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volume 4, issue 1, article 19, 2003.

Received 9 October, 2002;

accepted 29 January, 2003.

Communicated by:T.M. Mills

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Journal of Inequalities in Pure and Applied Mathematics

SOME INTEGRAL INEQUALITIES RELATED TO HILBERT’S

T.C. PEACHEY

School of Computer Science and Software Engineering Monash University

Clayton, VIC 3168, Australia.

EMail:[email protected]

c

2000Victoria University ISSN (electronic): 1443-5756 105-02

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Some Integral Inequalities Related to Hilbert’s

T.C. Peachey

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J. Ineq. Pure and Appl. Math. 4(1) Art. 19, 2003

http://jipam.vu.edu.au

Abstract

We prove some integral inequalities involving the Laplace transform. These are sharper than some known generalizations of the Hilbert integral inequality including a recent result of Yang.

2000 Mathematics Subject Classification:26D15.

Key words: Hilbert inequality, Laplace transform.

1. Introduction

By the Hilbert integral inequality we mean the following well-known result.

Theorem 1.1. Suppose p > 1, q = p/(p−1), functionsf and g to be non- negative and Lebesgue measurable on(0,∞)then

(1.1)

Z ∞

0

Z ∞

0

f(u)g(v) u+v dudv

≤B 1

p,1 q

Z ∞

0

f(u)^pdu

_p¹ Z ∞

0

g(v)^qdv ¹_q

,

whereB is the beta function. The inequality is strict unlessf org are null.

The constantB 1

p,¹_q

=πcosec π

p

is known to be the best possible. See [3, Chapter 9], for the history of this result.

Many authors on integral inequalities follow the practice of Hardy, Little- wood and Polya [3] of implicitly assuming all functions mentioned are measurable and non-negative. We intend to make such conditions explicit. Further, if one of the integrals on the right of (1.1) is infinite then the strict inequality gives the impression that the left side is finite. Recent authors such as Yang [7] avoid this by adding the conditions0 <R∞

0 f^p(u)du <∞,0 <R∞

0 g^q(v)dv < ∞to give the strict inequality. Hence it becomes convenient to introduce a class of functions that satisfy all the required conditions. We defineP(E)to be the class of functionsf :E →Rsuch that onE:

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1. f is measurable, 2. f is non-negative,

3. f is not null (hence positive on a set of positive measure soR

Ef >0) and 4. f is integrable (soR

Ef <∞).

Yang [7], [8] has recently found various inequalities related to that above.

One of these (in our notation) is

Theorem 1.2. Ifp > 1,q =p/(p−1),λ >2−min(p, q)and(u−a)^1−λf^p(u) and(v−a)^1−λg^q(v)are inP(a,∞)then

Z ∞

a

Z ∞

a

f(u)g(v)

(u+v−2a)^λdudv

< B

p+λ−2

p ,q+λ−2 q

Z ∞

a

(u−a)^1−λf(u)^pdu ¹_p

× Z ∞

a

(v−a)^1−λg(v)^qdv ¹_q

.

We will specifically write out the casea = 0since this case is equivalent to the complete theorem.

Theorem 1.3. If p > 1, q = p/(p−1), λ > 2−min(p, q), u^1−λf^p(u) and v^1−λg^q(v)are inP(0,∞)then

Z ∞

0

Z ∞

0

f(u)g(v) (u+v)^λ dudv

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< B

p+λ−2

p ,q+λ−2 q

Z ∞

0

u^1−λf(u)^pdu ¹_p

× Z ∞

0

v^1−λg(v)^qdv ¹_q

. To show that Theorem1.3 implies Theorem1.2, change the variablesuand v in Theorem1.3tos=u+aandt=v+aand then replacef(s−a),g(t−a) byφ(s)andψ(t)respectively. Thus Theorem1.3is equivalent to Theorem1.2.

Note also that for the caseλ= 1, Theorem1.3reduces to Theorem1.1.

This paper will consider a further generalization of this inequality.

