Tesis Johansson

DOCTORA L T H E S I S

Lule University of TechnologyDepartment of Mathematics

2007:53|: 02-5|: - -- 0753 --

2007:53

Carleman type inequalities and Hardy type inequalities

for monotone functions

Maria Johansson

Carleman type inequalities andHardy type inequalities for

monotone functions

by

Maria Johansson

Department of Mathematics

Lule University of Technology

971 87 Lule, Sweden

November 2007

SupervisorLars-Erik Persson,

Lule University of Technology, Sweden

Published 2007Printed in Sweden by University Printing Oce, Lule

To Hans, Emma and Nils

Abstract

This Ph.D. thesis deals with various generalizations of the inequalities byCarleman, Hardy and Plya-Knopp. In Chapter 1 we give an introductionand overview of the area that serves as a frame for the rest of the thesis. InChapter 2 we consider Carlemans inequality, which may be regarded as adiscrete version of Plya-Knopps inequality and also as a natural limitinginequality of the discrete Hardy inequality. We present several simple proofsof and remarks (e.g. historical) about this inequality. In Chapter 3 we givesome sharpenings and generalizations of Carlemans inequality. We discussand comment on these results and put them into the frame presented in theprevious chapter. In particular, we present some new proofs and results. InChapter 4 we prove a multidimensional Sawyer duality formula for radiallydecreasing functions and with general weights. We also state the correspond-ing result for radially increasing functions. In particular, these results implythat we can describe mapping properties of operators dened on cones ofsuch monotone functions. Moreover, we point out that these results can alsobe used to describe mapping properties of operators between some corre-sponding general weighted multidimensional Lebesgue spaces. In Chapter 5we give a new weight characterization of the weighted Hardy inequality fordecreasing functions and use this result to give a new weight characteriza-tion of the weighted Plya-Knopp inequality for decreasing functions and wealso give a new scale of weightconditions for characterizing the embeddingp(v) q(u) for the case 1 < p q

vi Abstract

we prove a weight characterization of Lp [0,)Lq[0,) boundedness of thegeneral Hardy operator (Hsf)(x) =

([0,x] f

sud) 1

s restricted to the cone ofmonotone functions f 0 for 0 < p, q, s < with positive Borel -nitemeasures , and . In Chapter 8 we present some new integral conditionscharacterizing the embedding p (v) q (w) , 0 < p, q , includingproofs also for the cases (i) p = , 0 < q < , (ii) q = , 1 < p < and(iii) p = q =. Only one condition is necessary for each case, which meansthat our conditions are dierent and simpler than other corresponding con-ditions in the literature. We even prove our results in a more general framenamely when the space q (w) is replaced by the more general space qu (w) .In our proof we use a technique of discretization and anti-discretization de-veloped by A. Gogatishvili and L. Pick, where they considered the oppositeembedding.

Preface

This Ph.D. thesis is written as a monograph. In particular, the contributionsof the author in the papers below are included and put into this generalframe.

M. Johansson, L.-E. Persson and A. Wedestig, Carlemans olikhet-historik, skrpningar och generaliseringar, Normat 51:3 (2003), 89-108(in Swedish).

M. Johansson, L.-E. Persson and A. Wedestig, Carlemans inequality:history, proofs and some new generalizations, J. Inequal. Pure Appl.Math. 3(4) (2003)(19 pages).

S. Barza, M. Johansson and L.-E. Persson, A Sawyer duality principlefor radially monotone functions in Rn, J. Inequal. Pure Appl. Math.6(2) (2005)(13 pages).

A. Gogatishvili, M. Johansson, C.A. Okpoti and L.-E. Persson, Char-acterizations of embeddings in Lorentz spaces, Bull. Austral. Math.Soc. 76 (2007), 69-92.

M. Johansson, V.D. Stepanov and E.P. Ushakova, Hardy inequalitywith three measures on monotone functions, to appear in Math. In-equal. Appl. (20 pages).

vii

viii Preface

M. Johansson, A unied approach to the Sawyer and Sinnamon char-acterizations of Hardys inequality for decreasing functions, Researchreport 6, Department of Mathematics, Lule University of Technology,(2007) (11 pages).

M. Johansson, A new characterization of the Hardy and its limit Polya-Knopp inequality for decreasing functions, Research report 7, De-partment of Mathematics, Lule University of Technology, (2007) (14pages), submitted.

See also [9], [32], [48], [49], [50], [51] and [52].

Acknowledgements

First of all I want to express my deepest gratitude to Professor Lars-ErikPersson, my main supervisor, for his guidance, support and encouragement.I consider myself fortunate to be one of his students.

Secondly, I want to express my gratitude to Professor Lech Maligrandaand Vladimir Stepanov, my co-supervisors, for their support and many fruit-ful discussions. Moreover , I want to Thank Dr. Sorina Barza and Dr. AnnaWedestig, my co-supervisors up to licentiate degree, for constant supportand generous advices.

I would also like to thank my co-authors Professor Lars-Erik Persson,Professor Vladimir Stepanov (Peoples Friendship University, Moscow, Rus-sia), Professor Amiran Gogatishvili (Mathematical Institute, Academy ofSciences, Czech Republic), Dr. Elena Ushakova (Computing Center FEBRAS, Khabarovsk, Russia), Dr. Sorina Barza (Karlstad University), Dr.Anna Wedestig and Dr. Christopher Okpoti (University of education, Win-neba, Ghana) , for their cooperation and also for their help with this Ph.D.thesis. Especially I want to thank Dr. Anna Wedestig for all her help andfor just being a good friend.

Furthermore, I thank everyone at the Department of Mathematics atLule University of Technology for the inspiring and challenging atmosphere.

Finally, I want to thank my husband Hans for his constant love andsupport, and I thank our children Emma and Nils for bringing so much joyinto our lives.

Lule, November 2007Maria Johansson

ix

x Acknowledgements

Conventions and notations

The notations R,Z,N stand for the eld of real numbers, the group of integersand the semigroup of natural numbers, respectively. If n N\{0} we denoteR

n+ := {x = (x1, ..., xn) : xi 0, i = 1, 2, ...n} and R+ := R1+.

Constants are always positive and denoted by C and may be dierentat dierent places. Throughout this thesis all functions are assumed to bemeasurable. Moreover, n Z+, 1 p < , p = pp1 (p = if p = 1),v (x) and u (x) are weights (positive and measurable functions on Rn) and

Lpv = Lpv (R

n)

=

{f : Rn R, measurable s.t.

(Rn

|f (x)|p v (x) dx) 1

p

0,then the set Dh,t = {x Rn : h (x) > t} is decreasing.

The symbol (c.f. (4.1)) means that the quotient of the right and lefthand sides is bounded from above and below by positive constants, whileexpressions as 0 = 0 are taken as zero. Other notations will beintroduced and explained when required.

xi

xii Conventions and notations

Contents

Abstract v

Preface vii

Acknowledgements ix

Conventions and notations xi

1 Introduction 1

1.1 On Hardys original inequalities and its pre-history . . . . . . 11.2 Some further developments of the Hardy inequality . . . . . . 41.3 On the limiting inequalities of the discrete Hardy inequality . 51.4 On the limiting inequalities of the continuous Hardy inequality 61.5 On Hardy and Plya-Knopp type inequalities on the cone of

decreasing functions . . . . . . . . . . . . . . . . . . . . . . . 81.6 On the connection between the one condition case and the

two conditions case . . . . . . . . . . . . . . . . . . . . . . . . 101.7 On multidimensional Hardy-type inequalities . . . . . . . . . 121.8 On the Hardy inequality with measures . . . . . . . . . . . . 13

2 Carlemans inequality - history and proofs 15

2.1 Some proofs of Carlemans inequality . . . 162.2 Plya-Knopps inequality . . . . . . . . . . . . . . . . . . . . . 242.3 Final remarks about Torsten Carleman and his work . . . . . 28

3 Carlemans inequality - some new sharpenings and general-izations 31

xiii

xiv Contents

4 A Sawyer duality principle for radially monotone functionsin Rn 414.1 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . 424.2 The Duality Principle for radially monotone functions . . . . 444.3 Further results and applications . . . . . . . . . . . . . . . . . 50

5 A new characterization of the Hardy inequality and its limitPlya-Knopp inequality for decreasing functions 595.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595.2 A new weight characterization of the weighted Hardy inequal-

ity for decreasing functions . . . . . . . . . . . . . . . . . . . 605.3 A characterization of the Plya-Knopp inequality for decreas-

ing functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 665.4 Another approach via an equivalence theorem . . . . . . . . . 68

6 A Unied approach to Sawyer and Sinnamon characteriza-tions of the Hardy inequality for decreasing functions 736.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.2 The main result . . . . . . . . . . . . . . . . . . . . . . . . . . 756.3 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . 81

7 Hardy inequality with three measures on monotone func-tions 837.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.2 Preliminary remarks . . . . . . . . . . . . . . . . . . . . . . . 867.3 The case 0 < p 1 . . . . . . . . . . . . . . . . . . . . . . . . 897.4 The case 1 < p, q

Chapter 1

Introduction

1.1 On Hardys original inequalities and its pre-

history

The dramatic pre-history until G.H. Hardy nally proved his famous inequal-ity (see (1.2) below) in his 1925 paper was recently described and analyzed inthe paper [61] by A. Kufner, L. Maligranda and L.-E. Persson. In particularwe have included some facts from this description in this section.

Let us rst consider the following standard forms of Hardys inequality.The discrete Hardy inequality : if p > 1 and {ak}1 is a sequence of

positive real numbers, then

n=1

(1

n

nk=1

ak

)p(

p

p 1)p

n=1

apn, (1.1)

and the continuous Hardy inequality : suppose that p > 1 and f is a non-negative p-integrable function on (0,). Then f is integrable over the in-terval (0, x) for all x > 0 and

0

1x

x0

f(t)dt

p dx ( pp 1

)p 0

fp(x)dx. (1.2)

We make the following remarks:

1

2 Carleman type inequalities and Hardy type inequalitiesfor monotone functions

1. The inequalities (1.1) and (1.2) are the standard forms of Hardysinequalities, which can be found in many textbooks in analysis and have beenhighlighted rst in the famous book [44] by G.H. Hardy, J.E. Littlewood andG. Plya.

