Equivalent integral conditions related to bilinear Hardy-type inequalities

M athematical I nequalities

& A pplications

With Compliments of the Author

MIA-22-106 1535–1548 Zagreb, Croatia Volume 22, Number 4, October 2019

Saikat Kanjilal, Lars-Erik Persson and Guldarya E. Shambilova

Equivalent integral conditions related to bilinear

Hardy-type inequalities

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Inequalities

& Applications

Volume 22, Number 4 (2019), 1535–1548 doi:10.7153/mia-2019-22-106

EQUIVALENT INTEGRAL CONDITIONS RELATED TO BILINEAR HARDY–TYPE INEQUALITIES

SAIKATKANJILAL, LARS-ERIKPERSSON∗ ANDGULDARYAE. SHAMBILOVA Dedicated to the 70th anniversary

of Professor Josip Peˇcari´c

(Communicated by I. Peri´c)

Abstract. Infinitely many, even scales of, equivalent conditions are derived to characterize the bilinear Hardy-type inequality under various ranges of parameters.

1. Introduction

Let Mdenote the set of all Lebesgue measurable functions on (a,b),−∞a<

b∞,M⁺⊂Mis the subset of all non-negative functions.

Let u,v,v_i,∈M⁺, 0<p,p_i,q∞,1p_i,i=1,2. Denote p_i:= _pi^p₋₁ⁱ for 1<

pi<∞,p_i:=1 for pi=∞ and p_i:=∞ for pi=1,i=1,2. We use the record A≈ B that means c1ABc2A with the constants c1,c2, depending only on irrelevant parameters.

We put

U(x) = _b

x u(t)dt, V_i(x) = _x

a v^1−p_i ⁱ(t)dt, i=1,2. It is well known that the best constantCfor the Hardy inequality

_b a

_x a f

q

u(x)dx ¹_q

C

_b a f^pv

¹_p

,f ∈M⁺, (1)

is equivalent to the Muckenhoupt constant in the case 1<pq<∞ A_M:= sup

a<x<bU^1q(x)V^p1(x)<∞

Mathematics subject classification(2010): 26D10, 46E35.

Keywords and phrases: Inequalities, the Hardy inequality, bilinear Hardy-type inequalities, scales of equivalent conditions.

∗Corresponding author.

c , Zagreb

Paper MIA-22-106 1535

or to the Mazya-Rozin constant for 0<q<p<∞,1<p<∞,¹_r :=¹_q−¹_p

B_MR:= _b

a U^rp(x)V^pr(x)u(x)dx ¹_r

<∞

and it can be replaced by different (equivalent) constants. For instance, the constants A_M andB_MR can be replaced by the following [4]:

A_T := sup

a<x<b

_x

a uV^{q 1}_q

V^−1p(x)<∞,

BPS:=

b a

_x

a uV^{q r}_p

V^q−rp(x)u(x)dx ¹_r

<∞.

More generally, it was discovered (see, e.g., [1], [5]) that these conditions are not unique and can be replaced by the scales of constants depending on a continuous parameter s>0 of a form:

A_M(s):= sup

a<x<b

_b

x uV^{q(p1−s) 1}_q

V^s(x)<∞, (2) A_T(s):= sup

a<x<b

_x

a uV^{q(p1+s) 1}_q

V^−s(x)<∞, (3)

B_MR(s):=

b a

_b

x uV^{q(p1−s) r}_q

V^rs−1(x)dV(x) ¹_r

<∞, (4)

B_PS(s):=

b a

_x

a uV^q(p1+s) _q^r

V^−rs−1(x)dV(x) ¹_r

<∞. (5)

Moreover, it yields that (see, e.g., [1], [3], [5])C≈AM(s)≈AT(s) for 1<p q<∞andC≈BMR(s)≈BPS(s)for 0<q<p<∞,p>1 ands>0.Concerning his- tory and references about such alternative conditions, even infinite scales of equivalent conditions, we refer to the book [3], Section 7, and the references therein e.g. [1], [5]

and the PhD theses [6] and [7].

