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volume 4, issue 3, article 61, 2003.

Received 12 December, 2002;

accepted 08 September, 2003.

Communicated by:B. Opi´c

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Journal of Inequalities in Pure and Applied Mathematics

SOME NEW HARDY TYPE INEQUALITIES AND THEIR LIMITING INEQUALITIES

ANNA WEDESTIG

Department of Mathematics Luleå University

SE- 971 87 Luleå SWEDEN.

EMail:annaw@sm.luth.se

c

2000Victoria University ISSN (electronic): 1443-5756 142-02

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Some New Hardy Type Inequalities and their Limiting

Inequalities Anna Wedestig

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J. Ineq. Pure and Appl. Math. 4(3) Art. 61, 2003

http://jipam.vu.edu.au

Abstract

A new necessary and sufficient condition for the weighted Hardy inequality is proved for the case1< p≤q <∞. The corresponding limiting Pólya-Knopp inequality is also proved for0< p≤q <∞. Moreover, a corresponding limiting result in two dimensions is proved. This result may be regarded as an endpoint inequality of Sawyer’s two-dimensional Hardy inequality. But here we need only one condition to characterize the inequality whereas in Sawyer’s case three conditions are necessary.

2000 Mathematics Subject Classification:26D15.

Key words: Inequalities, Weights, Hardy’s inequality, Pólya-Knopp’s inequality, Mul- tidimensional inequalities, Sawyer’s two-dimensional operator.

I thank Professor Lars-Erik Persson and the referee for some comments and good advice which has improved the final version of this paper.

1 Introduction. . . 3 2 A New Weight Characterization of Hardy’s Inequality. . . 8 3 A Weight Characterization of Pólya-Knopp’s Inequality . . . . 12 4 Weighted Two-Dimensional Exponential Inequalities . . . 15 5 Final Remarks and Proof . . . 25 5.1 On Minkowski’s integral inequalities . . . 25 5.2 On the conditions and the best constant in the Hardy

inequality (2.1): . . . 28 References

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1. Introduction

We are inspired by the clever Hardy-Pólya observation to the Hardy inequality, Z ∞

0

1 x

Z x 0

f(t)dt p

dx≤ p

p−1

pZ ∞ 0

f^p(x)dx, f ≥0, p >1, that by changing f to f¹^p and tending p → ∞ we obtain the Pólya-Knopp inequality

Z ∞ 0

Gf(x)dx≤e Z ∞

0

f(x)dx

with the geometric mean operator Gf(x) = exp

1 x

Z x 0

lnf(t)dt

.

Let 1 < p ≤ q < ∞. In particular, in [3] the authors tried to find a new condition for the weighted inequality

(1.1)

Z ∞ 0

exp

1 x

Z x 0

lnf(t)dt q

u(x)dx ¹_q

≤C Z ∞

0

f^p(x)v(x)dx ¹_p

by using the weighted Hardy inequality (1.2)

Z ∞ 0

Z x 0

f(t)dt q

u(x)dx ¹_q

≤C Z ∞

0

f^p(x)v(x)dx _p¹

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and replacing u(x) and f(x) by u(x)x^−q and f(x) by f^α(x) respectively in (1.2). Then, by replacingqwith _α^q andpwith ^p_α so that (1.2) becomes

Z ∞ 0

1 x

Z x 0

f^α(t)dt _α^q

u(x)dx

!¹_q

≤C Z ∞

0

f^p(x)v(x)dx ¹_p

and letting α → 0we obtain (1.1). The natural choice was of course to try to use the usual “Muckenhoupt” condition (see [4] and [6])

A_M = sup

x>0

Z ∞ x

u(t)dt

¹_q Z x 0

v(t)^1−p⁰dt _p¹0

<∞, p⁰ = p p−1, which, with the same substitutions, will be

A_M(α) = sup

x>0

Z ∞ x

u(t)t⁻^α^qdt

¹_q Z x 0

v^α−p^α (t)dt ^p−α_αp

<∞.