2. By restricting (1.2) to the class of step functions it can easily be provedthat (1.2) in fact implies (1.1). This important fact seems to have been rstmentioned by E. Landau (cf. [43, p. 154] and [61, Section 8], see also section1.3).

3. The constant(

pp1

)pin both inequalities (1.1) and (1.2) is sharp in

the sense that it can not be replaced by any smaller number.4. The inequalities (1.1) and (1.2) imply the following information:

If

n=1

apn

Introduction 3

in several papers and books e.g. [44], [62], [63] and [80], devoted only to thissubject.

Let us now present shortly the inequality that in fact was Hardys mainsource and motivation when he initiated the research that nally led himto the inequality (1.1). This was another famous inequality, which was dis-covered by D. Hilbert in the early 1900s. Here we present it in the mostbasic form: If

m=1 a

2m


the inequalities (1.2) and (1.1), e.g. E. Landau, G. Plya, M. Riesz and I.Schur. For example if the results of these persons had been published inanother way it could have changed the situation and it should not be unfairto call the discrete inequality (1.1) as the Riesz, or the Landau-Riesz or theHardy-Landau-Riesz inequality.

1.2 Some further developments of the Hardy in-

equality

Since 1925 the investigations and generalizations of Hardys inequality havedeveloped in several directions. Here we have to mention some books whichare important. The rst one is the classical Hardy-Littlewood-Plya book[44] from 1934. Moreover, in 1990 B. Opic and A. Kufner published thebook [80], which is completely concerned with the Hardy inequalities. Thenewest developments of Hardy-type inequalities can be found in [63] by A.Kufner and L.-E. Persson and in [62] by A. Kufner, L. Maligranda and L.-E.Persson. Here we also mention the following books on inequalities (relatedto this subject): [28], [29], [31] and [72].

There are also many papers written in this subject. For more informationsee the references in the books. Here we also have to mention some recentPh.D. theses devoted to this subject, namely those by S. Barza [7], A.A.Kalybay [55], M. Nassyrova [75], C.A. Okpoti [79], D.V. Prokhorov [89], E.Ushakova [101] and A. Wedestig [103].

Important developments have been made in, for example, the followingdirections:

Weighted forms of Hardy-type inequalities.

The mapping properties of the Hardy operator between weighted Lpor Lq spaces.

Compactness properties of the Hardy operator.

Weighted Hardy-type inequalities of weak type.

Weighted Hardy-type inequalities on the cone of monotone functionsor sequences.

Hardy-type inequalities in more general function spaces e.g. Orlicz,Lorentz or even Banach function spaces.

Introduction 5

Limiting (Carleman or Plya-Knopp type) inequalities of the Hardy-type inequalities (when p).

Hardy-type inequalities with more general integraloperators (describede.g. by an integralkernel satisfying an Oinarov condition).

Multidimensional Hardy-type inequalities (e.g. for cones or even moregeneral subsets of Rn).

Fractional order Hardy-type inequalities. Rened Hardy-type inequalities. Higher order Hardy-type inequalities. General three-weights Hardy-type inequalities. Hardy-type inequalities for negative indices. Mapping properties of the Hardy operator with variable limits and its

limiting (geometric mean) operator.

Various combinations of the developments above. Various applications e.g. mapping properties of the maximal function,

imbedding of Lorentz spaces, the theory of partial dierential equa-tions, signal processing, etc.

In the remaining part of this introduction we just comment more on somepoints in connection to which contributions have been made in this Ph.D.thesis.

1.3 On the limiting inequalities of the discrete Hardy

inequality

First we observe that by replacing ak in (1.1) by a1p

k and letting k weobtain the inequality

n=1

(n

k=1

ak

) e

n=1

an. (1.7)

This inequality was rst presented in 1922 in [14] by the Swedish mathemati-cian Torsten Carleman (1892-1942) and it is called Carlemans inequality.

Carleman discovered this inequality during his important work on quasian-alytical functions and he could hardly have imagined at that time that thisdiscovery would be an object for such great independent interest.

More information about Carlemans inequality is found in Chapter 2,where we present several proofs and remarks on (1.7). At the end of thechapter we also present some facts about T. Carleman and his work.

In Chapter 3 we give some examples of some generalizations of (1.7)and (1.9). We discuss and comment on these results and put them intothe frame presented in Chapter 2. We also include some new proofs andresults. In particular, we prove a new weight characterization of a generalweighted Carleman type inequality for the case 0 < p q < , i.e. weprove a necessary and sucient condition on the weight sequences {bk}k=1and {dk}k=1 so that the inequality

k=1

( ka1a2...ak)

q bk C(

k=1

apkdk

)1/p(1.8)

holds for some nite and positive constant C and for all sequences {ak}k=1of positive numbers.

1.4 On the limiting inequalities of the continuous

Hardy inequality

By making similar calculations as mentioned in the previous section, in theinequality (1.2) we obtain the corresponding limit inequality

0

exp

1x

x0

ln f(t)dt

dx < e 0

f(x)dx, (1.9)

where f(t) > 0 and it is sometimes called Knopps inequality with referenceto [54] (cf. Remark 2.12). However, it seems that it was G. Plya who rstdiscovered this inequality (see Remark 2.3). Therefore we prefer to call itPlya-Knopps inequality.

In Chapter 2 we prove that (1.9) implies (1.7). We also present someproofs of (1.9) and thereby some more proofs of (1.7).

Next we mention that in [84] L.-E. Persson and V.D. Stepanov proved aweight characterization for the inequality (1.14) by rst characterizing the

Introduction 7

Hardy inequality 0

1x

x0

f(t)dt

q u(x)dx

1q

C

0

fp(x)v(x)dx

1p

(1.10)

with a new weight criteria. Their result reads:

Theorem 1.1. Let 1 0

t0

u(x)xq

x0

v(y)1p

dy

q dx

1q t

0

v(x)1p

dx

1p

0

t 1

p

t0

w(x)dx

1q

characterization of the Hardy inequality for decreasing functions, and thenwe use this to give a weight characterization of the weighted Plya-Knoppinequality

0

exp1

x

x0

ln f(t)dt

q u(x)dx

1q

C

0

fp(x)v(x)dx

1p

(1.14)

for decreasing functions f 0, for the case 0 < p q }| t} .

It is well-known that the functional p,w is a norm if and only if w isdecreasing and p 1.

In order to be able to study the structure of the p (w)space it is impor-tant to study the boundedness of the Hardy operator f (t) = 1t

t0 f

(s) dsbetween weighted Lp-spaces, which means that the Hardy inequality

0

1x

x0

f(t)dt

p u(x)dx

1p

C

0

f(x)pu(x)dx

1p

holds for decreasing functions f (t). Recall that the rearrangement of theHardy-Littlewood maximal function Mf is equivalent to the Hardy operatorof the rearrangement of |f |. To be more precise, if

(Mf) (x) = supxQ

1

|Q|Q|f (z)| dz, x Rn,

Introduction 9

where Q is a cube in Rn with sides parallel to the coordinate axes and |Q|is its Lebesgue measure, then

(Mf) (x) 1t

t0

f (s) ds, t > 0.

Some historical remarks and further developments of this equivalence can befound in paper [6] (see also [62, p. 92]).

Hence, to prove that

M : p (v) q (u) , 1 < p, q

we mention L. Maligranda (see e.g. [26], [45], [57], [56], [68], [69], [70] and[71]) and for more references see also the Ph.D. Thesis [7] by S. Barza andthe new book [62] by A. Kufner, L. Maligranda and L.-E. Persson.

We also mention that in 1990 E. Sawyer [93] established the followingremarkable duality principle for weighted Lp spaces on the cone of decreasingfunctions:

Suppose that 1 < p < . Let g, v be positive measurable functions on(0,) with v locally integrable. Then

sup0f

0 f (x) g (x) dx(

0 fp (x) v (x) dx

) 1p

( 0

( x0

g (t) dt

)p ( x0

v (t) dt

)pv (x) dx

)1/p+

0 g (x) dx(

0 v (x) dx)1/p .

A very nice proof of this principle was presented by V.D. Stepanov in[99].

1.6 On the connection between the one condition

case and the two conditions case

In 1990 E. Sawyer also made a very important contribution to the study ofthe Hardy inequality for decreasing functions, namely he proved the follow-ing result by using a general approach (see [93, Theorem 2] and also [100,Theorem 2] for a simpler proof):

Theorem 1.3. Let 1 < p q 0

V 1

p (t)

t0

u(x)dx

1q

Introduction 11

supt>0

t

u(x)xqdx

1q t

0

xp

V p

(x)v(x)dx

1p

1.7 On multidimensional Hardy-type inequalities

There are few results concerning characterizations of the weights so thatmultidimensional Hardy-type inequalities hold in weighted Lp -spaces. Herewe just mention the following results, which have guided the investigationsin this thesis:

1. Hardy-type inequalities when the standard one-dimensional Hardy-operator is replaced by spherical means. One early result of this type is dueto P. Drbek, H.P. Heinig and A. Kufner [27] (see our Theorem 4.1). Thisidea is further developed and explained in the paper [24] by A. imeija,L.-E. Persson and A. Wedestig, where general cones in Rn are considered.

2. In the paper [93] E. Sawyer considered mapping properties of thetwo-dimensional Hardy operator H2 dened by

H2 (f (x1, x2)) =

x10

x20

f (t1, t2) dt1dt2.

