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CONVOLUTION INEQUALITIES IN LORENTZ SPACES ERLAN NURSULTANOV AND SERGEY TIKHONOV

Abstract. In this paper we study boundedness of the convolution operator in different Lorentz spaces. In particular, we obtain the limit case of the Young-O’Neil inequality in the classical Lorentz spaces. We also investigate the convolution operator in the weighted Lorentz spaces. Finally, norm inequalities for the potential operator are presented.

1. Introduction The Young convolution inequality of the form kf ∗ Kkp 6 kf kp kKk1 and in a more general form 1 1 1 = + q p r plays a very important role both in Harmonic Analysis and PDE. O’Neil [ON], Yap [Ya], and Blozinski [Bl3] studied the boundedness of the convolution operator Z (1.1) Af (y) = K(y − x)f (x)dx kf ∗ Kkq 6 kf kp kKkr ,

1+

D

in the Lorentz spaces. In particular, the following Young-O’Neil inequality was obtained: for 1 < p, q, r < ∞, 0 < h1 , h2 , h3 6 ∞, 1 + 1q = p1 + 1r , and 1 1 1 h1 = h2 + h3 , one has (1.2)

kAf kLqh1 (Ω) 6 ckf kLph2 (D) kKkLrh3 (Ω−D)

where Ω − D = {x − y : x ∈ Ω, y ∈ D}. In this paper we continue investigating the Young-type inequalities in different Lorentz spaces. The outline of the paper is as follows. In section 2 we study the boundedness of the operator A from Lph2 (Ω) into Lph1 (Ω), i.e., the limit case of the 2000 Mathematics Subject Classification. Primary 46E30; Secondary 44A35, 47G10. Key words and phrases. Young-O’Neil inequality, Lorentz spaces, Convolution, Riesz Potential Operator. 1

2

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Young-O’Neil inequality (p = q and r = 1). It is known (see [Bl2, Theorem 2]) that if Ω = Rn , h1 < h2 6 ∞ and K > 0, then A : Lph2 (Rn ) −→ Lph1 (Rn ) a.e.

implies A ≡ 0, i.e., K = 0. We show that in the case when Ω is of finite measure, the same problem has a nontrivial solution: for 1 6 p = q < ∞ one has k(K ∗ f )∗∗ kp,h1 6 ckf ∗∗ kp,h2 kK ∗∗ k1,h3 . The case p = q = ∞ is studied separately in Section 3. In this case we consider the L∞q -spaces ([BRS], [BMR]) and we obtain kK ∗ f kL∞,h1 6 2 kf kL∞,h2 kK ∗∗ kL1,h3 . Further, Section 4 is devoted to the general Young-O’Neil-type inequality in the weighted Lorentz spaces Λq (ω), Γq (ω), and S q (ω), i.e., ( ) Z ∞ 1/q
q q ∗ Λ (ω) = kf kΛq (ω) = f (t) ω(t) dt 0 h2 f dt for any non-increasing nonnegative function f on (0, ∞) (see, for example, [BS, p. 56]), and (2.3) follows.  Lemma 2.2. Let 1 < p < ∞ and 1 6 h < ∞. If f ∈ Lph (Ω, µ), then 1

lim t p f ∗∗ (t) = 0

t→0

1

lim t p f ∗∗ (t) = 0.

and

t→∞

Proof. It follows from Z s Z s 1 1 dt dt (t p f ∗∗ (t))h = 0 (t p f ∗∗ (t))h = lim lim s→∞ s/2 s→0 s/2 t t and Z

s

dt (t f ∗∗ (t))h t s/2

! h1 1

1 p

> cph s p f ∗∗ (s).



Theorem 2.1. Suppose measurable functions f, g, and K are defined on [0, 1]; then Z (2.4)

1

g ∗ (t)(K ∗ f )∗∗ (t)dt 6 2

1

Z

0

tg ∗∗ (t)f ∗∗ (t)K ∗∗ (t)dt.

0

If we also assume that lim t2 K ∗∗ (t)f ∗∗ (t)g ∗∗ (t) = 0,

(2.5)

t→+0

then Z (2.6)

1

g ∗ (t)(K ∗ f )∗∗ (t)dt 6 kgkL1 kf kL1 kKkL1

0

Z + 0

1

tg ∗∗ (t)f ∗∗ (t) (K ∗∗ (t) − K ∗ (t)) dt.

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

5

Proof. From (2.2) and the Hardy-Littlewood inequality [BS, p.44], we write Z

1

g ∗ (s)(K ∗ f )∗∗ (s)ds

0

Z

1

Z

1

Z 1 |K(y − x)| dydxds |e| e |e|=s 0 0 e⊂[0,1]  Z ∗∗ Z 1 Z 1 1 f ∗ (t) (t) dtds 6 g ∗ (s) sup |K(y − ·)| dy |e| e |e|=s 0 0 g ∗ (s) sup

6

|f (x)|

e⊂[0,1]

Z

1

g ∗ (s) sup

=

|e|=s e⊂[0,1]

0

Z 6

1

g ∗ (s)

0

1

Z 0

Z

1

f ∗ (t) sup |ω|=t ω⊂[0,1]

0

1 1 |e| |ω|

Z Z |K(y − x)| dxdydtds e

ω



 1 1 f ∗ (t)  sup sup |e| |ω| |e|=s |ω|=t e⊂[0,1] ω⊂[0,1]

Z Z |K(y − x)| dxdy  dtds. e

ω

We consider 1 1 Φ(s, t) = sup |e| |ω| |e|=s

Z Z |K(y − x)| dydx ω

|ω|=t

1 1 = sup |e|=s |e| |w|

|e|=s |w|=t

6 sup |e|=s |w|=t

1

Z

Z

0

1 1 |e| |w|

Z

1 1 |e| |w|

Z

1

χω (y) |K(y − x)| dydx

χe (x)

|ω|=t

= sup

e

0 1

|K(x)| |e ∩ (w + x)|dx 0 1

K ∗ (ξ)φ(ξ)dξ,

0

where φ(ξ) is the non-increasing rearrangement of the function |e ∩ (ω + x)|. Then φ(ξ) satisfies φ(s) 6 min(|e|, |w|) and Z

1

φ(ξ)dξ = |e||w|. 0

6

ERLAN NURSULTANOV AND SERGEY TIKHONOV

We assume that s = |e| < |w| = t, then the function φ0 (ξ) = φ(ξ)/|e| satisfies conditions of Lemma 2.1 with d = |w|. Then for s < t we have Z 1 1 K ∗ (ξ)φ0 (ξ)dξ Φ(s, t) 6 sup |e|=s,e⊂[0,1] |w| 0 |w|=t,w⊂[0,1]

1 |w|=t |w|

|w|

Z

K ∗ (ξ)dξ = K ∗∗ (t)

6 sup

0

As above, for s > t we write |e|

Z

1 Φ(s, t) 6 sup |e| |e|=s

K ∗ (ξ)dξ = K ∗∗ (s)

0

Therefore we have Z 1 Z ∗ ∗∗ g (s)(K ∗ f ) (s)ds 6 (2.7) 0

1

0

f ∗ (t)K ∗∗ (max{s, t})dt ds.

g (s)

0

Using this inequality, we get Z 1 Z g ∗ (s)(K ∗ f )∗∗ (s)ds 6

1

Z

∗

0

1

g ∗ (s)

1

Z

0

f ∗ (t)K ∗∗ (max(t, s))dtds

0 1

Z

∗

=

s

Z

∗∗

f ∗ (t)dtds

g (s)K (s) 0

0 1

Z

Z

∗

+

g (s) 0

1

f ∗ (t)K ∗∗ (t)dtds

s

1

Z

tK ∗∗ (t)g ∗∗ (t)f ∗∗ (t)dt.

