Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 3, Article 35, 2002
AN INEQUALITY IMPROVING THE SECOND 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 21 February, 2002
Communicated by A. Lupaş
A BSTRACT. 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.
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)
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)
(5− f (x)) (y) : = lim
, x, y ∈ X.
k→0−
k
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
080-01
2
S.S. D RAGOMIR
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) (t) dt ≤
.
g (a, b)
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) =
convex and thus the following limits exist
h
i
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 ;
1
2
kxk2 , x ∈ X is
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
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.
J. Inequal. Pure and Appl. Math., 3(3) Art. 35, 2002
http://jipam.vu.edu.au/
O N THE S ECOND H ERMITE -H ADAMARD I NEQUALITY
3
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
1
(2.1)
(β − γ)2 h0+ (γ) − (γ − α)2 h0− (γ)
2
Z β
≤ (γ − α) h (α) + (β − γ) h (β) −
h (t) dt
α
1
≤
(β − γ)2 h0− (β) − (γ − α)2 h0+ (α) .
2
The constant 12 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 β
Z β
0
(2.2)
(t − γ) h (t) dt = (γ − α) h (α) + (β − γ) h (β) −
h (t) dt
α
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 γ − t ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , we get
Z γ
1
(2.5)
(γ − t) h0 (t) dt ≤ (γ − α)2 h0− (γ)
2
α
and if we multiply (2.4) by t − γ ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , we also have
Z β
1
(2.6)
(t − γ) h0 (t) dt ≥ (β − γ)2 h0+ (γ) .
2
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 21 , i.e.,
(2.7) C (β − γ)2 h0+ (γ) − (γ − α)2 h0− (γ) ≤ (γ − α) h (α)+(β − γ) h (β)−
Z
β
h (t) dt
α
, k > 0, t ∈ [α, β], then
and take the convex function h0 (t) := k t − α+β
2
α+β
0
h0+
= k,
2
α+β
0
h0−
= −k,
2
J. Inequal. Pure and Appl. Math., 3(3) Art. 35, 2002
http://jipam.vu.edu.au/
4
S.S. D RAGOMIR
k (β − α)
= h0 (β) ,
2
Z β
1
h0 (t) dt =
k (β − α)2 ,
4
α
h0 (α) =
and the inequality (2.7) becomes, for γ = α+β
,
2
1
1
1
2
2
(β − α) k + (β − α) k ≤ k (β − α)2 ,
C
4
4
4
giving C ≤ 21 , which proves the sharpness of the constant 12 in the first inequality in (2.1).
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)
and
(2.9)
h0 (t) ≤ h0− (β) for a.e. t ∈ [γ, β] (γ may be equal to α) .
If we multiply (2.8) by γ − t ≥ 0, t ∈ [α, γ] and integrate on [α, γ] , then we deduce
Z γ
1
(2.10)
(γ − t) h0 (t) dt ≥ (γ − α)2 h0+ (α)
2
α
and if we multiply (2.9) by t − γ ≥ 0, t ∈ [γ, β], and integrate on [γ, β] , then we also have
Z β
1
(2.11)
(t − γ) 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
part of (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) (γ − α) f (α) + (β − γ) f (β) −
f (t) dt
α
≥ D (β − γ)2 f−0 (β) − (γ − α)2 f+0 (α) .
If we consider the convex function h0 given above, then we have h0− (β) = k, h0+ (α) = −k and
we get
by (2.12) applied for h0 and x = α+β
2
1
1
1
2
2
2
k (β − α) ≤ D k (β − α) + k (β − α) ,
4
4
4
giving D ≥ 21 , and the sharpness of the constant
1
2
is proved.
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
γ−α
β−γ
1
0
(2.13)
− γ h (γ) ≤
h (α) +
h (β) −
h (t) dt.
2
β−α
β−α
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
J. Inequal. Pure and Appl. Math., 3(3) Art. 35, 2002
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O N THE S ECOND H ERMITE -H ADAMARD I NEQUALITY
5
The following corollary provides some bounds for the difference
Z β
h (α) + h (β)
1
−
h (t) dt.
2
β−α α
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
h (α) + h (β)
1
≤
−
h (t) dt
2
β−α α
1 0
≤
h− (β) − h0+ (α) (β − α) .
8
1
The constant 8 is sharp in both inequalities.
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
(2.16)
(1 − s)2 (5+ f [(1 − s) a + sb]) (b − a) − s2 (5− f [(1 − s) a + sb]) (b − a)
2
Z 1
≤ (1 − s) f (a) + sf (b) −
f [(1 − t) a + tb] dt
0
1
≤
(1 − s)2 (5− f (b)) (b − a) − s2 (5+ f (a)) (b − a) .
