The necessary and sufficient conditions are found for the weight functionv, which provide the boundedness and compactness of the Riemann–Liouville operatorRα fromLp toLqv

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 functionv, which provide the boundedness and compactness of the Riemann–Liouville operatorRα fromL^p toL^qv. The criteria are also established for the weight functionw, which guarantee the boundedness and compactness of the Weyl operator Wα from L^pw

toL^q.

In this paper, the necessary and sufficient conditions are found for the weight function v (w), which provide the boundedness and compactness of the Riemann-Liouville transform Rαf(x) = Rx

0 f(t)

(x−t)^1−αdt (of the Weyl transformWαf(x) =R_∞

x f(t)

(t−x)^1−αdt) from L^p toL^q_v (fromL^p_w toL^q) when 1< p, q <∞, ¹_p < α <1 orα >1 (^q−_q¹ < α <1 orα >1).

A complete description of the weight pairs (v, w) providing the bounded- ness of the operatorsRα andWα fromL^p_w toL^q_v when 1< p < q <∞and 0< α <1 is given in [1]. For 1< p≤q <∞ andα >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 < ∞ andα >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.

Letv and wbe positive almost everywhere, locally integrable functions defined onR⁺.

1991Mathematics Subject Classification. 42B20, 46E40, 47G60.

Key words and phrases. Weight, Riemann–Liouville and Weyl operators, boundedness and compactness.

565

1072-947X/98/1100-0565$15.00/0 c1998 Plenum Publishing Corporation

Denote byL^p_v (1< p <∞) a class of all Lebesgue-measurable functions defined onR⁺ for which

kfkL^pv =Z^∞

0

|f(x)|^pv(x)dx‘¹_p

<∞.

First, let us recall some familiar results.

Theorem A ([5–10]). Let 1≤p≤q <∞. The inequality

’Z^∞

0

ŒŒ

ΠZx

0

f(t)dt

ŒŒ

Œ

q

v(x)dx

“¹_q

≤cZ^∞

0

|f(x)|^pw(x)dx‘¹_p

, (1)

where the positive constantc does not depend on f, is fulfilled iff

D= sup

t>0

Z^∞

t

v(x)dx‘¹_qZ^t

0

w^1−p0(x)dx‘_p0¹

<∞ 

p⁰= p p−1

‘ .

Moreover, if c is the best constant in (1), then c ≈ D (the symbol ≈ here denotes a two-sided inequality).

Theorem B ([10]). Let 1≤q < p <∞. Then inequality(1) holds iff

D1=

’Z^∞

0

”Z^∞

t

v(x)dx‘Z^t

0

w^1−p0(x)dx‘q−1•_p−^p_q

w^1−p0(t)dt

“^p_pq^−q

<∞.

Moreover, if c is the best constant in(1), thenc≈D1.

We also need Kolmogorov’s theorem formulated as follows (see, e.g., [11]):

Theorem C.Let 1< p, q <∞andK:L^p→L^q_v be an integral operator of the formKf(x) =R_∞

0 k(x, y)f(y)dy. If k kk(x,·)kL^p0kL^q_v <∞, then the operator K is compact.

Theorem 1. Let 1< p≤q <∞, _p¹ < α <1 orα >1. The inequality kRαfkL^qv ≤Akfk^Lp, (2) where the positive constantA does not depend onf, is fulfilled iff

B = sup

t>0B(t) = sup

t>0

Z^∞

t

v(x) x^(1−α)qdx‘_q¹

t^p01 <∞. (3)

Moreover, if Ais the best constant in (2), thenA≈B.

Proof. Sufficiency. Denoting I1αf(x) = R^x₂

0 f(t)

(x−t)^1−αdt and I2αf(x) = Rx

x 2

f(t)

(x−t)¹−αdt for f ∈ L^p we write Rαf as Rαf(x) = I1αf(x) +I2αf(x).

We obtain

kRαfk^qL^q_v≤c1

Z∞

0

|I1αf(x)|^qv(x)dx+c1

Z∞

0

|I2αf(x)|^qv(x)dx=S1+S2.

If 0< t < ^x₂, then (x−t)^α−1≤bx^α−1, where the positive constantbdepends only onα. Consequently, using Theorem A withw≡1, we have

S1≤c2

Z∞

0

v(x) x^(1−α)q

Z^x

0

|f(t)|dt‘q

dx≤c3B^qkfk^qL^p.

