(α,m)-CONVEX FUNCTIONS - Hacettepe Journal of Mathematics

Hacettepe Journal of Mathematics and Statistics
Volume 40 (2) (2011), 219 – 229
SOME NEW HADAMARD TYPE
INEQUALITIES FOR CO-ORDINATED
m-CONVEX AND (α, m)-CONVEX
FUNCTIONS
M. Emin Özdemir∗†, Erhan Set‡ and Mehmet Zeki Sarıkaya∗
Received 15 : 06 : 2010 : Accepted 21 : 11 : 2010
Abstract
In this paper, we establish some new Hermite-Hadamard type inequalities for m-convex and (α, m)-convex functions of 2-variables on the
co-ordinates.
Keywords: m-convex function, (α, m)-convex function, co-ordinated convex mapping,
Hermite-Hadamard inequality.
2010 AMS Classification: 26 A 51, 26 D 15.
Communicated by Alex Goncharov
1. Introduction
Let f : I ⊆ R → R be a convex mapping defined on the interval I of real numbers,
and a, b ∈ I with a < b. The following double inequality is well known in the literature
as the Hermite-Hadamard inequality [5]:
Z b
a+b
1
f (a) + f (b)
f
≤
f (x) dx ≤
.
2
b−a a
2
In [8], the notion of m-convexity was introduced by G.Toader as the following:
1.1. Definition. The function f : [0, b] → R, b > 0 is said to be m-convex, where
m ∈ [0, 1], if we have
f (tx + m (1 − t) y) ≤ tf (x) + m (1 − t) f (y)
for all x, y ∈ [0, b] and t ∈ [0, 1]. We say that f is m-concave if −f is m-convex.
∗Atatürk University, K. K. Education Faculty, Department of Mathematics, 25240 Campus,
Erzurum, Turkey. E-mail: emos@atauni.edu.tr
†
Corresponding Author.
‡
Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey.
E-mail: (E. Set) erhanset@yahoo.com (M. Z. Sarıkaya) sarikayamz@gmail.com
220
M. E. Özdemir, E. Set, M. Z. Sarıkaya
Denote by Km (b) the class of all m-convex functions on [0, b] for which f (0) ≤ 0.
Obviously, if we choose m = 1, Definition 1.1 recaptures the concept of standard convex
functions on [0, b].
In [6], S. S. Dragomir and G. Toader proved the following Hadamard type inequalities
for m-convex functions.
1.2. Theorem. Let f : [0, ∞) → R be an m-convex function with m ∈ (0, 1]. If 0 ≤ a <
b < ∞ and f ∈ L1 [a, b], then the following inequality holds:
Z b
b
a f (a) + mf m
f (b) + mf m
1
f (x) dx ≤ min
,
.
(1.1)
b−a a
2
2
Some generalizations of this result can be found in [2, 3].
1.3. Theorem. Let f : [0, ∞) → R be an m-convex differentiable function with m ∈
(0, 1]. Then for all 0 ≤ a < b the following inequality holds:
Z b
f (mb)
b−a ′
1
−
f (mb) ≤
f (x) dx
m
2
b−a a
(1.2)
(b − ma) f (b) − (a − mb) f (a)
≤
.
2 (b − a)
Also, in [5], Dragomir and Pearce proved the following Hadamard type inequality for
m-convex functions.
1.4. Theorem. Let f : [0, ∞) → R be an m-convex function with m ∈ (0, 1] and
0 ≤ a < b. If f ∈ L1 [a, b], then one has the inequality:
Z b
x
f (x) + mf m
a+b
1
(1.3)
f
≤
dx.
2
b−a a
2
In [7], the definition of (α, m)-convexity was introduced by V. G. Miheşan as the
following:
1.5. Definition. The function f : [0, b] → R, b > 0, is said to be (α, m)-convex, where
(α, m) ∈ [0, 1]2 , if we have
f (tx + m(1 − t)y) ≤ tα f (x) + m(1 − tα )f (y)
for all x, y ∈ [0, b] and t ∈ [0, 1].
α
Denote by Km
(b) the class of all (α, m)-convex functions on [0, b] for which f (0) ≤ 0.
If we take (α, m) = (1, m), it can be easily seen that (α, m)-convexity reduces to mconvexity, and for (α, m) = (1, 1), (α, m)-convexity reduces to the usual concept of
convexity defined on [0, b], b > 0.
