Journal of Inequalities in Pure and
Applied Mathematics
AN INEQUALITY IMPROVING THE FIRST HERMITE-HADAMARD
INEQUALITY FOR CONVEX FUNCTIONS DEFINED ON LINEAR
SPACES AND APPLICATIONS FOR SEMI-INNER PRODUCTS
volume 3, issue 2, article 31,
2002.
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
Received 16 November, 2001;
accepted 15 February, 2002.
Communicated by: C.P. Niculescu
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
082-01
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Abstract
An integral inequality for convex functions defined on linear spaces is obtained
which contains in a particular case a refinement for the first 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.
Contents
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
3
Applications for Semi-Inner Products . . . . . . . . . . . . . . . . . . . . 15
References
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1.
Introduction
Let X be a real linear space, a, b ∈ X, a 6= 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] → R and 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 that f is convex on [a, b] iff g (a, b) is convex on [0, 1], and
the following lateral derivatives exist and satisfy
0
(i) g±
(a, b) (s) = (5± f [(1 − s) a + sb]) (b − a), s ∈ (0, 1)
0
(ii) g+
(a, b) (0) = (5+ f (a)) (b − a)
0
(iii) g−
(a, b) (1) = (5− f (b)) (b − a)
where (5± f (x)) (y) are the Gâteaux lateral derivatives, we recall that
f (x + hy) − f (x)
,
(5+ f (x)) (y) := lim
h→0+
h
f (x + ky) − f (x)
, x, y ∈ X.
(5− f (x)) (y) := lim
k→0−
k
The following inequality is the well-known Hermite-Hadamard integral inequality for convex functions defined on a segment [a, b] ⊂ X :
Z 1
f (a) + f (b)
a+b
(HH)
f
≤
f [(1 − t) a + tb] dt ≤
,
2
2
0
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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which easily follows by the classical Hermite-Hadamard inequality for the convex function g (a, b) : [0, 1] → R
Z 1
1
g (a, b) (0) + g (a, b) (1)
g (a, b)
≤
g (a, b) (t) dt ≤
.
2
2
0
For other related results see the monograph on line [1].
Now, assume that (X, k·k) is a normed linear space. The function f0 (s) =
1
kxk2 , x ∈ X is convex and thus the following limits exist
2
h
i
2
2
(iv) hx, yis := (5+ f0 (y)) (x) = lim ky+txk2t−kyk ;
t→0+
(v) hx, yii := (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 norm k·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
that p, q ∈ {s, i} and p 6= q:
(a) hx, xip = kxk2 for all x ∈ X;
(aa) hαx, βyip = αβ hx, yip if α, β ≥ 0 and x, y ∈ X;
(aaa) hx, yip ≤ kxk kyk for all x, y ∈ X;
(av) hαx + y, xip = α hx, xip + hy, xip if x, y ∈ X and α ∈ R;
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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(v) h−x, yip = − hx, yiq for all x, y ∈ X;
(va) hx + y, zip ≤ kxk kzk + hy, zip for all x, y, z ∈ X;
(vaa) The mapping h·, ·ip is continuous and subadditive (superadditive) in the
first variable for p = s (or p = i);
(vaaa) The normed linear space (X, k·k) is smooth at the point x0 ∈ X\ {0} if
and only if hy, x0 is = hy, x0 ii for all y ∈ X; in general hy, xii ≤ hy, xis
for all x, y ∈ X;
(ax) If the norm k·k is induced by an inner product h·, ·i , then hy, xii = hy, xi =
hy, xis for all x, y ∈ X.
Applying inequality (HH) for the convex function f0 (x) = 12 kxk2 , one may
deduce the inequality
Z 1
x + y 2
kxk2 + kyk2
2
(1.1)
≤
k(1
−
t)
x
+
tyk
dt
≤
2 2
0
for any x, y ∈ X. The same (HH) inequality applied for f1 (x) = kxk , will
give the following refinement of the triangle inequality:
Z 1
x + y kxk + kyk
(1.2)
k(1 − t) x + tyk dt ≤
, x, y ∈ X.
2 ≤
2
0
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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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 → R be a convex function on [α, β]. Then for
any γ ∈ [α, β] one has the inequality
(2.1)
1
(β − γ)2 h0+ (γ) − (γ − α)2 h0− (γ)
2
Z β
≤
h (t) dt − (β − α) h (γ)
α
1
≤
(β − γ)2 h0− (β) − (γ − α)2 h0+ (α) .
2
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
The constant 12 is sharp in both inequalities.
The second inequality also holds for γ = α or γ = β.