Now we shall estimate S2. Using the H¨older inequality and the condition

1

p < α, we obtain

S2=c1

Z∞

0

v(x)ŒŒŒ Zx

x 2

f(t) (x−t)^1−αdtŒŒŒ

q

dx≤

≤c1

Z∞

0

v(x)Z^x

x 2

|f(t)|^pdt‘_p^qZ^x

x 2

dt (x−t)^(1−α)p0

‘_p0^q dx=

=c4

X

k∈Z 2Z^k+1

2^k

v(x)·x^{(α−1)q+p0q}Z^x

x 2

|f(t)|^pdt‘^q_p dx≤

≤c4

X

k∈Z

 ²Z^k+1

2^k−1

|f(t)|^pdt‘^q_p ²Z^k+1

2^k

v(x)·x^{(α−1)q+p0q}dx‘

≤

≤c5

X

k∈Z

 ²Z^k+1

2^k−1

|f(t)|^pdt‘_p^q²Z^k+1

2^k

v(x)·x^(α−1)qdx‘

·2^kqp0 ≤

≤c5B^qX

k∈Z

 ²Z^k+1

2^k−1

|f(t)|^pdt‘_p^q

≤c6B^qkfk^qL^p

which proves the sufficiency.

Necessity. Letf(x) =χ_(0,^t

2)(x). Note that if 0< y < ₂^t andx > t, then (x−y)^α−1 ≥ b1x^α−1, where the positive constant b1 depends only on α.

We have

kRαfkL^qv ≥

’Z^∞

t

v(x)Z^t2

0

dy (x−y)^1−α

‘q

dx

“¹_q

≥c7

Z^∞

t

v(x) x^(1−α)qdx‘¹_q

·t.

On the other hand, kfkL^p =c8t^1p and by virtue of inequality (2) we find thatB(t)≤c9Afor allt >0.

A most complicated proof of a similar theorem is given in [12] for the casep=q= 2.

Remark 1. Condition (3) is equivalent to the condition

Be= sup

k∈Z

²Z^k+1

2^k

v(x)

x^{(1−α)q−p0q} dx‘¹_q

<∞. (4)

Moreover,B≈B.e

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 ∈(2^m,2^m+1] for somem∈Z. We have

B(t)^q =Z^∞

t

v(x) x^(1−α)qdx‘

t^p0q ≤Z^∞

2^m

v(x) x^(1−α)qdx‘

2^(m+1)qp0 =

=c12^mqp0 X∞ k=m

 ²Z^k+1

2^k

v(x) x^(1−α)qdx‘

≤c22^mqp0 X∞ k=m

2^−kqp0

2Z^k+1

2^k

v(x)x^p0q x^(1−α)qdx≤

≤c2Be^q2^mqp0 X∞ k=m

2^−kqp0 ≤c3Be^q

and thereforeB≤c4B <e ∞.

By the duality argument and Theorem 1 we obtain

Theorem 2. Let 1 < p ≤ q < ∞, _q¹₀ < α < 1 or α > 1. For the inequality

kWαfkL^q ≤AkfkL^pw, (5) where the positive constantAdoes not depend onf, to be valid it is necessary and sufficient that

B = sup

t>0B(t) = sup

t>0

Z^∞

t

w^1−p0(x) x^(1−α)p0 dx‘_p0¹

t^1q <∞. (6)

Moreover, if Ais the best constant in inequality (5), thenA≈B.

We shall now consider the case 1< q < p <∞. Applying the integration by parts, we obtain

Lemma 1. Let 1< q < p <∞andube a locally integrable function on R+. Then the equality

Z^b

a

u(x)dx‘_p−^p_q

= p

p−q Zb

a

Z^b

x

u(t)dt‘_p−^q_q u(x)dx

holds, where0≤a < b <∞.

Theorem 3. Let 1< q < p <∞, _p¹ < α <1 orα >1. The inequality kRαfkL^qv ≤A1kfkL^p (7) is fulfilled iff

B1=

’Z^∞

0

Z^∞

x

v(t)

t^(1−α)qdt‘_p−^p_q

x^{(qp−−1)pq} dx

“^p_pq^−q

<∞. (8)

Moreover, if A1 is the best constant in inequality (7), thenA1≈B1. Proof. Sufficiency. In the notation introduced in the proof of Theorem 1 we have

kRαfk^q_L^q_v ≤S1+S2.

Using TheoremBwithw≡1 and the argument from the proof of Theorem 1, we obtain

S1≤c2B₁^qkfk^qL^p.