62 0

Thus, inequality (2.4) is verified. To prove (2.6), we use Z s Z s Z s Z ∗ ∗ ∗ ∗ g (s) f (t)dt + f (s) g (s)dt = g ∗ (t)dt 0

0

Therefore, (2.7) implies Z 1 Z g ∗ (s)(K ∗ f )∗∗ (s)ds 6 0

1

g ∗ (s)

0

Z =

0

= 0

0 f (t)dt . ∗

0

1

Z

f ∗ (t)K ∗∗ (max(t, s))dtds

0 1

  Z s Z s K ∗∗ (s) g ∗ (s) f ∗ (t)dt+f ∗ (s) g ∗ (t)dt ds

0

Z

s

0 1

K ∗∗ (s)

Z 0

s

g ∗ (t)dt

0

Z 0

s

0 ∗ f (t)dt ds.

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

7

Integrating by parts, we get 1

Z

∗

∗∗

s

Z

∗∗

g (s)(K ∗ f ) (s)ds 6 K (s) 1

s

Z

f ∗ (t)dt

− 0

0

Hence, using (K ∗∗ (s))0 = − and (2.5), we obtain

1 s2

Rs 0

0

Z

s

∗

f (t)dt 0

0

Z

1 g (t)dt

s

Z

∗

0 0

g ∗ (t)dt (K ∗∗ (s)) ds.

0


K ∗ (t)dt− 1s K ∗ (s) = − 1s (K ∗∗ (s)−K ∗ (s))

1

Z

g ∗ (t)(K ∗ f )∗∗ (t)dt 6 kgkL1 kf kL1 kKkL1

0

Z

1

+

tg ∗∗ (t)f ∗∗ (t) (K ∗∗ (t) − K ∗ (t)) dt.

0

The proof is complete.



Theorem 2.2. Let 1 6 p, q, r 6 ∞, 1 6 h1 , h2 , h3 6 ∞, and 1 1 1 +1= + , q p r

(2.8)

1 1 1 = + . h1 h2 h3

Suppose f ∈ Lp,h2 [0, 1] and K ∈ Lr,h3 [0, 1] are 1-periodic functions. a) (Young-O’Neil) If 1 < p, q, r < ∞, then K ∗ f ∈ Lq,h1 and kK ∗ f kLq,h1 6 ckf kLp,h2 kKkLr,h3 . b) If 1 < p = q < ∞ or p = q = h1 = h2 = h3 = ∞, then K ∗ f ∈ Lp,h1 and kK ∗ f kLp,h1 6 ckf kLp,h2 kK ∗∗ k1,h3 .

(2.9)

c) If p = q = 1 and 1 < h1 , h2 , h3 < ∞, then K ∗ f ∈ L1h1 and k(K ∗ f )∗∗ k1,h1 6 ckf ∗∗ k1,h2 kK ∗∗ k1,h3 . Proof. Let f ∗∗ ∈ Lp,h2 , K ∗∗ ∈ Lr,h3 , and g ∗∗ ∈ Lq0 ,h01 , where p, r, q, h1 , h2 , h3 satisfy (2.8). If p1 = q 0 , p2 = p, p3 = r, then p11 + p12 + p13 = 2 and h10 + 1 h2

+

1 h3

1

= 1. This implies that at least one of the parameters

differs from zero. Let us assume that

1 h01

1 1 1 h01 , h2 , h3

> 0. Then Lemma 2.2 implies

8

ERLAN NURSULTANOV AND SERGEY TIKHONOV

limt→0 t1/p1 g ∗∗ (t) = 0. Applying the H¨ older inequality twice, we get 1

1

1

t2 K ∗∗ (t)f ∗∗ (t)g ∗∗ (t) = (t p3 K ∗∗ (t))(t p2 f ∗∗ (t))(t p1 g ∗∗ (t)) Z t h03 !1/h03 Z t h02 !1/h02 ds 1 1 0 0 ds s p3 s p2 6 1−1/p 1−1/p 3 2 s s t t 0 0  1  kK ∗∗ kLp3 ,h3 kf ∗∗ kLp2 ,h2 t p1 g ∗∗ (t)   1 = CkK ∗∗ kLp3 ,h3 kf ∗∗ kLp2 ,h2 t p1 g ∗∗ (t) . Therefore, lim t2 K ∗∗ (t)f ∗∗ (t)g ∗∗ (t) = 0.

(2.10)

t→0

By Theorem 2.1 and the H¨ older inequality, we have Z 1 g(y)(K ∗ f )(y)dy kf ∗ KkLq,h1  sup kgkL

=1

q 0 ,h01

0

1

Z

g ∗ (y)(K ∗ f )∗∗ (y)dy

sup

6

kgkL

=1

q 0 ,h01

[ kgkL1 kf kL1 kKkL1

sup

6

kgkL

=1

q 0 ,h01

1

Z

0

 t (K ∗∗ (t) − K ∗ (t)) f ∗∗ (t)g ∗∗ (t)dt

+ 0

[ kgkL1 kf kL1 kKkL1

sup

6

kgkL

q 0 ,h01

Z

1

+

=1



  1 1 1 1 1 − 10 t p3 − h3 K ∗∗ (t) (t p2 − h2 f ∗∗ (t))(t p1 h1 g ∗∗ (t))dt

0

6c

sup kgkL

q 0 ,h01

=1

kK ∗∗ kr,h3 kf ∗∗ kp,h2 kg ∗∗ kq0 ,h01 .

Thus we derive (2.10)

kf ∗ KkLq,h1 6 c

sup kgkL

q 0 ,h01

=1

kK ∗∗ kr,h3 kf ∗∗ kp,h2 kg ∗∗ kq0 ,h01 .

In the case a), i.e., 1 < p, r, q, < ∞, we have kf ∗∗ kp,h2  kf kLp,h2 , kg kq0 ,h01  kgkLq0 ,h0 , kK ∗∗ kr,h3  kKkLr,h3 and inequality (2.10) implies ∗∗

1

kf ∗KkLq,h1 6 c

sup kgkL

q 0 ,h01

=1

kKkLr,h3 kf kLp,h2 kgkLq0 ,h0 = ckKkLr,h3 kf kLp,h2 . 1

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

9

b) Let now r = 1 and 1 < p = q < ∞, f ∈ Lph2 , K ∈ L1,h3 , g ∈ L . Then since at least one of the parameters h10 , h12 , h13 differs from 1 zero, by Lemma 2.2, we write (2.10) and, therefore, (2.10). Then because of kf ∗∗ kp,h2  kf kLp,h2 and kg ∗∗ kp0 ,h01  kgkLp0 ,h0 , we finally have p0 ,h01

1

kf ∗ KkLp,h1 6 c1

kK ∗∗ k1,h3 kf kLp,h2 kgkLp0 ,h0

sup kgkL

p0 ,h01

1

=1

= c1 kK ∗∗ k1,h3 kf kLp,h2 . For the case p = q = h1 = h2 = h3 = ∞, we simply write kf ∗ KkL∞ 6 ckf kL∞ kKk1 . c) Let q = p = r = 1 and 1 < h1 < ∞. Then using [CGMP, Th 4.1] and periodicity of functions, we get1 Z 1 ∗∗ g(y)(K ∗ f )(y)dy k(K ∗ f ) k1,h1  sup kgkL

∞h01

=1

0 1

Z =

f (x)(g ∗ K)(x)dx

sup kgkL

∞h01

=1

0

Z sup

6

kgkL

∞,h01

=1

1

f ∗ (s)(g ∗ K)∗∗ (s)ds.

