Directory UMM :Journals:Journal_of_mathematics:GMJ:
GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 6, 1998, 565-574
SOLUTION OF SOME WEIGHT PROBLEMS FOR THE
RIEMANN–LIOUVILLE AND WEYL OPERATORS
A. MESKHI
Abstract. The necessary and sufficient conditions are found for the
weight function v, which provide the boundedness and compactness
of the Riemann–Liouville operator Rα from Lp to Lqv . The criteria
are also established for the weight function w, which guarantee the
boundedness and compactness of the Weyl operator Wα from Lpw
to Lq .
In this paper, the necessary and sufficient conditions are found for the
weight function v (w), which provide the boundedness and compactness
R x f (t)
of the Riemann-Liouville transform Rα f (x) = 0 (x−t)
1−α dt (of the Weyl
R ∞ f (t)
p
q
transform Wα f (x) = x (t−x)1−α dt) from L to Lv (from Lpw to Lq ) when
1 < p, q < ∞, p1 < α < 1 or α > 1 ( q−1
q < α < 1 or α > 1).
A complete description of the weight pairs (v, w) providing the boundedp
to Lvq when 1 < p < q < ∞ and
ness of the operators Rα and Wα from Lw
0 < α < 1 is given in [1]. For 1 < p ≤ q < ∞ and α > 1 a similar problem
has been solved by many authors (see, e.g., [2, 3]).
The necessary and sufficient conditions for pairs of weights, which provide
the boundedness of the above-mentioned operators when 1 < q < p < ∞
and α > 1, are obtained in [4].
For 1 < q ≤ p < ∞ and 0 < α < 1, the two-weight problem for the
operators Rα and Wα remains unsolved and in this context the results
presented here are interesting.
Let v and w be positive almost everywhere, locally integrable functions
defined on R+ .
1991 Mathematics Subject Classification. 42B20, 46E40, 47G60.
Key words and phrases. Weight, Riemann–Liouville and Weyl operators, boundedness
and compactness.
565
c 1998 Plenum Publishing Corporation
1072-947X/98/1100-0565$15.00/0
566
A. MESKHI
Denote by Lpv (1 < p < ∞) a class of all Lebesgue-measurable functions
defined on R+ for which
kf k
p
Lv
=
Z∞
|f (x)|p v(x) dx
0
p1
< ∞.
First, let us recall some familiar results.
Theorem A ([5–10]). Let 1 ≤ p ≤ q < ∞. The inequality
Z∞ Zx
q1
q
p1
Z∞
≤c
|f (x)|p w(x)dx ,
f (t)dt v(x)dx
0
0
(1)
0
where the positive constant c does not depend on f , is fulfilled iff
D = sup
t>0
Z∞
v(x)dx
t
q1 Zt
0
1′
′
p
w1−p (x)dx
1. The inequality
kRα f kLvq ≤ Akf kLp ,
(2)
where the positive constant A does not depend on f , is fulfilled iff
q1 1
Z∞ v(x)
B = sup B(t) = sup
t p′ < ∞.
dx
x(1−α)q
t>0
t>0
t
Moreover, if A is the best constant in (2), then A ≈ B.
(3)
SOLUTION OF SOME WEIGHT PROBLEMS
567
R x f (t)
Proof. Sufficiency. Denoting I1α f (x) = 02 (x−t)
1−α dt and I2α f (x) =
R x f (t)
p
x
(x−t)1−α dt for f ∈ L we write Rα f as Rα f (x) = I1α f (x) + I2α f (x).
2
We obtain
kRα f kqLq
v
≤ c1
Z∞
Z∞
q
|I1α f (x)| v(x)dx+c1
0
|I2α f (x)|q v(x)dx = S1 +S2 .
0
If 0 < t < x2 , then (x−t)α−1 ≤ bxα−1 , where the positive constant b depends
only on α. Consequently, using Theorem A with w ≡ 1, we have
S1 ≤ c2
Z∞
Zx
v(x)
x(1−α)q
|f (t)|dt
0
0
q
dx ≤ c3 B q kf kqLp .