In [9], E. Set, M. Sardari, M. E. Ozdemir and J. Rooin proved the following Hadamard
type inequalities for (α, m)-convex functions.
1.6. Theorem. Let f : [0, ∞) → R be an (α,m)-convex function with (α, m) ∈ (0, 1]2 .
b
a
If 0 ≤ a < b < ∞ and f ∈ L1 [a, b] ∩ L1 m
,m
, then the following inequality holds:
b
α
Z f (x) + m(2 − 1)f x
a+b
1
m dx.
(1.4)
f
≤
2
b−a
2α
a
1.7. Theorem. Let f : [0, ∞) → R be an (α, m)-convex function with (α, m) ∈ (0, 1]2 .
If 0 ≤ a < b < ∞ and f ∈ L1 [a, b], then the following inequality holds:
Z b
b
a f (a) + mαf m
f (b) + mαf m
1
(1.5)
f (x)dx ≤ min
,
.
b−a a
α+1
α+1
Co-ordinated m-Convex and (α, m)-Convex Functions
221
1.8. Theorem. Let f : [0, ∞) → R be an (α, m)-convex function with (α, m) ∈ (0, 1]2 .
If 0 ≤ a < b < ∞ and f ∈ L1 [a, b], then the following inequality holds:
"
#
Z b
a
b
1
1 f (a) + f (b) + mαf m + mαf m
(1.6)
f (x)dx ≤
.
b−a a
2
α+1
Let us now consider a bidimensional interval ∆ =: [a, b] × [c, d] in R2 , with a < b and
c < d. A function f : ∆ → R is said to be convex on ∆ if the following inequality:
f (tx + (1 − t) z, ty + (1 − t) w) ≤ tf (x, y) + (1 − t) f (z, w)
holds, for all (x, y) , (z, w) ∈ ∆ and t ∈ [0, 1]. A function f : ∆ → R is said to be convex
on the co-ordinates on ∆ 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] and y ∈ [c, d]
(see [5, p. 317]).
Also, in [4], Dragomir proved the following similar inequalities of Hadamard’s type
for a co-ordinated convex mapping on a rectangle in the plane R2 .
1.9. Theorem. Suppose that f : ∆ → R is co-ordinated convex on ∆. Then one has the
inequalities:
a+b c+d
f
,
2
2
Z b Z d 1
c+d
1
a+b
1
f x,
dx +
f
, y dy
≤
2 b−a a
2
d−c c
2
Z bZ d
1
≤
f (x, y) dx dy
(b − a) (d − c) a c
(1.7)
Z b
Z b
1
1
1
≤
f (x, c) dx +
f (x, d) dx
4 b−a a
b−a a
Z d
Z d
1
1
f (a, y) dy +
f (b, y) dy
+
d−c c
d−c c
f (a, c) + f (a, d) + f (b, c) + f (b, d)
≤
4
The above inequalities are sharp.
For co-ordinated s-convex functions, another version of this result can be found in [1].
The main purpose of this paper is to establish new Hadamard-type inequalities for
functions of 2-variables which are m-convex or (α, m)-convex on the co-ordinates.
2. Inequalities for co-ordinated m-convex functions
Firstly, we can define co-ordinated m-convex functions as follows:
2.1. Definition. Consider the bidimensional interval ∆ := [0, b] × [0, d] in [0, ∞)2 . The
mapping f : ∆ → R is m-convex on ∆ if
f (tx + (1 − t) z, ty + m (1 − t) w) ≤ tf (x, y) + m (1 − t) f (z, w)
holds for all (x, y) , (z, w) ∈ ∆ with t ∈ [0, 1], b, d > 0, and for some fixed m ∈ [0, 1].
A function f : ∆ → R which is m-convex on ∆ is called co-ordinated m-convex on ∆
if the partial mappings
fy : [0, b] → R, fy (u) = f (u, y)
222
M. E. Özdemir, E. Set, M. Z. Sarıkaya
and
fx : [0, d] → R, fx (v) = f (x, v)
are m-convex for all y ∈ [0, d] and x ∈ [0, b] with b, d > 0, and for some fixed m ∈ [0, 1].
We also need the following Lemma for our main results.
2.2. Lemma. Every m-convex mapping f : ∆ ⊂ [0, ∞)2 → R is m-convex on the coordinates, where ∆ = [0, b] × [0, d] and m ∈ [0, 1].
Proof. Suppose that f : ∆ = [0, b] × [0, d] → R is m-convex on ∆. Consider the function
fx : [0, d] → R, fx (v) = f (x, v) , (x ∈ [0, b]) .