However, asα→0the first term tends to0. By making a suitable change in the conditionAM(α)the author was able to give a sufficient condition. In [7] this problem was solved in a satisfactory way by first proving a new necessary and sufficient condition for the weighted Hardy inequality (1.2), namely

A_P.S = sup

x>0

V(x)⁻¹^p Z x

0

u(t)V(t)^qdt ¹_q

<∞, whereV(x) = Rx

0 v(t)^1−p⁰dt, which was also already proved in the casep =q in [9], and the bounds for the best possible constantC in (1.2) are

AP.S ≤C ≤p⁰AP.S.

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Then, by using the limiting procedure described above, they obtained the following necessary and sufficient condition for the inequality (1.1) to hold for 0< p ≤q <∞:

D_P.S = sup

x>0

x⁻^p¹ Z x

0

w(t)dt ¹_q

<∞,

where

(1.3) w(t) =

exp

1 t

Z t 0

ln 1 v(y)dy

_p^q u(t) and

D_P.S ≤C ≤e¹^pD_P.S.

The lower bound was also proved in a lemma ([7, Lemma 1]) directly by using the following test function:

f(x) =t⁻^p¹χ_[0,t](x) + (xe)⁻^s^pt^s−1^p χ(t,∞)(x), s >1, t >0.

By omitting different intervals the following two lower bounds were pointed out:

D_P.S ≤C and

sup

s>1

(s−1)e^s) 1 + (s−1)e^s

¹_p sup

t>0

t^s−1^p

Z ∞ t

w(x) x

sq p

dx ¹_q

≤C.

Moreover, in [5] the following theorem was proved:

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Theorem 1.1. Let 0 < p ≤ q < ∞and u, v be weight functions. Then there exists a positive constant C < ∞ such that the inequality (1.1) holds for all f > 0if and only if there is as >1such that

(1.4) D_O.G =D_O.G(s, q, p) = sup

t>0

t^s−1^p Z ∞

t

w(x) x

sq p

dx ¹_q

<∞,

where w(x)is defined by (1.3). Moreover, ifC is the least constant for which (1.1) holds, then

sup

s>1

s−1 s

¹_p

D_O.G(s, q, p)≤C ≤inf

s>1e^s−1^p D_O.G(s, q, p).

We see that the condition (1.4) and the condition from Lemma 1 in [7] is the same but the lower bound is different. Since

(1.5)

(s−1)e^s) 1 + (s−1)e^s

¹_p

−

s−1 s

¹_p

=

s−1 s

¹_p

se^s se^s+ (1−e^s)

¹_p

−1

!

>0 for alls >1, we note that the lower bound from Lemma 1 in [7] is better than that from [5]. This suggests that the bounds for the best constantCin (1.1) with the condition (1.4) should be

(1.6) sup_s>1

(s−1)e^s) 1 + (s−1)e^s

_p¹

D_O.G(s, q, p)≤Cinf_s>1e^s−1^p D_O.G(s, q, p).

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In [5] the authors also proved that for the upper bound sshould bes = 1 + ^p_q. Thus, the best possible bounds for the constantCshould be

(1.7) sup

s>1

(s−1)e^s) 1 + (s−1)e^s

¹_p

D_O.G(s, q, p)≤C≤e¹^qD_O.G(1 + p q, q, p).

In Section 2 of this paper we prove a new necessary and sufficient condition for the weighted Hardy inequality (1.2) to hold (see Theorem 1). In Section 3 we make the limiting procedure described above in our new Hardy inequality and obtain condition (1.4) for the inequality ( 1.1) to hold and we also receive the expected bounds as in (1.6) or (1.7) (see Theorem 2). Finally, in Section 4 we prove that a two-dimensional version of the inequality (1.1) can be characterized by a two-dimensional version of the condition (1.4) and, moreover, the bounds corresponding to (1.6) or (1.7) hold (see Theorem 3). This result fits perfectly as an end point inequality of the Hardy inequalities proved by E. Sawyer ([8], Theorem 1) for the (rectangular) Hardy operator

H(f)(x1x2) = Z x

0

Z x 0

f(t1, t2)dt1dt2.