0

We again apply Theorem 2.1 and the H¨ older inequality:  k(K ∗ f )∗∗ k1,h1 6 c sup kf kL1 kf kL1 kKkL1 kgkL

∞h01

=1 1

Z

 tf (t)K (t) (g (t) − g (t)) dt ∗∗

+

∗∗

∗∗

∗

0

6c

sup kgkL

=1 ∞,h01

kf ∗∗ k1,h2 kKk∗1,h3  Z kgk1 + 0

1

! 10  0 h1 (g ∗∗ − g ∗ )h1  dt t

= ckf ∗∗ k1,h2 kK ∗∗ k1,h3 . The proof is complete.



1See the definition of the space L ∞h0 in the next section 1

10

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Let us give two examples showing sharpness of the results of Theorem 2.2. First, we prove that in inequality (2.9), the factor kK ∗∗ kL1,h3 could not be changed to kKkL1,h3 . That is, in general, for 1 6 p = q < ∞ and 1 1 1 h1 = h2 + h3 the inequality kK ∗ f kLp,h1 6 Ckf kLp,h2 kKkL1,h3

(2.11) does not hold.

1/p

Example 2.1. Let N ∈ N. We define f (t) = (min(N, 1/t)) min(N, 1/t). Then

and K(t) =

1/h2

kf kLp,h2  (ln N ) and 1/h3

kKkL1,h3  (ln N )

= (ln N )

1/h1 −1/h2

.

Define ( 0 ϕ(x) = (K ∗ f )(x)

for x ∈ [0, a) , for x ∈ [a, 1)

a=


1 p e +1 . N

Therefore if x ∈ [a, 1) we write Z x− N1 Z x− N1 ds 1 ds N 1/p ln(N x − 1) ϕ(x) > > = . 1 1 1 x−s (N x − 1)1/p s p (x − s) (x − N1 ) p N1 N  0 ξ p Noting ξln1/pξ = pξp−ln 1/p+1 < 0 for ξ ∈ [e , 1], we obtain that the function ln ξ ξ 1/p

is decreasing on [ep , 1]. Hence, ln N (t + a) − 1 ϕ (t) >
1/p t + a − N1




∗

for

t ∈ (0, 1 − a).

Using this, we get kf ∗

KkhL1p,h 1

Z

1

(t1/p ϕ∗ (t))h1

> 0

Z

1−a



t1/p ln(N (t + a) − 1) (t + a − N1 )1/p

h1

dt t

1−a



t1/p ln(N (t + a) − 1) (t + a − N1 )1/p

h1

dt t

> 0

Z

dt t

> a

> 2−

h1 p

Z a

1−a

(ln(N (t + a) − 1))h1 dt t 1−a

 lnh1 +1 (N (t + a) − 1) |a

 (ln N )h1 +1 .

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

11

Thus, (2.11) implies ln N 6 C. This contradiction concludes the proof. We now provide an example showing sharpness of (2.9) in the following sense: none of the parameters could be changed to make the inequality stronger. Corollary 2.2.1. Let 1 < p < ∞, 1 6 h1 , h2 6 ∞ and 1/h = (1/h1 − 1/h2 )+ . Then for the Ces` aro operator of 1-periodic function f , given by N

1 X Sk (f ; x) N +1 k=0 Z 1 f (y)FN (x − y)dy, =

σN (f ; x) =

FN (x)

0

=

1 sin2 π(2N − 1)x , N sin2 πx

we write 1/h

kσN kLp,h2 →Lp,h1  kFN∗∗ k1,h  (ln N )

.

Proof. By Theorem 2.2, we have kσN kLp,h2 →Lp,h1 6 CkFN∗∗ k1,h . Further, we estimate Z 1/N Z t  ∗ h dt ∗∗ kFN kL1,h 6 c FN (s)χ[0,1/2] (s) ds t 0 0 Z

1

+ 1/N

Z

1/N

Z

t

+

0

1/N



!h 1/h ∗ dt  FN (s)χ[0,1/2] (s) ds . t

Using the known inequalities FN (s) 6 cN for x > 0 and FN (s) 6 c/(N x2 ) for x ∈ (1/N, 1), we write !1/h Z 1/N Z 1 dt 1/h ∗ h h−1 6 c (ln N ) . kFN kL1,h 6 c N t dt + 0 1/N t Finally, we get kσN kLp,h2 →Lp,h1 6 c (ln N ) On the other hand, defining f0 (x) =

N X 1 2πikx e 1/p k k=1

1/h

.

12

ERLAN NURSULTANOV AND SERGEY TIKHONOV

and using Hardy-Littlewood’s theorem on Fourier coefficients for the Lorentz space [Se], we obtain h2 !1/h2 N  X 1 1/h 1/p 1  (ln N ) 2 kf0 kLp,h2  k 1/p k k k=1 h !1/h1 N  X N −k 1 1 1/h kσN (f0 )kLp,h1   (1 + ln N ) 1 . N k k=1

Therefore, we derive kσN kLp,h2 →Lp,h1 >

kσN (f0 )kLp,h1 1/h 1/h −1/h2 = (ln N ) ,  (ln N ) 1 kf0 kLp,h2

completing the proof.



3. Convolution in the Lorentz space of periodic functions: the case of p = q = ∞ Theorem 2.2 does not encompass the limit case p = q = ∞ (hi < ∞). It is clear that in this case the classical Lorentz space is trivial. We consider another scale of Lorentz spaces. Following Bennett et al. [BRS] (see also [BMR], [CGMP]), we define L∞h [0, 1] as follows ( L∞q [0, 1] =

f ∈ L1 [0, 1] : kf kL∞q [0,1] := kf kL1 [0,1] 1

Z + 0

)  q1 q (f ∗∗ − f ∗ ) dt 1. Without loss of generality, we can assume that K and f are from C ∞ . Put h ≡ K ∗ f . If Z 1 1/h1 h1 dt ∗∗ ∗ khk1 > (h (t) − h (t)) t 0 then khkL∞,h1 6 2khk1 6 2kf kL∞,h2 kK ∗∗ kL1,h3

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

15

and (3.2) is proved. Now suppose the reverse: 1

Z

∗∗

(h (t) − h (t))

khk1 6

(3.3)

∗

h1

0

1/h1

dt t

.

Then 1

Z khkL∞,h1 6 2

∗∗

∗

(h (t) − h (t))

dt t

h1

0 1

Z

1/h1

g(t) (h∗∗ (t) − h∗ (t)) dt

=2 0

where h −1

(h∗∗ (t) − h∗ (t)) 1 g(t) =  R1 h t 0 (h∗∗ (s) − h∗ (s)) 1

ds s

1/h01

The function g satisfies the following conditions: 1) 2) 3)

g(t) > 0, g(1) 6 1 (see (3.3)), 1/h01 R 1 h0 = 1, (tg(t)) 1 dt t 0

4)

g ∈ C ∞.

By A denote a collection of all functions satisfying 1) - 4). From (3.3), we get 1

Z kK ∗ f kL∞,h1 6 2 sup

g∈A

  ∗∗ ∗ g(t) (K ∗ f ) (t) − (K ∗ f ) (t) dt.

