Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 2, Article 31, 2002
AN INEQUALITY IMPROVING THE FIRST HERMITE-HADAMARD
INEQUALITY FOR CONVEX FUNCTIONS DEFINED ON LINEAR SPACES AND
APPLICATIONS FOR SEMI-INNER PRODUCTS
S.S. DRAGOMIR
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428
M ELBOURNE C ITY MC
V ICTORIA 8001, AUSTRALIA .
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
A BSTRACT. 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 HermiteHadamard inequality. Applications for semi-inner products on normed linear spaces are also
provided.
Key words and phrases: Hermite-Hadamard integral inequality, Convex functions, Semi-Inner Products.
2000 Mathematics Subject Classification. Primary 26D15, 26D10; Secondary 46B10.
1. I NTRODUCTION
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)
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
082-01
2
S.S. D RAGOMIR
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
a+b
f (a) + f (b)
(HH)
f
≤
f [(1 − t) a + tb] dt ≤
,
2
2
0
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) = 12 kxk2 , x ∈ X is
convex and thus the following limits exist
i
h
ky+txk2 −kyk2
(iv) hx, yis := (5+ f0 (y)) (x) = lim
;
2t
t→0+ h
i
2
2
(v) hx, yii := (5− f0 (y)) (x) = lim ky+sxk2s−kyk ;
s→0−
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;
(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) = 21 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
J. Inequal. Pure and Appl. Math., 3(2) Art. 31, 2002
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F IRST H ERMITE -H ADAMARD I NEQUALITY
FOR
C ONVEX F UNCTIONS
3
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.
2. T HE R ESULTS
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
Z β
1
2 0
2 0
(2.1)
(β − γ) h+ (γ) − (γ − α) h− (γ) ≤
h (t) dt − (β − α) h (γ)
2
α
1
≤
(β − γ)2 h0− (β) − (γ − α)2 h0+ (α) .
2
1
The constant 2 is sharp in both inequalities.
The second inequality also holds for γ = α or γ = β.
Proof. It is easy to see that for any locally absolutely continuous function h : (α, β) → R, we
have the identity
Z γ
Z β
Z β
0
0
(2.2)
(t − α) h (t) dt +
(t − β) h (t) dt = h (γ) −
h (t) dt
α
γ
α
0
for any γ ∈ (α, β) , where h 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)
h0 (t) ≤ h0− (γ) for a.e. t ∈ [α, γ]
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
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
and
Z β
1
h0 (t) dt = k (β − α)2 .
4
α
J. Inequal. Pure and Appl. Math., 3(2) Art. 31, 2002
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S.S. D RAGOMIR
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 ≤ 21 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
h0 (t) ≥ h0+ (α) for a.e. t ∈ [α, γ] (γ may be equal to β)
(2.8)
and
h0 (t) ≤ h0− (β) for a.e. t ∈ [γ, β] (γ may be equal to α) .
(2.9)
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
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 21 , i.e.,
Z β
(2.12)
h (t) dt − (β − α) h (γ) ≤ D (β − γ)2 h0− (β) − (γ − α)2 h0+ (α) .
α
, k > 0, t ∈ [α, β], then we have
If we consider the convex function h0 (t) = k t − α+β
2
h00− (β) = k, h00+ (α) = −k and by (2.12) applied for h0 in γ = α+β
we get
2
1
1
1
2
2
2
k (β − α) ≤ D k (β − α) + k (β − α) ,
4
4
4
giving D ≥
1
2
which proves the sharpness of the constant 12 in the second inequality in (2.1).
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
α+β
1
0
(2.13)
− γ h (γ) ≤
h (t) dt − h (γ) .
2
β−α α
Now, recall that the following inequality, which is well known in the literature as the HermiteHadamard inequality for convex functions, holds
Z β
α+β
1
h (α) + h (β)
(2.14)
h
≤
h (t) dt ≤
.
2
β−α α
2
The following corollary provides both a sharper lower bound for the difference,
Z β
1
α+β
h (t) dt − h
,
β−α α
2
which we know is nonnegative, and an upper bound.
J. Inequal. Pure and Appl. Math., 3(2) Art. 31, 2002
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F IRST H ERMITE -H ADAMARD I NEQUALITY
FOR
C ONVEX F UNCTIONS
5
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
1
α+β
≤
h (t) dt − h
β−α α
2
1 0
h (β) − h0+ (α) (β − α) .
≤
8 −
The constant 18 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,
0
h (x) =
 ex
if x > 0;
and h0− (0) = −1, h0+ (0) = 1. Also,
Z β
Z 0
Z
−x
h (t) dt =
e dx +
α
Now, if
(2.16)
If
α+β
2
α+β
2
α
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
− exp 0 ≤
β−α
2 1
≤
[exp (β) + exp (−α)] (β − α) .