We note that for the Hardy case E. Sawyer showed that three conditions were necessary to characterize the inequality but in our endpoint case only one condition is necessary and sufficient. In Section 5 we give some concluding remarks, shortly discuss the different weight condition for characterizing the Hardy inequality and prove a two-dimensional Minkowski inequality we needed for the proof of Theorem 3 but which is also of independent interest (see Proposition 1).

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2. A New Weight Characterization of Hardy’s Inequality

Our main theorem in this section reads:

Theorem 2.1. Let1< p≤q <∞,ands∈(1, p)then the inequality (2.1)

Z ∞ 0

Z x 0

f(t)dt q

u(x)dx ¹_q

≤C Z ∞

0

f^p(x)v(x)dx _p¹

holds for allf ≥0iff

(2.2) A_W(s, q, p) = sup

t>0

V(t)^s−1^p Z ∞

t

u(x)V(x)^q(^p−sp )dx ¹_q

<∞,

where V(t) = Rt

0 v(x)^1−p⁰dx. Moreover if C is the best possible constant in (2.1) then

(2.3) sup

1<s<p





p p−s

p

p p−s

p

+_s−1¹





1 p

AW(s, q, p)

≤C ≤ inf

1<s<p

p−1 p−s

_p¹0

A_W(s, q, p).

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Proof of Theorem2.1. Letf^p(x)v(x) = g(x)in (2.1). Then (2.1) is equivalent to

(2.4)

Z ∞ 0

Z x 0

g(t)¹^pv(t)⁻¹^pdt q

u(x)dx ¹_q

≤C Z ∞

0

g(x)dx ¹_p

. Assume that (2.2) holds. We have, by applying Hölder’s inequality, the fact thatDV(t) =v(t)^1−p⁰ =v(t)⁻^p

0

p and Minkowski’s inequality, Z ∞

0

Z x 0

g(t)¹^pv(t)⁻¹^pdt q

u(x)dx ¹_q

= Z ∞

0

Z x 0

g(t)^p¹V(t)^s−1^p V(t)⁻^s−1^p v(t)⁻¹^pdt q

u(x)dx ¹_q

≤ Z ∞

0

Z x 0

g(t)V(t)^s−1dt

^q_p Z x 0

V(t)⁻

(s−1)p0

p v(t)⁻^p

0 pdt

_p^q0

u(x)dx

!¹_q

=

p p−(s−1)p⁰

_p¹0 Z ∞ 0

Z x 0

g(t)V(t)^s−1dt _p^q

V(x)

p−(s−1)p0 p

q p0

u(x)dx

!¹_q

≤

p−1 p−s

_p¹0 Z ∞ 0

g(t)V(t)^s−1 Z ∞

t

V(x)^q(^p−s_p )u(x)dx ^p_q

dt

!¹_p

≤

p−1 p−s

_p¹0

AW(s, q, p) Z ∞

0

g(t)dt _p¹

.

Hence (2.4) and thus (2.1) holds with a constant satisfying the right hand side inequality in (2.3).

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Now we assume that (2.1) and thus (2.4) holds and choose the test function g(x) =

p p−s

p

V(t)^−sv(x)^1−p⁰χ_(0,t)(x) +V(x)^−sv(x)^1−p⁰χ_(t,∞)(x), wheretis a fixed number>0.Then the right hand side is equal to

Z t 0

p p−s

p

V(t)^−sv(x)^1−p⁰dx+ Z ∞

t

V(x)^−sv(x)^1−p⁰dx ¹_p

≤

p p−s

p

V(t)^1−s− 1

1−sV(t)^1−s ¹_p

Moreover, the left hand side is greater than Z ∞

t

Z t 0

p

p−sV(t)⁻^s^pv(y)^1−p⁰dy+ Z x

t

V(y)⁻^s^pv(y)^1−p⁰dy ^q

u(x)dx ¹q

= Z ∞

t

p p−s

q

V(x)(¹⁻^s^p)^qu(x)dx ¹_q

.

