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volume 3, issue 3, article 35, 2002.

Received 16 November, 2001;

accepted 21 February, 2002.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

AN INEQUALITY IMPROVING THE SECOND HERMITE-HADAMARD INEQUALITY FOR CONVEX FUNCTIONS DEFINED ON LINEAR SPACES AND APPLICATIONS FOR SEMI-INNER PRODUCTS

S.S. DRAGOMIR

School of Communications and Informatics Victoria University of Technology

PO Box 14428 Melbourne City MC 8001 Victoria, Australia

EMail:sever@matilda.vu.edu.au

URL:http://rgmia.vu.edu.au/SSDragomirWeb.html

c

2000Victoria University ISSN (electronic): 1443-5756 080-01

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An Inequality Improving the Second Hermite-Hadamard Inequality for Convex Functions

Defined on Linear Spaces and Applications for Semi-Inner

Products S.S. Dragomir

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Abstract

An inequality for convex functions defined on linear spaces is obtained which contains in a particular case a refinement for the second part of the celebrated Hermite-Hadamard inequality. Applications for semi-inner products on normed linear spaces are also provided.

2000 Mathematics Subject Classification: Primary 26D15, 26D10; Secondary 46B10.

Key words: Hermite-Hadamard integral inequality, Convex functions, Semi-Inner products.

1. Introduction

LetX be a real linear space,a, b∈X, a6=b and let[a, b] :={(1−λ)a+λb, λ ∈ [0,1]} be the segment generated by a and b. We consider the function f : [a, b] → Rand the attached function g(a, b) : [0,1] → R, g(a, b) (t) :=

f[(1−t)a+tb],t∈[0,1].

It is well known thatf is convex on[a, b]iffg(a, b)is convex on[0,1], and the following lateral derivatives exist and satisfy

(i) g_±⁰ (a, b) (s) = (5±f[(1−s)a+sb]) (b−a),s ∈(0,1) (ii) g₊⁰ (a, b) (0) = (5+f(a)) (b−a)

(iii) g₋⁰ (a, b) (1) = (5−f(b)) (b−a)

where(5±f(x)) (y)are the Gâteaux lateral derivatives, we recall that (5₊f(x)) (y) : = lim

h→0+

f(x+hy)−f(x) h

, (5−f(x)) (y) : = lim

k→0−

f(x+ky)−f(x) k

, x, y∈X.

The following inequality is the well known Hermite-Hadamard integral inequality for convex functions defined on a segment[a, b]⊂X :

(HH) f

a+b 2

≤ Z 1

0

f[(1−t)a+tb]dt≤ f(a) +f(b)

2 ,

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which easily follows by the classical Hermite-Hadamard inequality for the convex functiong(a, b) : [0,1]→R

g(a, b) 1

2

≤ Z 1

0

g(a, b) (t)dt≤ g(a, b) (0) +g(a, b) (1)

2 .

For other related results see the monograph on line [1].

Now, assume that(X,k·k)is a normed linear space. The functionf0(s) =

1

2kxk²,x∈X is convex and thus the following limits exist (iv) hx, yi_s:= (5₊f₀(y)) (x) = lim

t→0+

h_ky+txk2−kyk² 2t

i

; (v) hx, yi_i := (5−f₀(y)) (x) = lim

s→0−

h_ky+sxk2−kyk² 2s

i

;

for any x, y ∈ X. They are called the lower and upper semi-inner products associated to the normk·k.