He proved the remarkable result that it was necessary and sucient to havethree dierent integral conditions to characterize the weights u and v so thata Hardy-type inequality holds in the normal Lp (u) Lq (v) situation when1 < p q

Introduction 13

1.8 On the Hardy inequality with measures

Hardy was originally most interested in the discrete case but his nal result inthe 1925 paper was in the continuous case (but he was aware of the fact thatthe continuous case implies the discrete case in this unweighted situation).After that almost all development has been concerning the continuous case(see e.g. the book [62] and the references given there). However, fairlylate also the discrete case has been considered, by e.g. Bennett [13] and thePh.D. thesis by C. Okpoti [79] (and in the references given there). Thus, itis natural to consider the case with general measures. The rst result in thisdirection seems to be the following result by B. Muckenhoupt [74]: In 1972he proved that, in the case 1 p = q 0

( [r,)) 1p r

0

(dv

dx

)1pdx

1p


and the nal general three weights inequalityR

(,x]

fd

q

d (x)

1q

C

0

fpd

1p

. (1.23)

By using Sinnamons idea about level functions it is possible to characterize(1.22) and, furthermore, by [78, Theorem 2.1] we can in fact reduce (1.23)to (1.22).

In Chapter 7 we will study the inequality [0,)

[0,x]

fud

q

vd (x)

1q

C

0

fpwd

1p

where u, v and w are weights, for monotone functions and for all cases 0 0 and it is sometimes called Knopps inequality with referenceto [54] (cf. Remark 2.12). However it seems that it was G. Plya who rstdiscovered this inequality (see Remark 2.3). Therefore we prefer to call itPlya-Knopps inequality.

In Section 2.1 of this chapter we present several proofs of and remarkson (2.1). In Section 2.2 we prove that (2.2) implies (2.1) and present someproofs of (2.2) (and thus some more proofs of (2.1)).

15

Finally, we include some facts about Torsten Carleman and his work,which we have found, by studying [1], [58] and the recent article [67] by L.Maligranda. This description partly complements the information given byL. Grding in his book [30].

2.1 Some proofs of Carlemans inequality

Proof 1. (Rough sketch of Carlemans original proof)Carleman rst noted that the problem can be solved by nding maximum

of the expressionk

i=1

(a1a2...ai)1i

under the constraintk

i=1

ai = 1.

He then substituted ai = exi and obtained the simpler problem:Find maximum Mk for k = 1, 2, ... of

G =

ki=1

ex1+x2+...+xi

i

under the constraint

H =

ki=1

exi = 1.

This problem can be solved by using the Lagrange multiplier method. Unfor-tunately this leads to some technical calculations, which, of course Carlemancarried out in an elegant way. We leave out these calculations here, and onlyrefer to Carlemans paper [14], where all the details are presented. The resultis that Mk < e for all k Z+. Carleman then showed separately that theinequality (2.1) is strict when the sum on the left hand side converges.

Remark 2.1. In the same paper [14], Carleman proved that the inequality(2.1) does not hold in general for any constant c < e, i.e., that the constante is sharp.

Proof 2. (via Hardys inequality)

Carlemans inequality - history and proofs 17

The discrete version of Hardys inequality reads (see [43], [44])

k=1

(1

k

ki=1

ai

)p 1. (2.3)

Replace ai with a1/pi and note that by using that x = e

lnx and the denitionof the derivative we nd that(

1

k

ki=1

a1/pi

)p= exp

1

p

(ln

ki=1

a1/pi ln

ki=1

a0i

)

exp([

D(ln

ki=1

axi )

]x=0

)( when p)

= exp

([k

i=1

axi ln ai/

ki=1

axi

]x=0

)

= exp1

k

ki=1

ln ai =

(k

i=1

ai

)1/k

and we see that (2.3) leads to the non-strict inequality (2.1) since(p

p1

)p e when p . Observe that this method does not automati-cally prove that we have strict inequality in (2.2) and this has to be provedseparately (see for example our later proofs).

Remark 2.2. G.H. Hardy formulated his inequality (2.3) in 1920 in [42]and proved it 1925 in [43] but it seems that Carleman did not know about theinequality (2.3) at this time, since he does not refer to the simple connectionthat holds according to the proof above. This is somewhat remarkable sinceCarleman worked together with Hardy at that time, see for example theirjoint paper [15].

Remark 2.3. The above means that (2.1) may be considered as a limitinequality for the scale (2.3) of Hardy inequalities. This was pointed out byG.H. Hardy in 1925 in the paper [43, p. 156], but he pronounced that it wasG. Plya who made him aware of this interesting fact.

We now present two proofs which are based on variations of the arithmetic-geometric mean inequality (the AG-inequality).


Proof 3 Because of the AG-inequality the following holds for every i =1, 2, ..., every k and all ci > 0:(

ki=1

ai

)1/k=

(k

i=1

ci

)1/k( ki=1

ciai

)1/k(

ki=1

ci

)1/k1

k

ki=1

ciai. (2.4)

We now choose ci =(1+i)i

ii1, i = 1, 2, ..., k. Then(

k1

ci

)1/k= k + 1 (2.5)

and (2.4) and (2.5) gives that

k=1

ka1a2...ak

k=1

1

k (k + 1)

ki=1

ciai

=

i=1

ciai

k=i

1

k (k + 1)

=

i=1

ciaii

=

i=1

ai

(1 +

1

i

)i e

i=1

ai.

The strict inequality holds, since we cannot have equality at the same time inall terms of the inequality. This can only occur if ciai = c for some constant

c > 0 i.e. ai = c(

i1+i

)i1i , i = 1, 2, ... but this can not hold since

i=1

ai is

convergent (note that(

i1+i

)i e when i).Remark 2.4. This idea of proof was presented by G. Plya (see [86, p. 249])but here we have more closely followed the presentation which can be foundin L. Hrmanders book [47], p. 24.

Remark 2.5. We note that Proof 3 shows that for nite sums the inequality(2.1) holds even for some constant strictly less than e. More precise, forN = 1, 2, ...we have

Nk=1

ka1a2...ak

Nk=1

(1 +

1

k

)kak.

For historical reasons we also present another variant of Proof 3

Proof 4. We choose ci = i, i = 1, 2, ...k, in (2.4) and get that(k

i=1

ai

)1/k (k!)1/k 1

k

ki=1

iai. (2.6)

Moreover, it holds that

(k + 1)k

k!=

(1 +

1

1

)(1 +

1

2

)2...

(1 +

1

k

)k< ek (2.7)

and, by using this inequality and (2.6), we get that

k=1

ka1a2...ak

k=1

(k!)1/k1

k

ki=1

iai

k=1

e

k(k + 1)

ki=1

iai

= e

i=1

iai

k=1

1

k(k + 1)

= ei=1

ai.

The strict inequality can be proved in a similar manner as in Proof 3 (equality

demands that ak = ck but this contradicts the fact that

k=1

ak is convergent).

Remark 2.6. In the paper [42, p. 77], G.H. Hardy presented essentiallythis proof but he also pronounced that it was G. Knopp who pointed out thisproof to him.

Proof 5 (Carlesons proof)We rst note that we can assume that a1 a2 ... (because the sum at

the left hand side of (2.1) obviously becomes the greatest if the sequence {ai}is rearranged in decreasing order while the sum at the right hand side willbe the same for every rearrangement). Let m(x) be a polygon through the

points (0, 0) and (k,k1

log(1/ai)), k = 1, 2, ... The function m(x) is obviously

convex and because of that it holds that for every r > 1

m(rx) m(x) + (r 1)xm(x). (2.8)


Furthermorem(x) = log(1/ak), x (k 1, k) , (2.9)

and since m (0) = 0 and m is convex.

m(x)

x=

m(x)m(0)x

m(k)m(0)k

(2.10)

=m(k)

k=

1

k

ki=1

log(1/ai) for all x k.

We now make a substitution and use Hlders inequality and (2.8). Then wend, for every A > 0 and r > 1,

1

r

A0

em(x)/xdx 1r

rA0

em(x)/xdx

=

A0

em(rx)/rxdx

A

0

em(x)/rx((r1)/r)m(x)dx

A

0

em(x)/xdx

1/r A0

em(x)dx

(r1)

r

so thatA

0

em(x)/xdx r rr1A

0

em(x)dx.

We let A , r 1+ and note that then r rr1 e so that0

em(x)/xdx e0

em(x)dx. (2.11)

We now use (2.9) and (2.10) and get that

0

em(x)dx =

k=1

kk1

em(x)dx =

k=1

e log(1/ak) =

k=1

ak


respectively

0

em(x)/xdx =

k=1

kk1

em(x)/xdx

k=1

exp

(1

k

ki=1

log (1/ai)

)=

k=1

(k

i=1

a1/ki

)1/k.

The non-strict inequality (2.1) follows by using these estimates and (2.11).The strict inequality follows from the fact that we can not have equality in(2.10) at the same time for all x and k.

Remark 2.7. This is L. Carlesons proof (see [16]) and in fact he provedthat the even more general inequality (2.11) holds for every piecewise dif-ferentiable convex function m(x) on [0,] such that m(0) = 0. In fact,Carleson formulated and proved his inequality in the following slightly moregeneral form:

0

xpem(x)/xdx ep+10

xpem(x)dx, p > 1. (2.12)

Proof 6 (via Redheers inequality)R.M. Redheer proved in 1967 the following interesting inequality (see

[90] and also [91]):

nGn +

nk=1

k (bk 1)Gk n

k=1

akbkk (2.13)

which holds for all n = 1, 2, ... and all positive sequences {bk} and whereGk =

(k

i=1ai

)1/k.

In particular, we see that ifa) bk = 1, k = 1, 2, ... , then

Gn 1n

nk=1

ak = An,

i.e. (2.13) coincides with the AG-inequality


b) bk = 1 + 1k , k = 1, 2, ... ,then

nGn +

nk=1

Gk n

k=1

(1 +

1

k

)kak,

which implies that the non-strict inequality (2.1) follows when n. Thestrict inequality can also be proved by using the arguments in the followingproof of the inequality (2.13): We use the elementary inequality

1 + a (x 1) xa, x > 0, a > 1, (2.14)(a simple proof of (2.14) can be obtained by putting = 1a and replacing xwith xa in the following form of the AG-inequality: x11 x+(1 ) 1).We now use (2.14) with a = k and x = GkGk1 bk (k 2) to obtain that

1 + k

(Gk

Gk1bk 1

)(

GkGk1

bk

)k=

akGk1

bkk,

which can be written as

Gk1 + k (Gkbk Gk1) akbkk. (2.15)We now prove (2.13) by simple induction. First we observe that G1 = a1 sothat

G1 + (b1 1)G1 = a1b1i.e. (2.13) holds for n = 1. Assume that the inequality (2.13) holds forn = N Z+. Now us this induction assumption and (2.15) with k = N + 1and we get that

(N + 1)GN+1 +

N+1k=1

k (bk 1)Gk

= (N + 1) bN+1GN+1 +N

k=1

k (bk 1)Gk

= NGN +

Nk=1

k (bk 1)Gk + (N + 1) bN+1GN+1 NGN

N

k=1

akbkk + aN+1b

N+1N+1 =

N+1k=1

akbkk

i.e. (2.13) holds even for n = N+1 and, by the induction axiom, we concludethat (2.13) holds for all n Z+.