Hence, (2.4) implies that p

p−s

Z ∞ t

V(x)(¹⁻^p^s)^qu(x)dx ¹_q

≤C

p p−s

p

+ 1

s−1 ¹_p

V(t)^1−s^p i.e., that

p p−s

p

+ 1

s−1 −¹_p

V(t)^s−1^p Z ∞

t

≤C

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or, equivalently, that





p p−s

p

p p−s

p

+_s−1¹





1 p

V(t)^s−1^p Z ∞

t

≤C.

We conclude that (2.2) and the left hand side of the estimate of (2.3) hold. The proof is complete.

Remark 2.1. If we replace the interval(0,∞)in (2.1) with the interval(a, b), then, by modifying the proof above, we see that Theorem2.1 is still valid with the same bounds and the condition

A_W(s, q, p) = sup

t>0

V(t)^s−1^p Z b

t

u(x)V(x)^q(^p−s_p )dx

1 q

<∞,

whereV(t) = Rt

av(x)^1−p⁰dx.

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3. A Weight Characterization of Pólya-Knopp’s Inequality

In this section we prove that a slightly improved version of Theorem1.1can be obtained just as a natural limit of our Theorem2.1.

Theorem 3.1. Let0< p≤q <∞and s >1.Then the inequality (3.1)

Z ∞ 0

exp

1 x

Z x 0

lnf(t)

dt q

u(x)dx ¹_q

≤C Z ∞

0

f^p(x)v(x)dx ¹_p

holds for all f > 0 if and only if D_O.G(s, q, p) < ∞,where D_O.G(s, q, p)is defined by (1.4). Moreover, ifC is the best possible constant in (3.1), then

(3.2) sup

s>1

(s−1)e^s 1 + (s−1)e^s

_p¹

DO.G(s, q, p)≤C ≤e¹^qDO.G

1 + p

q, q, p

.

Remark 3.1. Theorem3.1is due to B. Opic and P. Gurka, but our lower bound in (3.2) is strictly better (see (1.5)). As mentioned before, other weight charac- terizations of (3.1) have been proved by L.E. Persson and V. Stepanov [7] and H.P. Heinig, R. Kerman and M. Krbec [1].

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Proof. If we in the inequality (3.1) replacef^p(x)v(x)withf^p(x)and letw(x) be defined as in (1.3), then we see that (3.1) is equivalent to

Z ∞ 0

exp

1 x

Z x 0

lnf(t)dt q

w(x)dx ¹_q

≤C Z ∞

0

f^p(x)dx ¹_p

. Further, by using Theorem 2.1 withu(x) = w(x)x^−q andv(x) = 1, we have that

(3.3)

Z ∞ 0

1 x

Z x 0

f(t)dt q

w(x)dx ¹_q

≤C Z ∞

0

f^p(x)dx ¹_p

holds for allf ≥0if and only if (3.4) A_W(s, q, p) = sup

t>0

t^s−1^p Z ∞

t

w(x) x

sq p

dx ¹_q

=D_O.G(s, q, p)<∞.

Moreover, ifC is the best possible constant in (3.3), then

(3.5) sup

1<s<p





p p−s

p

p p−s

p

+_s−1¹





1 p

D_O.G(s, q, p)

≤C ≤ inf

1<s<p

p−1 p−s

_p¹0

D_O.G(s, q, p).

Now, we replacef in (3.3) withf^α,0< α < p,and after that we replacepwith

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α andqwith _α^q in (3.3) – (3.5), we find that for1< s < _α^p (3.6)

Z ∞ 0

1 x

Z x 0

f^α(t)dt ^q_α

w(x)dx

!¹_q

≤C_α Z ∞

0

f^p(x)dx _p¹

holds for allf ≥0if and only if D_O.G

s, q α, p

α

=D^α_O.G(s, q, p)<∞.

Moreover, ifC_α is the best possible constant in (3.6), then

(3.7) sup

1<s<_α^p







p p−αs

^p_α

p p−αs

_α^p +_s−1¹







1 p

D_O.G(s, q, p)

≤C ≤ inf

1<s<^p_α

p−α p−αs

^p−α_αp

D_O.G(s, q, p).

We also note that 1

x Z x

0

f^α(t)dt _α¹

↓exp 1 x

Z x 0

lnf(t)dt, asα→0+.

