Selçuk J. Appl. Math. Selçuk Journal of Vol. 8. No.. pp. 101 - 112, 2007 Applied Mathematics
On Multilinear Fractional Integrals Generated by the Quasi-distance Hüseyin Yıldırım, Mehmet Zeki Sarıkaya and Umut Mutlu Ozkan
Department of Mathematics, Faculty of Science and Arts, Kocatepe University, Afyon-Turkey;
e-mail:hyildir@ aku.edu.tr,sarikaya@ aku.edu.tr,umut_ ozkan@ aku.edu.tr
Received : October 10, 2007
Summary. In R, we prove 1 × × boundedness for the multilinear
fractional integrals generated by the quasi-distance (1 )() =
Z
1( − 1)2( − 2)( − ) kk−
where the ’s are nonzero and distinct. We also prove multilinear versions of
two inequalities for fractional integrals and a multilinear Lebesgue differentiation theorem.
Key words:Fractional Integral, Riesz Potential and Quasi-distance. 2000 Mathematics Subject Classification. 31B10, 44A15 and 47B37 1. Introduction
Boundedness properties of the classical bilinear and multilinear fractional in-tegrals were studied by many authors. We refer to papers in [5-10, 12, 19]. In those studies, the kernels are homogeneous. But, in our paper the kernel is that nonhomogeneous. The importance here is, those kernel were studied by Grafakos and other’s are particular case of our kernel. The bi(sub)linear maximal function ( )() = sup 0 1 2 Z − |( + )( − )|
or the bilinear Hilbert transform ( )() =
Z
( + )( − )
map (R1) × 0(R1) → 1(R1) boundedness into 1 for the corresponding multilinear fractional integrals can be obtain.
We defined a quasimetric (a non-isotropic quasi-distance) in R by
k − k:= (|1− 1| 21 + | 2− 2| 22 + + | − | 2 )2 |1| where = (1 2 ) 0 = 1 2 , 0 and ¯ ¯1 ¯ ¯ = 1 1+ 1 2+ + 1
. This quasi-metric is named as non-isotropic quasi-distance see [4], [14-16]
and [21].
For ∈ R+ and ∈ R define = (
11 ) The quasi-metric has
the following properties:
1 kk= 0 ⇔ = = (0 0 0) 2 °°°° = () | 1 | kk 3 k + k≤ (kk+ kk) where = 2(+2max) 1 2|1| max= max{1 2 }
Now, we consider spherical coordinates by the following formulas : 1= ( cos 1) 1 = ( sin 1sin2 sin −1) We obtained that kk =
|1| It can be seen that the Jacobian
( ) of
this transformation is ( ) = |
1 |−1Ω
() where Ω() is the bounded
function, which only depends on angles 1 2 −1 It is clear that if =
= 1 then the quasi-metric is Euclidean metric on R.
Throughout this note, will denote an integer ≥ 2 and = 1 will
be fixed, distinct and nonzero real numbers. We are going to work in and
0 . We denote by the -tuples (1 ) and by the -linear
fractional integral operator generated by the quasi-distance defined as follows:
(1 )() =
Z
1( − 1)2( − 2)( − ) kk−
When = = 1 the operators are the multilinear fractional
integrals as studied in [6]. When = 1 and = = 1 the operators
are the usual fractional integrals as studied in [17]. When = 1 and
1= 1 the operator is the Riesz potential generated by the quasi-distance
which is studied in the [14-17], [21]. We also denote by ( ) the −linear maximal function ( )() = sup 0 (Ω| 1 |)−1 Z |1( − 1)| |( − )|
= () = { ∈ R: kk }
is the ball of centered at origin. It is trivial to check that for any positive 1 with harmonic mean 1 maps 1××into . If we denote
by ∗ the Hardy-Littlewood maximal function of , then is dominated by
the product (( 1 1 )∗) 1(( )∗)
and hence its boundedness follows from Hölder’s inequality and the
bound-edness of the Hardy-Littlewood maximal function. This argument breaks down when = 1 but a slight modification of it gives that maps into 1∞ in this
endpoint case. It is conceivable, however, that map into 1 since it carries −tuples of compactly supported functions into compactly supported functions. This problem remains unsolved. The × → boundedness of the bilinear Hilbert transform ( ) is more subtle and it remains unsolved even in the case 1
The aim of this paper is to show that boundedness of inhomogeneous version of multilinear fractional integrals generated by the Quasi-distance. Note that our result is the generalization of corresponding results in the homogeneous case, given in [6]. Our first result concerns the 1× × → boundedness of
for ≥ 1
Theorem 1: Let be the harmonic mean of 1 1 and let be such that 1
= 1 −
Then, maps 1××into for
+ ≤ ≤
(equivalently
1 ≤ ∞).
