In this paper, it is shown that a refinement of Hölder’s inequality can be established using the positive definiteness of the Gram matrix

A REFINEMENT OF HÖLDER’S INEQUALITY AND APPLICATIONS

XUEMEI GAO, MINGZHE GAO, AND XIAOZHOU SHANG DEPARTMENT OFMATHEMATICS ANDCOMPUTERSCIENCE

NORMALCOLLEGE, JISHOUUNIVERSITY

JISHOUHUNAN416000, PEOPLE’SREPUBLIC OFCHINA

mingzhegao@163.com

Received 10 February, 2007; accepted 10 June, 2007 Communicated by S.S. Dragomir

ABSTRACT. In this paper, it is shown that a refinement of Hölder’s inequality can be established using the positive definiteness of the Gram matrix. As applications, some improvements on Minkowski’s inequality, Fan Ky’s inequality and Hardy’s inequality are given.

Key words and phrases: Inner product space, Gram matrix, Variable unit-vector, Minkowski’s inequality, Fan Ky’s inequality, Hardy’s inequality.

2000 Mathematics Subject Classification. 26D15, 46C99.

1. INTRODUCTION

For convenience, we need to introduce the following notations which will be frequently used throughout the paper:

(a^r, b^s) =

∞

X

n=1

a^r_nb^s_n, kak_r =

∞

X

n=1

a^r_n

!¹_r

, kak₂ =kak,

(f^r, g^s) = Z ∞

0

f^r(x)g^s(x)dx, kfk_r= Z ∞

0

f^r(x)dx ¹_r

, kfk₂ =kfk, and

S_r(α, y) = α^r/2, y

kαk^−r/2_r ,

where a = (a₁, a₂, . . .)are sequences of real numbers, f : [0,∞) → [0,∞) are measurable functions andαandyare elements of an inner product spaceEof real sequences.

Leta= (a₁, a₂, . . .)andb = (b₁, b₂, . . .)be sequences of real numbers inRⁿ. Then Hölder’s inequality can be written in the form:

(1.1) (a, b)≤ kak_pkbk_q.

The equality in (1.1) holds if and only ifa^p_i =kb^q_i, i= 1,2, . . ., wherekis a constant.

The authors would like to thank the anonymous referee for valuable comments that have been implemented in the final version of this paper.

A Project Supported by scientific Research Fund of Hunan Provincial Education Department (06C657).

100-07

This inequality is important in function theory, functional analysis, Fourier analysis and ana- lytic number theory, etc. However, there are drawbacks in this inequality. For example, let

a= (a1, a2, . . . , an,0, . . . ,0), b = (0,0, . . . , bn+1, bn+2, . . . , b2n), a, b∈R²ⁿ. If we leta_i =b_j = 1, i = 1,2, . . . , n;j =n+ 1, n+ 2, . . . ,2n, and substitute them into (1.1), then we have0≤n. In this case, Hölder’s inequality is meaningless.

In the present paper we establish a new inequality that improves Hölder’s inequality and remedies the defect pointed out above. At the same time, some significant refinements for a number of the classical inequalities can be established. As space is limited, only several applications of the new inequality are given.

2. MAINRESULTS

Letαandβ be elements of an inner product spaceE.Then the inner product ofαandβ is denoted by(α, β)and the norm ofα is given bykαk = p

(α, α).In our previous papers ([1], [2]), the following result has been obtained by means of the positive definiteness of the Gram matrix.

Lemma 2.1. Letα, β andγ be three arbitrary vectors ofE. Ifkγk= 1, then (2.1) |(α, β)|² ≤ kαk²kβk²−(kαk |x| − kβk |y|)²,

wherex= (β, γ), y = (α, γ). The equality in (2.1) holds if and only ifαandβare linearly de- pendent, orγis a linear combination ofαandβ, andxy= 0butxandyare not simultaneously equal to zero.

For the sake of completeness, we give here a short proof of (2.1), which can also be found in [2].

Proof of Lemma 2.1. Consider the Gram determinant constructed by the vectorsα, βandγ:

G(α, β, γ) =

(α, α) (α, β) (α, γ) (β, α) (β, β) (β, γ) (γ, α) (γ, β) (γ, γ)

.