Now we shall estimate S2 . Using the H¨older inequality and the condition
1
p < α, we obtain
S2 = c1
Z∞
0
≤ c1
Z∞
v(x)
Zx
Zx
v(x)
x
2
p
|f (t)| dt
x
2
0
f (t)
q
dx ≤
dt
(x − t)1−α
pq Zx
= c4
k∈Z
(α−1)q+ pq′
v(x) · x
k+1
≤ c4
k∈Z
≤ c5
X
k∈Z
|f (t)| dt
pq 2Z
v(x) · x
2k
|f (t)|p dt
2k−1
≤ c5 B q
|f (t)|p dt
pq
dx ≤
k+1
p
2k−1
2Zk+1
Zx
x
2
2k
2
X Z
q
x
2
k+1
2
X Z
′
dt
p
dx =
′
(x − t)(1−α)p
X
k∈Z
pq
2Zk+1
2k−1
(α−1)q+ pq′
dx ≤
2Zk+1
2k
kq
v(x) · x(α−1)q dx · 2 p′ ≤
|f (t)|p dt
pq
≤ c6 B q kf kqLp
which proves the sufficiency.
Necessity. Let f (x) = χ(0, 2t ) (x). Note that if 0 < y < 2t and x > t, then
(x − y)α−1 ≥ b1 xα−1 , where the positive constant b1 depends only on α.
568
A. MESKHI
We have
kRα f kLvq ≥
Z∞
t
v(x)
t
Z2
0
q
dy
dx
1−α
(x − y)
On the other hand, kf kLp = c8 t
that B(t) ≤ c9 A for all t > 0.
1
p
q1
q1
Z∞ v(x)
≥ c7
· t.
dx
x(1−α)q
t
and by virtue of inequality (2) we find
A most complicated proof of a similar theorem is given in [12] for the
case p = q = 2.
Remark 1. Condition (3) is equivalent to the condition
k+1
e = sup
B
k∈Z
2Z
v(x)
x
2k
(1−α)q− pq′
q1
< ∞.
dx
(4)
e
Moreover, B ≈ B.
Indeed, the fact that (3) implies (4) follows from the proof of Theorem
1. Now let condition (4) be satisfied and t ∈ (0, ∞). Then t ∈ (2m , 2m+1 ]
for some m ∈ Z. We have
q
Z∞ v(x)
(m+1)q
Z∞ v(x)
q
′
p
dx
t
≤
dx
2 p′ =
B(t) =
x(1−α)q
x(1−α)q
2m
t
k+1
= c1 2
mq
p′
∞ 2Z
X
k=m
v(x)
x(1−α)q
dx ≤ c2 2
2k
mq
e q 2 p′
≤ c2 B
mq
p′
∞
X
2
− kq
p′
k=m
∞
X
2
− kq
p′
k=m
2Zk+1
q
v(x)x p′
dx ≤
x(1−α)q
2k
eq
≤ c3 B
e < ∞.
and therefore B ≤ c4 B
By the duality argument and Theorem 1 we obtain
1
q′
Theorem 2. Let 1 < p ≤ q < ∞,
inequality
< α < 1 or α > 1. For the
kWα f kLq ≤ Akf kLpw ,
(5)
where the positive constant A does not depend on f , to be valid it is necessary
and sufficient that
B = sup B(t) = sup
t>0
t>0
Z∞ w1−p′ (x)
(1−α)p′
x
t
dx
1′
p
1
t q < ∞.
(6)
SOLUTION OF SOME WEIGHT PROBLEMS
569
Moreover, if A is the best constant in inequality (5), then A ≈ B.
We shall now consider the case 1 < q < p < ∞. Applying the integration
by parts, we obtain
Lemma 1. Let 1 < q < p < ∞ and u be a locally integrable function on
R+ . Then the equality
Zb
u(x)dx
a
p
p−q
p
=
p−q
Zb Zb
a
u(t)dt
x
q
p−q
u(x)dx
holds, where 0 ≤ a < b < ∞.
1
p
Theorem 3. Let 1 < q < p < ∞,
< α < 1 or α > 1. The inequality
kRα f kLqv ≤ A1 kf kLp
(7)
is fulfilled iff
B1 =
Z∞ Z∞
v(t)
t(1−α)q
dt
x
0
p
p−q
x
(q−1)p
p−q
p−q
pq
< ∞.
dx
(8)
Moreover, if A1 is the best constant in inequality (7), then A1 ≈ B1 .
Proof. Sufficiency. In the notation introduced in the proof of Theorem 1
we have
q
kRα f kL
q ≤ S1 + S2 .
v
Using Theorem B with w ≡ 1 and the argument from the proof of Theorem
1, we obtain
S1 ≤ c2 B1q kf kqLp .