For the sake of completeness we list here some of the main properties of these mappings that will be used in the sequel (see for example [2]), assuming thatp, q ∈ {s, i}andp6=q:

(a) hx, xi_p =kxk² for allx∈X;

(aa) hαx, βyi_p =αβhx, yi_p ifα, β ≥0andx, y ∈X;

(aaa)

hx, yi_p

≤ kxk kykfor allx, y ∈X;

(av) hαx+y, xi_p =αhx, xi_p +hy, xi_p ifx, y ∈X andα∈R;

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(v) h−x, yi_p =− hx, yi_q for allx, y ∈X;

(va) hx+y, zi_p ≤ kxk kzk+hy, zi_pfor allx, y, z ∈X;

(vaa) The mapping h·,·i_p is continuous and subadditive (superadditive) in the first variable forp=s (orp=i);

(vaaa) The normed linear space(X,k·k) is smooth at the pointx0 ∈ X\ {0}if and only ifhy, x₀i_s = hy, x₀i_i for ally ∈ X; in generalhy, xi_i ≤ hy, xi_s for allx, y ∈X;

(ax) If the normk·kis induced by an inner producth·,·i,thenhy, xi_i =hy, xi= hy, xi_sfor allx, y ∈X.

Applying inequality (HH) for the convex functionf₀(x) = ¹₂kxk²,one may deduce the inequality

(1.1)

x+y 2

2

≤ Z 1

0

k(1−t)x+tyk²dt ≤ kxk²+kyk² 2

for any x, y ∈ X. The same (HH) inequality applied for f₁(x) = kxk,will give the following refinement of the triangle inequality:

(1.2)

x+y 2

≤ Z 1

0

k(1−t)x+tykdt≤ kxk+kyk

2 , x, y∈X.

In this paper we point out an integral inequality for convex functions which is related to the first Hermite-Hadamard inequality in (HH) and investigate its applications for semi-inner products in normed linear spaces.

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2. The Results

We start with the following lemma which is also of interest in itself.

Lemma 2.1. Let h : [α, β] ⊂ R→Rbe a convex function on[α, β]. Then for anyγ ∈[α, β]one has the inequality

1 2

(β−γ)²h⁰₊(γ)−(γ−α)²h⁰₋(γ) (2.1)

≤(γ−α)h(α) + (β−γ)h(β)− Z β

α

h(t)dt

≤ 1 2

(β−γ)²h⁰₋(β)−(γ −α)²h⁰₊(α) . The constant ¹₂ is sharp in both inequalities.

The second inequality also holds forγ =αorγ =β.

Proof. It is easy to see that for any locally absolutely continuous function h : (α, β)→R, we have the identity

(2.2) Z β

α

(t−γ)h⁰(t)dt= (γ−α)h(α) + (β−γ)h(β)− Z β

α

h(t)dt for anyγ ∈(α, β),whereh⁰ is the derivative ofhwhich exists a.e. on(α, β).

Sincehis convex, then it is locally Lipschitzian and thus (2.2) holds. More- over, for anyγ ∈(α, β), we have the inequalities

(2.3) h⁰(t)≤h⁰₋(γ) for a.e. t ∈[α, γ]

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and

(2.4) h⁰(t)≥h⁰₊(γ) for a.e. t∈[γ, β].

If we multiply (2.3) byγ−t≥0,t∈[α, γ]and integrate on[α, γ],we get (2.5)

Z γ

α

(γ−t)h⁰(t)dt ≤ 1

2(γ−α)²h⁰₋(γ)

and if we multiply (2.4) by t−γ ≥ 0, t ∈ [γ, β], and integrate on [γ, β],we also have

(2.6)

Z β

γ

(t−γ)h⁰(t)dt≥ 1

2(β−γ)²h⁰₊(γ).

Now, if we subtract (2.5) from (2.6) and use the representation (2.2), we deduce the first inequality in (2.1).

If we assume that the first inequality (2.1) holds with a constant C > 0 instead of ¹₂, i.e.,

(2.7) C

(β−γ)²h⁰₊(γ)−(γ−α)²h⁰₋(γ)

≤(γ−α)h(α) + (β−γ)h(β)− Z β

α

h(t)dt and take the convex functionh₀(t) :=k

t− ^α+β₂

,k >0,t∈[α, β], then h⁰₀+

α+β 2

= k, h⁰₀−

α+β 2

= −k,

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h0(α) = k(β−α)

2 =h0(β), Z β

α

h₀(t)dt = 1

4k(β−α)², and the inequality (2.7) becomes, forγ = ^α+β₂ ,

C 1

4(β−α)²k+1

4(β−α)²k

≤ 1

4k(β−α)²,

givingC ≤ ¹₂,which proves the sharpness of the constant ¹₂ in the first inequality in (2.1).