We conclude that (3.1) holds exactly when lim_α→0₊C_α < ∞ and this holds, according to (3.7), exactly when (3.4) holds. Moreover, whenα → 0₊ (3.7) implies that (3.2) holds, where we have inserted the optimal values= 1 +^p_q on the right hand side as pointed out in [5]. The proof is complete.

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4. Weighted Two-Dimensional Exponential Inequalities

In [8], E. Sawyer proved a two-dimensional weighted Hardy inequality for the case1< p≤q <∞.More exactly, he showed that for the inequality



 Z ∞

0

Z ∞ 0



 Z x

0 x2

Z

0

f(t_1,t₂)dt₁dt₂





q

w(x_1,x₂)dx₁dx₂





1 q

≤C Z ∞

0

Z ∞ 0

f^p(x₁, x₂)v(x₂, x₂)dx₁dx₂ ¹_p

to hold, three different weight conditions must be satisfied. Here we will show that when we consider the endpoint inequality of this Hardy inequality, we only need one weight condition to characterize the inequality. Our main result in this section reads:

Theorem 4.1. Let 0< p ≤ q < ∞,and let u, v andf be positive functions onR²+.If0< b₁, b₂ ≤ ∞,then

(4.1)

Z b1

0

Z b2

0

exp

1 x₁x₂

Z x1

0

Z x2

0

logf(y₁, y₂)dy₁dy₂ q

u(x₁, x₂)dx₁dx₂ ¹q

≤C Z b1

0

Z b2

0

f^p(x₁, x₂)v(x₁, x₂)dx₁dx₂ ¹p

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if and only if

(4.2) D_W(s_1,s₂, p, q) := sup

y1∈(0,b1) y2∈(0,b2)

y

s1−1 p

1 y

s2−1 p 2

Z b1

y1

Z b2

y2

x⁻

s1q p

1 x⁻

s2q p

2 w(x₁, x₂)dx₁dx₂

1 q

<∞,

wheres₁, s₂ >1and w(x₁, x₂) =

exp

1 x₁x₂

Z x1

0

Z x2

0

log 1

v(t₁, t₂)dt₁dt₂ _p^q

u(x₁, x₂) and the best possible constantCin (4.1) can be estimated in the following way:

sup

s1,s2>1

e^s¹(s₁−1) e^s¹(s₁−1) + 1

¹_p

e^s²(s₂−1) e^s²(s₂−1) + 1

¹_p

D_W(s₁, s₂, p, q) (4.3)

≤C

≤ inf

s1,s2>1e

s1+s2−2

p DW(s1, s2, p, q).

Remark 4.1. For the case p = q = 1, b₁ = b₂ = ∞ a similar result was recently proved by H. P. Heinig, R. Kerman and M. Krbec [1] but without the estimates of the operator norm (= the best constant C in (4.1)) pointed out in (4.3) here.

We will need a two-dimensional version of the following well-known Minkowski integral inequality:

(4.4)

Z b a

Φ(x) Z x

a

Ψ(y)dy r

dx ¹r

≤ Z b

a

Ψ(y) Z b

y

Φ(x)dx ¹r

dy.

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The following proposition will be required in the proof of Theorem4.1. Propo- sition4.2will be proved in Section5.

Proposition 4.2. Letr > 1, a₁, a₂, b₁, b₂ ∈ R, a₁ < b_1,a₂ < b₂ and letΦand Ψbe measurable functions on[a₁, b₁]×[a₂, b₂].Then

(4.5)

Z b1

a1

Z b2

a2

Φ(x1, x2) Z x1

a1

Z x2

a2

Ψ(y1, y2)dy1dy2

r

dx1dx2

¹r

≤ Z b1

a1

Z b2

a2

Ψ(y₁, y₂) Z b1

y1

Z b2

y2

Φ(x₁, x₂)dx₁d₂

1 r

dy₁dy₂. Proof of Theorem4.1. Assume that (4.2) holds. Letg(x₁, x₂) =f^p(x₁, x₂)v(x₁, x₂) in (4.1):

Z b1

0

Z b2

0

exp

1 x₁x₂

Z x1

0

Z x2

0

logg(y₁, y₂)dy₁dy₂ ^q_p

×

exp 1

x₁x₂ Z x1

0

Z x2

0

log 1

v(t₁, t₂)dt1dt2

^q_p

u(x1, x2)dx1dx2

!¹_q

≤C Z ∞

0

Z ∞ 0

g(x₁, x₂)dx₁dx₂ ¹_p

.