S.S. Dragomir
Proof. It is easy to see that for any locally absolutely continuous function h :
(α, β) → R, we have the identity
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Z
γ
Z
0
(t − α) h (t) dt +
(2.2)
α
β
0
Z
β
(t − β) h (t) dt = h (γ) −
γ
h (t) dt
α
for any γ ∈ (α, β) , where h0 is the derivative of h which exists a.e. on (α, β) .
Since h is convex, then it is locally Lipschitzian and thus (2.2) holds. Moreover, for any γ ∈ (α, β), we have the inequalities
(2.3)
0
h (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) by t − α ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , we get
Z γ
1
(2.5)
(t − α) h0 (t) dt ≤ (γ − α)2 h0− (γ)
2
α
and if we multiply (2.4) by β − t ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , we
also have
Z β
1
(2.6)
(β − t) h0 (t) dt ≥ (β − γ)2 h0+ (γ) .
2
γ
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
If we subtract (2.6) from (2.5) and use the representation (2.2), we deduce the
first inequality in (2.1).
Now, assume that the first inequality (2.1) holds with C > 0 instead of 12 ,
i.e.,
Z β
2 0
2 0
(2.7) C (β − γ) h+ (γ) − (γ − α) h− (γ) ≤
h (t) dt − (β − α) h (γ) .
α
, k > 0, t ∈ [α, β]. Then
Consider the convex function h0 (t) := k t − α+β
2
α+β
α+β
α+β
0
0
h0+
= k, h0−
= −k, h0
=0
2
2
2
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and
Z
β
α
1
h0 (t) dt = k (β − α)2 .
4
If in (2.7) we choose h = h0 , γ = α+β
, then we get
2
1
1
1
2
2
C
(β − α) k + (β − α) k ≤ k (β − α)2
4
4
4
which gives C ≤ 12 and the sharpness of the constant in the first part of (2.1) is
proved.
If either h0+ (α) = −∞ or h0− (β) = −∞, then the second inequality in (2.1)
holds true.
Assume that h0+ (α) and h0− (β) are finite. Since h is convex on [α, β] , we
have
(2.8)
h0 (t) ≥ h0+ (α) for a.e. t ∈ [α, γ] (γ may be equal to β)
S.S. Dragomir
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and
(2.9)
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
h0 (t) ≤ h0− (β) for a.e. t ∈ [γ, β] (γ may be equal to α) .
If we multiply (2.8) by t − α ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , then we
deduce
Z γ
1
(2.10)
(t − α) h0 (t) dt ≥ (γ − α)2 h0+ (α)
2
α
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and if we multiply (2.9) by β − t ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , then
we also have
Z β
1
(2.11)
(β − t) h0 (t) dt ≤ (β − γ)2 h0− (β) .
2
γ
Finally, if we subtract (2.10) from (2.11) and use the representation (2.2), we
deduce the second inequality in (2.1). Now, assume that the second inequality
in (2.1) holds with a constant D > 0 instead of 12 , i.e.,
Z
β
h (t) dt − (β − α) h (γ)
(2.12)
α
≤ D (β − γ)2 h0− (β) − (γ − α)2 h0+ (α) .
, k > 0, t ∈ [α, β], then
If we consider the convex function h0 (t) = k t − α+β
2
we have h00− (β) = k, h00+ (α) = −k and by (2.12) applied for h0 in γ = α+β
2
we get
1
1
1
2
2
2
k (β − α) ≤ D k (β − α) + k (β − α) ,
4
4
4
giving D ≥ 12 which proves the sharpness of the constant
equality in (2.1).
1
2
in the second in-
Corollary 2.2. With the assumptions of Lemma 2.1 and if γ ∈ (α, β) is a point
of differentiability for h, then
Z β
1
α+β
0
(2.13)
− γ h (γ) ≤
h (t) dt − h (γ) .
2
β−α α
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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Now, recall that the following inequality, which is well known in the literature as the Hermite-Hadamard inequality for convex functions, holds
Z β
α+β
1
h (α) + h (β)
≤
h (t) dt ≤
.
(2.14)
h
2
β−α α
2
The following corollary provides both a sharper lower bound for the difference,
Z β
α+β
1
h (t) dt − h
,
β−α α
2
which we know is nonnegative, and an upper bound.
Corollary 2.3. Let h : [α, β] → R be a convex function on [α, β]. Then we have
the inequality
α+β
1 0 α+β
0
(2.15)
0 ≤
h
− h−
(β − α)
8 +
2
2
Z β
α+β
1
≤
h (t) dt − h
β−α α
2
1 0
≤
h− (β) − h0+ (α) (β − α) .
8
1
The constant 8 is sharp in both inequalities.
