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International Mathematical Forum, 4, 2009, no. 47, 2327 - 2338
Hadamard-Type Inequalities for Product
Two Convex Functions on the Co-ordinates
M. A. Latif
Saudi Civi Aviation Academy
General Authority of Civil Aviation, P.O Box 15441
Jeddah 21444, Saudi Arabia
m amer [email protected]
M. Alomari1
School of Mathematical Sciences
Faculty of Science and Technology, Universiti
Kebangsaan Malaysia, Bangi, 43600, Selangor, Malaysia
Abstract. In this paper Hadamard-type inequalities for product of two convex func-
tions on the co-ordinates on the rectangle from the plane are established.
Mathematics Subject Classification: Primary 26D15; Secondary 26A51
Keywords: convex functions, Hadamard’s inequality, convex functions on the co-
ordinates
1. Introduction
Let f : I ⊆ R → R be a convex function and a, b ∈ I with a < b. Then the following
double inequality:
f
(a + b
2
)≤ 1
b − a
∫ b
af(x)dx ≤ f(a) + f(b)
2(1.1)
1Supported by the Grant: UKM-GUP-TMK-07-02-107.
2328 M. A. Latif and M. Alomari
is known as Hadamard’s inequality for convex mapping. For particular choice of the
function f in (1.1) yields some classical inequalities of means. Both inequalities in (1.1)
hold in reversed if f is concave. Dragomir [1] established the following similar inequality of
Hadamard’s type for convex functions on the co-ordinates on a rectangle from the plane
R2.
Theorem 1. Suppose f : Δ = [a, b] × [c, d] ⊆ [0,∞) → R is convex function on the
co-ordinates on Δ. Then one has the inequalities:
(1.2) f
(a + b
2,c + d
2
)≤ 1
(b − a)(d − c)
∫ b
a
∫ d
cf(x, y)dydx
≤ f(a, c) + f(b, c) + f(a, d) + f(b, d)4
The above inequalities are sharp.
In [5], Pachapatte established two Hadmard-type inequalities for product of convex
functions.
Theorem 2. Let f, g : [a, b] ⊆ R → [0,∞) be convex functions on [a, b], a < b. Then
1b − a
∫ b
af(x)g(x)dx ≤ 1
3M(a, b) +
16N(a, b)(1.2)
and
2f
(a + b
2
)g
(a + b
2
)≤ 1
b − a
∫ b
af(x)g(x)dx +
16M(a, b) +
13N(a, b)(1.3)
where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).
The main purpose of the present paper is establish new Hadamard-type inequalities like
those given above in the last theorem, but now for convex functions on the co-ordinates
on rectangle from the plane R2.
2. Preliminary Results
Let us consider a bidimensional interval Δ =: [a, b] × [c, d] in R2 with a < b and c < d.
A mapping f : Δ → R is said to be convex on Δ if the following inequality:
f(αx + (1 − α)z, αy + (1 − α)w) ≤ αf(x, y) + (1 − α)f(z,w)
holds, for all (x, y), (z,w) ∈ Δ and α ∈ [0, 1] .
A modification for convex functions which is so called co-ordinated convex function was
introduced by Dragomir in [1], as follows:
Hadamard-type inequalities for product of functions 2329
A function f : Δ → R is said to be convex on the co-ordinates on Δ if the partial
mappings fy : [a, b] → R, fy(u) = f(u, y) and fx : [c, d] → R, fx(v) = f(x, v) are convex
where defined for all x ∈ [a, b], y ∈ [c, d].
A formal definition for co-ordinated convex function may be stated as follows:
Definition 1. A function f : Δ → R+ will be called co-ordinated convex on Δ, for all
t, s ∈ [0, 1] and (x, y), (u, v) ∈ Δ, if the following inequality holds
f (tx + (1 − t) y, su + (1 − s)w)
≤ tsf (x, u) + s (1 − t) f (y, u) + t (1 − s) f (x,w) + (1 − t) (1 − s) f (y,w)
Clearly, every convex function is co-ordinated convex. Furthermore, there exist co-
ordinated convex function which is not convex, (see [1]).