If we let

w(x₁, x₂) =

exp 1

x₁x₂ Z x1

0

Z x2

0

log 1

v(t₁, t₂)dt₁dt₂ ^q_p

u(x₁, x₂),

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then we can equivalently write (4.1) as

(4.6)

Z b1

0

Z b2

0

exp

1 x₁x₂

Z x1

0

Z x2

0

logg(y₁, y₂)dy₁dy₂ ^q_p

w(x₁, x₂)dx₁dx₂

!¹_q

≤C Z b1

0

Z b2

0

g(x₁, x₂)dx₁dx₂

1 p

.

Lety₁ =x₁t₁ andy₂ =x₂t₂,then (4.6) becomes

(4.7)

Z b1

0

Z b2

0

exp

Z 1 0

logg(x₁t₁, x₂t₂)dt₁dt₂

q p

w(x₁, x₂)dx₁dx₂

!¹_q

≤C Z b1

0

Z b2

0

g(x1, x2)dx1dx2

¹p

.

By using the result

exp Z 1

0

Z 1 0

logt^s₁¹⁻¹t^s₂²⁻¹dt1dt2

^q_p

=e^−(s¹^+s²⁻²⁾^q^p and Jensen’s inequality, the left hand side of (4.7) becomes

e

s1+s2−2 p

Z b1

0

Z b2

0

exp

Z 1 0

log

t^s₁¹⁻¹t^s₂²⁻¹g(x1t1, x2t2) dt1dt2

^q_p

×w(x₁, x₂)dx₁dx₂ ¹_q

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≤e

s1+s2−2 p

Z b1

0

Z b2

0

Z 1 0

t^s₁¹⁻¹t^s₂²⁻¹g(x₁t₁, x₂t₂)dt₁dt₂ ^q_p

×w(x₁, x₂)dx₁dx₂ ¹_q

=e

s1+s2−2 p

Z b1

0

Z b2

0

Z x1

0

Z x2

0

y^s₁¹⁻¹y₂^s²⁻¹g(y₁, y₂)dy₁dy₂ _p^q

x^−s¹

q p

1

×x^−s²

q p

2 w(x₁, x₂)dx₁dx₂ ¹_q

.

Therefore, by also using Minkowski’s integral inequality (4.5) forp < q and Fubini’s theorem for p = q, we find that the left hand side in (4.6) can be estimated as follows:

≤e

s1+s2−2 p

Z b1

0

Z b2

0

y₁^s¹⁻¹y₂^s²⁻¹g(y₁, y₂)

× Z b1

y1

Z b2

y2

x^−s¹

q p

1 x^−s²

q p

2 w(x1, x2)dx1dx2

p q

dy1dy2

!¹_p

≤e

s1+s2−2

p D_W(s₁, s₂, q, p)· Z b1

0

Z b2

0

g(y₁, y₂)dy₁dy₂ ¹_p

.

Hence, (4.6) and, thus, (4.1) holds with a constant C satisfying the right hand side estimate in (4.3).

Now, assume that (4.1) holds. For fixedt1 andt2,0< t1 < b1,0< t2 < b2,

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we choose the test function g(x₁, x₂) = g₀(x₁, x₂)

