Báo cáo hóa học: "WEIGHTED WEAK-TYPE INEQUALITIES FOR GENERALIZED HARDY OPERATORS" ppt

Gogatishvili and Lang [3] characterized the pairs of weights for the strong- and weak-type p, q inequalities for the operator T in the case p ≤ q.. The goal of this paper is to character

GENERALIZED HARDY OPERATORS

A L BERNARDIS, F J MART´IN-REYES, AND P ORTEGA SALVADOR

Received 13 June 2006; Accepted 21 September 2006

We characterize the pairs of weights (v, w) for which the Hardy-Steklov-type operator

T f (x) = g(x)h(x)

s(x) K(x, y) f (y)d y applies L p(v) into weak-L q(w), q < p, assuming certain

monotonicity conditions ong, s, h, and K.

Copyright © 2006 A L Bernardis et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Let us consider the Hardy-Steklov-type operator defined by

T f (x) = g(x)

h(x)

s(x) K(x, y) f (y)d y, f ≥0, (1.1) whereg is a nonnegative measurable function, s and h are continuous and increasing

functions (x < y ⇒ s(x) ≤ s(y), h(x) ≤ h(y)) defined on an interval (a, b) such that s(x) ≤

h(x) for all x ∈(a, b), and the kernel K(x, y) defined on {(x, y) : x ∈(a, b) and s(x) ≤ y ≤

h(x) }satisfies

(i)K(x, y) ≥0,

(ii) it is increasing and continuous inx and decreasing in y,

(iii)K(x, z) ≤ D[K(x, h(y)) + K(y, z)] for y ≤ x and s(x) ≤ z ≤ h(y), where the

con-stantD > 1 is independent of x, y, and z.

Gogatishvili and Lang [3] characterized the pairs of weights for the strong- and weak-type (p, q) inequalities for the operator T in the case p ≤ q Actually, in [3] the au-thors deal with Banach functions spaces with some extra condition On the other hand, Chen and Sinnamon [2] have characterized the weighted strong-type inequality for 1< p,

q < ∞in terms of a normalizing measure In both papers, they work with more general functionss, h, and K.

Hindawi Publishing Corporation

Journal of Inequalities and Applications

Volume 2006, Article ID 62426, Pages 1 10

DOI 10.1155/JIA/2006/62426

The goal of this paper is to characterize the weighted weak-type inequalities in the case

q < p It is well known that strong-type inequalities for the operator T can be deduced

directly from the corresponding ones forg(x) =1, but this is not the case when we work with weak-type inequalities In [5] it was characterized the weighted weak-type inequality

in the case q < p for the operator T when s ≡0,h(x) = x, and K ≡1 The result was obtained for monotone functionsg In fact, in the proof of the result the authors used the

condition

inf

for any bounded setE, where α =infE and β =supE This property clearly holds if g is

monotone or if there existsx0such thatg is increasing in (a, x0] and decreasing in [x0,b).

In our result, we will assume (1.2) and the same condition for the functiong(x)K(x, y),

that is, for ally and every bounded set E y ⊂ { x : s(x) ≤ y ≤ h(x) },

inf

x ∈ E y



g(x)K(x, y)

x ∈(α y,β y)



g(x)K(x, y)

whereα y =infE yandβ y =supE y

Examples of Hardy-Steklov-type operators are the modified Riemann-Liouville oper-ators defined forα > 0 and η ∈ Rasx ηx

0(x − y) α f (y)d y or the more general version

x ηBx

Ax(x − y) α f (y)d y, with 0 < A < B ≤1 andx > 0; the modified logarithmic kernel

op-eratorsx ηx

0logβ(x/ y) f (y)d y, with β > 0 and η ∈ R; the Steklov operatorT f (x) =x x+1 −1 f ;

and the Riemann-Liouville operators, with general variable limitsh(x)

s(x)(x − y) α f (y)d y,

withs(x) ≤ h(x) ≤ x This last operator was studied in [6] in the case−1< α < 0.

As far as we know, our result is new even for the particular casesT f (x) = g(x)x

0K(x, y) f (y)d y and T f (x) =s(x) h(x) K(x, y) f (y)d y For this last operator, conditions (1.2) and (1.3) hold trivially becauseK(x, y) is increasing in x.