Equivalently, we can determine whether or not the function f is co-ordinated convex
by using the following lemma.
Lemma 1. Let f : Δ → R+. If f is twice differentiable then f is co-ordinated convex on
Δ iff for the functions fy : [a, b] → R, defined by fy (u) = f (u, y) and fx : [c, d] → R,
defined by fx (v) = f (x, v), we have f ′′x ≥ 0, and f ′′
y ≥ 0.
Proof. A straight forward using the elementary properties of convexity in one variable.
Proposition 1. If f, g : Δ → R are two co-ordinated convex functions on Δ and α ≥ 0,
then f + g and αf are co-ordinated convex on Δ.
Proof. Follows directly from Definition 1
Proposition 2. If f, g : Δ → R are two co-ordinated convex functions on Δ, then (fg)
is co-ordinated convex on Δ.
2330 M. A. Latif and M. Alomari
Proof.
f (αx + (1 − α) y, βu + (1 − β) v) g (αx + (1 − α) y, βu + (1 − β) v)
≤ [αβf (x, u) + α (1 − β) f (x, v) + (1 − α)βf (y, u) + (1 − α) (1 − β) f (y, v)]
× [αβg (x, u) + α (1 − β) g (x, v) + (1 − α) βg (y, u) + (1 − α) (1 − β) g (y, v)]
≤ α2β2f (x, u) g (x, u) + α2 (1 − β)2 f (x, v) g (x, v)
+ (1 − α)2 β2f (y, u) g (y, u) + (1 − α)2 (1 − β)2 f (y, v) g (y, v)
+ α2β (1 − β) [f (x, u) g (x, u) + f (x, v) g (x, v)]
+ αβ2 (1 − α) [f (x, u) g (x, u) + f (y, u) g (y, u)]
+ αβ (1 − α) (1 − β) [f (x, v) g (x, v) + f (y, u) g (y, u)
+f (x, u) g (x, u) + f (y, v) g (y, v)]
+ α (1 − α) (1 − β)2 [f (x, v) g (x, v) + f (y, v) g (y, v)]
+ β (1 − α)2 (1 − β) [f (y, v) g (y, v) + f (y, u) g (y, u)]
=[α2β2 + α2β (1 − β) + αβ2 (1 − α) + αβ (1 − α) (1 − β)
]f (x, u) g (x, u)
+[α2 (1 − β)2 + α2β (1 − β) + α (1 − α) (1 − β)2 + αβ (1 − α) (1 − β)
]f (x, v) g (x, v)
+[(1 − α)2 β2 + αβ2 (1 − α) + αβ (1 − α) (1 − β) + β (1 − α)2 (1 − β)
]f (y, u) g (y, u)
+[(1 − α)2 (1 − β)2 + α (1 − α) (1 − β)2 + β (1 − α)2 (1 − β) + αβ (1 − α) (1 − β)
]f (y, v) g (y, v)
= αβf (x, u) g (x, u) + α (1 − β) f (x, v) g (x, v)
+ (1 − α)βf (y, u) g (y, u) + (1 − α) (1 − β) f (y, v) g (y, v)
and therefore (fg) is co-ordinated convex on Δ.
Theorem 3. The function f : Δ → R+ is co-ordinated convex on Δ iff for distinct
x1, x2, x3 ∈ [a, b] such that x1 < x2 < x3 and distinct y1, y2, y3 ∈ [c, d] such that y1 < y2 <
y3, we have
(x3 − x2) [(y3 − y2) f (x1, y1) + (y2 − y1) f (x1, y3)]
+ (x2 − x1) [(y3 − y2) f (x3, y1) + (y2 − y1) f (x3, y3)]
− (x3 − x1) (y3 − y1) f (x2, y2) ≥ 0.
Proof. Let x1, x2, x3 ∈ [a, b] such that x1 < x2 < x3 and y1, y2, y3 ∈ [c, d] such that y1 <
y2 < y3. Setting α = x3−x2x3−x1
, x2 = αx1 + (1 − α) x3 and β = y3−y2
y3−y1, y2 = βy1 + (1 − β) y3.