Theorem 1.4. Ifp >1,q=p/(p−1),b >−¹_p,c >−¹_q andu^p−pb−2f^p(u)and v^q−qc−2g^q(v)are inP(0,∞)then

(1.2)

Z ∞

0

Z ∞

0

f(u)g(v) (u+v)^b+c+1dudv

< B

b+1 p, c+1

q

u^1−b−2/pf(u) _p

v^1−c−2/qg(v) _q. This reduces to Theorem1.3for the caseb = (λ+p−3)/p,c= (λ+q−3)/q.

Theorem1.4is itself a special case of the Hardy-Littlewood-Polya inequality, Proposition 319 of [3]. Recall that a functionK(u, v) is homogeneous of degree−1ifK(λu, λv) =λ⁻¹K(u, v)for allowedu,v,λ. The inequality is Theorem 1.5. If p >1, q=p/(p−1),φ^p andψ^q are inP(0,∞)andK(u, v) is positive on(0,∞)×(0,∞), homogeneous of degree−1with

Z ∞

0

K(u,1)u⁻¹^pdu=k

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then (1.3)

Z ∞

0

Z ∞

0

K(u, v)φ(u)ψ(v)dudv < kkφk_pkψk_q

and (1.4)

Z ∞

0

K(u, v)φ(u)du p

< kkφk_p.

The constantkis the best possible.

To obtain Theorem1.4substitute

K(u, v) = u^b+²^p⁻¹v^c+²^q⁻¹ (u+v)^b+c+1 andφ(u) =u^1−b−²^pf(u),ψ(v) = v^1−c−²^qg(v).

In summary, we have a chain of generalizations:

Th.1.1(Hilbert)⇐Th.1.2(Yang)⇔Th.1.3⇐Th.1.4⇐Th.1.5(HLP).

This paper presents a new inequality Theorem 2.1, sharper than Theorem 1.4, involving the Laplace transform. Logically the implications are

Th.2.2 Th.1.5(HLP)⇒Th.2.3

⇒Th.2.1⇒Th.1.4.

In Section2we state and prove this result. Section3considers the extension of our theory to the case of non-conjugate p andq. In that section, Theorem 1.1

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generalizes to Theorem3.1, Theorem1.5generalizes to Theorem3.2and The- orem2.1to Theorem3.4. The structure of the section is

Th.2.2 Th.3.2(Bonsall)⇒Th.3.3

⇒Th.3.4.

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2. A Sharper Inequality

Theorem 2.1. Suppose p > 1, q = p/(p − 1), b > −¹_p, c > −¹_q, sup- poseu^p−pb−2f^p(u)andv^q−qc−2g^q(v)areP(0,∞)andF, Gare the respective Laplace transforms of f andg, then

Z ∞

0

Z ∞

0

f(u)g(v)

(u+v)^b+c+1dudv

≤ 1

Γ(b+c+ 1)

s^bF(s)

pks^cG(s)k_q (2.1)

< B

b+ 1 p, c+1

q

u^1−b−2/pf(u) p

v^1−c−2/qg(v) q. (2.2)

To show that (2.1) gives a considerably sharper bound than (1.2) consider the casep=q= 2,b =c= 0andf(x) =g(x) =e^−x. ThenF(s) =G(s) = 1/(s+ 1)and

s^bF (s)

p =ks^cG(s)k_q = 1 while

kfk_p =kgk_q = 1

√2.

The right side of (2.1) is thus equal to1while that of (1.2) is ^π₂.

The caseb = 1− ²_p,c= 1− ²_q for Theorem2.1has been considered by the author in [6, Theorem 15].

The proof uses the following identity

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Theorem 2.2. Ifa >−1,f andg are non-negative and measurable on(0,∞) with respective Laplace transformsF andGthen

(2.3)

Z ∞

0

Z ∞

0

f(u)g(v)

(u+v)^a+1dudv= 1 Γ(a+ 1)

Z ∞

0

s^aF(s)G(s) ds.

The proof is a simple application of Fubini’s theorem, Z ∞

0

s^aF(s)G(s)ds = Z ∞

0

s^ads Z ∞

0

e^−suf(u)du Z ∞

0

e^−svg(v)dv

= Z ∞

0

f(u)du Z ∞

0

g(v)dv Z ∞

0

s^ae^−s(u+v)ds

= Z ∞

0

Z ∞

0

Γ(a+ 1)f(u)g(v) (u+v)^a+1 dudv which is equation (2.3).