Recently, M. Krepela [2] considered the following bilinear Hardy-type inequality _b

a

_x

a f

_{q x}

a g _q

u(x)dx ¹_q

C

_b

a f^p1v_{1 p}¹

1 _b

a g^p2v_{2 p}¹

2 ,f,g∈M⁺ (6) and characterized it for various range of parametersp1,p2,q. The results can be sum- marized in the following modified theorem:

THEOREMA. Let 0<q<∞,1<p₁,p₂<∞. Then the inequality (6) with the best constantCholds for every f,g∈M⁺,iff the following holds:

(i) If1<max(p1,p₂)q<∞, then C≈A₁, where A₁:= sup

a<x<bU^1q(x)V

p1

11(x)V

p1

22(x)<∞.

(ii) If1<p₁q<p₂<∞,_r¹₂ :=¹_q−_p¹₂,then C≈A₂, where

A₂:= sup

a<x<bV

p11

1 (x) _b

x U^rq2V

r2 q

2 dV_{2 r}¹

2.

(iii) If1<p2q<p1<∞,_r¹₁ :=¹_q−_p¹₁,then C≈A3, where

A₃:= sup

a<x<bV

1 p

22(x) _b

x U^rq1V

r1 q

1 dV_{1 r}¹

1.

(iv) Let 0<q<min(p₁,p₂)<∞,min(p₁,p₂)>1,_r¹_i := ¹_q−_p¹_i, i=1,2 and ¹_q

p11+_p¹₂.Then C≈A₄+A₅,where

A₄:= sup

a<x<bV

1 p

11(x) _b

x U^rq2V

r2 q

2 dV_{2 r}¹

2,

and

A₅:= sup

a<x<bV

p1

22(x) _b

x U^rq1V

r1 q

1 dV_{1 r}¹

1.

(v) Let0<q<min(p1,p2)<∞,min(p1,p2)>1,_r¹_i :=¹_q−_p¹_i,i=1,2, ¹_q>_p¹₁+_p¹₂ and _κ¹:=¹_q−_p¹₁−_p¹₂.Then C≈A₆+A₇, where

A₆:=

b a

_b

x U^rq2(y)V

r2 q

2 (y)dV2(y) _r^κ

2V

rκ

12(x)dV1(x) _κ¹

,

and

A7:=

b a

_b

x U^rq1(y)V

r1 q

1 (y)dV1(y) _r^κ

1V

rκ

21(x)dV2(x) _κ¹

.

Inspired by the works in [1] and [5] (see also [3]), in this paper we derive infinite many equivalent conditions to characterize the inequality (6) for each of the cases (i)- (v). More exactly, we prove that each of the Krepela constantsA₁−A₇can be replaced by infinite many other equivalent constants, even scales of constants.

2. The case1<max(p1,p₂)q<∞ We begin this section with the following:

THEOREM1. Let1<max(p₁,p₂)q<∞.Then C≈A⁽¹⁾₁ (s₁,s₂)≈A⁽²⁾₁ (s₁,s₂),

∀s₁,s₂>0,where

A⁽¹⁾₁ (s₁,s₂):= sup

a<x<bV₁^s1(x)V₂^s2(x)

b x uV^q(

p1 1−s1)

1 V^q(

p1 2−s2) 2

¹_q

and

A⁽²⁾₁ (s₁,s₂) = sup

a<x<bV₁^−s1(x)V₂^−s2(x)

x a uV^q(

p1 1+s1)

1 V^q(

p1 2+s2) 2

¹_q .

In this Theorem and also all other Theorems and proofs in this paper we let C denote the sharp constant in (6).

Proof. We have

C=sup

g sup

f

ab _t

afq _t

agqu(t)dt¹_q fp1,v1gp2,v2

(2)≈sup

g g⁻¹p2,v2 sup

a<x<b

b x

_y a g

q

u(y)V^q(

p11−s1) 1 (y)dy

¹_q V₁^s1(x)

(2)≈ sup

a<x<bV₁^s1(x)

⎛

⎝ sup

x<y<b

b y uV^q(

p11−s1)

1 V^q(

p12−s2) 2

¹_q V₂^s2(y)

⎞

⎠=A⁽¹⁾₁ (s1,s₂).