Hadamard-type inequalities for product of functions 2331
Therefore,
f (x2, y2) = f (αx1 + (1 − α)x3, βy1 + (1 − β) y3)
≤ αβf (x1, y1) + α (1 − β) f (x1, y3)
+ β (1 − α) f (x3, y1) + (1 − α) (1 − β) f (x3, y3)
=x3 − x2
x3 − x1
y3 − y2
y3 − y1f (x1, y1) +
x3 − x2
x3 − x1
y2 − y1
y3 − y1f (x1, y3)
+x2 − x1
x3 − x1
y3 − y2
y3 − y1f (x3, y1) +
x2 − x1
x3 − x1
y2 − y1
y3 − y1f (x3, y3)
and we write
(x3 − x2) [(y3 − y2) f (x1, y1) + (y2 − y1) f (x1, y3)]
+ (x2 − x1) [(y3 − y2) f (x3, y1) + (y2 − y1) f (x3, y3)]
− (x3 − x1) (y3 − y1) f (x2, y2) ≥ 0.
which completes the proof.
3. Main Results
In this section we obtain Hadamard’s type inequalities for product of co-ordinated
convex function. We begin with the following theorem.
Theorem 4. Let f, g : Δ = [a, b] × [c, d] ⊆ R2 → [0,∞) be convex functions on the
co-ordinates on Δ with a < b, c < d. Then
1(b − a)(d − c)
∫ b
a
∫ d
cf(x, y)g(x, y)dydx ≤ 1
9L(a, b, c, d) +
118
M(a, b, c, d) +136
N(a, b, c, d)
(3.1)
where
L(a, b, c, d) = f(a, c)g(a, c) + f(b, c)g(b, c) + f(a, d)g(a, d) + f(b, d)g(b, d),
M(a, b, c, d) = f(a, c)g(a, d) + f(a, d)g(a, c) + f(b, c)g(b, d) + f(b, d)g(b, c)
+ f(b, c)g(a, c) + f(b, d)g(a, d) + f(a, c)g(b, c) + f(a, d)g(b, d)
and N(a, b, c, d) = f(b, c)g(a, d) + f(b, d)g(a, c) + f(a, c)g(b, d) + f(a, d)g(b, c)
Proof. Since f, g : Δ = [a, b]×[c, d] ⊆ R2 → [0,∞) are convex functions on the co-ordinates
on Δ with a < b, c < d, therefore the partial mappings fy : [a, b] → [0,∞), fy(x) = f(x, y),
gy : [a, b] → [0,∞), gy(x) = g(x, y), fx : [c, d] → [0,∞), fx(y) = f(x, y) and gx : [c, d] →
2332 M. A. Latif and M. Alomari
[0,∞), gx(y) = g(x, y) are convex on [a, b] and [c, d] respectively for all x ∈ [a, b], y ∈ [c, d].
Now by applying (1.3) to fx(y)gx(y) on [c, d] we get
1d − c
∫ d
cfx(y)gx(y)dy ≤ 1
3[fx(c)gx(c) + fx(d)gx(d)]
+16
[fx(c)gx(d) + fx(d)gx(c)]
That is
1d − c
∫ d
cf(x, y)g(x, y)dy ≤ 1
3[f(x, c)g(x, c) + f(x, d)g(x, d)]
+16
[f(x, c)g(x, d) + f(x, d)g(x, c)]
Integrating over [a, b] and dividing both sides by b − a, we have
1(b − a)(d − c)
∫ b
a
∫ d
cf(x, y)g(x, y)dydx ≤ 1
3
[1
b − a
∫ b
af(x, c)g(x, c)dx
+1
b − a
∫ b
af(x, d)g(x, d)dx
]
+16
[1
b − a
∫ b
af(x, c)g(x, d)dx
+1
b − a
∫ b
af(x, d)g(x, c)dx
](3.2)
Now by applying (1.3) to each integral on R.H.S of (3.2) again, we get
1b − a
∫ b
af(x, c)g(x, c)dx ≤ 1
3[f(a, c)g(a, c) + f(b, c)g(b, c)]
+16
[f(a, c)g(b, c) + f(b, c)g(a, c)]
1b − a
∫ b
af(x, d)g(x, d)dx ≤ 1
3[f(a, d)g(a, d) + f(b, d)g(b, d)]