Remark 2.8. The proof above is somewhat more complicated than the otherproofs but it leads to a better result. In fact, this method of proving inequal-ities uses a well-known principle, which is sometimes referred to as Redhef-fers recursion principle (see [81]). This principle can also be used to improveseveral other classical inequalities.

Let a(n) = {a1, a2, ..., an} be a positive sequence (n = 1, 2, ...). We denethe powermeans Mr,n of a(n) in the following way

Mr,n = Mr,n

(a(n)

)=

(

1n

nk=1

ark

)1/r, r = 0,(

nk=1

ak

)1/n, r = 0.

Note that An = M1,n , Gn = M0,n and Hn = M1,n are the usual arithmetic,geometric and harmonic means, respectively. We also look at the followingsequence of powermeans:

M r,n = (Mr,1,Mr,2, ...,Mr,n) .

In 1996 B. Mond and J. Peari proved the following interesting inequalitybetween iterative powermeans, see [73]:

Ms,n (Mr,n) Mr,n (M s,n) , (2.16)

for every r s. We have equality if and only if a1 = ... = an. The proof in[73] is fairly complicated so we do not show it here. But we are convincedthat there is a more elementary proof and we leave that as an open questionto the reader. Our next proof is based on the Mond-Pearic result.

Proof 7 We use (2.16) with s = 1 and r = 0 and get that

1

n

nk=1

Gk (

nk=1

(1

k

ki=1

ai

))1/n. (2.17)

By using this inequality and the fact that

ki=1

ai n

i=1

ai , k n,

we nd thatn

k=1

Gk nnn!n

k=1

ak. (2.18)

We use our previous estimate (2.7) with k = n 1 and get thatnn

n!=

nn1

(n 1)! < en1 , i.e.,

nnn!

< e11/n.

By combining this inequality with (2.17) we nd that

nk=1

(k

i=1

ai

)1/k< e11/n

nk=1

ak. (2.19)

This non-strict inequality (2.1) follows when we let n . The fact thatthe inequality actually is strict follows from the fact that equality in (2.18)only can occur when all ai are equal, but this can not be true under our

assumption that

k=1

ak is convergent.

Remark 2.9. More information about how (2.16) can be used to prove andimprove inequalities can be found in the papers [25] and [22].

Remark 2.10. We note that if we, in the proof above, combine (2.17) withthe following variant of the AG-inequality(

nk=1

1

k

ki=1

ai

)1/n=

(1

n!

)1/n(a1 (a1 + a2) ... (a1 + a2 + ... + an))

1/n

(

1

n!

)1/n (na1 + (n 1) a2 + ... + an)n

we get the following strict improvement of (2.19):

nk=1

(k

i=1

ai

)1/k< e11/n

nk=1

(1 k 1

n

)ak. (2.20)

2.2 Plya-Knopps inequality

We begin by proving that (2.2) implies (2.1). As before we note that it isenough to prove (2.1) when {ak}1 is a decreasing sequence. Use (2.2) withthe function f(x) = ak , x [k 1, k), k = 1, 2, .... Then

0

f(x)dx =

k=1

ak (2.21)


and0

exp

1x

x0

ln f(t)dt

dx = k=0

k+1k

exp

1x

x0

ln f(t)dt

dx. (2.22)Furthermore, it yields that

10

exp

1x

x0

ln f(t)dt

dx = a1, (2.23)and, for k = 1, 2, ... ,(

k+1i=1

ai

) 1k+1

=

k+1k

exp

(1

k + 1

k+1i=1

ln ai

)dx (2.24)

k+1k

exp

(1

x

ki=1

ln ai +x k

xln ak+1)

)dx

=

k+1k

exp

1x

x0

ln f(t)dt

dx.The crucial inequality in (2.24) depends on the fact that the integrand

is a weighted arithmetic mean of the numbers ln ai, i = 1, 2, ..., k + 1, withweights 1x , ...,

1x (k 1 weights) and x(k1)k . Here k 1 x k and since

the sequence is decreasing the mean value is smallest for x = k, i.e., whenall weights = 1k . Now (2.1) follows by combining (2.21) - (2.24).

We now present some simple proofs of (2.2) (and thereby some moreproofs of (2.1)).

Proof 8 We note that the function m (x) = x0

ln f (t) dt fulls the condi-

tions to use Carlesons inequality (2.12) (here f (t) is the decreasing rear-rangement of the function f). Therefore, according to (2.12), it holds that

0

xp exp

1x

x0

ln f (t) dt

dx ep+1 0

xpf (x) dx, (2.25)

for every p > 1. Carlesons argument shows that we in fact have strictinequality in (2.25) and especially by using (2.25) for p = 0 we get Plya-Knopps inequality (2.2).


Remark 2.11. Carleson did not note this fact explicitly in his paper [16]since he obviously was mainly interested in giving a simple proof of the in-equality (2.1).

We now present two other proofs of (2.2) and thereby of (2.1) which, likeCarlesons proof, only are based on convexity arguments.

Proof 9 First we note that

exp

1x

x0

ln f(t)dt

(2.26)= exp

1x

x0

ln tf(t)dt 1x

x0

ln tdt

= exp

1x

x0

ln tf(t)dt

exp1

x

x0

ln tdt

.Furthermore, it yields that

1x

x0

ln tdt = lnx + 1 (2.27)

and, in view of Jensens inequality (or the AG-inequality),

exp

1x

x0

ln tf(t)dt

1x

x0

tf(t)dt. (2.28)

We integrate, use (2.26) - (2.28) and change the order of integration and ndthat

0

exp

1x

x0

ln f(t)dt

dx 0

e lnx+11

x

x0

tf(t)dt

dx= e

0

1

x2

x0

tf(t)dt

dx = e 0

tf(t)

t

1

x2dx = e

0

f(t)dt.

The strict inequality follows since equality in Jensens inequality requires that

tf(t) is constant a.e. but this can not occur since0

f(x)dx is convergent.

Remark 2.12. The proof above is partly related to Knopps original idea(see [59] p. 211). However Knopp worked with the interval [1, x] instead of[0, x] and hence Jensens inequality can not be used. Moreover, Knopp neverwrote out the inequality (2.2) explicitly even if it is sometimes referred in theliterature as this is the case, see for example [44, p. 250] and [72, p. 143].

Remark 2.13. By modifying the proof above we can easily prove even someweighted versions of (2.2) for instance the following

0

exp

1x

x0

ln f(t)dt

xpdx < e1 p

0

f(x)xpdx

for every p < 1 which is more general than (2.2), cf. also (2.25).

Proof 10 We rst note that if we replace f(t) by f(t)/t in (2.2), then by(2.27) the left hand side in (2.2) equals

0

exp

1x

x0

ln f(t)dtx

0

ln tdt

dx= e

0

exp

1x

x0

ln f(t)dt

dxx

.

Thus, (2.2) can equivalently be written in the maybe more natural form0

exp

1x

x0

ln f(t)dt

dxx

< 1

0

f(x)dx

x. (2.29)

In order to prove (2.29) we use the fact that the function f(u) = eu is convexand Jensens inequality :

0

exp

1x

x0

ln f(t)dt

dxx

0

1

x2

x0

f(t)dt

dx=

0

f(t)

t

1

x2dx

dt=

0

f(t)dt

t.

The strict inequality follows in the same way as in Proof 9.


Remark 2.14. There are some great similarities between Proof 3 of Carle-mans inequality and Proof 10 of Plya -Knopps inequality. After changingorder of summation and integration, respectively, in the concluding calcula-tions we get

k=i

1

k (k + 1)=

1

iand

t

1

x2dx =

1

t,

respectively, and this is crucial for the proofs. We also note that (2.29) doesnot follow from Carlesons inequality (2.12) since p = 1.

2.3 Final remarks about Torsten Carleman and his

work

Remark 2.15. A main reference concerning Torsten Carleman and hismathematics is of course the book [30] of L. Grding (see p. 233-276). Inthis book Carleman is described in the following way: With Torsten Carle-man (1892-1949) Sweden got their so far most outstanding mathematician.It is therefore not curious that Grding spent the next 30 pages to describeCarleman and his mathematical work and no other mathematician was giveneven close to so much space in the book. It is remarkable that (2.1) was notexplicitly mentioned in Grdings book, which can depend on the fact thathe (as well as Carleman himself) obviously regarded the inequality only as anecessary tool to prove his important main results concerning quasianalyticalfunctions. However, as we have seen in this chapter, Carlemans inequality(2.1) and its continuous variant (Plya-Knopps inequality (2.2)) has at-tracted a lot of attention and it is even mentioned in the title of a numberof papers. See our list of references, Chapter 4 in the book [72] (with 174references), Chapter 1 in the book [63] and the recently published paper [83](with 53 references). And the interest seems only to have increased duringthe very last years.

Remark 2.16. (About the person Torsten Carleman). There is a lot ofinteresting information in Grdings book [30] and some complementary in-formation can be found in [1]. However, the main reference in this con-nection is the recent biographic paper [67] by L. Maligranda. Tage GillisTorsten Carleman was born 8 July 1892. He defended his Ph.D. thesis 1917at Uppsala University. In 1923 he was appointed as full professor at theLund University. Shortly after this, and on an initiative of Professor GstaMittag-Leer, he was called as professor at the University of Stockholm, in


1924. He died 1949. Carleman was a remarkable person and there are manyremarks concerning him (see e.g. Professor B. Kjellbergs interesting andvery personal description in [58, p. 93]).