Example 2.1. Assume that −∞ < α < 0 < β < ∞ and consider the convex
function h : [α, β] → R, h (x) = exp |x| . We have

 −e−x if x < 0,
h0 (x) =
 x
e
if x > 0;
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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and h0− (0) = −1, h0+ (0) = 1. Also,
Z β
Z 0
Z
−x
h (t) dt =
e dx +
α
Now, if
(2.16)
α
α+β
2
β
ex dx = exp (β) + exp (−α) − 2.
0
6= 0, then by (2.15) we deduce the elementary inequality
α + β exp (β) + exp (−α) − 2
0 ≤
− exp β−α
2 1
≤
[exp (β) + exp (−α)] (β − α) .
8
If α+β
= 0 and if we denote β = a, a > 0, thus α = −a and by (2.15) we also
2
have
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
1
exp (a) − 1
1
a≤
− 1 ≤ a exp (a) .
2
a
2
S.S. Dragomir
The reader may produce other elementary inequalities by choosing in an
appropriate way the convex function h. We omit the details.
We are now able to state the corresponding result for convex functions defined on linear spaces.
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(2.17)
Theorem 2.4. Let X be a linear space, a, b ∈ X, a 6= b and f : [a, b] ⊂ X → R
be a convex function on the segment [a, b]. Then for any s ∈ (0, 1) one has the
inequality
(2.18)
1
(1 − s)2 (5+ f [(1 − s) a + sb]) (b − a)
2
−s2 (5− f [(1 − s) a + sb]) (b − a)
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Z
≤
1
f [(1 − t) a + tb] dt − f [(1 − s) a + sb]
0
≤
1
(1 − s)2 (5− f (b)) (b − a) − s2 (5+ f (a)) (b − a) .
2
The constant 21 is sharp in both inequalities.
The second inequality also holds for s = 0 or s = 1.
Proof. Follows by Lemma 2.1 applied for the convex function h (t) = g (a, b) (t) =
f [(1 − t) a + tb], t ∈ [0, 1], and the choices α = 0, β = 1, and γ = s.
Corollary 2.5. If f : [a, b] → R is as in Theorem 2.4 and Gâteaux differentiable
in c := (1 − λ) a + λb, λ ∈ (0, 1) along the direction (b − a), then we have the
inequality:
Z 1
1
− λ (5f (c)) (b − a) ≤
f [(1 − t) a + tb] dt − f (c) .
(2.19)
2
0
The following result related to the first Hermite-Hadamard inequality for
functions defined on linear spaces also holds.
Corollary 2.6. If f is as in Theorem 2.4, then
a+b
a+b
1
(2.20) 0 ≤
5+ f
(b − a) − 5− f
(b − a)
8
2
2
Z 1
a+b
≤
f [(1 − t) a + tb] dt − f
2
0
1
≤
[(5− f (b)) (b − a) − (5+ f (a)) (b − a)] .
8
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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The constant
1
8
is sharp in both inequalities.
Now, let Ω ⊂ Rn be an open and convex set in Rn .
If F : Ω → R is a differentiable convex function on Ω, then, obviously, for
any c̄ ∈ Ω we have
n
X
∂F (c̄)
∇F (c̄) (ȳ) =
· yi ,
∂xi
i=1
ȳ ∈ Rn ,
∂F
where ∂x
are the partial derivatives of F with respect to the variable xi
i
(i = 1, . . . , n) .
Using (2.18), we may state that
X
n
∂F λā + (1 − λ) b̄
1
(2.21)
−λ
· (bi − ai )
2
∂x
i
i=1
Z 1
≤
F (1 − t) ā + tb̄ dt − F (1 − λ) ā + λb̄
0
n
n
X
X
∂F b̄
∂F (ā)
2
≤ (1 − λ)
· (bi − ai ) − λ2
· (bi − ai )
∂x
∂x
i
i
i=1
i=1
for any ā, b̄ ∈ Ω and λ ∈ (0, 1) .
In particular, for λ = 21 , we get
Z 1
ā + b̄
(2.22)
0 ≤
F (1 − t) ā + tb̄ dt − F
2
0
!
n
∂F (ā)
1 X ∂F b̄
−
· (bi − ai ) .
≤
8 i=1
∂xi
∂xi
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
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Applications for Semi-Inner
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S.S. Dragomir
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In (2.22) the constant
1
8
is sharp.
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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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 products h·, ·ii and h·, ·is .
Proposition 3.1. For any x, y ∈ X and σ ∈ (0, 1) we have the inequalities:
(3.1)
(1 − σ)2 hy − x, (1 − σ) x + σyis − σ 2 hy − x, (1 − σ) x + σyii
Z 1
≤
k(1 − t) x + tyk2 dt − k(1 − σ) x + σyk2
0
2
2
≤ (1 − σ) hy − x, yii − σ hy − x, yis .