According to the positive definiteness of Gram matrix we haveG(α, β, γ)≥0, andG(α, β, γ) = 0if and only if the vectorsα, β andγ are linearly dependent.

Expanding this determinant and using the conditionkγk= 1we obtain G(α, β, γ) =kαk²kβk² −(α, β)²−

kαk²x²−2(α, β)xy+kβk²y²

≤ kαk²kβk²−(α, β)² −

kαk²x²−2|(α, β)xy|+kβk²y²

≤ kαk²kβk²−(α, β)² − {kαk |x| − kβk |y|}²

wherex= (β, γ)andy = (α, γ). It follows that the equality holds if and only if the vectorsα andβ are linearly dependent; or the vectorγ is a linear combination of the vectorαandβ, and

xy= 0butxandyare not simultaneously equal to zero.

Applying Lemma 2.1, we can now establish the following refinement of Hölder’s inequality.

Theorem 2.2. Leta_n,b_n ≥ 0, (n = 1,2, . . .), ¹_p + ¹_q = 1andp > 1. If0 <kak_p < +∞and 0<kbk_q <+∞, then

(2.2) (a, b)≤ kak_pkbk_q(1−r)^m,

where

r= (S_p(a, c)−S_q(b, c))², m= min 1

p,1 q

, kck= 1

and

a^p/2, c

b^q/2, c

≥0.

The equality in (2.2) holds if and only ifa^p/2and b^q/2 are linearly dependent; or if the vector c is a linear combination of a^p/2and b^q/2, and a^p/2, c

b^q/2, c

= 0, but the vector cis not simultaneously orthogonal toa^p/2 andb^q/2.

Proof. Firstly, we consider the casep6=q. Without loss of generality, we suppose thatp > q >

1. Since ¹_p + ¹_q = 1, we have p > 2. LetR = ^p₂, Q = _p−2^p , then _R¹ + _Q¹ = 1. By Hölder’s inequality we obtain

(a, b) =

∞

X

k=1

a_kb_k (2.3)

=

∞

X

k=1

a_kb^q/p_k b^1−q/p_k

≤

∞

X

k=1

a_kb^q/p_k R!_R¹ _∞ X

k=1

b^1−q/p_k Q!_Q¹

= a^p/2, b^q/22/p

kbk^q(1−2/p)_q .

The equality in (2.3) holds if and only ifa^p/2andb^q/2are linearly dependent. In fact, the equality in (2.3) holds if and only if for anyk, there existsc₀ (c₀ 6= 0)such that

a_kb^q/p_k R

=c₀

b^1−q/p_k Q

. It is easy to deduce thata^p/2_k =c0b^q/2_k .

Ifα, β andγ in (2.1) are replaced bya^p/2, b^q/2 andcrespectively, then we have

(2.4) a^p/2, b^q/22

≤ kak^p_pkbk^q_q(1−r),

wherer = (S_p(a, c)−S_q(b, c))². Substituting (2.4) into (2.3), we obtain after simplifications (2.5) (a, b)≤ kak_pkbk_q(1−r)^1p.

It is known from Lemma 2.1 that the equality in (2.5) holds if and only ifa^p/2andb^q/2 are lin- early dependent; or if the vectorcis a linear combination ofa^p/2andb^q/2, and a^p/2, c

b^q/2, c

= 0,but the vectorcis not simultaneously orthogonal toa^p/2 andb^q/2.

Note the symmetry ofpandq. The inequality (2.2) follows from (2.5).

Secondly, consider the casep= 2. By Lemma 2.1, we obtain:

(2.6) (a, b)≤ kak kbk(1−r)¹² ,

where r =

(a,c)

kak − ^(b,c)_kbk 2

, kck = 1 and (a, c) (b, c) ≥ 0.The equality in (2.6) holds if and only if a andb are linearly dependent, or the vector cis a linear combination ofa and b, and (a, c)(b, c) = 0, but(a, c)and(b, c)are not simultaneously equal to zero.

The proof of the theorem is thus completed.

Consider the example given in the Introduction. Let c = (c₁, c₂, . . . , c_2n), c ∈ R²ⁿ, where c_i = ^√1_n, i= 1,2, . . . , nandc_j = 0, j =n+ 1, n+ 2, . . . ,2n.It is easy to deduce thatkck= 1 andr = 1. Substituting them into (2.2), it follows that the equality is valid.