Chapter 3

Carlemans inequality - some new

sharpenings and generalizations

In this chapter we give some sharpenings and generalizations of (2.1) and(2.2). We discuss and comment on these results and put them into the framepresented in the previous chapter. We also include some new proofs andresults. In particular, we prove a new weight characterization of a generalweighted Carleman type inequality for the case 0 < p q

form (2.29) with the best constant 1. By using this observation, and bymodifying the proof, we nd that, in fact, the following more general theoremholds (cf. [53], Theorem 4.1):

Theorem 3.1. Let 0 < b . Let be a positive and strictly monotoneconvex function on (a, c), a < c . Then

b0

1x

x0

f(t)dt

dxx

1, (3.3)

where g (x) = f(x11/p

)x1/p. Note especially that Hardys inequality writ-

ten in the form (3.2) holds even when p = 1 but that the inequality (3.3) thenhas no meaning.

Remark 3.4. In particular, Theorem 3.1 implies the following improvementsof Plya-Knopps and Hardys inequalities for nite intervals (0, b) , b

Theorem 3.2. Let 0 0

x1/p

x0

w(s)ds

1/q

Carlemans inequality - some new sharpenings and gener-

alizations

35

Remark 3.6. For previous results of this type we also refer to the papers[2], [3], [4], [83], [104], [105] and the references found there. We note that

by using the estimate lk 0,(1 + 1k

)k< e and letting N we get (2.1)

as a special case of (3.5).

Remark 3.7. Renements of Carlemans inequality (2.1) with e replaced by(1 + 1k

)khas been known since at least 1967 (see [90] and [91] and compare

with our Proof 6). We also note that the factor 1 kN+1 in (3.5) means thatthe usual sum on the right hand side of the inequality has been replacedby the equivalent Cesarosum, i.e. we have calculated the arithmetic mean ofpartial sums. This mean value is of course strictly less than the usual sumsince the terms are positive.

Remark 3.8. In the paper [104] P. Yan and G. Sun proved that Carlemansinequality (2.1) can be improved in the following way:

k=1

(k

i=1

ai

)1/k< e

k=1

(1 +

1

k + 1/5

)1/2ak. (3.6)

This result easily follows from (3.5) by estimating the important factor(1 + 1k

)kin the following way: (

1 +1

k

)k e

(1 +

1

k + c

)1/2(3.7)

where c = 8e2

e24 0, 1802696 < 1/5. The inequality (3.7) does not hold for

numbers smaller than c. (See [53], Remark 12). This means that by using(3.5) we see that the Yan-Sun inequality (3.6) actually can be replaced withthe sharper inequality

k=1

(k

i=1

ai

)1/k+

k=1

bkk (k + 1)

< e

k=1

(1 +

1

k + c

)1/2ak.

Remark 3.9. The factor(1 + 1k

)khas also been of interest in some other

new papers. For example M. Gyllenberg and P. Yan recently proved in thepaper [39] that (

1 +1

k

)k= e

(1

n=1

an(1 + k)n

)where all an are positive and can be calculated recursively. For examplea1 =

12 , a2 =

124 , a3 =

148 etc. This answers an earlier question raised by

Yang (see [105]).

Remark 3.10. We have noted before that Carlesons inequality (2.12) givesboth (2.1) and (2.2) as special cases. Another inequality with that propertyhas recently been proved, namely the following see ([53], theorem 3.1):

B0

exp

1M(x)

x0

ln f(t)dM(t)

dM(x)+e

B0

(1 M(x)

M(x)

)f(x)dM(x)

eB

0

(1 M(x)

M(B)

)f(x)dM(x).

Here B R+,M(x) is a right continuous and increasing function on (0,)and M(x) is a special dened function with the property that M(x) M(x).By using this theorem with M(x) = x and B = we get (2.2) and by usingit with

M(x) =

{12 , 0 x 1k , k x k + 1, k = 1, 2, ...

we get a renement of (2.1).

In view of questions raised in connection to (3.4) it is natural to ask thefollowing connected to (2.1): Let 0 < p, q 0

N1/p

N+1k=1

(k

i=1

di

)q/kpbk

1/q


alizations

37

Proof. Assume that (3.9) holds. Let rst w1 = 0, and replace ak with akd1/pk

in (3.8). Then (3.8) is equivalent to k=1

(k

i=1

ai

)q/k( ki=1

di

)q/kpbk

1q

C(

k=1

apk

) 1p

or, if wk =(k

i=1 di

)q/kpbk,

I1/q :=

k=1

(k

i=1

ai

)q/kwk

1q

C(

k=1

apk

) 1p

. (3.10)

Now if {ak}k=1 is decreasing rearrangement of {ak}k=1, then obviously k=1

(k

i=1

ai

)q/kwk

1q

k=1

(k

i=1

ai

)q/kwk

1q

. (3.11)

Let f (x) = ak and w (x) = wk for x [k 1, k) , k = 1, 2, .... Then k=1

(k

i=1

ai

)q/kwk

1q

(3.12)

=

k=1

kk1

[exp

(k

i=1

log a 1

ki

)]qwkdx

=

k=1

kk1

[exp

(1

k

k1i=1

log ai +1

klog ak

)]qwkdx

k=1

kk1

[exp

(1

x

k1i=1

log ai +x (k 1)

xlog ak

)]qwkdx

=

k=1

kk1

exp1

x

x0

ln f(t)dt

q w(x)dx=

0

exp1

x

x0

ln f(t)dt

q w(x)dx1/q .


Moreover, it follows from Theorem 3.2 that if

D := supx>0

x1/p

x0

w(t)dt

1/q


alizations

39

If w1 = 0, then, by using what we just have proved and an elementaryinequality, we have

I1/q =

aq1w1 + k=2

(k

i=1

ai

)q/kwk

1/q (3.16) max

(1, 21/q1

)(w

1/q1 + C

)( k=2

apk

)1/p.

Therefore, by using (3.9), (3.13), (3.15) and (3.16), we conclude that (3.8)holds.

On the contrary, assume that (3.8) (and thus (3.10)) holds for all positivesequences. In particular, let ak = 1, k = 1, ..., N +1 and ak = 0, k > N +1.Then the left hand side in (3.10) can be estimated as follows:

k=1

(k

i=1

ai

)q/kwk

1/q N+1

k=1

(k

i=1

ai

)q/kwk

1/q = (N+1k=1

wk

)1/q.

For the right hand side we have(

k=1

apk

)1/p=

(N+1k=1

1

)1/p= (N + 1)1/p (2N)1/p

and in view of (3.8), it follows that

(2N)1/p(

N+1k=1

wk

)1/q C,

so that (3.9) holds. The proof is complete.

Remark 3.12. We note that our proof gives that if b1 = 0, then C B1, ormore exactly,

21/pB1 C e1/pB1.Remark 3.13. In [46] the weighted Carlemans inequality (3.8) was char-acterized with another condition than B1.

Chapter 4

A Sawyer duality principle for

radially monotone functions in Rn

As mentioned in the introduction a remarkable duality principle for positivedecreasing functions of one variable was proved by E. Sawyer in [93], and alsosome applications are well-known. Here we also refer to the useful proof andideas concerning this principle presented by V. Stepanov in [99]. Moreover,it is natural to look for extensions to functions of several variables. Suchgeneralizations were recently obtained in [8], [10] and [11]. For 0 < q 0

V1p ()

|x|

( |x|0

vn (s) ds

)pv (x) dx

1p

= sup>0

V1p ()

(

n1

tn1( t

0vn (s) ds

)pv (t) dtd

) 1p


= sup>0

V1p ()

(

( t0

vn (s) ds

)p (n1

tn1v (t) d

)dt

) 1p

= sup>0

V1p ()

(

( t0

vn (s) ds

)pvn (t) dt

) 1p

= sup>0

V1p ()

(

(d

dt

( t0

vn (s) ds

)p+1)1

1 pdt) 1

p

sup>0

1

(p 1) 1pV

1p ()

( 0

vn (t) dt

)p+1p

= sup>0

1

(p 1) 1pV

1p ()V

1p1

() =1

(p 1) 1p

A Sawyer duality principle for radially monotone func-

tions in Rn45

Remark 4.1. Theorem 4.1 for the case n = 1 is just Theorem 1 in [93].However, our proof below is based on the technique in [99] and our investi-gation in Section 4.1.

Proof. If f c > 0, then

C (g) supf=c

R

n+fg(

Rn+fpv

) 1p

=g1v

1p

1

= I1.

Moreover, we use the test function f (x) =|t|>|x| h(t)dt, where h (t) 0

(note that f is radially decreasing), Lemma 4.2 and usual duality in Lpv-space to nd that

C (g) = sup0f

Rn

f (x) g (x) dx(Rn

fp (x) v (x) dx) 1

p

suph0

Rn

(|x|


radially decreasing we haveRn

gf =

Rn

gfV1

V

=

0

n1

f (x) g (x)1

V (x)

(B(x)

v(t)dt

)dx

=

Rn

v (t)

(|x|>|t|

f (x) g (x)1

V (x)dx

)dt

Rn

v (t) f (t)

(|x|>|t|

g (x)1

V (x)dx

)dt. (4.7)

To estimate the inner integral we dene gn (s) and vn (s) analogously to(4.5), note that V (x) = V (|x|) and nd that

|x|>|t|g (x)

1

V (x)dx =

|t|

sn1n1

g (s)1

V (s)dsd

=

|t|

(n1

sn1g (s) d

)1

V (s)ds =

|t|

gn (s)1

V (s)ds

=

[1

V (s)

s0

gn (z) dz

]|t|

+

+

|t|

1

V 2 (s)

(n1

sn1v (s) d

)( s0

gn (z) dz

)ds

1V ()

0

gn (z) dz K1

+

|t|

1

V 2 (s)vn (s)

( s0

gn (z) dz

)ds

K2

.