+16
[f(a, d)g(b, d) + f(b, d)g(a, d)]
1b − a
∫ b
af(x, c)g(x, d)dx ≤ 1
3[f(a, c)g(a, d) + f(b, c)g(b, d)]
+16
[f(a, c)g(b, d) + f(b, c)g(a, d)]
1b − a
∫ b
af(x, d)g(x, c)dx ≤ 1
3[f(a, d)g(a, c) + f(b, d)g(b, c)]
+16
[f(a, d)g(b, c) + f(b, d)g(a, c)]
Hadamard-type inequalities for product of functions 2333
On substitution of these inequalities in (3.2) yields
1(b − a)(d − c)
∫ b
a
∫ d
cf(x, y)g(x, y)dydx ≤ 1
9[f(a, c)g(a, c) + f(b, c)g(b, c)]
+118
[f(a, c)g(b, c) + f(b, c)g(a, c)]
+19
[f(a, d)g(a, d) + f(b, d)g(b, d)]
+118
[f(a, d)g(b, d) + f(b, d)g(a, d)]
+118
[f(a, c)g(a, d) + f(b, c)g(b, d)]
+136
[f(a, c)g(b, d) + f(b, c)g(a, d)]
+118
[f(a, d)g(a, c) + f(b, d)g(b, c)]
+136
[f(a, d)g(b, c) + f(b, d)g(a, c)]
Therefore, by (3.2) we get the required result (3.1). Similarly, by applying (1.3) for
fy(x)gy(x) on [a, b], we get the same result.
Theorem 5. Let f, g : Δ = [a, b] × [c, d] ⊆ R2 → [0,∞) be convex functions on the
co-ordinates on Δ with a < b, c < d. Then
4f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 1
(b − a)(d − c)
∫ b
a
∫ d
cf(x, y)g(x, y)dydx
+536
L(a, b, c, d) +736
M(a, b, c, d) +29N(a, b, c, d)(3.3)
where L(a, b, c, d),M(a, b, c, d), and N(a, b, c, d) as in Theorem 4.
Proof. Now applying (1.4) to 2f(
a+b2 , c+d
2
)g
(a+b2 , c+d
2
), we get
2f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 1
b − a
∫ b
af
(x,
c + d
2
)g
(x,
c + d
2
)dx
+16
[f
(a,
c + d
2
)g
(a,
c + d
2
)+ f
(b,
c + d
2
)g
(b,
c + d
2
)]
+13
[f
(a,
c + d
2
)g
(b,
c + d
2
)+ f
(b,
c + d
2
)g
(a,
c + d
2
)](3.4)
2334 M. A. Latif and M. Alomari
and
2f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 1
d − c
∫ d
cf
(a + b
2, y
)g
(a + b
2, y
)dy
+16
[f
(a + b
2, c
)g
(a + b
2, c
)+ f
(a + b
2, d
)g
(a + b
2, d
)]
+13
[f
(a + b
2, c
)g
(a + b
2, d
)+ f
(a + b
2, d
)g
(a + b
2, c
)](3.5)
Adding (3.4) and (3.5) and multiplying both sides by 2, we get
8f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 2
b − a
∫ b
af
(x,
c + d
2
)g
(x,
c + d
2
)dx
+2
d − c
∫ d
cf
(a + b
2, y
)g
(a + b
2, y
)dy
+16
[2f
(a,
c + d
2
)g
(a,
c + d
2
)+ 2f
(b,
c + d
2
)g
(b,
c + d
2
)]
+13
[2f
(a,
c + d
2
)g
(b,
c + d
2
)+ 2f
(b,
c + d
2
)g
(a,
c + d
2
)]
+16
[2f
(a + b
2, c
)g
(a + b
2, c
)+ 2f
(a + b
2, d
)g
(a + b
2, d
)]
+13
[2f
(a + b
2, c
)g
(a + b
2, d
)+ 2f
(a + b
2, d
)g
(a + b
2, c
)](3.6)
Applying (1.4) to each term within the brackets we have
2f
(a,
c + d
2
)g
(a,
c + d
2
)≤ 1
d − c
∫ d
cf (a, y) g (a, y) dy