8
= 0 and if we denote β = a, a > 0, thus α = −a and by (2.15) we also have
1
exp (a) − 1
1
a≤
− 1 ≤ a exp (a) .
2
a
2
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.
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
1
(2.18)
(1 − s)2 (5+ f [(1 − s) a + sb]) (b − a) − s2 (5− f [(1 − s) a + sb]) (b − a)
2
Z 1
≤
f [(1 − t) a + tb] dt − f [(1 − s) a + sb]
(2.17)
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.
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S.S. D RAGOMIR
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
(2.19)
− λ (5f (c)) (b − a) ≤
f [(1 − t) a + tb] dt − f (c) .
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
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)
· yi ,
ȳ ∈ Rn ,
∇F (c̄) (ȳ) =
∂x
i
i=1
∂F
where ∂x
are the partial derivatives of F with respect to the variable xi (i = 1, . . . , n) .
i
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
2
≤ (1 − λ)
· (bi − ai ) − λ
· (bi − ai )
∂xi
∂xi
i=1
i=1
for any ā, b̄ ∈ Ω and λ ∈ (0, 1) .
In particular, for λ = 12 , we get
Z 1
ā + b̄
(2.22)
0 ≤
F (1 − t) ā + tb̄ dt − F
2
0
!
n
1 X ∂F b̄
∂F (ā)
≤
−
· (bi − ai ) .
8 i=1
∂xi
∂xi
In (2.22) the constant
1
8
is sharp.
3. A PPLICATIONS FOR S EMI -I NNER P RODUCTS
Let (X, k·k) be a real normed linear space. We may state the following results for the semiinner products h·, ·ii and h·, ·is .
J. Inequal. Pure and Appl. Math., 3(2) Art. 31, 2002
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F IRST H ERMITE -H ADAMARD I NEQUALITY
FOR
C ONVEX F UNCTIONS
7
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
≤ (1 − σ)2 hy − x, yii − σ 2 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) = 12 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
Z 1
(3.2) (1 − 2σ) [y − x, (1 − σ) x + σy] ≤
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 = 21 ,
1
[hy − x, y + xis − hy − x, y + xii ]
8
Z 1
x + y 2
2
≤
k(1 − t) x + tyk dt − 2 0
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
(3.3)
0 ≤
The second inequality also holds for σ = 0 or σ = 1.
We note that if the space is smooth, then we have
Z 1
1
[y − x, (1 − σ) x + σy]
(3.5)
−σ ·
≤
k(1 − t) x + tyk dt − k(1 − σ) x + σyk ,
2
k(1 − σ) x + σyk
0
and for σ = 21 , (3.4) will give the simple inequality
"*
+
*
+#
x+y
x+y
1
2 2 0 ≤
y − x, (3.6)
x+y − y − x, x+y 8
2
2
s
i
Z 1
x + y ≤
k(1 − t) x + tyk dt − 2 0
1
y
x
≤
y − x,
− y − x,
.
8
kyk i
kxk s
J. Inequal. Pure and Appl. Math., 3(2) Art. 31, 2002
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8
S.S. D RAGOMIR
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 − 2 8
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.
P
2
(1) Let `2 (K) , K = C, R; be the Hilbert space of sequences x = (xi )i∈N with ∞
i=0 |xi | <
∞. Then, by (3.7), we have the inequalities
! 12
2 ! 12
Z 1 X
∞
∞ X
xi + yi |(1 − t) xi + tyi |2
(3.8)
0 ≤
dt −
2 0
i=0
i=0
∞
1 X
·
|yi − xi |2 ,
8 i=0
P
P∞
2
2
for any x, y ∈ `2 (K) provided ∞
i=0 |xi | =
i=0 |yi | = 1.
(2) Let µ be a positive measure, L2 (Ω) the Hilbert space of µ−measurable
on Ω
R functions
2
with 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
12
Z 1 Z
2
(3.9)
0 ≤
|(1 − λ) f (t) + λg (t)| dµ (t) dλ
≤
0
Ω
! 12
Z f (t) + g (t) 2
dµ (t)
−
2
Ω
Z
1
· |f (t) − g (t)|2 dµ (t)
≤
8 Ω
R
R
for any f, g ∈ L2 (Ω) provided Ω |f (t)|2 dµ (t) = Ω |g (t)|2 dµ (t) = 1.
R EFERENCES
[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)
J. Inequal. Pure and Appl. Math., 3(2) Art. 31, 2002
http://jipam.vu.edu.au/