The notation is standard:w(E) denotes the integral

E w; if 1 < p < ∞, thenp denotes the conjugate exponent ofp defined by 1/ p + 1/ p =1, andL q, ∞(w) will denote the space

of measurable functions f such that

f q, ∞;w =sup

λ>0

λ

w

x :f (x)> λ 1/q

2 Statement and proof of the result

In the next theorem we state the result of this article

Theorem 2.1 Let s and h be increasing continuous functions defined on an interval (a, b) satisfying s(x) ≤ h(x) for x ∈(a, b) Let K(x, y) be defined on {(x, y) : x ∈(a, b) and s(x) ≤

y ≤ h(x) } satisfying (i), (ii), (iii) and let g be a nonnegative function defined on (a, b) satis-fying ( 1.2 ) and ( 1.3 ) Let q, p, and r be such that 0 < q < p, 1 < p < ∞ , and 1/r =1/q −1/ p.

Let w and v be nonnegative measurable functions defined on (a, b) and (s(a), h(b)), respec-tively The following statements are equivalent.

(i) There exists a positive constant C such that



w

x ∈(a, b) : T f (x) > λ 1/q ≤ C

λ

h(b)

1/ p

(2.1)

for all f ≥ 0 and all positive real number λ.

(ii) The functions

Φ1(x) =sup inf

t ∈(c,d)



g(t)K

t, h(c)

d

1/ ph(c) s(d) v1− p

1/ p 

where the supremum is taken over all the numbers c, c, and d such that a ≤ c ≤ c < x < d ≤ b and s(d) ≤ h(c) and

Φ2(x) =sup inf

t ∈(c,d) g(t)

d

1/ ph(c)

s(d) K p (c, y)v1− p (y)d y

1/ p 

where the supremum is taken over all the numbers c and d such that a ≤ c < x < d ≤ b and s(d) ≤ h(c), belong to L r, ∞(w).

Let us observe that ifg ≡1, we get thatΦ1≤Φ2 Then, in this case, the weighted weak-type inequality (i) is equivalent toΦ2∈ L r, ∞(w) On the other hand, if K ≡1, then

Φ1=Φ2and we recover [1, Theorem 1.9]

To prove the theorem we will use the following lemma (see [1, Lemma 1.4] for the proof)

Lemma 2.2 Let a and b be real numbers such that a < b Let s, h : (a, b) → R be increas-ing and continuous functions such that s(x) ≤ h(x) for all x ∈(a, b) Let {(a j,b j)} j be the connected components of the open setΩ= { x ∈(a, b) : s(x) < h(x) } Then

(a) (s(a j),h(b j))∩(s(a i),h(b i))= ∅ for all j = i,

(b) for every j there exists a (finite or infinite) sequence { m k j } of real numbers such that:

(i)a j ≤ m k j < m k+1 j ≤ b j for all k and j;

(ii) (a j,b j)=k(m k j,m k+1 j ) a.e for all j;

(iii)s(m k+1 j )≤ h(m k j ) for all k and j and s(m k+1 j )= h(m k j ) if a j < m k j < m k+1 j < b j Proof of Theorem 2.1 (i) ⇒(ii) First, we will prove thatΦ1∈ L r, ∞(w), that is, we will

prove that

sup

λ>0

λ

w

x ∈(a, b) :Φ1(x) > λ 1/r < ∞ (2.4)

Letλ > 0 and S λ = { x ∈(a, b) :Φ1(x) > λ } For everyz ∈ S λthere existc z,c z, andd z, with

a ≤ c z ≤ c z < z < d z ≤ b such that s(d z)≤ h(c z) and

λ < inf

t ∈(c,d)



g(t)K

t, h

c z

d z

1/ ph(c z)

s(d)v1− p

1/ p

Let᏷⊂ S λbe a compact set Then there exist (c z1,d z1), , (c z k,d z k) which cover᏷ We may assume without loss of generality thatk

j =1χ(c z j,d z j)≤2 k

j =1 (c z j,d z j) Let f : (s(a), h(b)) →

Rdefined by

f (y) =

⎛

⎜

⎝

k



j =1

v − p (y)χ(s(d z j),h(c z j))(y)

 inft ∈(c z j,d z j)



g(t)K

t, h

c z j

h(c z j)

s(d z j)v1− p p

⎞

⎟

⎠

1/ p

Ifz ∈(z j,d z j), then (s(d z j),h(c z j))⊂(s(z), h(z)) and since K(z, y) is decreasing in y, we

get that

T f (z) ≥ g(z)

h(c z j)

s(d z j) K(z, y) f (y)d y ≥ g(z)K

z, h

c z j

h(c z j)