Despite the simplicity of the proof of (2.3) and the obvious relation with Hilbert’s inequality, we can find no explicit reference to this identity in the literature before the case a = 0in [6]. That case appears implicitly in Hardy’s proof of Widder’s inequality, [2]. Mulholland [5] used the discrete analogue, namely

∞

X

m=0

∞

X

n=0

a_mb_n m+n+ 1 =

Z 1

0

∞

X

m=0

a_mx^m

∞

X

n=0

a_nxⁿdx,

to prove the series case of Hilbert’s inequality.

Theorem2.1also depends on the following bound on Laplace transforms.

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Theorem 2.3. Suppose p >1, α+¹_p > 0, x^p−pα−2f^p(x)isP(0,∞)andF is the Laplace transform off, then

ks^αF(s)k_p <Γ

α+ 1 p

x^1−α−²^pf(x) p. This is also a corollary of Theorem1.5. For if we make

K(u, v) =e⁻^u^vu^α+²^p⁻¹v^−α−²^p andφ(u) =u^−α−²^p⁺¹f(u) then we obtain from (1.4)

v^−α−²^pF 1

v

p

<Γ

α+1 p

x^1−α−²^pf(x) p

and a substitution s = 1/v in the left side gives the theorem. We can find no direct mention of this result although the last two formulae in Propositions 350 of [3] are special cases.

We may now prove Theorem2.1. Sinceb+c >−1, Theorem2.2gives Z ∞

0

Z ∞

0

f(u)g(v)

(u+v)^b+c+1dudv = 1 Γ(b+c+ 1)

Z ∞

0

s^b+cF (s)G(s)ds.

Then by Hölder’s inequality Z ∞

0

Z ∞

0

f(u)g(v)

(u+v)^b+c+1dudv≤ 1 Γ(b+c+ 1)

s^bF(s)

pks^cG(s)k_q

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which is equation (2.1). Applying Theorem2.3 to each norm, the right side of this is less than

Γ(b+ ¹_p)Γ(c+ ¹_q) Γ(b+c+ 1)

u^1−b−²^pf(u) p

v^1−c−²^qg(v) q

and this is (2.2).

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3. Non-Conjugate Parameters

Here we consider inequalities of the type considered in Section 2 but where q 6=p/(p−1). (Henceforth we will usep⁰ andq⁰ for the respective conjugates ofpandq.) The earliest investigation of this type seems to be Theorem 340 of Hardy, Littlewood and Polya, [3]. In our notation this is

Theorem 3.1. Suppose p > 1, q > 1, ¹_p + ¹_q ≥ 1 and λ = 2− ¹_p − ¹_q (so 0< λ≤1); supposef^p andg^qare inP(0,∞)then

Z ∞

0

Z ∞

0

f(u)g(v)

(u+v)^λ dudv < Ckfk_pkgk_q, whereCdepends onpandqonly.

Hardy, Littlewood and Polya did not give a specific value for the constantC.

An alternative proof by Levin, [4] established thatC = B^λ

1 λp⁰,_λq¹0

suffices but the paper did not decide whether this was the best possible constant. This question remains open.

Just as Theorem1.5 generalized Theorem 1.1 to a general kernel, Bonsall [1] has generalized the above result.

Theorem 3.2. Supposep > 1,q >1, ¹_p+¹_q ≥1,λ= 2−¹_p−¹_q;φ^p andψ^qare inP(0,∞)andK(u, v)is positive on(0,∞)×(0,∞), homogeneous of degree

−1with

Z ∞

0

K(x,1)x^−1/(λq⁰⁾dx =k

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then (3.1)

Z ∞

0

Z ∞

0

K^λ(u, v)φ(u)ψ(v)dudv < k^λkφk_pkψk_q and

(3.2)

Z ∞

0

K^λ(u, v)φ(u)du q⁰

< k^λkφk_p.