Similarly, by using (3) twice, wefind thatC≈A⁽²⁾₁ (s₁,s₂)so the proof is complete.

REMARK1. The Krepela constantA₁ appears just as a point(_p¹ 1,_p¹

2)on ourfirst scale of equivalent constants, i.e. A⁽¹⁾₁ (_p¹

1,_p¹ 2) =A1.

Theorem 1 gives two scales of equivalent conditions, namely A⁽¹⁾₁ (s1,s2) <∞ andA⁽²⁾₁ (s₁,s₂) <∞,characterizing the inequality (6). Below, we prove that also these scales of equivalent conditions are not unique and can in fact be replaced by several more equivalent conditions for the inequality (6) to hold. Unlike in Theorem1, here we do not link any of the conditions with the inequality (6). Rather it is provided that each of these conditions is equivalent to the conditionA₁<∞.

THEOREM2. Let1<max(p₁,p₂)q<∞. Then A₁≈A⁽³⁾₁ (s)≈A⁽⁴⁾₁ (s)

≈A⁽⁵⁾₁ (s)≈A⁽⁶⁾₁ (s1,s2),∀s,s1,s2>0, where A₁:= sup

a<x<bU^1/q(x)V₁^1/p1(x)V₂^1/p2(x),

A⁽³⁾₁ (s):= sup

a<x<b

b x u(t)V

1/p 1/q+s1

1 (t)V

1/p 1/q+s2

2 (t)dt 1/q+s

U^−s(x),

A⁽⁴⁾₁ (s):= sup

a<x<b

x a u(t)V

1/p 1/q−1s

1 (t)V

1/p 1/q−2s

2 (t)dt 1/q−s

U^s(x), 1/q>s,

A⁽⁵⁾₁ (s):= sup

a<x<b

b x u(t)V

1/p1 1/q−s

1 (t)V

1/p2 1/q−s

2 (t)dt 1/q−s

U^s(x), 1/q<s and

A⁽⁶⁾₁ (s1,s2):=sup

a<x<b

x

av^1−p₁ ¹(t)U^1/1/2qp1+s1(t)dt _1/p

1+s1

x

av^1−p₂ ²(t)U^1/1/2qp2+s2(t)dt _1/p

2+s2

×V₁^−s1(x)V₂^−s2(x).

Proof. Note that the functionU is decreasing while V₁ andV₂ are increasing.

First we prove thatA₁≈A⁽³⁾₁ (s). We have

U^1/q(x)V₁^1/p1(x)V₂^1/p2(x) =U^1/q+s(x)V₁^1/p1(x)V₂^1/p2(x)U^−s(x)

= _b

x u(t)dt _1/q+s

V₁^1/p1(x)V₂^1/p2(x)U^−s(x)

=

b x u(t)V

1/p 1/q+1s

1 (x)V

1/p 1/q+2s

2 (x)dt 1/q+s

U^−s(x)

b x u(t)V

1/p 1/q+1s

1 (t)V

1/p 1/q+2s

2 (t)dt 1/q+s

U^−s(x).

Hence,

a<x<bsupU^1/q(x)V₁^1/p1(x)V₂^1/p2(x) sup

a<x<b

b x u(t)V

1/p 1/q+s1

1 (t)V

1/p 1/q+s2

2 (t)dt 1/q+s

U^−s(x).

(7) Moreover,

b x u(t)V

1/p 1/q+s1

1 (t)V

1/p 1/q+s2

2 (t)dt 1/q+s

U^−s(x)

sup

x<t<bU^1/q(t)V₁^1/p1(t)V₂^1/p2(t) _b

x u(t)U^−1/q+s1/q (t)dt 1/q+s

U^−s(x)

= sup

x<t<bU^1/q(t)V₁^1/p1(t)V₂^1/p2(t)

1/q+s s

U^1/q+ss (x) _1/q+s

U^−s(x)

1/q+s s

_1/q+s

a<t<bsup U^1/q(t)V₁^1/p1(t)V₂^1/p2(t). (8) The equivalenceA₁≈A⁽³⁾₁ (s) now follows by taking supremum in (8) and using (7).