s(d z j) f (y)d y ≥1. (2.7) Therefore,k

j =1( z j,d z j)⊂ { x ∈(a, b) : T f (x) ≥1} Applying the weighted weak-type in-equality and (2.5) we obtain



k

j =1 (c z j,d z j)w ≤ C

⎛

⎜

⎝

k



j =1

h(c z j)

s(d z j) v1− p

 inft ∈(c z j,d z j)



g(t)K

t, h

c z j

h(c z j)

s(d z j)v1− p p

⎞

⎟

⎠

q/ p

⎛

⎜

⎝

k



j =1

1 inft ∈(c z j,d z j)



g(t)K

t, h

c z j ph(c z j)

s(d z j)v1− p p −1

⎞

⎟

⎠

q/ p

λ q

k

j =1

d z j

c z j w

q/ p

≤ λ C q



k

j =1 (c z j,d z j)w

q/ p

.

(2.8)

The last inequality implies thatλ(

᏷w)1/r ≤ C for any compact set᏷⊂ S λwhich implies (2.4) The proof of (2.4) for the functionΦ2follows in a similar way applying (i) to the

function

f (y) =

⎛

⎜

⎝

k



j =1

K p 

c z j,y v − p

(y)χ(s(d z j),h(c z j))(y)

 inft ∈(c z j,d z j)g(t)h(c z j)

s(d z j)K p 

c z j,t v1− p

(t)dtp

⎞

⎟

⎠

1/ p

(ii)⇒(i) Let{ a N } ∞

N =1be sequences in (a, b) such that

lim

In order to prove (i) it will suffice to show that

w

x ∈a N,b N :T f (x) > λ ≤ C

for all nonnegative function f bounded with compact support such thath(b)

and with a constantC independent of N, λ, and f

Let us fixN ∈ N Observe that ifO λ = { x ∈(a N,b N) :T f (x) > λ }andU = { x ∈(a, b) :

Φ1(x) ≤ λ q/r,Φ2(x) ≤ λ q/r }, then

w

O λ ≤ w

O λ ∩ U +w

x ∈(a, b) :Φ1(x) > λ q/r

+w

x ∈(a, b) :Φ2(x) > λ q/r

≤ w

O λ ∩ U +

Φ1r

r, ∞;w

Φ2r

r, ∞,;w

λ q

(2.12)

Therefore, the implication will be proved if we establish that w(O λ ∩ U) ≤ C/λ q Let (a j,b j) and{ m k j }be the sequences given by the lemma for the setΩN = { x ∈(a N,b N) :

s(x) < h(x) } Then, for fixedj,

w

O λ ∩ U ∩a j,b j =

k

w

O λ ∩ U ∩m k j,m k+1 j (2.13)

Ifx ∈(m k j,m k+1 j ), sinces(m k+1 j )≤ h(m k j), we get that

T f (x) = g(x)

s(m j k+1)

s(x) K(x, y) f (y)d y + g(x)

h(m j

k)

s(m k+1 j )K(x, y) f (y)d y

+g(x)

h(x)

h(m k j)K(x, y) f (y)d y = T1j,k f (x) + T2j,k f (x) + T3j,k f (x).

(2.14)

It is clear that

w

O λ ∩ U ∩m k j,m k+1 j ≤ w

E1 +w

E2 +w

whereE  = { x ∈(m k j,m k+1 j )∩ U : T  j,k f (x) > λ/3 }, =1, 2, 3

First, notice that the property (iii) of the kernel K implies

K(x, y) ≤ D

K

x, h

m k j +K

forx ∈(m k j,m k+1 j ) andy ∈(s(m k+1 j ),h(m k j))

In order to estimatew(E1) let us observe that

T1

j,k f (x) ≤ Dg(x)K

x, h

m k j

s(m k+1 j )

s(x) f (y)d y

+Dg(x)

s(m j k+1)

m k j,y f (y)d y = DT1,1j,k f (x) + DT1,2j,k f (x).