If eitherh⁰₊(α) =−∞orh⁰₋(β) =−∞, then the second inequality in (2.1) holds true.

Assume that h⁰₊(α) andh⁰₋(β)are finite. Since his convex on [α, β],we have

(2.8) h⁰(t)≥h⁰₊(α) for a.e. t∈[α, γ] (γmay be equal toβ) and

(2.9) h⁰(t)≤h⁰₋(β) for a.e.t ∈[γ, β] (γ may be equal toα).

If we multiply (2.8) by γ−t ≥ 0, t ∈ [α, γ]and integrate on [α, γ],then we deduce

(2.10)

Z γ

α

(γ−t)h⁰(t)dt≥ 1

2(γ−α)²h⁰₊(α)

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and if we multiply (2.9) byt−γ ≥ 0, t ∈ [γ, β], and integrate on[γ, β],then we also have

(2.11)

Z β

γ

(t−γ)h⁰(t)dt ≤ 1

2(β−γ)²h⁰₋(β).

Finally, if we subtract (2.10) from (2.11) and use the representation (2.2), we deduce the second part of (2.1).

Now, assume that the second inequality in (2.1) holds with a constantD >0 instead of ¹₂, i.e.,

(2.12) (γ −α)f(α) + (β−γ)f(β)− Z β

α

f(t)dt

≥D

(β−γ)²f₋⁰ (β)−(γ−α)²f₊⁰ (α) . If we consider the convex function h0 given above, then we haveh⁰₋(β) = k, h⁰₊(α) = −kand by (2.12) applied forh₀ andx= ^α+β₂ we get

1

4k(β−α)² ≤D 1

4k(β−α)²+ 1

4k(β−α)²

, givingD≥ ¹₂,and the sharpness of the constant¹₂ is proved.

Corollary 2.2. With the assumptions of Lemma2.1and ifγ ∈(α, β)is a point of differentiability forh, then

(2.13)

α+β 2 −γ

h⁰(γ)

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≤

γ−α β−α

h(α) +

β−γ β−α

h(β)− 1 β−α

Z β

α

h(t)dt.

Now, recall that the following inequality, which is well known in the litera- ture as the Hermite-Hadamard inequality for convex functions, holds

(2.14) h

α+β 2

≤ 1 β−α

Z β

α

h(t)dt≤ h(α) +h(β)

2 .

The following corollary provides some bounds for the difference h(α) +h(β)

2 − 1

β−α Z β

α

h(t)dt.

Corollary 2.3. Leth: [α, β]→Rbe a convex function on[α, β]. Then we have the inequality

0 ≤ 1

8

h⁰₊

α+β 2

−h⁰₋

α+β 2

(β−α) (2.15)

≤ h(α) +h(β)

2 − 1

β−α Z β

α

h(t)dt

≤ 1 8

h⁰₋(β)−h⁰₊(α)

(β−α). The constant ¹₈ is sharp in both inequalities.

We are now able to state the corresponding result for convex functions defined on linear spaces.

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Theorem 2.4. LetXbe a linear space,a, b∈X,a 6=bandf : [a, b]⊂X →R be a convex function on the segment[a, b]. Then for anys ∈ (0,1)one has the inequality

1 2

(1−s)²(5₊f[(1−s)a+sb]) (b−a) (2.16)

−s²(5−f[(1−s)a+sb]) (b−a)

≤(1−s)f(a) +sf(b)− Z 1

0

f[(1−t)a+tb]dt

≤ 1 2

(1−s)²(5₋f(b)) (b−a)−s²(5₊f(a)) (b−a) . The constant ¹₂ is sharp in both inequalities.