The following theorem provides a similar result to Theorem 2.2.

Theorem 2.3. Letf(x), g(x)≥ 0 (x∈ (0,+∞)), ¹_p +¹_q = 1andp >1. If0 <kfk_p <+∞

and0<kgk_q <+∞, then

(2.7) (f, g)≤ kfk_pkgk_q(1−r)^m,

where

r = (S_p(f, h)−S_q(g, h))², m = min 1

p,1 q

, khk= 1, i.e. khk=

Z ∞ 0

h²(x)dx ¹₂

= 1 and

f^p/2, h

g^q/2, h

≥0.

The equality in (2.3) holds if and only if f^p/2and g^q/2 are linearly dependent; or the vectorh is a linear combination of f^p/2and g^q/2, and f^p/2, h

g^q/2, h

= 0, but the vector h is not simultaneously orthogonal tof^p/2andg^q/2.

Its proof is similar to that of Theorem 2.2. Hence it is omitted.

3. APPLICATIONS

3.1. A Refinement of Minkowski’s Inequality. We firstly give a refinement of Minkowski’s inequality for the discrete form.

Theorem 3.1. Leta_k, b_k ≥0,p > 1. If0<kak_p <+∞and0<kbk_p <+∞, then

(3.1) ka+bk_p <

kak_p+kbk_p

(1−r)^m, where

ka+bk_p =

∞

X

k=1

(a_k+b_k)^p

!¹_p , r = min{r(a), r(b)}, m= min

1

p,1− 1 p

,

r(x) =

( x^p/2, c

kxk^p/2_p − (a+b)^p/2, c ka+bk^p/2_p

)2

, x=a, b;

(a+b)^p/2, c

=

∞

X

k=1

(a_k+b_k)^p/2c_k, andcis a variable unit-vector.

Proof. Letm= minn

1

p,1− ¹_po ,

ka+bk_p =

∞

X

k=1

(a_k+b_k)^p

!¹_p .

By Theorem 2.2, we have (3.2)

∞

X

k=1

a_k(a_k+b_k)^p−1 ≤ kak_p

∞

X

k=1

(a_k+b_k)^p

!1−¹

p

(1−r(a))^m and

(3.3)

∞

X

k=1

b_k(a_k+b_k)^p−1 ≤ kbk_p

∞

X

k=1

(a_k+b_k)^p

!1−¹_p

(1−r(b))^m, where

r(x) =

( x^p/2, c

kxk^p/2_p − (a+b)^p/2, c ka+bk^p/2_p

)2

, x=a, b,

ka+bk^p/2_p =

∞

X

k=1

(a_k+b_k)^p

!¹₂ ,

(a+b)^p/2, c

=

∞

X

k=1

(a_k+b_k)^p/2c_k, andcis a variable unit-vector.

Adding (3.5) and (3.3) we obtain, after simplifying:

(3.4) ka+bk_p ≤ kak_p(1−r(a))^m+kbk_p(1−r(b))^m.

Let r = min{r(a), r(b)}, then the inequality (3.1) follows. This completes the proof of

Theorem 3.1.

If we choose a unit-vector csuch that itsith component is 1 and the rest is zero, i.e. c = (0,0, . . . ,0, 1

(i)

,0, . . .), then

r(x) =

( x^p/2_i

kxk^p/2_p − (a_i+b_i)^p/2 ka+bk^p/2_p

)2

x=a, b.

Similarly, we can establish a refinement of Minkowski’s integral inequality.

Theorem 3.2. Letf(x), g(x)≥0,p >1. If0<kfk_p <+∞and0<kgk_p <+∞, then

(3.5) kf +gk_p <

kfk_p+kgk_p

(1−r)^m, where

kf +gk_p = Z ∞

0

(f(x) +g(x))^pdx ¹_p

, r= min{r(f), r(g)}, m = min

1

p,1− 1 p

,

r(t) =

( t^p/2, h

ktk^p/2_p − (f+g)^p/2, h kf+gk^p/2_p

)2

, t=f, g,

(f +g)^p/2, h

= Z ∞

0

(f(x) +g(x))^p/2h(x)dx,

andhis a variable unit-vector, i.e.

khk= Z ∞

0

h²(x)dx ¹₂

= 1.