The second inequality in (3.1) also holds for σ = 0 or σ = 1.
The proof is obvious by Theorem 2.4 applied for the convex function f (x) =
kxk2 , x ∈ X.
If the space is smooth, then we may put [x, y] = hx, yii = hx, yis for each
x, y ∈ X and the first inequality in (3.1) becomes
1
2
(3.2)
(1 − 2σ) [y − x, (1 − σ) x + σy]
Z 1
≤
k(1 − t) x + tyk2 dt − k(1 − σ) x + σyk2 .
0
An interesting particular case one can get from (3.1) is the one for σ = 12 ,
(3.3)
1
[hy − x, y + xis − hy − x, y + xii ]
8
Z 1
x + y 2
2
≤
k(1 − t) x + tyk dt − 2 0
0≤
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
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S.S. Dragomir
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≤
1
[hy − x, yii − hy − x, xis ] .
4
The inequality (3.3) provides a refinement and a counterpart for the first
inequality (1.1).
If we consider now two linearly independent vectors x, y ∈ X and apply
Theorem 2.4 for f (x) = kxk, x ∈ X, then we get
Proposition 3.2. For any linearly independent vectors x, y ∈ X and σ ∈ (0, 1) ,
one has the inequalities:
1
2 hy − x, (1 − σ) x + σyiσ
2 hy − x, (1 − σ) x + σyii
(3.4)
(1 − σ)
−σ
2
k(1 − σ) x + σyk
k(1 − σ) x + σyk
Z 1
≤
k(1 − t) x + tyk dt − k(1 − σ) x + σyk
0
1
2 hy − x, yii
2 hy − x, xis
(1 − σ)
−σ
.
≤
2
kyk
kxk
The second inequality also holds for σ = 0 or σ = 1.
We note that if the space is smooth, then we have
(3.5)
1
[y − x, (1 − σ) x + σy]
−σ ·
2
k(1 − σ) x + σyk
Z 1
≤
k(1 − t) x + tyk dt − k(1 − σ) x + σyk ,
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
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S.S. Dragomir
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and for σ = 12 , (3.4) will give the simple inequality
+#
"*
+
*
x+y
x+y
1
2 2 (3.6)
− y − x, 0≤
y − x, x+y x+y 8
2
2
i
s
Z 1
x + y ≤
k(1 − t) x + tyk dt − 2 0
y
x
1
≤
y − x,
− y − x,
.
8
kyk i
kxk s
The inequality (3.6) provides a refinement and a counterpart for the first inequality in (1.2).
Moreover, if we assume that (H, h·, ·i) is an inner product space, then by
(3.6) we get for any x, y ∈ H with kxk = kyk = 1 that
Z 1
x + y 1
≤ ky − xk2 .
(3.7)
0≤
k(1 − t) x + tyk dt − 8
2
0
1
8
The constant is sharp.
Indeed, if H = R, ha, bi = a · b, then taking x = −1, y = 1, we obtain
equality in (3.7).
We give now some examples.
1. Let `2 (K) , K = C, R; be the Hilbert space of sequences x = (xi )i∈N with
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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P∞
i=0
|xi |2 < ∞. Then, by (3.7), we have the inequalities
Z
(3.8)
∞
X
1
0≤
0
! 21
|(1 − t) xi + tyi |2
∞ X
xi + y i 2
2 dt −
i=0
! 12
i=0
∞
1 X
≤ ·
|yi − xi |2 ,
8 i=0
for any x, y ∈ `2 (K) provided
P∞
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 on
R Ω with2 complex values that are 2−integrable on Ω, i.e., f ∈
L2 (Ω) iff Ω |f (t)| dµ (t) < ∞. Then, by (3.7), we have the inequalities
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
1
Z
(3.9)
Z
0≤
0
Ω
! 12
Z f (t) + g (t) 2
dµ (t)
2
Ω
−
≤
1
·
8
21
2
|(1 − λ) f (t) + λg (t)| dµ (t) dλ
Z
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|f (t) − g (t)|2 dµ (t)
Ω
for any f, g ∈ L2 (Ω) provided
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2
R
Ω
|f (t)| dµ (t) =
2
R
Ω
|g (t)| dµ (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
Topics
on
Hermite-Hadamard
Inequalities
and
Applications,
RGMIA Monographs,
Victoria University,
2000. (ONLINE:
http://rgmia.vu.edu.au/monographs)
An Inequality Improving the
First Hermite-Hadamard
Inequality for Convex Functions
Defined on Linear Spaces and
Applications for Semi-Inner
Products
S.S. Dragomir
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