1
p
Applying the H¨older inequality twice and the fact that
S2 ≤ c1
Z∞ Zx
p
|f (t)| dt
x
2
x
2
0
pq Zx
k+1
= c3
2
x
X Z Z
k∈Z
2k
|f (t)|p dt
x
2
≤ c3
k∈Z
2k−1
′
dt
p
v(x)dx =
′
(x − t)(1−α)p
q
pq
(α−1)q+ pq′
v(x)x
dx ≤
k+1
k+1
2
X Z
p
|f (t)| dt
< α, we have
pq 2Z
2k
v(x)x
(α−1)q+ pq′
dx ≤
570
A. MESKHI
k+1
k+1
≤ c3
p−q
p
p−q
pq X 2Z
p
(α−1)q+ pq′
p
≤
dx
v(x)x
|f (t)| dt
X 2Z
k∈Z
k∈Z
2k
2k
k+1
≤ c4 kf kqLp
X 2Z
k∈Z
(α−1)q+pq′
v(x)x
dx
2k
p−q
p
p−q
p
= c4 kf kqLp
X
S2k
k∈Z
p−q
p
.
By Lemma 1, we find for S2k that
k+1
S2k ≤ 2
(k+1)qp
p′ (p−q)
2Z
v(x)x(α−1)q dx
2k
kqp
≤ c5 2 p′ (p−q)
2Zk+1
2k
≤ c5
2Zk+1
2Zk+1
v(t)
t
dt
(1−α)q
x
p
p−q
≤
q
p−q
v(x)
dx ≤
x(1−α)q
q
p−q
Z∞ v(t)
q(p−1)
v(x)
dt
· x p−q dx.
(1−α)q
(1−α)q
t
x
2k
x
Using integration by parts we get
S2 ≤
q
c6 kf kL
p
Z∞ Z∞
x
0
=
c7 kf kqLp
Z∞ Z∞
p−q
q
p−q
p
q(p−1)
v(x)
p−q
dx
=
dt
·x
(1−α)q
(1−α)q
t
x
v(t)
v(t)
t(1−α)q
dt
x
0
p
p−q
x
p(q−1)
p−q
p−q
p
= c7 kf kqLp B1q
dx
and finally we obtain inequality (7).
Necessity. Let p1 < α < 1 and v0 (t) = v(t) · χ(a,b) (t), w0 (t) = χ(a,b) (t),
where 0 < a < b < ∞, and let
x
q−1
1 Z
Z∞ v (t) p−q
p−q
0
f (x) =
dt
w0 (x).
w
(t)dt
0
t(1−α)q
x
0
Then we have
kf k
Lp
=
Zb Z∞
a
x
v0 (t)
t(1−α)q
dt
x
p Z
p−q
0
w0 (t)dt
(q−1)p
p−q
dx
p1
< ∞.
SOLUTION OF SOME WEIGHT PROBLEMS
571
On the other hand,
kRα f kLqv =
Z∞
v(x)
0
0
≥ c8
Z∞
Zx
v(x)
x(1−α)q
0
Zx
f (t)dt
0
q q1
f (t)
≥
dt dx
(x − t)1−α
q
dx
q1
≥ c9
Z∞
v(x)
x(1−α)q
×
0
x
t
q Z Z
q−1
q q1
Z∞ v (y)
p−q
p−q
0
w0 (t)dt dx
dy
≥
w0 (y)dy
×
y (1−α)q
x
≥ c10
0
0
Z∞
0
= c11
x
q1
Z∞
q Z
(p−1)q
p−q
v0 (y)
v0 (x)
p−q
dx =
w0 (y)dy
dy
x(1−α)q
y (1−α)q
x
Z∞ Z∞
x
0
= c11
t
0
q1
Zx
p
(q−1)p
v0 (t) p−q
p−q
w0 (t)dt
=
w0 (x)dx
dt
(1−α)q
0
Zb Z∞
v0 (t)
t(1−α)q
dt
x
a
x
p Z
p−q
0
q1
(q−1)p
p−q
w0 (t)dt
dx .
From inequality (7) we have
Zb Z∞
a
x
x
q−p
p Z
p−q
(q−1)p
pq
p−q
dx
dt
≤ c12 A1 ,
w
(t)dt
0
t(1−α)q
v0 (t)
0
where c12 does not depend on a and b. By Fatou’s lemma we finally obtain
condition (8). The case α > 1 is proved similarly.