Hence the inner integral can be estimated by K1 + K2 and by substitutingthis into (4.7) and applying Hlders and Minkowskis inequalities we getthat

Rn

fg Rn

fv (K1 + K2) (

Rn

fpv

) 1p(

Rn

(K1 + K2)p v

) 1p

(

Rn

fpv

) 1p

((Rn

Kp

1 v

) 1p

+

(Rn

Kp

2 v

) 1p

).


tions in Rn47

Moreover,

(Rn

Kp

1 v

) 1p

=

(Rn

(1

V () 0

gn (z) dz

)pv (x) dx

) 1p

=1

V ()

( 0

n1

sn1g (s) dds

)(Rn

v (x) dx

) 1p

= v1+1p

1 g1 = v 1

p

1 g1 = I1,

and, according to Lemma 4.2, (Rn

Kp

2 v

) 1p

=

Rn

( |t|

1

V 2 (s)vn (s)

( s0

gn (z) dz

)ds

)pv (t) dt

1p

=

Rn

( |t|

n1

sn1v (s)

V 2 (s)

( s0

n1

zn1g (z) dzd

)dds

)pv

1p

=

Rn

(Rn\B(t)

v (x)

V 2 (x)

B(x)

g (y) dy

)pv (t) dt

1p

p(

Rn

V (t)p

(B(|x|)

g

)v (t) dt

) 1p

= pI2.

The upper bound of C (g) follows by combining the last estimates and theproof is complete.

Remark 4.2. According to our proof we see that the duality constant C (g)in (4.6) can in fact be estimated in the following more precise way:

max (I1, I2) C (g) I1 + pI2.

In particular we have the following useful information:


Corollary 4.4. Let the assumptions in Theorem 4.3 be satised andRn

v =. Then

I2 supfr

Rn

fg(Rn

fpv) 1

p

pI2. (4.8)

The proof above is self-contained and not depending directly on the one-dimensional result (only on similar arguments as V.D. Stepanov used whenhe proved this case). Here we give another shorter proof where we directlyuse the (Sawyer) onedimensional result.

Proof. Make the following changes of variables

t = s and x = y, (4.9)

where s, y (0,) and , n1 . By using that f (s) = f (s) since fis radial, we get:

Rn

fg(Rn

fpv) 1

p

=

0

n1

f (s) g (s) sn1dds(0

n1

fp (s) v (s) sn1dds)1/p

=

0 f (s) G (s) ds(

0 fp (s) V (s) ds

)1/p ,and hence, by using the (Sawyer) one-dimensional result we nd that

Rn

fg(Rn

fpv) 1

p

(4.10)

( 0

V (s) ds

) 1p(

0G (s) ds

)+

0

( s0 G (y) dy

)p( s

0 V (y) dy)p V (s) ds

1p

:= I.

Moreover,

tions in Rn49

I =

( 0

n1

v (s) sn1d ds

) 1p(

0

n1

g (s) sn1dds

)

+

0

( s0

n1

g (y) yn1ddy)p

( s0

n1

v (y) yn1ddy)p

n1

v (s) sn1dds

1p

=

(Rn

v (t) dt

) 1p(

Rn

g (t) dt

)

+

Rn

(B(t) g (x) dx

)p(

B(t) v (x) dx)p v (t) dt

1p

(4.11)

The proof follows by combining (4.10) and (4.11).

For completeness and our applications we also state the correspondingresult for radially increasing functions in Rn:

Theorem 4.5. Suppose that v is a weight on Rn and 1 < p < . If fis a positive radially increasing function on Rn and g a positive measurablefunction on Rn, then

D (g) := supfr

Rn

fg(Rn

fpv) 1

p

I1 + I3,

where

I1 = v1p

1 g1 ,and

I3 =

(Rn

G1 (t)p V1 (t)

p v (t) dt

) 1p

,

with G1 (t) =Rn\B(t) g (x) dx and V1 (t) =

Rn\B(t) v (x) dx.

Proof. We now use Theorem 4.1 (i) (instead of (ii) as in the proof of Lemma4.2) and obtain as in the proof of (4.4):

Rn

v(x)

(B(x)

f(y)dy

)pdx p

Rn

fp(x)V p1 (x) v1p(x)dx.

By using this estimate the proof follows similarly as the proof of Theorem4.3 so we leave out the details.

Remark 4.3. In fact, similar to Remark 4.2 and Corollary 4.4, we nd that

max (I1, I3) D (g) I1 + pI3

and if, in addition to the assumptions in Theorem 4.5,Rn

v = , then

I3 supfr

Rn

fg(Rn

fpv) 1

p

pI3.

4.3 Further results and applications

Let T be an integral operator dened on the cone of functions f : Rn R,which are radially decreasing (0 < f r) and let T be the adjoint operator.Then our results imply the following useful duality result:

Theorem 4.6. Let 1 < p, q

tions in Rn51

supfr

(Rn

(Tf (x))q u (x) dx) 1

q

(Rn

gq

(x) uq

q (x) dx

) 1q

(Rn

fp (x) v (x) dx) 1

p

= c

(Rn

gq

(x) u1q

(x) dx

) 1q

.

On the contrary assume that (4.13) holds for all g 0. Then, by usingCorollary 4.4 again, we have that

pc

(Rn

(g (x))q

(u (x))1q

dx

) 1p

p

Rn

(B(x)

T g (t) dt

)pV p

(x) v (x) dx

1p

Rn

f (x)T g (x) dx(Rn

fp (x) v (x) dx) 1

p

=

Rn

Tf (x) g (x) dx(Rn

fp (x) v (x) dx) 1

p

for each xed 0 < f r. Therefore we nd thatRn

h (x)Tf (x) u1q (x) dx pc

(Rn

fp (x) v (x) dx

) 1p

, (4.14)

where

h (x) =g (x)u

1q (x)(

Rn

(g (x)u

1q (x)

)qdx

) 1q

.

Since hLq = 1 we obtain (4.12) by taking supremum in (4.14) and usingusual duality in Lp-spaces.

Remark 4.4. By modifying the proof above we see that a similar dual-ity result also holds for positive radially increasing functions. In fact, inthis case we just need to replace

B(x)by

Rn\B(x) and V (x) by V1 (x) =

Rn\B(x) v (x) dx in (4.13).

For example when T is the identity operator we obtain the followingresult:

Corollary 4.7. Let 1 < p q 0

(B()

v(x)dx

) 1p(

B()u(x)dx

) 1q

0

(Rn\B()

v(x)dx

) 1p(

Rn\B()u(x)dx

) 1q


tions in Rn53

(i) with f replaced by g, q by p, p by q, W by vV p

and U by u1q

. Infact, then (4.16) is equivalent to (note that (1 q) (1 q) = 1 )

sup>0

(|x|>

vV p

) 1p(

|x|0

(|x|

Proof. Since Tf (x) =B(x) f (t) dt its conjugate T

is dened by T g (x) =Rn\B(x) g (t) dt. Assume rst that (4.18) and (4.19) hold. We note that,

according to Theorem 4.6, (4.17) for 0 < f r is equivalent to (4.13) forarbitrary g 0. Moreover, to be able to characterize weights for which(4.13) is satised, we rst compute:

B(x)T g =

B(x)

(|y|>|z|

g (y) dy

)dz

=

B(x)

( |z|

n1

tn1g (t) ddt

)dz =

B(x)

( |z|

gn (t) dt

)dz

=

|x|0

n1

sn1(

sgn (t) dt

)dds

= |n1|( |x|

0sn1

( |x|s

gn (t) dt

)ds +

|x|0

|x|

gn (t) dtds

)

= |n1| |x|0

( t0

sn1ds

)gn (t) dt + |n1|

|x|0

ds

|x|

gn (t) dt

=

|x|0

(n1

d

t0

sn1ds

n1

tn1g (t) d

)dt

+

( |x|0

n1

dds

) |x|

n1

tn1g (t) ddt

=

B(x)

|B (y)| g (y) dy I1

+ |B (x)|Rn\|B(x)|

g (y) dy I2

= I1 (x) + I2 (x) .

This means that (4.17) holds if and only if

(Rn

(I1 (x) + I2 (x))p V p

(x) v (x) dx

) 1p

(4.20)

c(

Rn

gq

(x)U1q

(x) dx

) 1q

.

Moreover, by Theorem 4.1 (i) with q replaced by p and p replaced by q, we


tions in Rn55

obtain that (Rn

(I1 (x))p V p

(x) v (x) dx

) 1p

(4.21)

=

Rn

(B(x)

|B (y)| g (y) dy)p

V p

(x) v (x) dx

1p

c(

Rn

(|B (x)| g (x))q u (x)1q

|B (x)|q dx) 1

q

= c

(Rn

g (x)q

u (x)1q

dx

) 1q

,

which holds because, according to (4.18),

sup>0

Rn\B()

v (x)

(B(x)

v (y) dy

)pdx

1p

B()

(u (y)1q

|B (y)|q)1q

dy

1q

sup>0

(B() v (y) dy

) 1p

(p 1) 1p

(B()

u (y) |B (y)|q dy) 1

q

0

(B()

|B (x)|p V p (x) v (x) dx) 1

p(

Rn\B()u (x) dx

) 1q

Now assume that (4.17) holds i.e. that (4.20) holds. Then, in particular,(Rn

(I1 (x))p V p

(x) v (x) dx

) 1p c

(Rn

gq

(x)u1q

(x) dx

) 1q

(4.23)

and, by using Theorem 4.1 (i) and arguing as above, we nd that (4.18)holds. Moreover, (8.13) holds also with I1 replaced by I2 so that, by usingTheorem 4.1 (ii) and again arguing as in the suciently part, we see that(4.19) holds. The proof is complete.

Remark 4.7. For the case when also the weights are radially decreasing orincreasing some of our results can be written in a more suitable form. Herewe only state the following consequence of Proposition 4.8:

Corollary 4.9. Let 1 < p q < and let f (x) and B (x) be positive andradially decreasing in Rn. Then the Hardy inequality(

Rn

(1

|B (x)|B(x)

f (y) dy

)q|B (x)|b dx

) 1q

c(

Rn

fp(x) |B (x)|a dx) 1

p

holds if and only if 1 < a < p 1, 1 < b < q 1 anda + 1

p=

b + 1

q. (4.24)

Proof. Apply Proposition 4.8 with v (x) = |B (x)|a and u (x) = |B (x)|b. Wenote that some straightforward calculations give that(

B()v (x) dx

) 1p(

B()u (x) dx

) 1q

(a+1)n

p+ (b+1)n

q (4.25)

whenever a > 1, b > 1 and(B()

|B (x)|p V p (x) v (x) dx) 1

p(

Rn\B()u(x)dx

) 1q

(4.26)

ananp+n

p+ bnnq+n

q = n( a+1

p+ b+1

q

)whenever a < p 1 and b < q 1.