Its proof is similar to that of Theorem 3.1. Hence it is omitted.

Remark 3.3. The variable unit-vectorhcan be chosen in accordance with our requirements. For example, we may choosehsuch that

h(x) =

s 2 π(1 +x²). 3.2. A Strengthening of Fan Ky’s Inequality.

Theorem 3.4. LetA, BandCbe three positive definite matrices of ordern,0≤λ≤1. Then (3.6) |A|^λ|B|^1−λ ≤ |λA+ (1−λ)B|



1−





|AC|¹⁴

¹₂(A+C)

1 2

− |BC|¹⁴

¹₂ (B +C)

1 2





2



m

, where|C|=πⁿ, m= min{λ,1−λ}.

Proof. Whenλ= 0,1, the inequality (3.3) is obviously valid. Hence we need only consider the case0< λ <1.

IfDis a positive definite matrix of ordern, then it is known from [4] that

(3.7) J_n =

Z +∞

−∞

· · · Z +∞

−∞

e^−(x,Dx)dx= π^n/2

|D|¹², wherex= (x₁, x₂, . . . , x_n), anddx=dx₁dx₂· · ·dx_n.

LetF (x) =e^−λ(x,Ax) andG(x) = e−(1−λ)(x,Bx). Ifp = ¹_λ andq = _1−λ¹ , according to (3.4) and (2.7) we have

π^n/2

|λA+ (1−λ)B|¹² (3.8)

= Z +∞

−∞

· · · Z +∞

−∞

F (x)G(x)dx

≤

Z +∞

−∞

· · · Z +∞

−∞

F^p(x)dx

¹_pZ +∞

−∞

· · · Z +∞

−∞

G^q(x)dx ¹_q

(1−r)^m

= π^n/2(1−r)^m

|A|^λ|B|^1−λ1₂, where

r=

S¹

λ(F, H)−S ¹

1−λ(G, H) 2

=

F^2λ1 , H kFk⁻

1 2λ 1 λ

−

G^2(1−λ)1 , H kGk⁻

1 2(1−λ) 1 1−λ

, whereH =e^−12(x,Cx),C is a positive definite matrix of ordern, and

kHk=

Z +∞

−∞

· · · Z +∞

−∞

H²(x)dx ¹₂

= 1.

By the definition of the variable unit-vector H, it is easy to deduce that |C| = πⁿ. Hence we have

F^2λ1 , H

= Z +∞

−∞

· · · Z +∞

−∞

F^2λ1 (x)H(x)dx

= π^n/2

¹₂ (A+C)

1 2

=

( |C|

¹₂(A+C)

)¹₂

and

kFk^1/2λ_1/λ =

Z +∞

−∞

· · · Z +∞

−∞

F^1/λ(x)dx ¹₂

=

( π^n/2

|A|^1/2 )¹₂

= |C|

|A|

¹₄ , whence

S_1/λ(F, H) = |AC|¹⁴

¹

2(A+C)

1 2

. Similarly,

S_1/(1−λ)(G, H) = |BC|¹⁴

¹₂(B+C)

1 2

, therefore we obtain

(3.9) r=





|AC|¹⁴

¹₂(A+C)

1 2

− |BC|¹⁴

¹₂(B+C)

1 2





2

.

It follows from (3.8) and (3.9) that the inequality (3.3) is valid.

3.3. An Improvement of Hardy’s Inequality. We give firstly a refinement of Hardy’s inequal- ity for the discrete form.

Theorem 3.5. Letan ≥0, βn = _n¹ Pn

k=1ak, ¹_p +¹_q = 1andp >1. If0<kak_p <+∞, then

(3.10) kβk_p ≤

p p−1

kak_p(1−r)^m, where

r= (a^p/2, c)

kak^p/2_p − (β^p/2, c) kβk^p/2_p

!2

, cis a variable unit-vector andm= minn

1 p,¹_qo

.