(2.17)

Then,w(E1)≤ w(E1,1) +w(E1,2), where

E1, =

x ∈m k j,m k+1 j ∩ U : T1,j,k  f (x) > λ

6D

 ,  =1, 2. (2.18)

Let us select an increasing sequence{ x i } i,x i ∈(m k j,m k+1 j ), such thatx0= m k jand

s(m j k+1)

s(x i)

LetE i1,1= E1,1∩(x i,x i+1),α1

i =infE1,1i , andβ1

i =supE1,1i IfE1,1i = ∅, lett ∈ E1,1i Using the property of the sequence{ x i } iwe have

λ

6D ≤4g(t)K

t, h

m k j

s(x i+2)

Now, by using (1.3) and H¨older inequality we get

λ

6D ≤4 inf

t ∈(α1

i,β1

i)



g(t)K

t, h

m k j

s(x i+2)

s(x i+1)v1− p

1/ p s(x i+2)

s(x i+1) f p v

1/ p

Now, multiplying by (β1

i

α1

i w)1/ p and using the inequalities s(β1i)≤ s(x i+1) ands(x i+2)≤

s(m k+1 j )≤ h(m k j) we get that

λ

6D

β1

i

α1

i

w

1/ p

≤4Φ1(x)

s(x i+2)

s(x i+1) f p v

1/ p

≤4 q/r

s(x i+2)

s(x i+1) f p v

1/ p

wherex is any element of E i1,1; and summing up ini we obtain

w

E1,1 ≤ C

λ q

s(m j k+1)

To estimatew(E1,2), we select an increasing sequence{ z i } i,z i ∈(m k j,m k+1 j ) such thatz0=

m k j and

s(m j k+1)

m k j,y f (y)d y =

s(z i)

m k j,y f (y)d y. (2.24)

As before, let E1,2i = E1,2∩( i,z i+1), α2

i =infE i1,2, and β2

i =supE1,2i If E i1,2= ∅, then H¨older inequality and (1.2) give

λ

6D ≤4 inf

t ∈(α2

i,β2

i)g(t)

s(z i+2)

s(z i+1)K p 

m k j,t v1− p

(t)dt

1/ p s(z i+2)

s(z i+1) f p v

1/ p

Notice thats(β2

i)≤ s(z i+1),m k j ≤ α2

i, ands(z i+2)≤ s(m k+1 j )≤ h(m k j)≤ h(α2

i) Then multi-plying by (β2

i

α2

i w)1/ pboth members of the above inequality we get

λ

6D

β2

i

α2

i

w

1/ p

≤4Φ2(x)

s(z i+2)

s(z i+1) f p v

1/ p

≤4 q/r

s(z i+2)

s(z i+1) f p v

1/ p

wherex is any element of E1,2i Now, summing up ini and putting together with (2.23)

we obtain

w

E1 ≤ C

λ q

s(m j k+1)

To estimatew(E2) we proceed in a similar way In fact, by using (2.16) we get that

T2

j,k f (x) ≤ Dg(x)K

x, h

m k j

h(m k j)

s(m j k+1)f (y)d y

+Dg(x)

h(m j

k)

s(m k+1 j )K

m k j,y f (y)d y = DT2,1j,k f (x) + DT2,2j,k f (x),

(2.28)

which implies thatw(E2)≤ w(E2,1) +w(E2,2), where the setsE2,, =1, 2 are defined as the setsE1,withT2,j,k  f instead of T1,j,k  f Now, the estimates of w(E2,1) andw(E2,2) follow

as in the previous cases obtaining

w

E2 ≤ C

λ q

h(m j

k)

Actually, the estimations are easier because we do not need to split the setsE2, For the estimation ofw(E3) let us define the function

H(x) =

h(x)

h(m k j)K(x, y) f (y)d y. (2.30)

Sinceh is continuous and K is continuous in the first variable, we may select a decreasing

sequence { x i } i in (m k j,m k+1 j ) such thatx0= m k+1 j andH(x i)=h(x i)

h(m k j)K(x i,y) f (y)d y =

(D + 1) − i H(m k+1 j ) We claim that

H

x i ≤(D + 1)4



K

x i+2,h

x i+3 h(x i+3)

h(m k j) f (y)d y +

h(x i+2)

h(x i+3)K

x i+2,y f (y)d y



.

(2.31)

In fact, first notice that

H(x i)=(D + 1)2

h(x i+2)

h(m k j) K

x i+2,y f (y)d y

=(D + 1)2

h(x i+3)

h(m k j) K

x i+2,y f (y)d y +

h(x i+2)

h(x i+3)K

x i+2,y f (y)d y

(2.32)

Now, applying property (iii) ofK we get that

H

x i ≤ D(D + 1)2

!

K

x i+2,h

x i+3 h(x i+3)

h(m k j) f (y)d y +

h(x i+3)

h(m k j) K

x i+3,y f (y)d y

"

+ (D + 1)2

h(x i+2)

h(x i+3)K

x i+2,y f (y)d y

≤(D + 1)3

!

