =
( cosh x+cosh y cosh z cos
sinh xsinh y , if xy 6 = 0;
0 , if xy = 0;
g (x; y; z) = 1 cosh 2 x cosh 2 y cosh 2 z + 2 cosh x cosh y cosh z cos ;
t + =
( t , if t > 0;
0 , if t 0;
and
M ; = 8 <
:
2
2( +1)
p ( ) ( +
12) , if > ;
0 , if = :
In [2], we have
F ; (T h f)( ) = ; (h) F ; (f )( ) and F ; ( ; f )( ) = i F ; (f )( ): (3) For 2 1 , we introduce the Bessel normalized function of the …rst kind de…ned by
j (x) = ( + 1) X 1 n=0
( 1) n ( x 2 ) 2n
n! (n + + 1) ; x 2 R : Moreover, we see that
x lim ! 0
j (x) 1
x 2 6 = 0; (4)
by consequence, there exists C 1 > 0 and > 0 satisfying
j j (x) 1 j C 1 j x j 2 for j x j : (5) By Lemma 9 in [4], we obtain the following Lemma 1.2.
LEMMA 1.2. Let 2 1 ; 6 = 2 1 . Then for j v j , there exists a positive constant C 2 such that
1 ' +i ; (x) C 2 j 1 j ( x) j :
Denote by L m p ( ; ); 1 < p 2; m = 0; 1; 2; :::, the class of functions f 2 L p ; ( R ) that have generalized derivatives satisfying m ; f 2 L p ; ( R ), i.e.,
L m p ( ; ) = n
f 2 L p ; ( R ) : m ; f 2 L p ; ( R ) o
; where 0 ; f = f and m ; f = ; ( m ; 1 f ) for m = 1; 2; :::.
2 Dini Lipschitz Condition
Denote N h by
N h = T h + T h 2I;
where I is the unit operator in the space L p ; ( R ).
DEFINITION 2.1. Let f 2 L m p ( ; ), and de…ne k N h m ; f (x) k p; ; C h
(log h 1 ) ; 0 and m = 0; 1; 2; :::;
i.e.,
k N h m ; f (x) k p; ; = O h (log 1 h ) ;
for all x in R and for all su¢ ciently small h; C being a positive constant. Then we say that f satis…es a Jacobi-Dunkl Dini Lipschitz of order , or f belongs to Lip( ; ; p).
DEFINITION 2.2. If
k N h m ; f (x) k p; ; h
(log
h1)
! 0; as h ! 0;
i.e.,
k N h m ; f (x) k p; ; = O h (log 1 h ) ;
then f is said to belong to the little Jacobi-Dunkl Dini Lipschitz class lip( ; ; p).
REMARK. It follows immediately from these de…nitions that lip( ; ; p) Lip( ; ; p):
THEOREM 2.3. Let > 1 . If f 2 Lip( ; ; p), then f 2 lip(1; ; p).
PROOF. For x 2 R , h small and f 2 Lip( ; ; p); we have k N h m
; f (x) k p; ; C h (log 1 h ) : Then
(log 1
h ) k N h m ; f (x) k p; ; Ch : Therefore,
(log 1 h )
h k N h m
; f (x) k p; ; Ch 1 ; which tends to zero with h ! 0. Thus
(log 1 h )
h k N h m
; f (x) k p; ; ! 0; h ! 0:
Then f 2 lip(1; ; p).
THEOREM 2.4. If < , then Lip( ; 0; p) Lip( ; 0; p) and lip( ; 0; p) lip( ; 0; p):
PROOF. We have 0 h 1 and < , then h h . Then the proof of the
theorem is immediate.
3 New Results on Dini Lipschitz Class
LEMMA 3.1. For f 2 L m p ( ; ), then Z
R
2 q qm j ' ; (h) 1 j q jF ; f ( ) j q d ( )
1 q
K k N h m ; f k p; ; ;
where 1 p + 1 q = 1 and m = 0; 1; 2; ::::
PROOF. From (3), we have
F ; ( m ; f )( ) = i m m F ; (f )( ); m = 0; 1; 2; :::: (6) By using formulas (3) and (6), we conclude that
F ; (N h m ; f )( ) = i m ( ( ; ) (h) + ( ; ) ( h) 2) m F ; (f )( );
Since
( ; )
(h) = ' ; (h) + i
4( + 1) sinh(2h)' +1; +1 (h);
( ; )
( h) = ' ; ( h) i
4( + 1) sinh(2h)' +1; +1 ( h);
and ' ; is even, we see that F ; (N h m
; f )( ) = 2i m (' ; (h) 1) m F ; (f )( ):
Now by formula (2), we have the result.
THEOREM 3.2. Let > 2. If f belongs to the Jacobi-Dunkl Dini Lipschitz class, i.e.,
f 2 Lip( ; ; p); > 2; 0:
Then f is null almost everywhere on R .
PROOF. Assume that f 2 Lip( ; ; p). Then we have k N h m
; f(x) k p; ; C h
(log h 1 ) ; 0:
From Lemma 3.1, we get Z
R
qm j ' ; (h) 1 j q jF ; f ( ) j q d ( ) K q C q 2 q
h q (log h 1 ) q : In view of Lemma 1.3, we conclude that
Z
R j 1 j ( h) j q qm jF ; f ( ) j q d ( ) K q C q 2 q C 2 q
h q
(log h 1 ) q :
Then R
R j 1 j ( h) j q qm jF ; f ( ) j q d ( ) h 2q
K q C q 2 q C 2 q
h q 2q (log 1 h ) q : Since > 2; we have
h lim ! 0
h q 2q (log 1 h ) q = 0:
Thus
h lim ! 0
Z
R
j 1 j ( h) j
2 h 2
q
2q qm
jF ; f ( ) j q d ( ) = 0;
and also from the formula (4) and Fatou theorem, we obtain Z
R 2q qm
jF ; f ( ) j q d ( ) = 0:
Hence 2 m F ; f ( ) = 0 for all 2 R , and so f (x) is the null function.