Then for t, m ∈ [0, 1] and v1 , v2 ∈ [0, d], we have
fx (tv1 + m (1 − t) v2 ) = f (x, tv1 + m (1 − t) v2 )
= f (tx + (1 − t) x, tv1 + m (1 − t) v2 )
≤ tf (x, v1 ) + m (1 − t) f (x, v2 )
= tfx (v1 ) + m (1 − t) fx (v2 ) .
Therefore, fx (v) = f (x, v) is m-convex on [0, d]. The fact that fy : [0, b] → R, fy (u) =
f (u, y) is also m-convex on [0, b] for all y ∈ [0, d] goes likewise, and we shall omit the
details.
2.3. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is an m-convex function on the
co-ordinates on ∆. If 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞ with m ∈ (0, 1], then one has
the inequality:
Z dZ b
1
f (x, y) dx dy
(d − c) (b − a) c a
(2.1)
1
1
≤
min {v1 , v2 } +
min {v3 , v4 } ,
4 (b − a)
4 (d − c)
where
v1 =
Z
b
f (x, c) dx + m
a
Z
b
Z
b
f
a
Z
b
d
x,
m
dx
c
dx
f x,
m
a
a
Z d
Z d
b
v3 =
f (a, y) dy + m
f
, y dy
m
c
c
Z d
Z d a v4 =
, y dy.
f (b, y) dy + m
f
m
c
c
v2 =
f (x, d) dx + m
Proof. Since f : ∆ → R is co-ordinated m-convex on ∆ it follows that the mapping
gx : [0, d] → R, gx (y) = f (x, y) is m-convex on [0, d] for all x ∈ [0, b]. Then by the
inequality (1.1) one has:
(
)
Z d
d
c
gx (c) + mgx m
gx (d) + mgx m
1
gx (y) dy ≤ min
,
,
d−c c
2
2
or
1
d−c
Z
d
f (x, y) dy ≤ min
c
(
where 0 ≤ c < d < ∞ and m ∈ (0, 1].
f (x, c) + mf x,
2
d
m
c
f (x, d) + mf x, m
,
2
)
,
Co-ordinated m-Convex and (α, m)-Convex Functions
223
Dividing both sides by (b − a) and integrating this inequality over [a, b] with respect
to x, we have
(2.2)
Z bZ d
1
f (x, y) dy dx
(b − a) (d − c) a c
Z b
Z b 1
m
d
≤ min
f (x, c) dx +
f x,
dx,
2 (b − a) a
2 (b − a) a
m
Z b
Z b 1
m
c
f (x, d) dx +
f x,
dx
2 (b − a) a
2 (b − a) a
m
Z b
Z b 1
d
=
min
f (x, c) dx + m
f x,
dx,
2 (b − a)
m
a
a
Z b
Z b c
dx
f (x, d) dx + m
f x,
m
a
a
where 0 ≤ a < b < ∞.
By a similar argument applied to the mapping gy : [0, b] → R, gy (x) = f (x, y) with
0 ≤ a < b < ∞, we get
1
(d − c) (b − a)
(2.3)
≤
Z
d
b
f (x, y) dx dy
Z d
Z
min
f (a, y) dy + m
c
1
2 (d − c)
Z
a
c
Z
c
d
d
f
b
,y
m
dy,
f (b, y) dy + m
c
Z
d
f
c
a , y dy .
m
Summing the inequalities (2.2) and (2.3), we get the inequality (2.1).
2.4. Corollary. With the above assumptions, and provided that the partial mappings
fy : [0, b] → R, fy (u) = f (u, y)
and
fx : [0, d] → R, fx (v) = f (x, v)
are differentiable on (0, b) and (0, d), respectively, we have the inequalities
1
(b − a) (d − c)
(2.4)
and
Z
b
Z
d
f (x, y) dy dx
1
d
≤
min (b − ma) f (b, c) + mf b,
4 (b − a)
m
d
− (a − mb) f (a, c) + mf a,
,
m
h
c i
(b − ma) f (b, d) + mf b,
m
h
c i − (a − mb) f (a, d) + mf a,
,
m
a
c
224
M. E. Özdemir, E. Set, M. Z. Sarıkaya
1
(d − c) (b − a)
(2.5)
Z
d
Z
b
f (x, y) dx dy
1
b
≤
min (d − mc) f (a, d) + mf
,d
4 (d − c)
m
b
,c ,
− (c − md) f (a, c) + mf
m
h
a i
(d − mc) f (b, d) + mf
,d
m
h
a i − (c − md) f (b, c) + mf
,c
.