Note that in the case = 1, the corresponding range of is the smaller interval 1 (equivalently − ∞).
When = 1 = = 1 , the following theorem has been proved by
Hirschman [12] for periodic function and Hedberg [10] for positive functions. Furthermore, this theorem has been proved by Grafakos [6] for = =
1
Theorem 2: Let be positive real numbers and let 1 be their harmonic
mean. Then for 1 and 0 1
(1.1) k k ≤ k k
Y
kk1−
where 1 =+1−
In the endpoint case = Trudinger [20] for = 1 and Strichartz [19] for other proved exponential integrability of when = 1 = = 1 .
Hedberg [10] and Grafakos [6] gave a proof of Theorem 3 below when = 1 = = 1 , and = = 1 , respectively.
By −1we denote the area of the unit sphere −1 The factor in the
expo-nent below is a normalizing factor and should be there by homogeneity.
Theorem 3: Let =
be the harmonic mean of 1 1. Let be a
ball of radius in R and let
∈ () be supported in Then for any
1, there exists a constant 0() depending on the ’s and on
such that (1.2) R exp µ −1 ³ (12) k1k1kk ´ −¶ ≤ 0()| 1 | where =Q | |
All the comments in this paragraph refer to the case = 1 Hempel et al. [11] (for = 1) and later Adams [1] (for all ) showed that inequality (12) cannot hold if 1 Moser [13] showed exponential integrability of
1 −1 −1 ³ |()| k∇k ´ −1
suggesting that Theorem 3 be true in the endpoint case = 1 = =
1 (Use formula (18), p. 125 in [17] to show that Moser’s result follows from an improved Theorem 3 with = 1 = = 1 ) In fact, Adams [1]
provide inequality (12) in the endpoint case = 1 = = 1 and also
deduced the sharp constants for Moser’s exponential inequality for higer order derivatives. Chang and Marshall [2] proved a similar sharp exponential inequal-ity concerning the Dirichlet integral. Later, Grafakos [6] provide inequalinequal-ity (12) in the case =³1 − | 1 | ´ − 1 = = 1
When 1, Theorem 3 does not hold either, while the case = 1 remains open when ≥ 2
2. Proof of Theorems
Proof of Theorem 1: We will adapt to our paper the proof given by Grafakos [6] in the homogeneous case. We denote by || the measure of the ball
and by the characteristic function of the set We also use the notation 0= −1 for ≥ 1
We consider first the case ≥ 1. In this case we will show maps 1 ×
× → ∞ The required result, when 1, follows from an application
of the Marcinkiewicz interpolation theorem. Without loss of generality we can assume that ≥ 0 and that kk = 1 Fix a 0 and define 0 by
−1 à −1 ( − ) ¯ ¯1 ¯ ¯ 0+ ¯¯1 ¯ ¯ !1 0 −|1| = 2
where −1 and are as in Theorem 3. Hölder’s inequality and our choice of
(2.1) ∞ () = R kk 1( − 1)( − ) kk− ≤ ⎛ ⎝ R kk |1( − 1)( − )| ⎞ ⎠ 1 × ⎛ ⎝ R kk kk(−)0 0 kk ⎞ ⎠ 1 0 ≤ Q | |− k( − )k () µ −1 (−) | 1 |0+| 1 | ¶1 0 −| 1 | = 2 Let 0 () = R kk≤ 1( − 1)( − ) kk− .We compute its norm : (2.2) ° ° °0 ° ° ° ≤ ° ° ° ° ° µR µ Q ¶ kk−kk≤ ¶1 × ³Rkk−kk≤ ´1 0 ° ° ° ° ≤ |1|01 µR Rµ Q ¶ kk−kk≤ ¶1 ≤ | 1 | 1 0 ⎛ ⎝Q k k R kk≤ kk−kk≤ ⎞ ⎠ 1 ≤ | 1 | 1 0+ | 1 | 1 = | 1 | By (2.1) the set { : ∞ ()