0

For g ∈ A, Z

1

1

Z   g(t) h∗∗ (t) − h∗ (t) dt = −

0

 0 tg(t) h∗∗ (t) dt

0

1 Z = −tg(t)h∗∗ (t) + 0

= −g(1)h∗∗ (1) +

1

 0   tg(t) h∗∗ (t) dt

0

Z

1



0   tg(t) h∗∗ (t) dt

0

Z 6 0

1



tg(t)

0 

 h∗∗ (t) dt.

16

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Hence, 1

Z kK ∗ f kL∞,h1 6 sup

g∈A

(3.4)



0 tg(t) h∗∗ (t) dt

0 1

Z =

sup g∈A, (tg(t))0 >0



0 tg(t) h∗∗ (t) dt.

0

We now use Lemma 3.1: 1

Z



0 tg(t) (K ∗ f )∗∗ (t)dt

0 1

Z 6

Z t h (tg(t))0 f ∗ (1)kKk1 + K ∗∗ (t) (−f ∗ (s))0 s ds

0

0

Z

1

+

i (−f ∗ (s))0 sK ∗∗ (s)ds dt

t

Z

∗

1

(tg(t))0 K ∗∗ (t)

= f (1)g(1)kKk1 + 0

Z +

t

(−f ∗ (s))0 s dsdt

0

1

(−f ∗ (t))0 sK ∗∗ (t)

0

= f ∗ (1)g(1)kKk1 +

Z

t

Z

(sg(s))0 dsdt

0

Z

1

K ∗∗ (t)

t

Z

0

(−f ∗ (s))0 sds

Z

0

t

0 (sg(s))0 ds dt

0

Z t Z t  1 ∗ ∗∗ ∗ 0 0 = f (1)g(1)kKk1 + K (t) (−f (s)) sds (sg(s)) ds 0

1

Z

t−1 (K ∗∗ (t) − K ∗ (t))

+ 0

Z

0

t

0

Z t  (−f ∗ (s))0 sds (sg(s))0 ds dt.

0

0

Integrating by parts, we write Z

t

(−f ∗ (s))0 sds = −f ∗ (t)t +

0

Z

t

f ∗ (s)ds = t (f ∗∗ (t) − f ∗ (t))

0

and therefore Z

1



0 tg(t) (K ∗ f )∗∗ (t)dt 6 g(1)kKk1 kf k1

0

Z + 0

1

t (K ∗∗ (t) − K ∗ (t)) (f ∗∗ (t) − f ∗ (t)) g(t)dt.

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

17

By the H¨ older inequality and the definition of A, we obtain ( Z 1 1/h2 h dt kK ∗ f kL∞,h1 6 2 sup g(1)kKk1 kf k1 + (f ∗∗ (t) − f ∗ (t)) 2 t g∈A 0 1/h3 Z 1 1/h01 ) Z 1 h01 dt h3 dt ∗∗ ∗ (tg(t)) [t(K (t)−K (t))] × t t 0 0 6 2 kf kL∞,h2 kK ∗∗ kL1,h3 . Let now h1 = 1. Then Z 1 K(x − y)f (y)dy kf ∗ KkL∞1 = kf ∗ KkL∞ = sup x∈[0,1] 0 Z t Z 1 1 K ∗ (s)ds K ∗ (t)f ∗ (t)dt = f ∗ (t) 6 0

0

0

Z

1

(−f ∗ (t))0 tK ∗∗ (t)dt

+ 0

0 (−f (s)) sds K ∗∗ (t)dt = f (1)kKk1 + 0 0 Z t  1 ∗ ∗ 0 = f (1)kKk1 + (−f (s)) sds K ∗∗ (t) 0 0  Z 1 Z t  dt + (−f ∗ (s))0 sds K ∗∗ (t) − K ∗ (t) t 0 0 Z 1  = f ∗ (1)kKk1 + (−f ∗ (s))0 sds kKkL1 Z

∗

1

Z

t

∗

0

0 1

Z

t

  dt + (−f (s)) sds K ∗∗ (t) − K ∗ (t) t 0 0 Z 1 = kf k1 kKk1 + (f ∗∗ − f ∗ )(t)(K ∗∗ (t) − K ∗ (t))dt Z

∗

0

0

6 kf kL∞h2 kK ∗∗ kL1h2 The proof is complete.



4. Convolution in the weighted Lorentz spaces Let µ be the Lebesgue measure and Ω, D be measurable subsets of Rn . For functions f and K defined on Ω and D − Ω we consider the convolution Z Af (y) = K(y − x)f (x)dx. Ω

18

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Theorem 4.1. Let f , g, and K be measurable functions on Ω, D, and D−Ω of Rn . One has ∞

Z 0

∞ g ∗ (s)(Af )∗∗ (s)ds 6 t2 f ∗∗ (t)g ∗∗ (t)K ∗∗ (t) 0 Z ∞ + tf ∗∗ (t)g ∗∗ (t)(K ∗∗ (t) − K ∗ (t))dt; 0

Z (4.1)

∞ ∗

Z

∗∗

∞

tf ∗∗ (t)g ∗∗ (t)K ∗∗ (t)dt;

g (s)(Af ) (s)ds 6 2 0

0

and Z

∞

(4.2)

∗

Z

∗∗

∞

g (s)(Af ) (s)ds 6

Z

∗

∞

g (t)

0

0

f ∗ (s)K ∗∗ (|t − s|)dsdt.

0

Proof. It is similar to the proof of Theorem 2.1. Indeed, the same technique that we used implies the following inequality: Z (4.3)

∞

g ∗ (s)(Af )∗∗ (s)ds 6

Z

0

0

∞

g ∗ (s)

Z

∞

f ∗ (t)K ∗∗ (max(s, t))dt ds.

0

Then Z 0

∞

∞

Z s g ∗ (s) f ∗ (t)K ∗∗ (s)dtds 0 0 Z ∞ Z ∞ + g ∗ (s) f ∗ (t)K ∗∗ (t)dtds 0 s Z s 0 Z ∞ Z s = K ∗∗ (s) f ∗ (t)dt g ∗ (t)dt ds 0 0 0 ∞ 2 ∗∗ ∗∗ ∗∗ = t f (t)g (t)K (t) 0 Z s  Z ∞ Z s 0 ∗∗ ∗ ∗ − (K (s)) f (t)dt g (t)dt ds 0 0 0 ∞ = t2 f ∗∗ (t)g ∗∗ (t)K ∗∗ (t) 0 Z ∞ + sf ∗∗ (s)g ∗∗ (s) (K ∗∗ (s) − K ∗ (s)) ds,

g ∗ (s)(Af )∗∗ (s)ds 6

Z

0

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

19

The second inequality follows from Z ∞ Z ∞ Z s g ∗ (s)(Af )∗∗ (s)ds 6 g ∗ (s)K ∗∗ (s) f ∗ (t)dtds 0 0 Z ∞0 Z ∞ ∗ f ∗ (t)K ∗∗ (t)dtds g (s) + s Z ∞ 0 Z s ∗ ∗∗ = g (s)K (s) f ∗ (t)dtds 0 0 Z ∞ Z t + f ∗ (t)K ∗∗ (t) g ∗ (s)dsdt 0 0 Z ∞ 62 tf ∗∗ (t)g ∗∗ (t)K ∗∗ (t)dt. 0 ∗∗

∗∗

Using K (max(s, t)) 6 K (|t − s|), (4.3) implies inequality (4.2). The proof is now complete.  Now we are going to study the Young-O’Neil inequality in weighted Lorentz spaces Λq (ω) and Γq (ω). We remark (see [Sa]) that Λq (ω) = Γq (ω) if and only if the weight ω satisfies the Bp -condition, that is, Z ∞ Z ω(t) C x dt 6 ω(t)dt. tq xq 0 x We also recall the definition of the associated space   Z   E 0 = kf kE 0 = sup f g dµ(x) < ∞ .   g∈E, X kgkE =1

It is known ([CRS, 2.4], [KM]) that (4.4)

0

Γq (˜ ω ) = (Λq (ω))

0

q > 1,

where (4.5)

0

0

ω e (t) = tq W (t)−q ω(t),

and Z W (x) =

x

ω(t)dt. 0

In the case of the classical example ω(t) = tq/p−1 , 1 < p, q < ∞, we have 0 0 0 0 kf kΛq (ω) = kf kΓq (ω) = kf kLpq , ω e (t) = tq /p −1 , i.e., (Lpq )0 = Lp q .