+16
[f (a, c) g (a, c) + f (a, d) g (a, d)]
+13
[f (a, c) g (a, d) + f (a, d) g (a, c)]
2f
(b,
c + d
2
)g
(b,
c + d
2
)≤ 1
d − c
∫ d
cf (b, y) g (b, y) dy
+16
[f (b, c) g (b, c) + f (b, d) g (b, d)]
+13
[f (b, c) g (b, d) + f (b, d) g (b, c)]
Hadamard-type inequalities for product of functions 2335
2f
(a,
c + d
2
)g
(b,
c + d
2
)≤ 1
d − c
∫ d
cf (a, y) g (b, y) dy
+16
[f (a, c) g (b, c) + f (a, d) g (b, d)]
+13
[f (a, c) g (b, d) + f (a, d) g (b, c)]
2f
(b,
c + d
2
)g
(a,
c + d
2
)≤ 1
d − c
∫ d
cf (b, y) g (a, y) dy
+16
[f (b, c) g (a, c) + f (b, d) g (a, d)]
+13
[f (b, c) g (a, d) + f (b, d) g (a, c)]
2f
(a + b
2, c
)g
(a + b
2, c
)≤ 1
b − a
∫ b
af(x, c)g(x, c)dx
+16
[f (a, c) g (a, c) + f (b, c) g (b, c)]
+13
[f (a, c) g (b, c) + f (b, c) g (a, c)]
2f
(a + b
2, d
)g
(a + b
2, d
)≤ 1
b − a
∫ b
af(x, d)g(x, d)dx
+16
[f (a, d) g (a, d) + f (b, d) g (b, d)]
+13
[f (a, d) g (b, d) + f (b, d) g (a, d)]
2f
(a + b
2, c
)g
(a + b
2, d
)≤ 1
b − a
∫ b
af(x, c)g(x, d)dx
+16
[f (a, c) g (a, d) + f (b, c) g (b, d)]
+13
[f (a, c) g (b, d) + f (b, c) g (a, d)]
2f
(a + b
2, d
)g
(a + b
2, c
)≤ 1
b − a
∫ b
af(x, d)g(x, c)dx
+16
[f (a, d) g (a, c) + f (b, d) g (b, c)]
+13
[f (a, d) g (b, c) + f (b, d) g (a, c)]
2336 M. A. Latif and M. Alomari
Substituting these inequalities in (3.6) and simplifying we have:
8f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 2
b − a
∫ b
af
(x,
c + d
2
)g
(x,
c + d
2
)dx
+2
d − c
∫ d
cf
(a + b
2, y
)g
(a + b
2, y
)dy
+16
1d − c
∫ d
cf (a, y) g (a, y) dy +
16
1d − c
∫ d
cf (b, y) g (b, y) dy
+13
1d − c
∫ d
cf (a, y) g (b, y) dy +
13
1d − c
∫ d
cf (b, y) g (a, y) dy
+16
1b − a
∫ b
af(x, c)g(x, c)dx +
16
1b − a
∫ b
af(x, d)g(x, d)dx
+13
1b − a
∫ b
af(x, c)g(x, d)dx +
13
1b − a
∫ b
af(x, d)g(x, c)dx
+118
L(a, b, c, d) +19M(a, b, c, d) +
29N(a, b, c, d)(3.7)
Now by applying (1.4) to 2f(
a+b2 , y
)g
(a+b2 , y
), integrating over [c, d] and dividing both
sides by d − c we get
2d − c
∫ d
cf
(a + b
2, y
)g
(a + b
2, y
)dy ≤ 1
(b − a)(d − c)
∫ b
a
∫ d
cf (x, y) g (x, y) dydx
+16
1d − c
∫ d
cf (a, y) g (a, y) dy +
16
1d − c
∫ d
cf (b, y) g (b, y) dy
+13
1d − c
∫ d
cf (a, y) g (b, y) dy +
13
1d − c
∫ d
cf (b, y) g (a, y) dy(3.8)
Now by applying (1.4) to 2f(x, c+d
2
)g
(x, c+d
2
), integrating over [a, b] and dividing both
sides by b − a we get
2b − a
∫ d
cf
(x,
c + d
2
)g
(x,
c + d
2
)dx ≤ 1
(b − a)(d − c)
∫ b
a
∫ d
cf (x, y) g (x, y) dydx
+16
1b − a
∫ b
af(x, c)g(x, c)dx +