The second inequality also holds fors = 0ors= 1.

Proof. Follows by Lemma2.1applied for the convex function h(t) =g(a, b) (t) = f[(1−t)a+tb], t ∈[0,1], and for the choicesα= 0,β = 1, andγ =s.

Corollary 2.5. Iff : [a, b]→Ris as in Theorem2.4and Gâteaux differentiable inc:= (1−λ)a+λb,λ∈(0,1)along the direction(b−a), then we have the inequality:

(2.17) 1

2−λ

(5f(c)) (b−a)

≤(1−λ)f(a) +λf(b)− Z 1

0

f[(1−t)a+tb]dt.

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The following result related to the second Hermite-Hadamard inequality for functions defined on linear spaces also holds.

Corollary 2.6. Iff is as in Theorem2.4, then 1

8

5+f

a+b 2

(b−a)− 5−f

a+b 2

(b−a) (2.18)

≤ f(a) +f(b)

2 −

Z 1

0

f[(1−t)a+tb]dt

≤ 1

8[(5−f(b)) (b−a)−(5₊f(a)) (b−a)]. The constant ¹₈ is sharp in both inequalities.

Now, letΩ⊂Rⁿbe an open convex set inRⁿ.

IfF : Ω → Ris a differentiable convex function on Ω,then, obviously, for any¯c∈Ωwe have

∇F (¯c) (¯y) =

n

X

i=1

∂F(¯c)

∂x_i ·yi, y¯∈Rⁿ, where _∂x^∂F

i are the partial derivatives of F with respect to the variable x_i (i= 1, . . . , n).

Using (2.16), we may state that 1

2 −λ ⁿ

X

i=1

∂F λ¯a+ (1−λ) ¯b

∂x_i ·(b_i−a_i) (2.19)

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≤(1−λ)F(¯a) +λF ¯b

− Z 1

0

F

(1−t) ¯a+t¯b dt

≤ 1 2

"

(1−λ)²

n

X

i=1

∂F ¯b

∂x_i ·(b_i−a_i)−λ²

n

X

i=1

∂F(¯a)

∂x_i ·(b_i−a_i)

#

for any¯a,¯b∈Ωandλ∈(0,1).

In particular, forλ= ¹₂, we get 0 ≤ F (¯a) +F ¯b

2 −

Z 1

0

F

(1−t) ¯a+t¯b dt (2.20)

≤ 1 8

n

X

i=1

∂F ¯b

∂x_i − ∂F(¯a)

∂x_i

!

·(bi−ai). In (2.20) the constant ¹₈ is sharp.

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3. Applications for Semi-Inner Products

Let(X,k·k)be a real normed linear space. We may state the following results for the semi-inner productsh·,·i_i andh·,·i_s.

Proposition 3.1. For anyx, y ∈X andσ ∈(0,1)we have the inequalities:

(1−σ)²hy−x,(1−σ)x+σyi_s−σ²hy−x,(1−σ)x+σyi_i (3.1)

≤(1−σ)kxk² +σkyk²− Z 1

0

k(1−t)x+tyk²dt

≤(1−σ)²hy−x, yi_i−σ²hy−x, yi_s. The second inequality in (3.1) also holds forσ= 0orσ= 1.

The proof is obvious by Theorem2.4applied for the convex functionf(x) =

1

2kxk²,x∈X.

If the space is smooth, then we may put[x, y] = hx, yi_i = hx, yi_s for each x, y ∈X and the first inequality in (3.1) becomes

(3.2) (1−2σ) [y−x,(1−σ)x+σy]

≤(1−σ)kxk²+σkyk²− Z 1

0

k(1−t)x+tyk²dt.