Next, we prove that A₁≈A⁽⁴⁾₁ (s). Fixx∈(a,b)and definey∈(x,b)such that _y

x u(t)dt= ^b

y u(t)dt. (9)

Then U^1/q(x) =

_b

x u(t)dt _1/q

= _y

x u(t)dt+ ^b

y u(t)dt _1/q

=2^1/q _b

y u(t)dt _1/q

=2^1/q _b

y u(t)dt

_{1/q−s b}

y u(t)dt _s

=2^1/q _y

x u(t)dt _1/q−s

U^s(y).

We have

U^1/q(x)V₁^1/p1(x)V₂^1/p2(x) =2^1/q _y

x u(t)dt _1/q−s

U^s(y)V₁^1/p1(x)V₂^1/p2(x)

2^1/q

y x u(t)V

1/p 1/q−1s

1 (t)V

1/p 1/q−2s

2 (t)dt 1/q−s

U^s(y)

2^1/q

y a u(t)V

1/p1 1/q−s

1 (t)V

1/p2 1/q−s

2 (t)dt 1/q−s

U^s(y).

Hence,

U^1/q(x)V₁^1/p1(x)V₂^1/p2(x)2^1/q sup

a<y<b

y a u(t)V

1/p 1/q−1s

1 (t)V

1/p 1/q−2s

2 (t)dt 1/q−s

U^s(y), (10) and the estimateA₁<cA⁽⁴⁾₁ (s)follows.

Moreover, since 1/q−s>0, we have

x a u(t)V

1/p 1/q−1s

1 (t)V

1/p 1/q−2s

2 (t)dt 1/q−s

U^s(x)

sup

a<t<xU^1/q(t)V₁^1/p1(t)V₂^1/p2(t) _x

a u(t)U^{−1/1/qq−s}(t)dt _1/q−s

U^s(x)

sup

a<t<bU^1/q(t)V₁^1/p1(t)V₂^1/p2(t)

− ^x

a U^{−1/q−s1/q} (t)dU(t) _1/q−s

U^s(x)

= sup

a<t<bU^1/q(t)V₁^1/p1(t)V₂^1/p2(t)

1/q−s s

_1/q−s

U^−1/qs−s(x)−U^−1/qs−s(a) _1/q−s

U^s(x)

1/q−s s

_1/q−s sup

a<t<bU^1/q(t)V₁^1/p1(t)V₂^1/p2(t). (11) By now just taking supremum in (11) and using (10), we obtain the equivalenceA₁≈ A⁽⁴⁾₁ (s).

Next, we prove that A₁≈A⁽⁵⁾₁ (s). Since 1/q−s<0, we get U^1/q(x)V₁^1/p1(x)V₂^1/p2(x) =U^1/q−s(x)U^s(x)V₁^1/p1(x)V₂^1/p2(x)

= _b

x u(t)dt 1/q−s

U^s(x)V₁^1/p1(x)V₂^1/p2(x)

b x u(t)V

1/p 1/q−1s

1 (t)V

1/p 1/q−2s

2 (t)dt 1/q−s

U^s(x).

Therefore

a<x<bsup U^1/q(x)V₁^1/p1(x)V₂^1/p2(x) (12)

a<x<bsup

b x u(t)V

1/p 1/q−1s

1 (t)V

1/p 1/q−2s

2 (t)dt 1/q−s

U^s(x). (13) Nowfix x∈(a,b)and definey∈(x,b)such that

_y

x u(t)dt= ^b

y u(t)dt.