20

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Theorem 4.2. Let ω ∈ Bq . We have (4.6)

kK ∗ f kΓq (ω)  kK ∗ f kΛq (ω) 6 Ckf kΓs1 (ω1 ) kKkΓs2 (ω2 )

for 1/q = 1/s1 + 1/s2 , 1 < q < ∞, 1 6 s1 , s2 6 ∞, and Z t ω1 (t)1/s1 ω2 (t)1/s2 (4.7) ω(x) dx 6 c ω(t) . ω(t)1/s1 +1/s2 0 Proof. We first remark that (4.7) can be rewritten as follows 0

t 6 cω e (t)1/q ω1 (t)1/s1 ω2 (t)1/s2 .

(4.8)

Then by Theorem 4.1 ( see (4.1)) and H¨ older’s inequality, we write Z  Z g(y) f (x)K(x − y)dx dy R R Z ∞h ih i ih 0 6 2c g ∗∗ (t)e ω (t)1/q f ∗∗ (t)ω1 (t)1/s1 K ∗∗ (t)ω2 (t)1/s2 dt 0 ∞

Z 6 2c 0

 q10 Z q 0 g (t) ω e (t) dt ∗∗

∞

 s1 s1 1 f (t) ω1 (t) dt ∗∗

0

Z

∞

∗∗

K (t)

s2

 s1 2 ω2 (t) dt

0

6 2ckgkΓq0 (eω) kf kΓs1 (ω1 ) kKkΓs2 (ω2 ) . Considering the supremum over g such that kgkΓq0 (eω) = 1, we obtain kK ∗ f k(Γq0 (eω))0 6 2c kf kΓs1 (ω1 ) kKkΓs2 (ω2 ) .
00 Finally, since ω ∈ Bp , we have Λq (ω) = Λq (ω) and then (4.6) follows from (4.4).  Corollary 4.2.1. Under the hypotheses of Theorem 4.2, if additionally, ω1 ∈ Bs1 and ω2 ∈ Bs2 , then kK ∗ f kΛq (ω) 6 Ckf kΛs1 (ω1 ) kKkΛs2 (ω2 ) . Examples. 1. For ω(t) = tq/h−1 , ω1 (t) = ts1 /p−1 , ω2 (t) = ts2 /r−1 and 1/q = 1/s1 + 1/s2 , inequality (4.7) is equivalent to hq t1−1/h 6 c t1/p+1/r . Hence, for 1 + 1/h = 1/p + 1/r and for 1 < p, r, h < ∞, we have kK ∗ f kh,q 6 CkK ∗∗ kp,s1 kf ∗∗ kr,s2 6 CkKkp,s1 kf kr,s2 , (see (2.1)).

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

21

2. Let ω(t) = tq/h−1 ξ1Aq (t), ω1 (t) = ts1 /p−1 ξ2Bs1 (t), ω2 (t) = ts2 /r−1 ξ3Ds2 (t), where 1/q = 1/s1 + 1/s2 , 1 + 1/h = 1/p + 1/r, and ξi are slowly oscillating functions. Then inequality (4.7) is equivalent to ξ1A (x) 6 Cξ2B (x)ξ3D (x), i.e., in this case we obtain the Young-O’Neil-type inequality for the LorentzZygmund spaces [BS, p. 253]. Further we study the convolution inequality for periodic functions. Suppose ω, ω1 , ω2 are weights on [0, 1] and 0 < q < ∞; then by definition, ) ( 1 Z 1

S q (ω) =

q

(f ∗∗ (s) − f ∗ (s))q ω(s)ds

kf kS q (ω) = kf k1 + 0

Theorem 4.3. Let ω2 ∈ L1 . We have kK ∗ f kS q (ω) 6 Ckf kS s1 (ω1 ) kKkS s2 (ω2 ) for 1/q = 1/s1 + 1/s2 , 1 < q < ∞, 1 6 s1 , s2 6 ∞, and tω 1/q (t) 6 Cω1 1/s1 (t)ω2 1/s2 (t).

(4.9)

Proof. Since ω2 ∈ L1 we have kKk1 6 CkKkΓs2 (ω2 ) . In the case when (h = K ∗ f ) Z 1  q1 ∗∗ ∗ q , kf k1 6 (f (s) − f (s)) ω(s)ds 0

we write Z khkSq (ω) 6 2 Z =2

1

1/q (h (t) − h (t)) ω(t)dt ∗∗

∗

q

0 1

g(t) (h∗∗ (t) − h∗ (t)) dt

0

where

 q−1 ω(t) h∗∗ (t) − h∗ (t) g(t) =  1/q0 . R1 ∗∗ (s) − h∗ (s))q ω(s)ds (h 0

We remark that the function g satisfies the following conditions: 1) g(t) > 0, 2) g(1) 6 ω(1), 1/q0   R 1 g(t) q0 ω(t)dt = 1, 3) ω(t) 0 4)

g ∈ C ∞.

kAkLp1 →Lq∞ Z Z 1 1 ν(y) µ−1 (x)|x − y|γ−n dxdy  sup 1/q 0 |w|1/p |e|>0,|w|>0 |e| e w Z Z 1 ν(y) µ−1 (x)|x − y|γ−n dxdy. > sup 1 1 +p |e|=|w|>0 q0 e w |w| 1/26|w|/|e|62 Let us now show the accuracy of the right-hand side inequality. We denote M ∗ := {e ∈ Rn : 0 < |e| < ∞} and take e, w ∈ M ∗ . Suppose also that |e| > |w|. Then there exists an integer M > 1 such that (M − 1)|w| < |e| 6 M |w|, where p, q are any numbers satisfying 1 < p < n/γ, 1/q < 1/p. γ Let K(x) = |x|γ−n . Noting K ∗∗ (|t − s|)  |t − s| n −1 and using inequality (4.2) from theorem 4.1, we estimate Z Z 1 1 ν(y) µ−1 (x)|x − y|γ−n dxdy |e|1/q0 |w|1/p e w Z |e| Z |w| C ∗ 6 ν (t) (µ−1 )∗ (s)|t − s|γ/n−1 dsdt ((M − 1)|w|)1/q0 |w|1/p 0 0 C 6 (M |w|)1/q0 |w|1/p

Z

|w| −1 ∗

(µ 0

) (s)

M Z X k=1

k|w|

(k−1)|w|

ν ∗ (t)|t − s|γ/n−1 dtds.