Moreover, according to the estimates (4.25) and (4.26), the condition(4.24) (and only this) gives nite supremum. The proof is complete.

tions in Rn57

Remark 4.8. It is easy to see that Theorem 4.1 is true also if Rn is replacedby Rn+ or even some more general cone in R

n since the proof essentiallyconsists of using polar coordinates and carrying over the problem to the (well-known) one dimensional question. Therefore, by modifying our proofs, we seethat all our results in this chapter indeed holds also when Rn is replaced byR

n+ or even general cones in R

n as dened in [24].

By a general cone C in Rn we mean a subset of Rn dened as follows:Let be a measurable subset of the unit sphere Sn1. Then

C = {x Rn : x = s, 0 < s

Chapter 5

A new characterization of the

Hardy inequality and its limit

Plya-Knopp inequality for

decreasing functions

5.1 Introduction

As mentioned in the introduction E. Sawyer [93] (see Theorem 1.3) provedthe following theorem by performing a general approach for general operators(In [100] V.D. Stepanov gave a direct proof, which also gave good estimatesof the equivalence constants).

Theorem 5.1. Let 1 < p q 0

V 1

p (t)

t0

u(x)dx

1q


and

A1 = A1(p, q, u, v) := (5.3)

supt>0

t

u(x)xqdx

1q t

0

xp

V p

(x)v(x)dx

1p

A new characterization of the Hardy inequality and its

limit Plya-Knopp inequality for decreasing functions

61

Theorem 5.2. Let 1 0

y0

x0

tp

V p

(t)v(t)dt

q u(x)xqdx

1q y

0

tp

V p

(t)v(t)dt

1p


=

0

x

h(y)dy +1

x

x0

yh(y)dy

q u(x)dx

1q

0

f q (x)u(x)dx

1q

+

0

1x

x0

yh(y)dy

q u(x)dx

1q

:=

0

f q(x)u(x)dx

1q

+ F1.

The rst term on the right hand side can be estimated by using Lemma 5.3and, thus,

F A0

0

fp(x)v(x)dx

1p

+ F1. (5.9)

Let us estimate the second term F1. We have, setting H = hV, where Vis dened by (5.4),

F q1 =

0

1x

x0

yh(y)dy

q u(x)dx = 0

1x

x0

yH(y)

V (y)dy

q u(x)dx.Moreover, by integrating by parts and setting

y0

H(t)dt = G(y), we get that

x0

yH(y)

V (y)dy =

xG(x)

V (x)

x0

G(y)

V (y)dy +

x0

yv(y)

V 2(y)G(y)dy

xG(x)V (x)

+

x0

yv(y)

V 2(y)G(y)dy.

Hence, according to Minkowskis inequality,

F1

0

(G(x)

V (x)

)qu(x)dx

1/q +

0

1x

x0

yv(y)

V 2(y)G(y)dy

q u(x)dx1/q(5.10)

:= F2 + F3.



63

Moreover, by integrating by parts and using an approximation argument, wend that

F q2 =

0

(G(x)

V (x)

)qu(x)dx q

0

G(x)qv(x)

V (x)q+1

x0

u(t)dt

dx.Hence, by applying condition (5.2), we have that

F q2 qAq00

G(x)qv(x)

V (x)q+1q/pdx. (5.11)

Moreover,

G(y) =

y0

h(x)

x0

v(t)dtdx =

y0

v(t)

yt

h(x)dx

dt (5.12)

y0

v(t)

t

h(x)dx

dt y0

f(t)v(t)dt.

Consequently, by inserting (5.12) into (5.11), we get that

F q2 qAq00

x0

f(t)v(t)dt

q v(x)V (x)q+1q/p

dx.

If we apply Theorem 1.1 to the function f (t) v (t) instead of f (t) and theweights V (x)

qpq1

v (x) xq instead of u (x) and v (x)1p instead of v (x) wend that the condition (1.11) becomes

AqPS = supt>0

t0

V (x)q/p1qv(x)

x0

v(1p)(1p)dy

q dx

t0

v(x)(1p)(1p)dx

qp

= supt>0

t0

V (x)q/p1v(x))dx

t0

v(x)dx

qp

=p

q,


and, thus,

F q2 p(p)q

Aq0

0

fp(x)v(x)dx

qp

. (5.13)

Denotingyv(y)

V 2(y)G(y) = (y)

andxp

V p

(x)v(x) = (x)1p

(5.14)

and, again applying Theorem 1.1, (5.5) and (5.12), we nd that

F q3 =

0

1x

x0

(y)dy

q u(x)dx (p)q Aq2

0

p(x)(x)dx

qp

=(p)q

Aq2

0

Gp(x)v(x)V (x)pdx

qp

(p)q Aq2

0

x0

f(y)v(y)dy

p v(x)V (x)pdx

qp

(p)2q Aq2

0

f(x)pv(x)dx

qp

. (5.15)

Here we in the last step apply Theorem 1.1 with q = p, f (x) replaced byf (x) v (x), v (x) replaced by v1p (x) and u (x) replaced by v (x)V p (x)xp

so that the condition (1.11) APS = 1. By combining the estimates (5.9),(5.10), (5.13) and (5.15) we nd that (5.1) holds with a constant C satisfying

C (1 + p1/qp

)A0 +

(p)2

A2),

i.e. the upper estimate in (5.6) holds.For the necessity, assume that the inequality (5.1) holds for all f 0.

Consider, for xed y > 0, the decreasing testfunction

fy(s) =

ys

tp

V (t)p1v(t)dt

1p

[0,y](s).



65

Applying this function to the right hand side of (5.1) and changing order ofintegration we obtain that

0

fy(s)pv(s)ds

1/p = y

0

ys

tp

V (t)p1v(t)dt

v(s)ds

1p

=

y0

tp

V (t)p1

t0

v(s)ds

dt

1p

=

y0

tp

V (t)p

v(t)dt

1p

. (5.16)

For the left hand side of (5.1) we have that 0

1x

x0

f(s)ds

q u(x)dx

1q

=

0

1x

x0

ys

tp

V (t)p1v(t)dt

1p

ds

q

u(x)dx

1q

y0

x0

ys

tp

V (t)p1v(t)dt

1p

ds

q

u(x)xqdx

1q

. (5.17)

For the inner integral it yields that

x0

ys

tp

V (t)p1v(t)dt

1p

ds

x

0

sp

p

ys

V (t)p1v(t)dt

1p

ds

1p

x0

sp

p

xs

yt

V (z)p1v(z)dz

1p

V (t)p1v(t)dt

ds := I


Now, by changing the order of integration, we nd that

I =1

p

x0

V (t)p1v(t)

t0

sp

p ds

yt

V (z)p1v(z)dz

1p

dt

=1

pp

x0

V (t)p1v(t)tp

yt

V (z)p1v(z)dz

1p

dt

1p (p)

1p

x0

V (t)p

v(t)tp

dt. (5.18)

Therefore, by combining (5.16) and (5.17) with (5.18), we obtain that y0

tp

V (t)p

v(t)dt

1p y

0

x0

tp

V p

(t)v(t)dt

q u(x)xqdx

1q

p (p) 1p CHence C A2

p(p)1p. Moreover, since f(x) 1x

x0

f(t)dt when f is decreasing,

it follows that A0 C, by choosing ft (x) = [0,t] (x) and taking supremum.Hence also the lower estimate of (5.6) holds and the proof is complete.

Remark 5.1. The crucial part of the proof above is to use the results in[84], [99] and the technique in [100] combined with nding explicitly the cor-responding test functions. For the estimates C max (A0, A2) we have foundthat

max

(A0, A2

1

p (p)1/p

) C

(1 + p1/qp

)A0 +

(p)2

A2) (5.19)

but we strongly believe that these estimates can be improved.

5.3 A characterization of the Plya-Knopp inequal-

ity for decreasing functions

In this Section we will give a weight characterization of the Plya-Knoppinequality (1.14) for decreasing functions. In fact, the following Plya-Knopptype inequality may be regarded as a limit result of our Theorem 5.2:

67

Theorem 5.4. Let 0 < p q < . Let u (x) and v (x) be weight functionsand assume that v (x) is decreasing. Then the inequality

0

exp1

x

x0

ln f(t)dt

q u(x)dx

1q

C

0

fp(x)v(x)dx

1p

(5.20)

holds for all f 0 if and only if

DPS = DPS(p, q, w) := supt>0

t 1

p

t0

w(x)dx

1q

1

k (r)DPS, (5.22)

where

k (r) =(1 + r

pqr r +

(r)2) rp

,

Proof. Assume that (5.21) holds and let g(x) = f(x)v(x)1p . Then as v(x) is

decreasing and f(x) is decreasing so is g(x). Moreover, if w(x) be dened asin (1.13), then (5.20) is equivalent to

0

exp1

x

x0

ln g(t)dt

q w(x)dx

1q

C

0

gp(x)dx

1p

, (5.23)

and with g(t) = h(t)rp equivalent to

0

exp1

x

x0

lnh(t)dt

qrp

w(x)dx

1q

C

0

hr(x)dx

1p

. (5.24)

Note that if r > 1 and 1 < p q < , then we have 1 < r qrp < . Byapplying Jensens inequality we nd that if the inequality

0

x0

h(t)dt

qrp

w(x)xqps dx

1q

C

0

hr(x)dx

1p

(5.25)


holds for all decreasing functions h (x) , then also (5.24) holds with the sameconstant C. Now we replace p by r and q by qrp , u (x) with w(x) and letv(x) = 1 in the inequality (5.1) and apply Theorem 5.2. This means thatthe inequality (5.25) holds if and only if

(A0(r,

qr

p,w, 1)

) rp

= DPS(p, q, w) = supt>0

t 1

p

t0

w(x)dx

1q

0

t1p

t0

w(x)dx

1q

69

and 0

x

f(t)

tdt

p xpV (x)pv(x)dx

1p

C

0

f q

(x)u(x)1q

dx

1q

,

(5.27)where V is dened by (5.4). Then, by using the Hardy inequality and theso called Muckenhoupt condition to characterize the inequalities (5.26) and(5.27), E. Sawyer received the before mentioned result in Theorem 5.1.