ThenU(x) =2U(y)and using this fact, we obtain

b xu(t)V

1/p 1/q−s1

1 (t)V

1/p 1/q−s2

2 (t)dt 1/q−s

U^s(x)

y xu(t)V

1/p 1/q−s1

1 (t)V

1/p 1/q−s2

2 (t)dt 1/q−s

U^s(x)

_y

xu(t)dt _1/q−s

V₁^1/p1(y)V₂^1/p2(y)U^s(x)=

_b y u(t)dt

_1/q−s

V₁^1/p1(y)V₂^1/p2(y)U^s(x)

=U^1/q−s(y)V₁^1/p1(y)V₂^1/p2(y)2^sU^s(y) =2^sU^1/q(y)V₁^1/p1(y)V₂^1/p2(y) 2^s sup

a<y<bU^1/q(y)V₁^1/p1(y)V₂^1/p2(y). (14) We now take supremum in (14) and use (13) and the equivalenceA₁≈A⁽⁵⁾₁ (s)follows.

Finally, we prove thatA₁≈A⁽⁶⁾₁ (s₁,s₂). It yields that U^1/q(x)V₁^1/p1(x)V₂^1/p2(x)

=U^1/q(x)V₁^1/p1+s1(x)V₂^1/p2+s2(x)V₁^−s1(x)V₂^−s2(x)

=U^1/q(x) _x

a v^1−p₁ ¹(t)dt

_{1/p1+s1 x}

a v^1−p₂ ²(t)dt

_1/p2+s2

V₁^−s1(x)V₂^−s2(x)

x

av^1−p₁ ¹(t)U

1/2q 1/p

1+s1(t)dt _1/p

1+s1

x

a v^1−p₂ ²(t)U

1/2q 1/p

2+s2(t)dt _1/p

2+s2

V₁^−s1(x)V₂^−s2(x).

(15) On the other hand

x

av^1−p₁ ¹(t)U^1/1/2qp1+s1(t)dt _1/p

1+s1

x

av^1−p₂ ²(t)U^1/1/2qp2+s2(t)dt _1/p

2+s2

V₁^−s1(x)V₂^−s2(x)

=

x a U

1/2q 1/p

1+s1(t)dV1(t)

_1/p1+s1

x a U

1/2q 1/p

2+s2(t)dV2(t)

_1/p2+s2

V₁^−s1(x)V₂^−s2(x)

U

1/2q 1/p

1+s1(x)V₁(x) _1/p

1+s1 U

1/2q 1/p

2+s2(x)V₂(x) _1/p

2+s2

V₁^−s1(x)V₂^−s2(x)

=U^1/2q(x)V₁^1/p1+s1(x)U^1/2q(x)V₂^1/p2+s2(x)V₁^−s1(x)V₂^−s2(x)

=U^1/q(x)V₁^1/p1(x)V₂^1/p2(x). (16)

The equivalenceA₁≈A⁽⁶⁾₁ (s₁,s₂) now follows by just combining (15) and (16). The proof is complete.

3. The cases1<p1q<p2<∞and 1<p2q<p1<∞ We begin this section with the following:

THEOREM3. Let1<p₁q<p₂<∞and_r¹

2 :=¹_q−_p¹₂.Then C≈A⁽¹⁾₂ (s1,s2)≈ A⁽²⁾₂ (s₁,s₂)≈A⁽³⁾₂ (s₁,s₂)≈A⁽⁴⁾₂ (s₁,s₂),s₁,s₂>0,where

A⁽¹⁾₂ (s₁,s₂) = sup

a<x<bV₁^s1(x)

⎛

⎝ ^b

x

b y uV^q(

p1 1−s1)

1 V^q(

p1 2−s2) 2

^r_q²

V₂^r2s2−1(y)dV₂(y)

⎞

⎠

r12

,

A⁽²⁾₂ (s1,s2) = sup

a<x<bV₁^−s1(x)

⎛

⎝ ^x

a

y a uV^q(

1 p

1+s1)

1 V^q(

1 p

2+s2) 2

^r_q²

V₂^−r2s2−1(y)dV2(y)

⎞

⎠

r12

,

A⁽³⁾₂ (s₁,s₂) = sup

a<x<bV₁^s1(x)

⎛

⎝ ^b

x

y x uV^q(

p11−s1)

1 V^q(

p12+s2) 2

^r_q²

V₂^−r2s2−1(y)dV₂(y)

⎞

⎠

r12