24

ERLAN NURSULTANOV AND SERGEY TIKHONOV

We divide the last expression into two terms: Z |w| Z 2|w| C −1 ∗ (µ ) (s) ν ∗ (t)|t − s|γ/n−1 dtds (M |w|)1/q0 |w|1/p 0 0 Z |w| M Z k|w| X C −1 ∗ (µ + ) (s) ν ∗ (t)|t − s|γ/n−1 dtds (M |w|)1/q0 |w|1/p 0 k=3 (k−1)|w| =: I1 + I2 . By changing the variables t → t + (k − 1)|w|, Z k|w| Z |w| γ/n−1 ∗ γ/n−1 ν (t)|t−s| dt. dt = ν ∗ (t+(k−1)|w|) t+(k−1)|w|−s (k−1)|w|

0

Further, we estimate γ/n−1  γ/n−1  γ/n−1 6 (k − 1)|w| − |t − s| 6 (k − 2)|w| (t − s) + (k − 1)|w| and then Z Z |w| γ/n−1 dt 6 ν ∗ (t+(k−1)|w|) t+(k−1)|w|−s

0

0

|w|



ν ∗ (t) 1−γ/n dt. (k − 2)|w|

This yields C I2 6 (M |w|)1/q0 |w|1/p

Z

|w| −1 ∗

(µ 0

Z

|w|

) (s) 0

M 1 ν ∗ (t) X dt ds. |w|1−γ/n k=3 (k − 2)1−γ/n

0

Noting that 1/q > 1 − 1/p + γ/n, we get M −1/q

0

M X

0

(k − 2)γ/n−1 6 CM γ/n−1/q 6 C.

k=3

Summing up the estimates for I1 and I2 , we finally have Z Z 1 1 ν(y) µ−1 (x)|x − y|γ−n dxdy |e|1/q0 |w|1/p e w Z |w| Z 2|w| 1 −1 ∗ 6C (µ ) (s) ν ∗ (t)|t − s|γ/n−1 dtds (M |w|)1/q0 |w|1/p 0 0 ! Z |w| Z |w| 1 −1 ∗ ∗ γ/n−1 + (µ ) (s) ν (t)|t − s| dtds |w|1/p+1/q0 0 0 Z |e| Z |w| 1 ∗ ν (t) (µ−1 )∗ (s)|t − s|γ/n−1 dsdt. 6C sup 1 +1 1/26|w|/|e|62 |w| q0 p 0 0 If |ω| > |e|, then we use similar estimates (we also use γ/n < 1/p).

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

25

Thus we obtain the following estimates kAkLp,1 (Rn )→Lq,∞ (Rn ) 6 CJ, where J=

sup 1/26|w|/|e|62

Z

1 |w|

1 q0

1 +p

|e|

Z

∗

|w|

(µ−1 )∗ (s)|t − s|γ/n−1 dsdt.

ν (t) 0

0

We note that the expression on the right-hand side is independent on all pairs (p, q) such that (1/p−1/q) is a constant. Therefore, if J < ∞, then the operator A is a (p, q) weak-type operator for 1/p−1/q = C and 1 < p < γ/n. Further, there exist two pairs (p0 , q0 ) and (p1 , q1 ) such that 1 < p0 < p < p1 < ∞, 1 < q0 < q < q1 < ∞, and 1/p0 − 1/q0 = 1/p1 − 1/q1 = 1/p − 1/q. Therefore, if the right-hand side of (5.2) is bounded, then A : Lp0 ,1 → Lq0 ,∞

and A : Lp1 ,1 → Lq1 ,∞ ,

where the norms are controlled by CJ. Then by the interpolation theorem, we write A : Lp(θ) → Lq(θ) and kAk 6 C J for 1/p(θ) = (1 − θ)/p0 + θ/p1 ,

1/q(θ) = (1 − θ)/q0 + θ/q1 .

We have p(θ) = p for some 0 < θ < 1. Since 1/p0 − 1/q0 = 1/p1 − 1/q1 = 1/p(θ) − 1/q(θ), in this case q(θ) coincides with q. The proof is now complete.  Note that in the case when weights satisfy some regular conditions, the left-hand side and the right-hand side integrals are equivalent. Then Theorem 5.1 implies the following relation for the norm of the potential operator. Corollary 5.1.1. Let 1 < p 6 q 6 ∞, 1/q 6 1/p − γ. Suppose weights µ and ν satisfy Z c µ(x)dx, µ∗ (t) 6 t 2t 6|x|6t Z (5.3) c ν ∗ (s) 6 ν(x)dx, s 2s 6|x|6s then kIγ kLp (R,µ−1 )→Lq (R,ν) 

sup |e| 1 2 6 |w| 62

1

Z

1 1

|e|

Z

1

|e| q0 |w| p

0

0

|w|

µ∗ (t)ν ∗ (s) dsdt. |t − s|1−γ

26

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Remark. Any monotone or quasi-monotone (i.e., there exists τ > 0 such that µ(t)t−τ is monotone) µ and ν satisfy (5.3). Proof. By Theorem 5.1, it is sufficient to prove the inequality Z |e| Z |w| ∗ 1 1 µ (t)ν ∗ (s) dsdt sup 1 1 |t − s|1−γ |e| 1 0 |e| q0 |w| p 0 2 6 |w| 62 Z Z 1 1 µ(x)ν(y) 6 C sup dxdy. 1 1 1−γ |e| 1 |e| q0 |w| p e w |x − y| 2 6 |w| 62 Using (5.3), we have R Z |e|Z |w| R t Z |e| Z |w| ∗ µ(η)ν(ξ)dηdξ s µ (t)ν ∗ (s) 6|η|6t 2 2 2 6|ξ|6s dsdt 6 c dsdt |t − s|1−γ ts|t − s|1−γ 0 0 0 0 Z 2|η|Z 2|ξ| Z |e| Z |w| dsdt µ(η)ν(ξ) =C dηdξ. 1−γ |ξ| ts|t − s| |η| −|e| −|w| Now let 0 < ξ < η. If ξ > η4 , then Z 2ξ Z 2η Z ∞Z ∞ Z ∞Z ∞ dtds dtds dtds 1 6 = 1−γ 1−γ 1−γ ξ ts|t − s| |t − s| ts η |t − s|1−γ ts η ξ ξ η 1 η Z ∞Z ∞ dtds cγ cγ 1 = 1−γ 6 . 6 1−γ 1−γ ts 1 η |t − s| η |η − ξ|1−γ 1 4 If ξ < η4 , then Z 2ξ Z ξ

η

2η

dtds 1 6 1−γ |t − s| ts |η − 2ξ|1−γ =

2ξ

Z

Z

ξ

η

2η

dtds ts

c (ln 2)2 c . = 1−γ 6 |η − η2 |1−γ η |η − ξ|1−γ

Similarly, in the case of 0 < η 6 ξ, one has Z 2ξ Z 2η cγ dtds 6 . 1−γ |t − s| ts |η − ξ|1−γ ξ η Thus, Z

2|η|

|η|

and

|e|

|w|

Z

2|ξ|

|ξ|

cγ dsdt 6 |t − s|1−γ ts |η − ξ|1−γ

µ∗ (t)ν ∗ (s) dtds 6 cγ |t − s|1−γ 0 0 which concludes the proof. Z

Z

Z

|e|

−|e|

Z

|w|

−|w|

µ(ξ)ν(η) dξdη, |ξ − η|1−γ 

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

27

6. Convolution in multidimensional Lorentz spaces Let vectors p = (p1 , . . . , pn ), q = (q1 , . . . , qn ) be such that 0 < pj , qj 6 ∞, j = 1, . . . , n and if pj = ∞, then qj = ∞. Then for two vectors p and q and for a given rearrangement ∗ = (j1 , j2 , · · · , jn ) of the sequence (1, 2, · · · , n) we define the following functional Φpq∗ (ϕ)  Z =

∞

Z ...