By the duality argument and Theorem 3 we have
Theorem 4. Let 1 < q < p < ∞,
1
q′
< α < 1 or α > 1. The inequality
p ,
kWα f kLq ≤ A1 kf kLw
(9)
where the positive constanr A1 does not depend f , holds iff
B1 =
Z∞ Z∞
0
x
w1−p (t)
dt
t(1−α)p′
′
q(p−1)
p−q
x
q
p−q
dx
p−q
pq
< ∞.
(10)
Moreover, if A1 is the best constant in inequality (9), then A1 ≈ B 1 .
Let us now investigate the compactness of the operators Rα and Wα .
572
A. MESKHI
Theorem 5. Let 1 < p ≤ q < ∞, p1 < α < 1 or α > 1. The operator
Rα is compact from Lp to Lqv iff condition (3) and the condition
lim B(t) = lim B(t) = 0
t→∞
t→0
is satisfied.
Proof. Sufficiency. Let 0 < a < b < ∞. We write Rα f as
Rα f = χ[0,a] Rα (f · χ(0,a) ) + χ(a,b) Rα (f · χ(0,b) ) + χ[b,∞) Rα (f · χ(0, b ) ) +
2
+χ[b,∞) Rα (f · χ( b ,∞) ) = P1α f + P2α f + P2α f + P4α f.
2
R∞
For P2α f we have P2α f (x) = χ(a,b) (x) 0 k1 (x, y)f (y)dy, with k1 (x, y) =
(x − y)α−1 for y < x and k1 (x, y) = 0 for y ≥ x. Consequently
Zb
v(x)
a
q′
Z∞
Zb
′
p
(k1 (x, y))p dy
dx ≤
v(x)
x(1−α)q
a
0
q
dx b p′ < ∞
and by Theorem C we conclude that P2α is compact from Lp to Lqv .
In a similar manner we show that P3α is compact too.
Using Theorem 1 for the operators P1α and P4α , we obtain
kP1α k ≤ c1 sup B(t) and kP4α k ≤ c2 sup B(t).
0 1. The operator
Rα is compact from Lp to Lqv iff condition (8) is satisfied.
Proof. The sufficiency is proved as in proving Theorem 5 while the necessity
follows from Theorem 3.
By the duality argument we have
Theorem 8. Let 1 < q < p < ∞, q1′ < α < 1 or α > 1. The operator
Wα is compact from Lpw to Lq iff condition (10) is fulfilled.
In [13, 14] the necessary and sufficient conditions are found for the operators Rα and Wα to be compact when 1 < p ≤ q < ∞ and α = 1.
An analogous problem for α > 1 was investigated in [15].
Remark 2. In Theorems 1 and 5 it suffices to consider v as a measurable
almost everywhere, positive function. The same assumption can be made
for w in Theorems 2 and 6.
Acknowledgement
The author expresses his gratitude to Professor V. Kokilashvili and also
to A. Gogatishvili for helpful remarks.
574
A. MESKHI
References
1. I. Genebashvili, A. Gogatishvili, and V. Kokilashvili, Solution of twoweight problems for integral transforms with positive kernels. Georgian
Math. J. 3(1996), No. 1, 319–342.
2. H. P. Heinig, Weighted inequalities in Fourier analysis. Nonlinear
Analysis, Function Spaces and Appl. 4, Proc. Spring School, 1990, 42–85,
Teubner-Verlag, Leipzig, 1990.
3. K. F. Andersen and H. P. Heinig, Weighted norm inequalities for
certain integral operators. SIAM J. Math. Anal. 14(1983), No. 4, 834–
844.
4. V. D. Stepanov, Two-weighted estimates for Riemann–Liouville integrals. Report No. 39, Math. Inst., Czechoslovak Acad. Sci., 1988.
5. B. Muckenhoupt, Hardy’s inequality with weights. Studia Math.
44(1972), 31–38.
6. G. Talenti, Osservazioni sopra una classe di disuguaglianze. Rend.
Sem. Mat. Fis. Milano 39(1969), 171–185.
7. G. Tomaselli, A class of inequalities. Boll. Un. Mat. Ital. 21(1969),
622–631.
8. V. M. Kokilashvili, On Hardy’s inequalities in weighted spaces. (Russian) Soobshch. Akad. Nauk Gruzin. SSR 96(1979), 37–40.