16
1b − a
∫ b
af(x, d)g(x, d)dx
+13
1b − a
∫ b
af(x, c)g(x, d)dx +
13
1b − a
∫ b
af(x, d)g(x, c)dx(3.9)
Hadamard-type inequalities for product of functions 2337
Adding (3.7) and (3.8) we have
2b − a
∫ d
cf
(x,
c + d
2
)g
(x,
c + d
2
)dx +
2d − c
∫ d
cf
(a + b
2, y
)g
(a + b
2, y
)dy
≤ 2(b − a)(d − c)
∫ b
af (x, y) g (x, y) dxdy
+16
1d − c
∫ d
cf (a, y) g (a, y) dy +
16
1d − c
∫ d
cf (b, y) g (b, y) dy
+13
1d − c
∫ d
cf (a, y) g (b, y) dy +
13
1d − c
∫ d
cf (b, y) g (a, y) dy
+16
1b − a
∫ b
af(x, c)g(x, c)dx +
16
1b − a
∫ b
af(x, d)g(x, d)dx
+13
1b − a
∫ b
af(x, c)g(x, d)dx +
13
1b − a
∫ b
af(x, d)g(x, c)dx(3.10)
Therefore from (3.7) and (3.10), we get
8f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 2
(b − a)(d − c)
∫ b
a
∫ d
cf (x, y) g (x, y) dxdy
+13
1d − c
∫ d
cf (a, y) g (a, y) dy +
13
1d − c
∫ d
cf (b, y) g (b, y) dy
+23
1d − c
∫ d
cf (a, y) g (b, y) dy +
23
1d − c
∫ d
cf (b, y) g (a, y) dy
+13
1b − a
∫ b
af(x, c)g(x, c)dx +
13
1b − a
∫ b
af(x, d)g(x, d)dx
+23
1b − a
∫ b
af(x, c)g(x, d)dx +
23
1b − a
∫ b
af(x, d)g(x, c)dx
+118
L(a, b, c, d) +19M(a, b, c, d) +
29N(a, b, c, d)(3.11)
By using (1.3) to each of the integral in (3.9) and simplifying we get
8f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 2
(b − a)(d − c)
∫ b
a
∫ d
cf (x, y) g (x, y) dydx
+518
L(a, b, c, d) +718
M(a, b, c, d) +49N(a, b, c, d)
Dividing both sides by 2
4f
(a + b
2,c + d
2
)g
(a + b
2,c + d
2
)≤ 1
(b − a)(d − c)
∫ b
a
∫ d
cf (x, y) g (x, y) dxdy
+536
L(a, b, c, d) +736
M(a, b, c, d) +29N(a, b, c, d)
Which is (3.2) and this completes the proof of the theorem.
2338 M. A. Latif and M. Alomari
Acknowledgement 1. The first author is thankful for professor Mohammad Alomari for
his guidance and valuable suggestions for the improvement of the present paper.
References
[1] S. S. Dragomir, On Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from
the plane, Taiwanese Journal of Mathematics, 5 (2001), 775 - 788.
[2] S. S. Dragomir, Selected Topics on Hermite-Hadamard Inequalities and Applications,
http://rgmia.vu.edu.au/SSDragomirWeb.html.
[3] S. S. Dragomir, J. Pecaric and L. E. Persson, Some inequalities of Hadamard type, Soochow J. Math.,
21 (1995), 335–241.
[4] D. S. Mitrinovic, J. Pecaric and A. M. Fink, Classical and new inequalities in analysis, Kluwer Aca-
demic, Dordrecht, 1993.
[5] B.G. Pachpatte, On some inequalities for convex functions, RGMIA Res. Rep. Coll., 6(E), 2003.
Received: February, 2009