0

0

∞

 q1 1 qj1  qqj2 jn 1 j1 p dt dt j1 jn  t 1 . . . tnpn ϕ(t1 , . . . , tn ) , ... 1 tj1 tjn


1/q R∞ where the expression 0 (G(s))q ds is understood as sups>0 G(s) for s q = ∞. Let m = (m1 , · · · , mn ) ∈ Nn and let f (x1 , · · · , xn ) be a measurable function on Rm = Rm1 × · · · × Rmn . Applying decreasing rearrangement according to x1 ∈ Rm1 , · · · , xn ∈ Rmn with other variables being fixed we will construct a function f ∗1 ,··· ,∗n (t1 , · · · , tn ). This function is called a decreasing rearrangement of f in Rm . The Lorentz space Lpq∗ (Rm ) is defined as the collection of measurable functions on Rm such that kf kLpq∗ (Rm ) = Φpq∗ (f ∗1 ···∗n ) < ∞. In the case m = (1, · · · , 1) and ∗ = (1, · · · , n), this space was introduced by Blozinski [Bl1]. He also obtained corresponding convolution inequalities. In a general case, this definition can be found in [Nu]. Below we prove the Young-O’Neil-type inequality for this case. Theorem 6.1. Let m = (m1 , · · · mn ) ∈ Nn and let 1 < q, p, r < ∞, 1/q + 1 = 1/p + 1/r, 1 6 h, h1 , h2 6 ∞, and 1/h = 1/h1 + 1/h2 . Let also ∗ = (j1 , · · · , jn ) be a rearrangement of (1, 2, · · · , n). If K ∈ Lr,h∗1 (Rm ) and g ∈ Lp,h∗2 (Rm ), then for the convolution Z (K ∗ g)(y) =

K(x − y)g(x)dx Rm

one has (6.1)

kK ∗ gkLq,h∗ (Rm ) 6 CkKkLr,h∗ (Rm ) kgkLp,h∗ (Rm ) . 1

2

28

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Proof. First, we show that for all measurable functions f , g and K on Rm one has (6.2) Z

Z

n

f (y) Rm

∞

Z

!t1 · · · tn f ∗∗ (t1 , · · · , tn )

···

g(x)K(x−y) dx dy 6 2 Rm

∞

Z 0

0

g ∗∗ (t1 , · · · , tn )K ∗∗ (t1 , · · · , tn )dt1 · · · dtn , where t1

Z

1 f ∗∗ (t1 , · · · , tn ) = Qn

tn

Z ···

i=1 ti

0

f ∗1 ···∗n (s1 , · · · , sn )ds1 · · · dsn .

0

Using (4.1), Z Z f (y) g(x)K(x − y) dx dy m Rm Z Z Z R Z ··· ··· = m2 m2 Rmn Rmn Z R  Z R f (y1 , y) K(x1 − y1 , x − y)g(x1 , x)dx1 dy1 dx dy Rm1 Rm1 Z Z Z Z ··· ··· 62 Rm2 Rm2 Rmn Rmn Z ∞  t1 K ∗∗ (t1 , x − y)g ∗∗ (t1 , x)f ∗∗ (t1 , y) dt1 dx dy 0

Z =2

∞

Z

Z

f (∗∗)1 (t1 , y)

···

t1

Rm2

Rmn

0

Z ··· Rmn

!

Z K

(∗∗)1

(t1 , x − y)g

(∗∗)1

(t1 , x) dt1 dx dy,

Rm2

where (x) = (x2 , · · · , xn ) and dx = dx2 · · · dxn . Continuing this process, we will obtain the required inequality. Secondly, inequality (6.2) implies Z Z f (y) K(x − y)g(x)dx dy kK ∗ gkLq,h∗ (Rm ) = sup kf kL

(R q0 h0 ∗

m ) =1

Rm

Z

6 2n

sup

kf kL

(R q0 h0 ∗

m ) =1

0

Z

= 2n

sup

kf kL

(R q0 h0 ∗

m ) =1

0

Rm

∞

Z ∞ · · · tf ∗∗ (t)g ∗∗ (t)K ∗∗ (t)dt1 · · · dtn 0

∞Z ∞ · · · tf ∗∗ (t)g ∗∗ (t)K ∗∗ (t)dtj1 · · · dtjn . 0

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

29

Then the H¨ older inequality implies kK ∗ gkLq,h∗ (Rm ) 6 2n

sup kf kL

(R q0 h0∗

m ) =1

kf ∗∗ kLq0 h0 ∗ (Rm ) kK ∗∗ kLr,h∗ (Rm ) kg ∗∗ kLp,h∗ (Rm ) 1

2

Using Hardy’s inequality, one can see for 1 < p < ∞ that kψkLp,h∗ (Rm )  kψ ∗∗ kLp,h∗ (Rm ) := Φp,h∗ (ψ ∗∗ ). Thus, we derive kK ∗ gkLq,h∗ (Rm )

6 ckKkLr,h∗ (Rm ) kgkLp,h∗ (Rm ) . 1

2

The proof is complete.



We note that in the case of m = (1, · · · , 1) and ∗ = (1, · · · , n) the O’Neiltype inequality (6.1) was proved by Blozinski [Bl1]. Remarks. 1. The classical Young-O’Neil inequality for the convolution Af = K ∗ f in the case 1 < p = (p, · · · , p) < q = (q, · · · , q) < ∞ and 1/r = 1 + 1/q − 1/p is given by kAkLp (Rn )→Lq (Rn ) 6 ckKkLr∞ (Rn ) . Also, Stepanov [St] proved the following inequality (6.3)

kAkLp (Rn )→Lq (Rn ) 6 ck · · · kKkLr∞ (Rx1 ) · · · kLr∞ (Rxn ) .

On the other hand, inequality (6.1) yields (6.4)

kAkLp (Rn )→Lq (Rn ) 6 ckKkL(r,··· ,r),(∞,··· ,∞) (Rn ) ,

which is an improvement. Indeed, suppose n = 2, then kKkL(r,r),(∞,∞) (R2 ) = 6 =

sup t1 >0,t2 >0

sup t1 >0,t2 >0

sup t1 >0,t2 >0

1/r 1/r

t1 t2 K ∗1 ∗2 (t1 , t2 ) 1/r 1/r−1

t1 t2

t2

Z

K ∗1 ∗2 (t1 , s2 )ds2 Z sup K ∗1 (t1 , x2 )dx2 0

1/r 1/r−1

t1 t2

|e|=t2

1/r−1 1/r−1 t2 sup t1 >0,t2 >0 |e|=t2

=

e

Z

sup t1

Z |K(x1 , x2 )|dx1 dx2 .

sup e |w|=t1

w

From the definition of supremum, for a.e. x2 ∈ e there exists a measurable set w(x2 ) (in general, depending on x2 ) such that |w(x2 )| = t1 and Z Z sup |K(x1 , x2 )|dx1 6 2 |K(x1 , x2 )|dx1 . |w|=t1

w

w(x2 )

30

ERLAN NURSULTANOV AND SERGEY TIKHONOV

Therefore, Z

1 1−1/r µΩ>0 (µΩ)

kKkL(r,r),(∞,∞) (R2 ) 6 2 sup

|K(x)|dµ  kKkLr,∞ (R2 ) Ω

where µ is the Lebesgue measure, x = (x1 , x2 ). Let us also verify that (6.4) implies (6.3). Indeed, Z 1/r 1/r−1 kKkL(r,r),(∞,∞) (R2 ) 6 sup t1 t2 sup K ∗1 (t1 , x2 )dx2 t1 >0,t2 >0