9. J. S. Bradley, Hardy inequality with mixed norms. Canad. Math.
Bull. 21(1978), 405–408.
10. V. G. Maz’ya, Sobolev spaces. Springer, Berlin, 1985.
auser,
11. H. K¨onig, Eigenvalue distribution of compact operators. Birkh¨
Boston, 1986.
12. J. Newman and M. Solomyak, Two-sided estimates on singular values for a class of integral operators on the semi-axis. Integral Equations
Operator Theory 20(1994), 335–349.
13. S. D. Riemenschneider, Compactness of a class of Volterra operators.
Tˆ
ohoku Math. J. (2) 26(1974), 385–387.
14. D. E. Edmunds, W. D. Evans, and D. J. Harris, Approximation
numbers of certain Volterra integral operators. J. London Math. Soc. (2)
37(1988), 471–489.
15. V. D. Stepanov, Weighted inequalities of Hardy type for higher
derivatives, and their applications. Soviet Math. Dokl. 38(1989), 389–393.
(Received 25.12.1997)
Author’s address:
A. Razmadze Mathematical Institute
Georgian Academy of Sciences
1, M. Aleksidze St., Tbilisi 380093
Georgia
SOLUTION OF SOME WEIGHT PROBLEMS FOR THE
RIEMANN–LIOUVILLE AND WEYL OPERATORS
A. MESKHI
Abstract. The necessary and sufficient conditions are found for the
weight function v, which provide the boundedness and compactness
of the Riemann–Liouville operator Rα from Lp to Lqv . The criteria
are also established for the weight function w, which guarantee the
boundedness and compactness of the Weyl operator Wα from Lpw
to Lq .
In this paper, the necessary and sufficient conditions are found for the
weight function v (w), which provide the boundedness and compactness
R x f (t)
of the Riemann-Liouville transform Rα f (x) = 0 (x−t)
1−α dt (of the Weyl
R ∞ f (t)
p
q
transform Wα f (x) = x (t−x)1−α dt) from L to Lv (from Lpw to Lq ) when
1 < p, q < ∞, p1 < α < 1 or α > 1 ( q−1
q < α < 1 or α > 1).
A complete description of the weight pairs (v, w) providing the boundedp
to Lvq when 1 < p < q < ∞ and
ness of the operators Rα and Wα from Lw
0 < α < 1 is given in [1]. For 1 < p ≤ q < ∞ and α > 1 a similar problem
has been solved by many authors (see, e.g., [2, 3]).
The necessary and sufficient conditions for pairs of weights, which provide
the boundedness of the above-mentioned operators when 1 < q < p < ∞
and α > 1, are obtained in [4].
For 1 < q ≤ p < ∞ and 0 < α < 1, the two-weight problem for the
operators Rα and Wα remains unsolved and in this context the results
presented here are interesting.
Let v and w be positive almost everywhere, locally integrable functions
defined on R+ .
1991 Mathematics Subject Classification. 42B20, 46E40, 47G60.
Key words and phrases. Weight, Riemann–Liouville and Weyl operators, boundedness
and compactness.
565
c 1998 Plenum Publishing Corporation
1072-947X/98/1100-0565$15.00/0
566
A. MESKHI
Denote by Lpv (1 < p < ∞) a class of all Lebesgue-measurable functions
defined on R+ for which
kf k
p
Lv
=
Z∞
|f (x)|p v(x) dx
0
p1
< ∞.
First, let us recall some familiar results.
Theorem A ([5–10]). Let 1 ≤ p ≤ q < ∞. The inequality
Z∞ Zx
q1
q
p1
Z∞
≤c
|f (x)|p w(x)dx ,
f (t)dt v(x)dx
0
0
(1)
0
where the positive constant c does not depend on f , is fulfilled iff
D = sup
t>0
Z∞
v(x)dx
t
q1 Zt
0
1′
′
p
w1−p (x)dx
1. The inequality
kRα f kLvq ≤ Akf kLp ,
(2)
where the positive constant A does not depend on f , is fulfilled iff
q1 1
Z∞ v(x)
B = sup B(t) = sup
t p′ < ∞.
dx
x(1−α)q
t>0
t>0
t
Moreover, if A is the best constant in (2), then A ≈ B.