Since this theorem is more than that claimed by Bonsall we will repeat and extend his proof, using the standard methods of [3] Section 9.3.

Sincep⁰λandq⁰λare conjugate, K^λ(u, v)φ(u)ψ(v)

=

K(u, v)^p¹⁰ψ^p^q⁰(v)u v

_p⁻¹0q0λ

K(u, v)^q¹⁰φ^q^p⁰(u)v u

_p⁻¹0q0λ

×

φ^p(1−λ)(u)ψ^q(1−λ)(v)

=F₁F₂F₃ say.

Then since _p¹0 +_q¹0 + (1−λ) = 1, Holder’s inequality gives Z ∞

0

Z ∞

0

K(u, v)^λφ(u)ψ(v)dudv≤ Z ∞

0

Z ∞

0

F₁^p⁰dudv

_p¹0 Z ∞

0

Z ∞

0

F₂^q⁰dudv _q¹0

× Z ∞

0

Z ∞

0

F₃^1/(1−λ)dudv (1−λ)

=I

1 p0

1 I

1 q0

2 I₃^1−λ.

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Here

I₁ = Z ∞

0

Z ∞

0

K(u, v)ψ^q(v)u v

− ¹

q0λ

dudv

= Z ∞

0

Z ∞

0

K(x,1)ψ^q(v)x⁻^q¹⁰^λdxdv=kkψk^q_q.

Similarly

I₂ = Z ∞

0

Z ∞

0

K(1, x)φ^p(v)x⁻^p¹⁰^λdxdu=kkφk^p_p and

I₃ = Z ∞

0

Z ∞

0

φ(u)^pψ(v)^qdudv=kφk^p_pkψk^q_q. Substituting theseI_i, gives

(3.3)

Z ∞

0

Z ∞

0

K^λ(u, v)φ(u)ψ(v)dudv ≤k^λkφk_pkψk_q.

Note that equality in (3.3) can only occur if one of theF_i is null or all are effectively proportional, see [3, Proposition 188]. The first possibility would contradict one of the hypotheses; the alternative implies that for almost allv,

K(u, v)φ^p(u)v u

− ¹

p0λ

=K(u, v)ψ^q(v)u v

− ¹

q0λ

for almost all u. For such a v, φ^p(u) = Au⁻¹ for some positive constant A, contradictingkφk_p <∞. Thus the inequality in (3.3) is strict giving (3.1).

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Finally we note that Z ∞

0

ψ(v)dv Z ∞

0

K^λ(u, v)φ(u)du < k^λkφk_pkψk_q

for all ψ ∈ L_q. Equation (3.2) follows by the converse of Hölder’s inequality, [3, Proposition 191].

We next extend Theorem2.3to non-conjugatepandq.

Theorem 3.3. Supposep > 1,q >1, ¹_p+¹_q ≥1,λ = 2−¹_p−¹_q andα+_q¹0 >0;

supposeu^{p(λ−α−2/q}⁰⁾f^p(u)is inP(0,∞)andF is the Laplace transform of f, then

s^αF(s)

q⁰ < λ^α+^q¹⁰ Γ^λ α

λ + 1 λq⁰

u^λ−α−^q²⁰f(u) p. This follows from Theorem3.2by substituting

K(u, v) =e^−u/vu(α/λ)+(2/λq⁰)−1

v−(α/λ)−(2/λq⁰)

and

φ(u) =u^−α−(2/q⁰^)+λf(u) givingk = Γ

α λ +_λq¹0

and Z ∞

0

K^λ(u, v)φ(u)du=v^−α−^q²⁰F λ

v

and then equation (3.2) gives the required inequality.

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The caseq = p⁰ givesλ = 1and reduces to Theorem 2.3. Another known case occurs ifq =p(which requires1< p≤2) andα= 0givingλ= 2/p⁰ and

kFk_p⁰ <

2π p⁰

_p¹0

kfk_p, which is Proposition 352 of [3].