2
1
The constant 2 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] ,
and for the choices α = 0, β = 1, and γ = s.
t ∈ [0, 1] ,
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
(2.17)
− λ (5f (c)) (b − a) ≤ (1 − λ) f (a) + λf (b) −
f [(1 − t) a + tb] dt.
2
0
The following result related to the second Hermite-Hadamard inequality for functions defined
on linear spaces also holds.
Corollary 2.6. If f is as in Theorem 2.4, then
1
a+b
a+b
(2.18)
5+ f
(b − a) − 5− f
(b − a)
8
2
2
Z 1
f (a) + f (b)
≤
−
f [(1 − t) a + tb] dt
2
0
1
≤ [(5− f (b)) (b − a) − (5+ f (a)) (b − a)] .
8
1
The constant 8 is sharp in both inequalities.
J. Inequal. Pure and Appl. Math., 3(3) Art. 35, 2002
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6
S.S. D RAGOMIR
Now, let Ω ⊂ Rn be an open 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 , ȳ ∈ Rn ,
∂x
i
i=1
∂F
are the partial derivatives of F with respect to the variable xi (i = 1, . . . , n).
where ∂x
i
Using (2.16), we may state that
X
n
∂F λā + (1 − λ) b̄
1
(2.19)
−λ
· (bi − ai )
2
∂xi
i=1
Z 1
≤ (1 − λ) F (ā) + λF b̄ −
F (1 − t) ā + tb̄ dt
0
"
#
n
n
X
X
∂F
b̄
∂F
(ā)
1
(1 − λ)2
· (bi − ai ) − λ2
· (bi − ai )
≤
2
∂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
F (ā) + F b̄
0 ≤
−
F (1 − t) ā + tb̄ dt
2
0
!
n
X
∂F b̄
1
∂F (ā)
≤
−
· (bi − ai ) .
8 i=1
∂xi
∂xi
(2.20)
In (2.20) 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 .
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
2
2
≤ (1 − σ) kxk + σ kyk −
k(1 − t) x + tyk2 dt
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) = 21 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
(3.2)
2
2
Z
(1 − 2σ) [y − x, (1 − σ) x + σy] ≤ (1 − σ) kxk + σ kyk −
1
k(1 − t) x + tyk2 dt.
0
J. Inequal. Pure and Appl. Math., 3(3) Art. 35, 2002
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O N THE S ECOND H ERMITE -H ADAMARD I NEQUALITY
7
An interesting particular case one can get from (3.1) is the one for σ = 12 ,
1
[hy − x, y + xis − hy − x, y + xii ]
8
Z 1
kxk2 + kyk2
≤
−
k(1 − t) x + tyk2 dt
2
0
1
≤
[hy − x, yii − hy − x, xis ] .
4
0 ≤
(3.3)
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 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 + σyis
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
≤ (1 − σ) kxk + σ kyk −
k(1 − t) x + tyk dt
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)
Z 1
[y − x, (1 − σ) x + σy]
1
−σ ·
≤ (1 − σ) kxk + σ kyk −
k(1 − t) x + tyk dt
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 (3.6)
0 ≤
y − x, x+y − y − x, x+y 8
2
2
i
s
Z 1
kxk + kyk
≤
−
k(1 − t) x + tyk dt
2
0
1
y
x
≤
y − x,
− y − x,
.
8
kyk i
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 any
x, y ∈ H with kxk = kyk = 1 that
Z 1
1
(3.7)
0≤1−
k(1 − t) x + tyk dt ≤ ky − xk2 .
8
0
The constant 18 is sharp.
Indeed, if we choose H = R, ha, bi = a · b, x = −1, y = 1, then we get equality in (3.7).
We give now some examples.
J. Inequal. Pure and Appl. Math., 3(3) Art. 35, 2002
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8
S.S. D RAGOMIR
(1) Let `2 (K) , K = C, R; be the Hilbert space of sequences x = (xi )i∈N with
∞. Then, by (3.7), we have the inequalities
! 21
Z 1 X
∞
(3.8)
0 ≤ 1−
|(1 − t) xi + tyi |2
dt
0
≤
1
·
8
∞
X
P∞
i=0
|xi |2 <
i=0
|yi − xi |2 ,
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
0 ≤ 1−
|(1 − λ) f (t) + λg (t)| dµ (t) dλ
(3.9)
0
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(3) Art. 35, 2002
http://jipam.vu.edu.au/