(3)
SOLUTION OF SOME WEIGHT PROBLEMS
567
R x f (t)
Proof. Sufficiency. Denoting I1α f (x) = 02 (x−t)
1−α dt and I2α f (x) =
R x f (t)
p
x
(x−t)1−α dt for f ∈ L we write Rα f as Rα f (x) = I1α f (x) + I2α f (x).
2
We obtain
kRα f kqLq
v
≤ c1
Z∞
Z∞
q
|I1α f (x)| v(x)dx+c1
0
|I2α f (x)|q v(x)dx = S1 +S2 .
0
If 0 < t < x2 , then (x−t)α−1 ≤ bxα−1 , where the positive constant b depends
only on α. Consequently, using Theorem A with w ≡ 1, we have
S1 ≤ c2
Z∞
Zx
v(x)
x(1−α)q
|f (t)|dt
0
0
q
dx ≤ c3 B q kf kqLp .
Now we shall estimate S2 . Using the H¨older inequality and the condition
1
p < α, we obtain
S2 = c1
Z∞
0
≤ c1
Z∞
v(x)
Zx
Zx
v(x)
x
2
p
|f (t)| dt
x
2
0
f (t)
q
dx ≤
dt
(x − t)1−α
pq Zx
= c4
k∈Z
(α−1)q+ pq′
v(x) · x
k+1
≤ c4
k∈Z
≤ c5
X
k∈Z
|f (t)| dt
pq 2Z
v(x) · x
2k
|f (t)|p dt
2k−1
≤ c5 B q
|f (t)|p dt
pq
dx ≤
k+1
p
2k−1
2Zk+1
Zx
x
2
2k
2
X Z
q
x
2
k+1
2
X Z
′
dt
p
dx =
′
(x − t)(1−α)p
X
k∈Z
pq
2Zk+1
2k−1
(α−1)q+ pq′
dx ≤
2Zk+1
2k
kq
v(x) · x(α−1)q dx · 2 p′ ≤
|f (t)|p dt
pq
≤ c6 B q kf kqLp
which proves the sufficiency.
Necessity. Let f (x) = χ(0, 2t ) (x). Note that if 0 < y < 2t and x > t, then
(x − y)α−1 ≥ b1 xα−1 , where the positive constant b1 depends only on α.
568
A. MESKHI
We have
kRα f kLvq ≥
Z∞
t
v(x)
t
Z2
0
q
dy
dx
1−α
(x − y)
On the other hand, kf kLp = c8 t
that B(t) ≤ c9 A for all t > 0.
1
p
q1
q1
Z∞ v(x)
≥ c7
· t.
dx
x(1−α)q
t
and by virtue of inequality (2) we find
A most complicated proof of a similar theorem is given in [12] for the
case p = q = 2.
Remark 1. Condition (3) is equivalent to the condition
k+1
e = sup
B
k∈Z
2Z
v(x)
x
2k
(1−α)q− pq′
q1
< ∞.
dx
(4)
e
Moreover, B ≈ B.
Indeed, the fact that (3) implies (4) follows from the proof of Theorem
1. Now let condition (4) be satisfied and t ∈ (0, ∞). Then t ∈ (2m , 2m+1 ]
for some m ∈ Z. We have
q
Z∞ v(x)
(m+1)q
Z∞ v(x)
q
′
p
dx
t
≤
dx
2 p′ =
B(t) =
x(1−α)q
x(1−α)q
2m
t
k+1
= c1 2
mq
p′
∞ 2Z
X
k=m
v(x)
x(1−α)q
dx ≤ c2 2
2k
mq
e q 2 p′
≤ c2 B
mq
p′
∞
X
2
− kq
p′
k=m
∞
X
2
− kq
p′
k=m
2Zk+1
q
v(x)x p′
dx ≤
x(1−α)q
2k
eq
≤ c3 B
e < ∞.
and therefore B ≤ c4 B
By the duality argument and Theorem 1 we obtain
1
q′
Theorem 2. Let 1 < p ≤ q < ∞,
inequality
< α < 1 or α > 1. For the
kWα f kLq ≤ Akf kLpw ,
(5)
where the positive constant A does not depend on f , to be valid it is necessary
and sufficient that
B = sup B(t) = sup
t>0
t>0
Z∞ w1−p′ (x)
(1−α)p′
x
t
dx
1′
p
1
t q < ∞.
(6)
SOLUTION OF SOME WEIGHT PROBLEMS
569
Moreover, if A is the best constant in inequality (5), then A ≈ B.