=t⁻¹₁ t⁻¹₂ χ_(0,t₁₎(x₁)χ_(0,t₂₎(x₂) +t⁻¹₁ χ_(0,t₁₎(x₁)e^−s²t^s₂²⁻¹

x^s₂² χ_(t₂,∞)(x₂) +e^−s¹t^s₁¹⁻¹

x^s₁¹ χ_(t₁,∞)(x₁)t⁻¹₂ χ_(0,t₂₎(x₂) + e^−(s¹^+s²⁾t^s₁¹⁻¹t^s₂²⁻¹

x^s₁¹x^s₂² χ_(t₁,∞)(x₁)χ_(t₂,∞)(x₂). Then the right side of (4.6) yields

Z b1

0

Z b2

0

g0(y1, y2)dy1dy2

¹p

= Z t1

0

Z t2

0

t⁻¹₁ t⁻¹₂ dy1dy2+ Z t1

0

Z b2

t2

t⁻¹₁ e^−s²t^s₂²⁻¹

y₂^s² dy1dy2

+ Z b1

t1

Z t2

0

t⁻¹₂ e^−s¹t^s₁¹⁻¹

y^s₁¹ dy₁dy₂+ Z b1

t1

Z b2

t2

e^−(s¹^+s²⁾t^s₁¹⁻¹t^s₂²⁻¹

y₁^s¹y₂^s² dy₁dy₂

1 p

= 1 + e^−s²

s₂−1 1− t₂

b₂

s2−1!

+ e^−s¹

s₁−1 1− t₁

b₁

s1−1!

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+ e^−s¹e^−s²

(s1−1) (s2−1) 1− t₁

b1

s1−1! 1−

t₂ b2

s2−1!!¹_p

≤

1 + e^−s²

s₂−1 + e^−s¹

s₁−1+ e^−s¹e^−s² (s₁−1) (s₂ −1)

₁

p,

i.e.,

(4.8)

Z b1

0

Z b2

0

g₀(y₁, y₂)dy₁dy₂ ¹p

≤

e^s¹(s1 −1) + 1 e^s¹(s₁−1)

¹_p

e^s²(s2−1) + 1 e^s²(s₂−1)

_p¹ .

Moreover, for the left hand side in (4.6) we have

(4.9)

Z b1

0

Z b2

0

w(x₁, x₂)

exp 1

x₁x₂ Z x1

0

Z x2

0

logg(y₁, y₂)dy₁dy₂ _p^q

dx₁dx₂

!¹_q

≥ Z b1

t1

Z b2

t2

w(x₁, x₂)

exp 1

x₁x₂ Z x1

0

Z x2

0

logg(y₁, y₂)dy₁dy₂ _p^q

dx₁dx₂

!¹_q .

With the functiong₀(y₁, y₂)we get that exp

1 x₁x₂

Z x1

0

Z x2

0

logg0(y1, y2)dy1dy2

= exp (I1+I2+I3+I4),

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where I₁ = 1

x1x2

Z t1

0

Z t2

0

log t⁻¹₁ t⁻¹₂

dy₁dy₂ =− t₁t₂ x1x2

logt₁− t₁t₂ x1x2

logt₂,

I₂ = 1 x₁x₂

Z t1

0

Z x2

t2

log

t⁻¹₁ e^−s²t^s₂²⁻¹ y^s₂²

dy₁dy₂

=−t₁

x₁logt₁ + t₁t₂

x₁x₂ logt₁+ (s₂−1) t₁

x₁ logt₂+ t₁t₂

x₁x₂logt₂ −s₂t₁

x₁ logx₂, I₃ = 1

x₁x₂ Z x1

t1

Z t2

0

log

t⁻¹₂ e^−s¹t^s₁¹⁻¹ y^s₁¹

dy₁dy₂

=−t2

x₂logt₂ + t1t2

x₁x₂ logt₂+ (s₁−1)t2

x₂ logt₁+ t1t2

x₁x₂ logt₁−s₁t2

x₂ logx₁, and

I₄ = 1 x₁x₂

Z x1

t1

Z x2

t2

log

e^−(s¹^+s²⁾t^s₁¹⁻¹t^s₂²⁻¹ y^s₁¹y^s₂²

dy₁dy₂

= (s₁−1) logt₁−(s₁−1) t₂

x₂ logt₁− t₁t₂ x₁x₂ logt₁ + (s₂−1) logt₂−(s₂−1) t₁

x1

logt₂− t₁t₂ x1x2

logt₂

−s₁logx₁+ t₁

x₁ logt₁+s₁t₂ x₂ logx₁

−s2logx2+ t₂

x₂ logt2+s2

t₁

x₁ logx2.