Proof. Firstly, we estimate the difference of the following two terms:

β_n^p− p

p−1β_n^p−1a_n=β_n^p− p

p−1(nβ_n−(n−1)β_n−1)β_n^p−1 (3.11)

=β_n^p

1− np p−1

+ (n−1)p

p−1 (β_n^p)^p−1β_n−1^p _p¹ .

Applying the arithmetic-geometric mean inequality to the second term on the right-hand side of (3.11) we get

(3.12) (β_n^p)^p−1β_n−1^p _p¹

≤ 1

p (p−1)β_n^p +β_n−1^p .

It follows from (3.11) and (3.12) that β_n^p− p

p−1β_n^p−1an≤β_n^p

1− np p−1

+(n−1)

p−1 (p−1)β_n^p +β_n−1^p

= 1

p−1 (n−1)β_n−1^p −nβ_n^p . Summing the above inequality with respect ton, we have

N

X

n=1

β_n^p − p p−1

N

X

n=1

β_n^p−1an ≤ − 1

p−1(N β_N^p)≤0.

Hence

N

X

n=1

β_n^p ≤ p p−1

N

X

n=1

β_n^p−1a_n. LettingN → ∞, we get

(3.13)

∞

X

n=1

β_n^p ≤ p p−1

∞

X

n=1

β_n^p−1a_n.

Applying the inequality (2.2) to the right-hand side of (3.13) we obtain p

p−1

∞

X

n=1

a_nβ_n^p−1 ≤ p p−1

∞

X

n=1

a^p_n

!¹_{p ∞} X

n=1

β_n^(p−1)q

!¹_q

(1−r)^m (3.14)

= p

p−1kak_p

kβk^p_p¹_q

(1−r)^m,

wherer = (S_p(a, c)−S_q(β^p−1, c))²,cis a variable unit-vector andm = minn

1 p,¹_qo

. We obtain from (3.13) and (3.14) after simplification

(3.15) kβk_p ≤

p p−1

kak_p(1−r)^m. It is easy to deduce that

Sp(a, c) = a^p/2, c

kak^p/2_p and Sq(β^p−1, c) = (β^(p−1)q/2, c)

kβ^p−1k^q/2_q = (β^p/2, c) kβk^p/2_p . Hence

r=

a^p/2, c

kak^−p/2_p − β^p/2, c

kβk^−p/2_p 2

,

wherecis a variable unit-vector. The proof of the theorem is completed.

A variable unit-vectorccan be chosen in accordance with our requirements. For example, we may choosec∈R^∞such thatc= (1,0,0, . . .). Obviously,kck= 1and

r=a^p₁

kak^−p/2_p − kβk^−p/2_p 2

.

Similarly, we can establish a refinement of Hardy’s integral inequality.

Theorem 3.6. Letf(x)≥ 0, g(x) = ¹_xRx

0 f(t)dt, ¹_p + ¹_q = 1 andp >1. If0< R∞

0 f(t)dt <

+∞,then

(3.16) kgk_p < p

p−1kfk_p(1−r)^m,

where

r= (f^p/2, h)

kfk^p/2_p − (g^p/2, h) kgk^p/2_p

!2

, his a variable unit-vector, i.e.

khk= Z ∞

0

h²(t)dt ¹₂

= 1 and m= min 1

p,1 q

. Proof. Using integration by parts and then applying (2.2) we obtain that

kgk^p_p = Z ∞

0

g^p(t)dt= p

p−1 f, g^p−1 (3.17)

≤ p

p−1kfk_p g^p−1

q(1−r)^m

= p

p−1kfk_pkgk^p−1_p (1−r)^m, where r = (S_p(f, h)−S_q(g^p−1, h))², m = minn

1 p,¹_qo

and h is a variable unit-vector. It is easy to deduce that

S_p(f, h) = f^p/2, h

kfk^p/2_p and S_q g^p−1, h

= g^p/2, h kgk^p/2_p .

It follows that the inequality (3.16) is valid. The theorem is thus proved.

A variable unit-vector hcan be chosen in accordance with our requirements. For example, we may choosehsuch thath(x) = e^−x/2. Obviously, we then have

khk= Z ∞

0

h²(t)dt ¹₂

= 1.

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[4] E.F. BECKENBACKANDR. BELLMAN, Inequalities, 2nd ed., Springer, 1965.