We shall now consider the case 1 < q < p < ∞. Applying the integration
by parts, we obtain
Lemma 1. Let 1 < q < p < ∞ and u be a locally integrable function on
R+ . Then the equality
Zb
u(x)dx
a
p
p−q
p
=
p−q
Zb Zb
a
u(t)dt
x
q
p−q
u(x)dx
holds, where 0 ≤ a < b < ∞.
1
p
Theorem 3. Let 1 < q < p < ∞,
< α < 1 or α > 1. The inequality
kRα f kLqv ≤ A1 kf kLp
(7)
is fulfilled iff
B1 =
Z∞ Z∞
v(t)
t(1−α)q
dt
x
0
p
p−q
x
(q−1)p
p−q
p−q
pq
< ∞.
dx
(8)
Moreover, if A1 is the best constant in inequality (7), then A1 ≈ B1 .
Proof. Sufficiency. In the notation introduced in the proof of Theorem 1
we have
q
kRα f kL
q ≤ S1 + S2 .
v
Using Theorem B with w ≡ 1 and the argument from the proof of Theorem
1, we obtain
S1 ≤ c2 B1q kf kqLp .
1
p
Applying the H¨older inequality twice and the fact that
S2 ≤ c1
Z∞ Zx
p
|f (t)| dt
x
2
x
2
0
pq Zx
k+1
= c3
2
x
X Z Z
k∈Z
2k
|f (t)|p dt
x
2
≤ c3
k∈Z
2k−1
′
dt
p
v(x)dx =
′
(x − t)(1−α)p
q
pq
(α−1)q+ pq′
v(x)x
dx ≤
k+1
k+1
2
X Z
p
|f (t)| dt
< α, we have
pq 2Z
2k
v(x)x
(α−1)q+ pq′
dx ≤
570
A. MESKHI
k+1
k+1
≤ c3
p−q
p
p−q
pq X 2Z
p
(α−1)q+ pq′
p
≤
dx
v(x)x
|f (t)| dt
X 2Z
k∈Z
k∈Z
2k
2k
k+1
≤ c4 kf kqLp
X 2Z
k∈Z
(α−1)q+pq′
v(x)x
dx
2k
p−q
p
p−q
p
= c4 kf kqLp
X
S2k
k∈Z
p−q
p
.
By Lemma 1, we find for S2k that
k+1
S2k ≤ 2
(k+1)qp
p′ (p−q)
2Z
v(x)x(α−1)q dx
2k
kqp
≤ c5 2 p′ (p−q)
2Zk+1
2k
≤ c5
2Zk+1
2Zk+1
v(t)
t
dt
(1−α)q
x
p
p−q
≤
q
p−q
v(x)
dx ≤
x(1−α)q
q
p−q
Z∞ v(t)
q(p−1)
v(x)
dt
· x p−q dx.
(1−α)q
(1−α)q
t
x
2k
x
Using integration by parts we get
S2 ≤
q
c6 kf kL
p
Z∞ Z∞
x
0
=
c7 kf kqLp
Z∞ Z∞
p−q
q
p−q
p
q(p−1)
v(x)
p−q
dx
=
dt
·x
(1−α)q
(1−α)q
t
x
v(t)
v(t)
t(1−α)q
dt
x
0
p
p−q
x
p(q−1)
p−q
p−q
p
= c7 kf kqLp B1q
dx
and finally we obtain inequality (7).
Necessity. Let p1 < α < 1 and v0 (t) = v(t) · χ(a,b) (t), w0 (t) = χ(a,b) (t),
where 0 < a < b < ∞, and let
x
q−1
1 Z
Z∞ v (t) p−q
p−q
0
f (x) =
dt
w0 (x).
w
(t)dt
0
t(1−α)q
x
0
Then we have
kf k
Lp
=
Zb Z∞
a
x
v0 (t)
t(1−α)q
dt
x
p Z
p−q
0
w0 (t)dt
(q−1)p
p−q
dx
p1
< ∞.