1/r−1

6 sup t2 t2 >0

|e|=t2

Z

e t1 >0

|e|=t2

1/r−1

t2 >0

1/r

sup t1 K ∗1 (t1 , x2 )dx2

sup t2

Z

= sup t2



0

e

kK(·, x2 )kLr∞ (Rx1 )

1/r−1 6 kKkLr∞ (Rx1 ) L (R ) sup t2 r∞ x2 t2 >0

= c kKkLr∞ (Rx1 ) Lr∞ (Rx )

 ∗2

Z

t2

(s2 )ds2 1/r

s2 ds2 0

2

Now we show that for the function −1/r

K(x1 , x2 ) = (|x1 ||x2 |)

,

10 t1 t2 K ∗1 ∗2 (t1 , t2 ) < ∞. Let us prove that supt>0 t1/r K ∗ (t) = ∞. Indeed, for ε > 0 we define a set eε = {(x1 , x2 ) : ε 6 x1 6 1, 0 6 x2 6 ε/x1 }. Then Z 1 K(x1 , x2 )dx1 dx2  | ln ε|1/r → ∞ as ε → 0, |eε |1−1/r eε and therefore kKkLr∞  sup |e|>0

1 |e|1−1/r

Z K(x)dx = ∞. e

2. The statement of this theorem in the limit case is not true, that is, if for some i0 , we have pi0 = 1, then (6.1) does not generally hold. The counter-example can be constructed as in [Bl2]. We present an analogue of Young-O’Neil inequality for the limit case. Let vectors p = (p1 , . . . , pn ), q = (q1 , . . . , qn ) be such that 0 < pj , qj 6 ∞, j = 1, . . . , n and if pj = ∞, then qj = ∞. The Lorentz space Lpq∗ ([0, 1]m )

CONVOLUTION INEQUALITIES IN LORENTZ SPACES

31

is defined as the collection of measurable 1-periodic (on each variable) functions f such that kf kLpq∗ ([0,1]m ) q 1  qj1 jn qqjj2 Z 1 Z 1 1 1 p dtj1 1 dtjn  pn ∗1 ,··· ,∗n 1  = ··· (t1 , . . . , tn ) < ∞. ··· t1 · · · tn f tj1 tjn 0 0 Theorem 6.2. Let m = (m1 , · · · mn ) ∈ Nn and let 1 < p 6 q < ∞, 1/q + 1 = 1/p + 1/r, 1 6 h, h1 , h2 6 ∞, and 1/h = 1/h1 + 1/h2 . Let also ∗ = (j1 , · · · , jn ) be a rearrangement of (1, 2, · · · , n). Then for the convolution Z (K ∗ g)(y) = K(x − y)g(x)dx [0,1]m

one has kK ∗ gkLq,h∗ ([0,1]m ) 6 CkK ∗∗ kLr,h∗ ([0,1]m ) kg ∗∗ kLp,h∗ ([0,1]m ) . 1

2

The proof is similar to the proof of Theorem 6.1 (using inequality (4.2)). Acknowledgements The research was supported by INTAS (05-1000008-815), the RFFI (Grants N 06-01-00268, N 08-01-00302), Russian Foundation for the Humanities (Grants N 08-06-00144a), the Leading Scientific Schools (Grant NSH2787.2008.1), and Scuola Normale Superiore. The paper was started when the authors were staying at the Centre de Recerca Matem´atica (Barcelona) in Spring 2006. References [BMR] J. Bastero, M. Milman, F. Ruiz, A Note on L(∞, q) Spaces and Sobolev Embeddings, Indiana Univ. Math. J., 52 (2003), 1215–1230. [BRS] C. Bennett, R. DeVore, R. Sharpley, Weak-L∞ and BMO, Ann. Math. (2), 113(3) (1981), 601–611. [BS] C. Bennett, R. Sharpley, Interpolation of Operators, Academic Press, 1988. [Bl1] A. P. Blozinski, On a convolution theorem for Lp,q spaces Trans. Amer. Math. Soc., 164 (1972), 255–265. [Bl2] A. P. Blozinski, Convolution of L(p, q) functions, Proc. Amer. Math. Soc., 32(1) (1972), 237–240. [Bl3] A. P. Blozinski, Multivariate rearrangements and Banach function spaces with mixed norms Trans. Amer. Math. Soc., 263(1) (1981), 149–167. [CGMP] M. J. Carro, A. Gogatishvili, J. Martin, L. Pick, Functional properties of rearrangement invariant spaces defined in terms of oscillations, J. Func. Anal., 227 (2005), 375–404.

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[CRS] M. J. Carro, J. A. Raposo, J. Soria, Recent Developments in the Theory of Lorentz Spaces and Weighted Inequalities, Memoirs of the American Mathematical Society 187 (2007). [EKM] D. Edmunds, V. Kokilashvili, A. Meskhi, Bounded and compact integral operators., Kluwer Academic Publishers, 2002. [KM] A. Kami´ nska, L. Maligranda, On Lorentz spaces Γp,w , Israel J. Math., 140 (2004), 285–318. [Ke] R. A. Kerman, Convolution theorems with weights, Trans. Amer. Math. Soc., 280(1) (1983), 207–219. [KK] V Kokilashvili, M. Krbec, Weighted inequalities in Lorentz and Orlicz spaces, World Scientific, River Edge, 1991. [KN] A. G. Kostyuchenko, E. D. Nursultanov, On integral operators in Lp -spaces Fundam. Prikl. Mat., 5(2) (1999), 475–491. [MW] B. Muckenhoupt, R. Wheeden, Weighted norm inequalities for fractional integrals Trans. Amer. Math. Soc., 192 (1974), 261–274. [Nu] E. D. Nursultanov, On the coefficients of multiple Fourier series in Lp -spaces, Izv. Math., 64(1) (2000), 93–120 ; translation from Izv. Ross. Akad. Nauk, Ser. Mat. 64(1) (2000), 95–122. [NT] E. D. Nursultanov, S. Tikhonov, Net spaces and boundedness of integral operators. Submitted. Online: http://www.crm.cat [ON] R. O’Neil, Convolution operators and L(p, q) spaces, Duke Math. J., 30 (1963), 129–142. [Sa] E. Sawyer, Boundedness of classical operators on classical Lorentz spaces, Studia Math., 96(2) (1990), 145–158. [Se] E. M. Semenov, Interpolation of linear operators and estimates of Fourier coefficients, Dokl. Akad. Nauk SSSR, 176 (1967), 1251–1254. [So] S. L. Sobolev, Some applications of functional analysis in mathematical physics, Providence, RI, 1991. [SW] E. Stein, G. Weiss, Introduction to Fourier analysis on Euclidean spaces, Princeton, N. J., 1971. [St] V. V. Stepanov, Some questions of the theory of convolution integral operators., Vladivostok, 2000. [Ya] L. Y. H. Yap, Some remarks on convolution operators and l(p, q) spaces, Duke Math. J., 36 (1969), 647–658. E. Nursultanov Kazakh Branch of Moscow State University Munatpasova, 7 010010 Astana Kazakhstan E-mail address: [email protected] S. Tikhonov Scuola Normale Superiore Piazza dei Cavalieri, 7 56126 Pisa Italy E-mail address: [email protected]