A third case of interest can be obtained by substituting α = _p¹0 − _q¹0 and introducingr=q⁰ withr≥p. Then Theorem3.3gives

x¹⁻¹^p⁻¹^rF(x) r

< λ^p¹⁰Γ^λ 1

p⁰λ

kfk_p

whereλ= _p¹0+¹_r. This is given in [3] as Proposition 360 except that the constant λ^p¹⁰Γ^λ

1 p⁰λ

is not specified there.

We can now extend Theorem2.1to the case of non-conjugate parameters.

Theorem 3.4. Suppose p > 1, q > 1, ¹_p + ¹_q ≥ 1, q ≤ r ≤ p⁰, b + _r¹0 > 0 andc+¹_r >0;u^p(1/p⁰^−1/r⁰^−b)f^p(u)andv^q(1/q⁰^−1/r−c)g^q(v)are inP(0,∞)and supposeF,Gare the Laplace transforms off andgrespectively, then

Z ∞

0

Z ∞

0

f(u)g(v)

(u+v)^b+c+1dudv ≤ 1 Γ(b+c+ 1)

s^bF(s)

r⁰ks^cG(s)k_r (3.4)

< C

u^p¹⁰⁻^r¹⁰^−bf(u) p

v^q¹⁰⁻¹^r^−cg(v) q, (3.5)

whereβ = _p¹0 +_r¹0,γ = _q¹0 + ¹_r and C =β^b+^r¹⁰γ^c+¹^rΓ^β

b β + 1

r⁰β

Γ^γ c

γ + 1 rγ

Γ⁻¹(b+c+ 1).

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The proof of (3.4) proceeds as that of (2.1) except that pandq are replaced by r⁰ and r respectively. We then apply Theorem 3.3 to ks^bF(s)kr⁰. In that theorem replaceαbyb,qbyrandλ= _p¹0 +_q¹0 byβ = _p¹0 +_r¹0, giving

ks^bF(s)k_r⁰ < β^b+^r¹⁰ Γ^β b

β + 1 βr⁰

u^p¹⁰⁻^r¹⁰^−bf(u) p

.

The condition ¹_p +¹_q ≥ 1is satisfied becauser ≤ p⁰. Alternatively if in Theo- rem3.3we replaceαbyc,pbyqandq⁰ byrwithr≥qwe obtain

ks^cG(s)k_r < γ^c+¹^r Γ^γ c

γ + 1 γr

u^q¹⁰⁻¹^r^−cg(u) q

.

Applying these to (3.4) gives (3.5) and completes the proof.

The question arises as to whether the constant C in (3.5) is better than Levin’sB^λ

1 λp⁰,_λq¹0

in the caseb = _p¹0 − _r¹0,c= _q¹0 −¹_r. For that case C =β^p¹⁰γ^q¹⁰Γ^β

1 βp⁰

Γ^γ

1 γq⁰

Γ⁻¹(λ)

whereβ = _p¹0+_r¹0 andγ = _q¹0+¹_r. Experiments with Maple suggest that Levin’s constant is the better one.

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References

[1] F.F. BONSALL, Inequalities with non-conjugate parameters, Quart. J.

Math., 2 (1951), 135–150.

[2] G.H. HARDY, Remarks in addition to Dr Widder’s note on inequalities, J.

London Math. Soc., 4 (1929), 199–202.

[3] G.H. HARDY, J.E. LITTLEWOODANDG. POLYA, Inequalities, (Second Edition) CUP, 1952.

[4] V. LEVIN, On the two parameter extension and analogue of Hilbert’s in- equality, J. London Math. Soc., 11 (1936), 119–124.

[5] H.P. MULHOLLAND, Note on Hilbert’s double series theorem, J. London Math. Soc., 3 (1928), 197–199.

[6] T.C. PEACHEY, A framework for proving Hilbert’s inequality and related results, RGMIA Research Report Collection, Victoria University of Technol- ogy, 2(3) (1999), 265–274.

[7] BICHENG YANG, On new generalizations of Hilbert’s inequality, J .Math.

Anal. Appl., 248 (2000), 29–40.

[8] BICHENG YANG, On Hardy-Hilbert’s integral inequality, J .Math. Anal.

Appl., 261 (2001), 295–306.

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