Recently the weighted Hardy inequality has been characterized with somenew conditions (see e.g. [64] and [102]). More generally, nowadays we knowthat the Hardy inequality (5.1) for 1 < p q < holds for all f 0 ifand only if just one of the following (innite many equivalent) conditions issatised (see [33]):

Theorem 5.5. Let 1 < p q

holds for all measurable functions f 0 if and only if any of the quantitiesAi(s) is nite. Moreover, for the best constant C in (5.29) we have C Ai(s), i = 1, 2, 3, 4.

This gives us the possibility to characterize the weighted Hardy inequal-ity for decreasing functions with some new conditions, by just using thetechnique of E. Sawyer described above. We can therefore formulate thefollowing generalization of our Theorem 5.2:

Theorem 5.6. Let 1 < p q < . Then the inequality (5.1) holds for allf 0 if and only if for any s, r, 0 < s, r



71

E2(r) := sup0


where (x) is dened by (5.14) so we can again use Theorem 5.5 now withs, u (x) and v (x) replaced by r, u (x)xqand (x) respectively, and we getthe second bunch of conditions in Theorem 5.6. The proof is complete.

Remark 5.4. By using Theorem 5.6 and arguing as in the proof of Theo-rem 5.4 we can get a number of other characterizations of the Plya-Knoppinequality (5.20) then that is stated in Theorem 5.4.

Chapter 6

A Unied approach to Sawyer and

Sinnamon characterizations of the

Hardy inequality for decreasing

functions

6.1 Introduction

Let us rst recall that in the Sawyer characterization of Hardys inequalityfor decreasing functions two independent conditions were used (see Theorem1.3 = Theorem 5.1 and c.f. also our alternative descriptions in Theorem 5.2and 5.6).

On the contrary, in the next result by G. Sinnamon [97] only one condi-tion is necessary to characterize the corresponding Hardy type inequality.

Theorem 6.1. Let 1 < p q


holds for all decreasing f 0 if and only if

A3 := supt>0

t0

u (x)

x0

v (y) dy

q dx

1q t

0

v (x) dx

1p

A Unified approach to Sawyer and Sinnamon characteri-

zations of the Hardy inequality for decreasing functions

75

Remark 6.2. The equivalence of the conditions A0 < and A1 < forv (x) 1 follows by using Theorem 5.5 with u (x) = u (x)xq and A1 withs = 1p and A2 with s = 1p .

In Section 6.2 we state and prove the main result of this Chapter (The-orem 6.2), which generalize the above mentioned result of Sinnamon (seeCorollary 6.3) and also put some new light on the result of Sawyer for thecase when the weight v (x) is increasing. Section 6.3 is reserved for someconcluding remarks.

6.2 The main result

Our main result reads:

Theorem 6.2. Let 1 < p q < and let u (x) , v (x) and (x) be weightfunctions on (0,), where (x) is decreasing. Then the inequality

0

x0

f (t) (t) v (t) dt

q u (x) dx

1q

C

0

fp (x) v (x) dx

1p

(6.4)

holds for all C < and decreasing f if and only if one of the followingconditions holds for some s > 0:

D (s) := supt>0

t

u (x)

x0

v (y)p

(y) dy

q(

1ps)dx

1q

(6.5)

t0

v (x)p

(x) dx

s 0

t

0

v (x)p

(x)

x

u (y) dy

p(

1qs)dx

1p

(6.6)

t

u (x) dx

s

E (s) := supt>0

t

0

u (x)

x0

v (y)p

(y) dy

q(

1p

+s)dx

1q

(6.7)

t0

v (x)p

(x) dx

s 0

t

v (x)p

(x)

x

u (y) dy

p(

1q+s)dx

1p

(6.8)

t

u (x) dx

s 0, the conditions (6.5) - (6.8) are equivalent. In particular,now we assume that (6.5) holds for some s, 0 < s < 1p . Since f is decreasingwe can write

fp (x) =

x

h (y) dy, where h 0.

Thus, by using the Fubini theorem, we nd that (6.4) can equivalently bewritten as

0

x0

t

h (y) dy

1p

v (t) (t) dt

q

u (x) dx

1q

(6.9)

C

0

h (y)

y0

v (x) dx

dy

1p

Now, we dene

V (t) =

x0

v (t)p

(t) dt and C0 =(

1

1 sp) 1

p

.



77

By using Hlders inequality, Minkowskis generalized integral inequality andchanging order of integration we nd that

0

( x0

( t

h (y) dy

) 1p

v (t) (t) dt

)qu (x) dx

1q

=

0

( x0

( t

h (y) dy

) 1p

(t) V s (t) Vs (t) v

1p (t) v

1p (t) dt

)qu (x) dx

1q

0

( x0

( t

h (y) dy

)V sp (t) v (t)

) qp

( x0

V sp

(t)p (t) v(t)dt

) qp

u (x) dx

) 1q

C0

0

( x0

( t

h (y) dy

)V sp (t) v (t) dt

) qp

Vq(

1ps)

(x) u (x) dx

1q

C0

0

( t

h (y) dy

)V sp (t) v (t)

t

Vq(

1ps)

(x) u (x) dx

pq

dt

1p

C0

0

h (y)

y0

V sp (t) v (t)

t

Vq(

1ps)

(x) u (x) dx

pq

dt

dy

1p

C0supt0

V s (t)

t

Vq(

1ps)

(x)u (x) dx

1q

0

h (y)

( y0

v (t) dt

)dy

1p

=

(1

1 sp) 1

p

D1 (s)

0

h (y)

( y0

v (t) dt

)dy

1p

=

(1

1 sp) 1

p

D (s)

0

fp (t) v (t) dt

1p

.

Hence (6.9) and, thus, (6.4) holds.Now we assume that (6.4) holds for some C < and choose the test-

function

f (x) =

(p

p (ps + 1))

V(

1p+s)

(t)p

p (x)(0,t) (x) +

V(

1p+s)

(x)p

p (x)(t,) (x)

where t > 0 is xed. Then the integral on the right hand side of (6.4) isequal to

=

t0

(p

p (1 s) 1)p

V (1+ps) (t)p (x) v (x) dx (6.10)

+

t

V (1+ps) (x)p (t) v (x) dx

1p

=

((p

p (1 s) 1)p

V ps (t) +1

psV ps (t)

) 1p

=

((p

p (1 s) 1)p

+1

ps

) 1p

V s (t) .

Moreover, the left hand side of (6.4) is greater than t

x0

f (t) (t) v (t) dt

q u (x) dx

1q

=

[since

p

p+ 1 = p

]

=

t

t0

(p

p (1 s) 1)

V(

1p+s)

(t)p (y) v (y) dy +

xt

V(

1p+s)

(y)p (y) v (y) dy

q u (x) dx

1q

=

t

((p

p (1 s) 1)

V

(1ps)

(t) +

(p

p (1 s) 1)

V

(1ps)

(x)

79

(p

p (1 s) 1)

V

(1ps)

(t)

)qu (x) dx

) 1q

=

(p

p (1 s) 1)

t

Vq(

1ps)

(x) u (x) dx

1q

.

By combining this estimate with (6.10) we nd that, in view of (6.4),

(p

p (1 s) 1)

t

Vq(

1ps)

(x)u (x) (x) dx

1q

C ((

p

p (1 s) 1)p

+1

ps

) 1p

V s (t)

i.e. that(

pp(1s)1

)p(

pp(1s)1

)p+ 1ps

1p

t

Vq(

1ps)

(x) u (x) dx

1q

V s (t) C 0 we nd that (6.5) holds for 0 < s < 1p .Hence, according to Theorem 5.5, all conditions (6.5) - (6.8) holds for alls > 0. The proof is complete.

By applying Theorem 6.2 with 1 we obtain the following version ofthe before mentioned result by G. Sinnamon:

Corollary 6.3. Let 1 0

t

u (x)

x0

v (y) dy

q(

1ps)dx

1q x

0

v (t) dt

s


D1 (s) = supt>0

t

0

v (x)

x

u (y) dy

p(

1qs)dx

1p

t

u (x) dx

s 0

t

0

u (x)

x0

v (y) dy

q(

1p

+s)dx

1q x

0

v (t) dt

s 0

t

v (x)

x

u (y) dy

p(

1q+s)dx

1p

t

u (x) dx

s



81

F 1 (s) = supt>0

t

0

v1p

(x)

x

u (y) yqdy

p(

1qs)dx

1p

t

u (x) xqdx

s 0

t

0

u (x) xq

x0

v1p

(y) dy

q(

1p

+s)dx

1q

t0

v1p

(x) dx

s 0

t

v1p

(x)

x

u (y) yqdy

p(

1q+s)dx

1p

t

u (x)xqdx

s

Remark 6.6. According to our proof of Theorem 6.2 we see that the suf-cient part of the Theorem holds in general i.e. without the additional as-sumption that (x) is decreasing.

Remark 6.7. Note that condition F1 (s) < , with 0 < s < 1p coincideswith that obtained by Wedestig in [102] to characterize (6.11) for all functionsf 0. The explanation of this phenomenon is that among all rearrangementsof the function f the integral

0

fp (x) v (x) dx with v (x) non-decreasing, will

be the smallest possible when f is rearranged in decreasing order. It followsthat if v (x) is increasing, then (6.11) for all functions is equivalent to (6.11)restricted to the cone of decreasing functions.

Remark 6.8. In view of Sinnamons result that (6.1) and (6.3) are in factequivalent our result also gives directly new results concerning embedding be-

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