SOLUTION OF SOME WEIGHT PROBLEMS
571
On the other hand,
kRα f kLqv =
Z∞
v(x)
0
0
≥ c8
Z∞
Zx
v(x)
x(1−α)q
0
Zx
f (t)dt
0
q q1
f (t)
≥
dt dx
(x − t)1−α
q
dx
q1
≥ c9
Z∞
v(x)
x(1−α)q
×
0
x
t
q Z Z
q−1
q q1
Z∞ v (y)
p−q
p−q
0
w0 (t)dt dx
dy
≥
w0 (y)dy
×
y (1−α)q
x
≥ c10
0
0
Z∞
0
= c11
x
q1
Z∞
q Z
(p−1)q
p−q
v0 (y)
v0 (x)
p−q
dx =
w0 (y)dy
dy
x(1−α)q
y (1−α)q
x
Z∞ Z∞
x
0
= c11
t
0
q1
Zx
p
(q−1)p
v0 (t) p−q
p−q
w0 (t)dt
=
w0 (x)dx
dt
(1−α)q
0
Zb Z∞
v0 (t)
t(1−α)q
dt
x
a
x
p Z
p−q
0
q1
(q−1)p
p−q
w0 (t)dt
dx .
From inequality (7) we have
Zb Z∞
a
x
x
q−p
p Z
p−q
(q−1)p
pq
p−q
dx
dt
≤ c12 A1 ,
w
(t)dt
0
t(1−α)q
v0 (t)
0
where c12 does not depend on a and b. By Fatou’s lemma we finally obtain
condition (8). The case α > 1 is proved similarly.
By the duality argument and Theorem 3 we have
Theorem 4. Let 1 < q < p < ∞,
1
q′
< α < 1 or α > 1. The inequality
p ,
kWα f kLq ≤ A1 kf kLw
(9)
where the positive constanr A1 does not depend f , holds iff
B1 =
Z∞ Z∞
0
x
w1−p (t)
dt
t(1−α)p′
′
q(p−1)
p−q
x
q
p−q
dx
p−q
pq
< ∞.
(10)
Moreover, if A1 is the best constant in inequality (9), then A1 ≈ B 1 .
Let us now investigate the compactness of the operators Rα and Wα .
572
A. MESKHI
Theorem 5. Let 1 < p ≤ q < ∞, p1 < α < 1 or α > 1. The operator
Rα is compact from Lp to Lqv iff condition (3) and the condition
lim B(t) = lim B(t) = 0
t→∞
t→0
is satisfied.
Proof. Sufficiency. Let 0 < a < b < ∞. We write Rα f as
Rα f = χ[0,a] Rα (f · χ(0,a) ) + χ(a,b) Rα (f · χ(0,b) ) + χ[b,∞) Rα (f · χ(0, b ) ) +
2
+χ[b,∞) Rα (f · χ( b ,∞) ) = P1α f + P2α f + P2α f + P4α f.
2
R∞
For P2α f we have P2α f (x) = χ(a,b) (x) 0 k1 (x, y)f (y)dy, with k1 (x, y) =
(x − y)α−1 for y < x and k1 (x, y) = 0 for y ≥ x. Consequently
Zb
v(x)
a
q′
Z∞
Zb
′
p
(k1 (x, y))p dy
dx ≤
v(x)
x(1−α)q
a
0
q
dx b p′ < ∞
and by Theorem C we conclude that P2α is compact from Lp to Lqv .
In a similar manner we show that P3α is compact too.
Using Theorem 1 for the operators P1α and P4α , we obtain
kP1α k ≤ c1 sup B(t) and kP4α k ≤ c2 sup B(t).
0 1. The operator
Rα is compact from Lp to Lqv iff condition (8) is satisfied.
Proof. The sufficiency is proved as in proving Theorem 5 while the necessity
follows from Theorem 3.
By the duality argument we have
Theorem 8. Let 1 < q < p < ∞, q1′ < α < 1 or α > 1. The operator
Wα is compact from Lpw to Lq iff condition (10) is fulfilled.
In [13, 14] the necessary and sufficient conditions are found for the operators Rα and Wα to be compact when 1 < p ≤ q < ∞ and α = 1.
An analogous problem for α > 1 was investigated in [15].
Remark 2. In Theorems 1 and 5 it suffices to consider v as a measurable
almost everywhere, positive function. The same assumption can be made
for w in Theorems 2 and 6.
Acknowledgement
The author expresses his gratitude to Professor V. Kokilashvili and also
to A. Gogatishvili for helpful remarks.
574
A. MESKHI
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(Received 25.12.1997)
Author’s address:
A. Razmadze Mathematical Institute
Georgian Academy of Sciences
1, M. Aleksidze St., Tbilisi 380093
Georgia