2 } is empty. This fact together with
Chebyshev’s inequality and (2.2) gives
{ : () } ≤ { : 0 () 2 } + { : ∞ () 2 } ≤ 2− ° ° °0 ° ° ° ≤ −| 1 | = −
which is the required weak type estimate for
We now do the case + ≤ ≤ 1. The corresponding range of ’s is 1 ≤ ≤ − Assume that = 2 and that 1≥ 2 1 Also assume that = 1
(2.3) k(1 2)k1 = R R 1( − 1)2( − 2) kk− = R1() R 2( − (2− 1) ) kk− ≤ |2− 1|−[ (1 2)] | 1 |(−)R 1()(2) () ≤ 12k1k1k2k01 where (1 2) = max{|2− 1| 1 | 2− 1| } 12= |2− 1| −[ ( 1 2)] | 1 |(−)
Note that = 1 implies 10 1 + = 1 2 Since 1 2
we have the following
inequality [17]
k(1 2)k1 ≤ 12k1k1k2k2
The case of general 1 follows by interpolating between the endpoint case = 1 and the case of close to ∞ Suppose now that the Theorem is true for − 1 ≥ 3. We will show that it is true for Again we first do the case = 1 We may assume without loss of generality that 1≥ ≥ 1 Now,
(2.4) k k1 = R R 1( − 1)( − ) kk− = R1()R2( − (2− 1) )( − (− 1) ) kk− = Q 6=1| − 1|−[ (1 )] | 1 |(−)R 1()(2 ) () ≤ 1k1k1k(2 )k01
where (1 ) has properties of (1 2) Define 1 by 11 = 1−11 Since
= 1 we have 10 1 +
=
1
1 We can apply the induction hypothesis only
provided + ≤ 1 ≤ This inequality follows from the identity 1 + = 1
which relates and = 1 From our induction hypothesis we have the following inequality for (24)
k k1 ≤ 12
Q
kk.
The case ≥ 1 follows by interpolation.
Proof of Theorem 2. As in the proof of Theorem 1, fix ≥ 0 such that
kk1 = 1. Let 0. Then ( )() = ⎛ ⎜ ⎝ R kk |1| + R kk≥ |1| ⎞ ⎟ ⎠Q( − ) kk− = 1+ 2.
For the 1we have 1 = ∞ P =0 R (2−−1)|1|≤kk (2−) |1| Q ( − ) kk− ≤ P∞ =0 (2−−1)| 1 |(−) R (2−−1)|1|≤kk (2−) |1| Q ( − ) = P∞ =0 (2−−1)(−) | 1 |(2 −+1)|1| (2−+1)|1| × ⎛ ⎜ ⎝ R (2−−1)|1|≤kk (2−) |1| Q ( − ) ⎞ ⎟ ⎠ = 1 |1|() ∞ P =0 2−| 1 | ≤ 2 | 1 |().
For the 2we have
2 = R kk≥ |1| Q ( − ) kk−kk (−1) ≤ (−1)| 1 | R kk≥ |1| Q ( − ) kk− = (−1)|1| ( ) ().
Thus, there is the following inequality for any 0
( ) () ≤ ³ |1|() + (−1)|1| ( ) () ´ . Minimizing the right-hand side with respect to , we see that its minimum is reached at min = 3( ()) − |1| ( ()) |1|
and easy evaluations give
( ) () ≤ [( ())][ ()]1−.
Hölder’s inequality with exponents 1 = 1 ( 1−)+ 1 will give (2.5) k k ≤ ° ° °( ) ° ° ° ° ° °[()]1−°°° 1− ≤ k kk()k 1−
From the following inequality for the Maximal function on (by the bound-dedness of the maximal function on ) with 1
= + 1− kk1− ≤ Q k k1− ,
there is the following inequality for (2.5)
k k ≤ k k
Q
k
k1− ≤ k k
Proof of Theorem 3. A simple dilation argument shows that if we know Theorem 3 for a specific value of = 0 with a constant 01() on the right
hand side of (21), then we also know it for all other values of with constant 1 0() ³ 0 ´|1 | . We select 0= 1 where = 2 min ||
−1and we will assume
that the radius of is 0. Furthermore, we can assume that the ’s satisfy
≥ 0 and kk = 1.