Applying the H¨older inequality twice and the fact that _p¹ < α, we have

S2≤c1

Z∞

0

Z^x

x 2

|f(t)|^pdt‘^q_pZ^x

x 2

dt (x−t)^(1−α)p0

‘_p0^q

v(x)dx=

=c3

X

k∈Z 2Z^k+1

2^k

Z^x

x 2

|f(t)|^pdt‘^q_p

v(x)x^{(α−1)q+p0q}dx≤

≤c3

X

k∈Z

 ²Z^k+1

2^k−1

|f(t)|^pdt‘_p^q²Z^k+1

2^k

v(x)x^{(α−1)q+p0q}dx‘

≤

≤c3

 X

k∈Z 2Z^k+1

2^k

|f(t)|^pdt‘^q_p’ X

k∈Z

 ²Z^k+1

2^k

v(x)x^{(α−1)q+p0q}dx‘_p−^p_q“^p−_p^q

≤

≤c4kfk^qL^p

’ X

k∈Z

²Z^k+1

2^k

v(x)x^{(α−1)q+p0q}dx‘_p−^p_q“^p−_p^q

=c4kfk^qL^p

 X

k∈Z

S2k

‘^p−_p^q .

By Lemma 1, we find forS2k that

S2k ≤2

(k+1)qp p0(p−q)

 ²Z^k+1

2^k

v(x)x^(α−1)qdx‘_p−^p_q

≤

≤c52^p0(pkqp−q)

2Z^k+1

2^k

 ²Z^k+1

x

v(t)

t^(1−α)qdt‘_p−^q_q v(x) x^(1−α)qdx≤

≤c5 2Z^k+1

2^k

Z^∞

x

v(t)

t^(1−α)qdt‘_p−^q_q v(x)

x^(1−α)q ·x^{q(pp−−q1)}dx.

Using integration by parts we get

S2≤c6kfk^qL^p

’Z^∞

0

Z^∞

x

v(t)

t^(1−α)qdt‘_p−^q_q v(x)

x^(1−α)q ·x^{q(pp−−q1)}dx

“^p−_p^q

=

=c7kfk^qL^p

’Z^∞

0

Z^∞

x

v(t)

t^(1−α)qdt‘_p−^p_q

x^{p(qp−−q1)}dx

“^p−_p^q

=c7kfk^qL^pB₁^q

and finally we obtain inequality (7).

Necessity. Let ¹_p < α < 1 and v0(t) =v(t)·χ(a,b)(t),w0(t) =χ(a,b)(t), where 0< a < b <∞, and let

f(x) =Z^∞

x

v0(t)

t^(1−α)qdt‘_p−¹_qZ^x

0

w0(t)dt‘^q_p⁻₋¹_q w0(x).

Then we have

kfkL^p=

’Z^b

a

Z^∞

x

v0(t)

t^(1−α)qdt‘_p−^p_qZ^x

0

w0(t)dt‘^(q_p⁻₋^1)p_q dx

“_p¹

<∞.

On the other hand,

kRαfkL^qv =

’Z^∞

0

v(x)Z^x

0

f(t) (x−t)^1−αdt‘q

dx

“¹_q

≥

≥c8

’Z^∞

0

v(x) x^(1−α)q

Z^x

0

f(t)dt‘q

dx

“¹_q

≥c9

’Z^∞

0

v(x) x^(1−α)q ×

אZ^∞

x

v0(y)

y^(1−α)qdy‘_p−^q_qZ^x

0

Z^t

0

w0(y)dy‘^q_p⁻₋¹_q

w0(t)dt‘q

dx

“¹_q

≥

≥c10

’Z^∞

0

v0(x) x^(1−α)q

Z^∞

x

v0(y)

y^(1−α)qdy‘_p−^q_qZ^x

0

w0(y)dy‘^(p_p⁻₋^1)q_q dx

“¹_q

=

=c11

’Z^∞

0

Z^∞

x

v0(t)

t^(1−α)qdt‘_p−^p_qZ^x

0

w0(t)dt‘^(q_p⁻₋^1)p_q

w0(x)dx

“¹_q

=

=c11

’Z^b

a

Z^∞

x

v0(t)

t^(1−α)qdt‘_p−^p_qZ^x

0

w0(t)dt‘^(q_p⁻₋^1)p_q dx

“¹_q .

From inequality (7) we have

’Z^b

a

Z^∞

x

v0(t)

t^(1−α)qdt‘_p−^p_qZ^x

0

w0(t)dt‘^(q_p⁻₋^1)p_q dx

“^q_pq^−p

≤c12A1,

wherec12 does not depend onaandb. By Fatou’s lemma we finally obtain condition (8). The caseα >1 is proved similarly.