K

x i+2,h

x i+3 h(x i+3)

h(m k j) f (y)d y +

h(x i+2)

h(x i+3)K

x i+2,y f (y)d y

"

D + 1 H



x i ,

(2.33)

and the claim follows Now, we have

w

E3 ≤

i ≥0



w

E3,1i +w

where

E3,1i = x ∈x i+1,x i ∩ U : g(x)K

x i+2,h

x i+3 h(x i+3)

h(m k j) f (y)d y > λ

6(D + 1)4

 ,

E3,2i = x ∈(x i+1,x i)∩ U : g(x)

h(x i+2)

h(x i+3)K

x i+2,y f (y)d y > λ

6(D + 1)4



.

(2.35)

Working as in previous cases we have



i ≥0

w

E3,2i ≤ C

λ q

h(m j k+1)

In order to estimate

i ≥0w(E3,1i ) we will apply the ideas of [4, Lemma 1] Let{ u s }be the decreasing sequence in (m k j,m k+1 j ) defined byu 0= m k+1 j and

h(u

s)

h(m k j)f =2− s

h(m j k+1)

and let{ u n }be the subsequence of{ u s }defined byu0= u 0and if [u s+1,u s)∩ { x i } = ∅, then we delete the termu s+1 of{ u s } LetE# 3,1

n ={ i ≥0:u n+1 ≤ x i+3 <u n } E3,1i ,α#n =infE# 3,1

n , and

#

β n =supE# 3,1

n Ifu s+1 = u n+1 ≤ x i+3 < u n, by the construction of the sequences we get that

x i+3 ≤ u sandu n+2 ≤ u s+2, then

h(x i+3)

h(m k j) f ≤

h(u

s)

h(m k j)f =4

h(u

s+1)

h(u s+2) f ≤4

h(u n+1)

h(u n+2) f (2.38)

Let us assume thatE# 3,1

n = ∅ By the above inequalities and the monotonicity ofK we have

for allt ∈ # E3,1

λ

6(D + 1)4≤4g(t)K

t, h

x i+3 h(u s+1)

h(u

s+2) f ≤4g(t)K

t, h

u n+1

h(u n+1)

h(u n+2) f (2.39)

Now, multiplying by (β#n

#

α n w)1/ p, applying H¨older inequality, and using thats( β#n)≤ h(u n+2)

we get that

λ

6(D + 1)4

β#n

#

α n

w

1/ p

≤4Φ1(x)

h(u n+1)

h(u n+2) f p v

1/ p

≤4 q/r

h(u n+1)

h(u n+2) f p v

1/ p

, (2.40)

wherex is any point in E# 3,1

n Then



i ≥0

w

E3,1i =

n



{ i ≥0:u n+1 ≤ x i+3 <u n }

w

E3,1i

n

w #E3,1

n

β#n

#

α n w

λ q



n

h(u n+1)

λ q

h(m j k+1)

h(m k j) f p v.

(2.41)

Putting together the estimations ofw(E1),w(E2), andw(E3) we have

w

O λ ∩ U ∩m k j,m k+1 j ≤ C

λ q

h(m j k+1)

Summing up ink in the above inequality and by (2.13) we get that

w

O λ ∩ U ∩a j,b j ≤ C

λ q

h(b j)

Keeping in mind the lemma and summing up in j we obtain the desired inequality. 

Acknowledgments

This research has been partially supported by Spanish goverment Grant MTM2005-8350-C03-02 The first author was supported in part by CAI+D-UNL and CONICET The second and third authors were supported by Junta de Andaluc´ıa Grant FQM 354

References

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Hardy-Steklov operators, to appear in Canadian Journal of Mathematics.

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A L Bernardis: IMAL-CONICET, G¨uemes 3450, Santa Fe 3000, Argentina

E-mail address:bernard@ceride.gov.ar

F J Mart´ın-Reyes: Departamento de An´alisis Matem´atico, Facultad de Ciencias,

Universidad de M´alaga, 29071 M´alaga, Spain

E-mail address:martin reyes@uma.es

P Ortega Salvador: Departamento de An´alisis Matem´atico, Facultad de Ciencias,

Universidad de M´alaga, 29071 M´alaga, Spain

E-mail address:ortega@anamat.cie.uma.es

Hardy- Steklov operators, to appear in Canadian Journal of Mathematics.

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