Now fix ∈ . The same argument as in Theorem 2 with = 1 gives
(2.6) () ≤ | 1 |() + R kk≥ |1| Q ( − ) kk−
since all are supported in the ball and ∈ the integral in (26) is over
the set n : |1| ≤ kk
≤ 0= 1
o
. Hölder’s inequality with exponents 1 2 and − gives (2.7) R |1|≤kk ≤1 Q ( − ) kk(−) ≤Qk( − )k() ⎛ ⎜ ⎝ R |1|≤kk ≤1 kk−|1| ⎞ ⎟ ⎠ − ≤ µ Q | | ¶−1⎛ ⎝−1 1 R |1| −|1||1|−1 ⎞ ⎠ = −1 µ −1ln 1 |1| ¶− . Combining (26) and (27) we get
(2.8) () ≤ | 1 |() + −1 µ −1 ln 1 |1| ¶− . The choice = 1 gives
() ≤ ()
for all ∈ and therefore the selection = () =
h () () i |1 | will
satisfy ≤ 1 for all ≤ 1. (28) now implies
() ≤ | 1 | () + −1 à −1 ln à ( ()) | 1 |( ()) !!− . Algebraic manipulation of the above gives
(2.9) −1 h³ 1 − | 1 | ´ () i − ≤ ln à ( ()) | 1 |( ()) ! .
We exponentiate (29) and we integrate over the set 1= { ∈ : () ≥ 1}
to obtain R 1 exp ∙ −1 ³³ 1 − | 1 | ´ () ´ −¸ ≤ 1 | 1 | R 1 ( ()) ( ()) ≤ 1 |1| R 1 ( ()) ≤ 2 |1|
The last inequality follows from the boundedness of the maximal function of on . The integral of same exponential over the set 2 = − 1 is
estimated trivially by R 2 exp ∙ −1 ³³ 1 − | 1 | ´ () ´ −¸ ≤ exp¡−1 −¢|2| ≤ 3Ω |1 | 0 = 4
Adding the integrals above 1and 2we obtain the required inequality with
a constant 01() = max(2 4)(1 + (1 − − )− )
where =³1 − | 1 |
´ −
. The constant 0() in the statement of
Theo-rem 3 is then 01() −|1 | 0 = 01()| 1 |.
We obtain the following.
Corollary. Let , , and be as in Theorem 3, then (1 2 )
is in (
) for every 0. In fact,
k(1 2 )k() ≤
Q
k k
for some constant C depending on and the ’s.
The corollary follows since exponential integrability of implies
integrabil-ity to any power . (Here 1 is fixed). 3. A Multilinear Differentiation Theorem
We end this study by proving the following multilinear Lebesgue differentiation theorem.
Let ∈ (R) and suppose that the harmonic mean of 1 2 is ≥ 1
Then lim →0 () = lim→0 1 Ω |1| R kk≤ |1| 1( − 1)( − ) = 1()() a.e.
The case = 1 is a consequence of weak type inequality |{ ∈ R: () }| ≤
k1k1 kk
which is easily obtained from
|{ ∈ R: () }| ≤ P =1 ¯ ¯ ¯n ∈ R : ( )∗() ³ −1 ´o¯¯ ¯ ≤ P =1 ³ −1 ´− kk
after minimizing over all 1 2 0 (Take 0= ). The standard argument
presented in [18], p.61, will prove that the sequence { ()}0 is Cauchy for
almost all and therefore it converges in this case 0= = = 1 2
Since for continuous 1 it converges to the value of their product at the
point ∈ R, to deduce the general case it will suffice to show that { ()}0
converges to the product of the ’s in the norm as → 0 Setting ( ()) =
k( ) − 1k ≤ 1 Ω |1| R kk≤ |1| ° ° ° ° ° Q − Q ° ° ° ° ° ≤ 1 Ω |1| R kk≤ |1| P =1 ° °− ° ° Q 6=k k → 0
as kk→ 0 since the last integrand is a continuous function of which
van-ishes at the origin. The last inequality above follows by adding and subtracting 2 − 2 suitable terms and applying Hölder’s inequality times.
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