By the duality argument and Theorem 3 we have

Theorem 4. Let 1< q < p <∞, _q¹0 < α <1or α >1. The inequality kWαfkL^q ≤A1kfkL^pw, (9) where the positive constanr A1 does not depend f, holds iff

B1=

’Z^∞

0

Z^∞

x

w^1−p0(t)

t^(1−α)p0 dt‘^q(p_p−⁻_q¹⁾ x^p−qqdx

“^p_pq^−q

<∞. (10)

Moreover, if A1 is the best constant in inequality (9), thenA1≈B1. Let us now investigate the compactness of the operatorsRαandWα.

Theorem 5. Let 1< p ≤q <∞, ¹_p < α <1 or α >1. The operator Rα is compact fromL^p toL^q_v iff condition (3)and the condition

tlim→0B(t) = lim

t→∞B(t) = 0 is satisfied.

Proof. Sufficiency. Let 0< a < b <∞. We writeRαf as

Rαf =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

2,∞)) =P1αf+P2αf+P2αf+P4αf.

For P2αf we have P2αf(x) = χ_(a,b)(x)R_∞

0 k1(x, y)f(y)dy, with k1(x, y) = (x−y)^α−1 fory < xandk1(x, y) = 0 fory≥x. Consequently

Zb

a

v(x)Z^∞

0

(k1(x, y))^p0dy‘_p0^q

dx≤Z^b

a

v(x) x^(1−α)qdx‘

b^p0q <∞

and by TheoremC we conclude thatP2α is compact fromL^p toL^q_v. In a similar manner we show thatP3α is compact too.

Using Theorem 1 for the operatorsP1αandP4α, we obtain kP1αk ≤c1 sup

0<t<a

B(t) and kP4αk ≤c2sup

t>^b₂

B(t).

Consequently

kRα−P2α−P3αk ≤ kP1αk+kP4αk ≤c1 sup

0<t<aB(t) +c2sup

t>^b₂

B(t)→0

asa→0 andb→ ∞.

Thus the operatorRαis compact, since it is a limit of compact operators.

The sufficiency is proved.

Necessity. Note that the fact B < ∞ follows from Theorem 1. Thus we need to prove the remaining part. Letft(x) =χ_(0,t)(x)t^−1/p. Then the sequenceftis weakly convergent to 0. Indeed, assuming that ϕ∈L^p0, we obtain

ŒŒ

Œ Z∞

0

ft(x)ϕ(x)dxŒŒŒ≤Z^t

0

|ϕ(x)|^p0dx‘_p0¹

→0 as t→0.

On the other hand, we have

kRαftkL^qv ≥

’Z^∞

t

v(x)Z^t

0

ft(y) (x−y)^1−αdy‘q

dx

“¹_q

≥

≥c3

Z^∞

t

v(x) x^(1−α)qdx‘¹_q

t^p01 =c3B(t).

Using the fact that a compact operator maps a weakly convergent sequence into a strongly convergent form, we find thatB(t)→0 ast→0.

Keeping in mind that the operator Wα is compact from L^q_v⁰₁₋q0 to L^p0 and arguing as above, we prove the remaining part of the theorem.

In [12] a similar theorem is proved for the casep=q= 2.

Since the operatorWα is compact fromL^p_w to L^q iff the operatorRα is compact fromL^q0 toL^p_w⁰1−p0, by Theorem 5 we obtain

Theorem 6. Let 1 < q < p < ∞, _q¹₀ < α < 1 or α > 1. The operator Wα is compact fromL^p_w toL^q iff condition (6)and the condition

tlim→0B(t) = lim

t→∞B(t) = 0 are fulfilled.

Theorem 7. Let 1< q < p < ∞, ¹_p < α <1 or α >1. The operator Rα is compact fromL^p toL^q_v 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 < ∞, _q¹0 < α < 1 or α > 1. The operator Wα is compact fromL^p_w toL^q iff condition (10)is fulfilled.

In [13, 14] the necessary and sufficient conditions are found for the oper- ators RαandWαto be compact when 1< p≤q <∞andα= 1.

An analogous problem for α >1 was investigated in [15].

Remark 2. In Theorems 1 and 5 it suffices to considerv as a measurable almost everywhere, positive function. The same assumption can be made forwin Theorems 2 and 6.

Acknowledgement

The author expresses his gratitude to Professor V. Kokilashvili and also to A. Gogatishvili for helpful remarks.

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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