Analog of the Theorem 3.2, we obtain this theorem.
THEOREM 3.3. Let f 2 L m p ( ; ). If f belong to lip(2; 0; p), i.e., k N h m ; f (x) k p; ; = O(h 2 ); as h ! 0:
Then f is null almost everywhere on R .
THEOREM 3.4. Let f belong to Lip( ; ; p). Then Z
j j r
qm jF ; f ( ) j q d ( ) = O r q
(log r) q ; as r ! 1 ; where 1 p + 1 q = 1 and m = 0; 1; 2; :::.
PROOF. Let f 2 Lip( ; ; p). Then k N h m
; f (x) k p; ; = O h
(log h 1 ) ; as h ! 0:
From Lemma 3.1, we have Z
R
qm j ' ; (h) 1 j q jF ; f ( ) j q d ( ) K q
2 q k N h m
; f (x) k q p; ; : By (5) and Lemma 1.2, we get
Z
2h
j j
hqm j 1 ' ; (h) j q jF ; (f )( ) j q d ( )
C 1 q C 2 q Z
2h
j j
hqm j h j 2q jF ; (f )( ) j q d ( ):
From 2h j j h ; we have
2h
2 2 2
h
2 2 ;
which implies that
2 h 2
2
4
2 h 2 : Take h 3 , then we have 2 h 2 C 3 = C 3 ( ). So
Z
2h
j j
hqm j 1 ' ; (h) j q jF ; (f )( ) j q d ( )
C 1 q C 2 q C 3 q Z
2h
j j
hqm jF ; (f )( ) j q d ( ):
Note that there exists then a positive constant C 4 such that Z
2h
j j
hqm jF ; (f )( ) j q d ( ) C 4
Z
R
qm j 1 ' ; (h) j q jF ; (f )( ) j q d ( )
= O h q
(log h 1 ) q : So we obtain Z
r j j 2r
qm jF ; (f )( ) j q d ( ) C 5
r q (log r) q ; where C 5 is a positive constant. Now, we have
Z
j j r
qm jF ; (f )( ) j q d ( ) = X 1 i=0
Z
2
ir j j 2
i+1r
qm jF ; (f )( ) j q d ( )
C 5
r q
(log r) q + (2r) q
(log 2r) q + (4r) q (log 4r) q + C 5 r q
(log r) q + (2r) q
(log r) q + (4r) q (log r) q + C 5 r q
(log r) q 1 + 2 q + (2 q ) 2 + (2 q ) 3 + K r q
(log r) q ;
where K = C 5 (1 2 q ) 1 since 2 q < 1. Consequently Z
j j r
qm jF ; f ( ) j q d ( ) = O r q
(log r) q ; as r ! 1 :
Thus, the proof is …nished.
COROLLARY 3.5. Let f 2 L m p ( ; ). If k N h m ; f k p; ; = O h
(log h 1 ) ; as h ! 0;
then Z
j j r jF ; (f )( ) j q d ( ) = O r qm q
(log r) q ; as r ! 1 : where 1 p + 1 q = 1 and m = 0; 1; 2; :::.
DEFINITION 3.6. A function f 2 L m p ( ; ) is said to be in the ( ; )-Jacobi-Dunkl Dini Lipschitz class, denoted by Lip( ; ; p), if
k N h m ; f (x) k p; ; = O (h)
(log 1 h ) ; as h ! 0; 0;
where
(1) (t) a continuous increasing function on [0; 1 ), (2) (0) = 0,
(3) (ts) = (t) (s) for all t; s 2 [0; 1 ).
THEOREM 3.7. Let f belong to Lip( ; ; p). Then Z
j j r
qm jF ; f ( ) j q d ( ) = O (r q )
(log r) q ; as r ! 1 ; where 1 p + 1 q = 1 and m = 0; 1; 2; :::.
PROOF. Let f 2 Lip( ; ; p). Then
k N h m ; f (x) k p; ; = O (h)
(log h 1 ) ; as h ! 0:
From Lemma 3.1, we have Z
R
qm j ' ; (h) 1 j q jF ; f ( ) j q d ( ) K q
2 q k N h m ; f (x) k q p; ; : By (5) and Lemma 1.2, we get
Z
2h
j j
hqm j 1 ' ; (h) j q jF ; (f )( ) j q d ( )
C 1 q C 2 q Z
2h
j j
hj h j 2q qm jF ; (f )( ) j q d ( ):
From 2h j j h we have
2h
2 2 2
h
2 2 ;
which implies that
2 h 2
2
4
2 h 2 : Take h 3 , then we have 2 h 2 C 3 = C 3 ( ). So
Z
2h
j j
hqm j 1 ' ; (h) j q jF ; (f )( ) j q d ( )
C 1 q C 2 q C 3 q Z
2h
j j
hqm jF ; (f )( ) j q d ( ):
Note that there exists then a positive constant C 4 such that Z
2h