An interesting particular case one can get from (3.1) is the one forσ = ¹₂, 0≤ 1

8[hy−x, y+xi_s− hy−x, y +xi_i] (3.3)

≤ kxk²+kyk²

2 −

Z 1

0

k(1−t)x+tyk²dt

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≤ 1

4[hy−x, yi_i− hy−x, xi_s].

The inequality (3.3) provides a refinement and a counterpart for the second inequality in (1.1).

If we consider now two linearly independent vectors x, y ∈ X and apply Theorem2.4forf(x) = kxk,x∈X,then we get

Proposition 3.2. For any linearly independent vectorsx, y ∈Xandσ ∈(0,1), one has the inequalities:

1 2

(1−σ)² hy−x,(1−σ)x+σyi_s

k(1−σ)x+σyk −σ²hy−x,(1−σ)x+σyi_i k(1−σ)x+σyk

(3.4)

≤(1−σ)kxk+σkyk − Z 1

0

k(1−t)x+tykdt

≤ 1 2

(1−σ)² hy−x, yi_i

kyk −σ²hy−x, xi_s kxk

. The second inequality also holds forσ = 0orσ = 1.

We note that if the space is smooth, then we have (3.5)

1 2−σ

· [y−x,(1−σ)x+σy]

k(1−σ)x+σyk

≤(1−σ)kxk+σkyk − Z 1

0

k(1−t)x+tykdt

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and forσ = ¹₂, (3.4) will give the simple inequality

0 ≤ 1

8

"*

y−x,

x+y 2

^x+y₂

+

s

−

* y−x,

x+y 2

^x+y₂

+

i

# (3.6)

≤ kxk+kyk

2 −

Z 1

0

k(1−t)x+tykdt

≤ 1 8

y−x, y kyk

i

−

y−x, x kxk

s

.

The inequality (3.6) provides a refinement and a couterpart of the second inequality in (1.2).

Moreover, if we assume that (H,h·,·i) is an inner product space, then by (3.6) we get for anyx, y ∈Hwithkxk=kyk= 1that

(3.7) 0≤1−

Z 1

0

k(1−t)x+tykdt ≤ 1

8ky−xk². The constant ¹₈ is sharp.

Indeed, if we chooseH = R, ha, bi = a·b, x = −1, y = 1, then we get equality in (3.7).

We give now some examples.

1. Let`²(K), K=C,R; be the Hilbert space of sequencesx= (x_i)_i∈

Nwith

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P∞

i=0|x_i|² <∞.Then, by (3.7), we have the inequalities 0 ≤ 1−

Z 1

0

∞

X

i=0

|(1−t)x_i+ty_i|²

!¹₂ (3.8) dt

≤ 1 8·

∞

X

i=0

|y_i−x_i|²,

for anyx, y ∈`²(K)providedP∞

i=0|xi|² =P∞

i=0|yi|² = 1.

2. Let µ be a positive measure, L₂(Ω) the Hilbert space of µ−measurable functions onΩwith complex values that are2−integrable on Ω,i.e., f ∈ L2(Ω)iffR

Ω|f(t)|²dµ(t)<∞.Then, by (3.7), we have the inequalities 0 ≤ 1−

Z 1

0

Z

Ω

|(1−λ)f(t) +λg(t)|²dµ(t) ¹₂

dλ (3.9)

≤ 1 8·

Z

Ω

|f(t)−g(t)|²dµ(t) for anyf, g ∈L2(Ω)providedR

Ω|f(t)|²dµ(t) = R

Ω|g(t)|²dµ(t) = 1.

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References

[1] I. CIORANESCU, Geometry of Banach Spaces, Duality Mappings and Nonlinear Problems, Kluwer Academic Publishers, Dordrecht, 1990.

[2] S.S. DRAGOMIR and C.E.M. PEARCE, Selected Top-

ics on Hermite-Hadamard Inequalities and Applications, RGMIA Monographs, Victoria University, 2000. (ONLINE:

http://rgmia.vu.edu.au/monographs)