m
c
a
Proof. Since the partial mappings
fx : [0, d] → R, fx (v) = f (x, v)
are differentiable on [0, d], by the inequality (1.2) we have
Z b
1
(b − ma) f (b, c) − (a − mb) f (a, c)
f (x, c) dx ≤
,
(b − a) a
2 (b − a)
Z b d
d
(b − ma) f (b, m
) − (a − mb) f (a, m
)
1
d
f x,
dx ≤
,
(b − a) a
m
2 (b − a)
Z b
1
(b − ma) f (b, d) − (a − mb) f (a, d)
f (x, d) dx ≤
, and
(b − a) a
2 (b − a)
Z b c
c
(b − ma) f (b, m
) − (a − mb) f (a, m
)
1
c
f x,
dx ≤
.
(b − a) a
m
2 (b − a)
Hence, using (2.2), we get the inequality (2.4).
Analogously, Since the partial mappings
fy : [0, b] → R, fy (u) = f (u, y)
are differentiable on [0, b], using the inequality (2.3), we get the inequality (2.5). The
proof is completed.
2.5. Remark. Choosing m = 1 in (2.4) or (2.5), we get the relationship between the
third and fourth inequalities in (1.7).
2.6. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is an m-convex function on the
co-ordinates on ∆. If 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞, m ∈ (0, 1] with fx ∈ L1 [0, d]
and fy ∈ L1 [0, b], then one has the inequality:
Z b Z d 1
c+d
1
a+b
f x,
dx +
f
, y dy
b−a a
2
d−c
2
"Z Z c
y
b
d
f (x, y) + mf x, m
1
≤
dy dx
(2.6)
(b − a) (d − c) a c
2
#
Z dZ b
x
f (x, y) + mf m
,y
dx dy .
+
2
c
a
Proof. Since f : ∆ → R is co-ordinated m-convex on ∆ it follows that the mapping
gx : [0, d] → R, gx (y) = f (x, y) is m-convex on [0, d] for all x ∈ [0, b]. Then by the
Co-ordinated m-Convex and (α, m)-Convex Functions
inequality (1.3) one has:
Z d
gx (y) + mgx
c+d
1
gx
≤
2
d−c c
2
y
m
225
dy,
or
f
x,
c+d
2
≤
1
d−c
Z
d
c
y
f (x, y) + mf x, m
dy
2
for all x ∈ [0, b]. Integrating this inequality on [a, b], we have
Z b 1
c+d
f x,
dx
b−a a
2
(2.7)
Z bZ d
y
f (x, y) + mf x, m
1
≤
dy dx.
(b − a) (d − c) a c
2
By a similar argument applied to the mapping gy : [0, b] → R, gy (x) = f (x, y), we get
Z d 1
a+b
f
, y dy
d−c c
2
(2.8)
Z dZ b
x
f (x, y) + mf m
,y
1
≤
dx dy.
(d − c) (b − a) c a
2
Summing the inequalities (2.7) and (2.8), we get the inequality (2.6).
2.7. Remark. Choosing m = 1 in (2.6), we get the second inequality of (1.7).
3. Inequalities for co-ordinated (α, m)-convex functions
3.1. Definition. Consider the bidimensional interval ∆ := [0, b] × [0, d] in [0, ∞)2 . The
mapping f : ∆ → R is (α, m)-convex on ∆ if
(3.1)
f (tx + (1 − t) z, ty + m (1 − t) w) ≤ tα f (x, y) + m (1 − tα ) f (z, w)
holds for all (x, y) , (z, w) ∈ ∆ and (α, m) ∈ [0, 1]2 , with t ∈ [0, 1].
A function f : ∆ → R which is (α, m)-convex on ∆ is called co-ordinated (α, m)-convex
on ∆ if the partial mappings
fy : [0, b] → R, fy (u) = f (u, y)
and
fx : [0, d] → R, fx (v) = f (x, v)
are (α, m)-convex for all y ∈ [0, d] and x ∈ [0, b] with some fixed (α, m) ∈ [0, 1]2 .
Note that for (α, m) = (1, 1) and (α, m) = (1, m), one obtains the class of co-ordinated
convex and of co-ordinated m-convex functions on ∆, respectively.
3.2. Lemma. Every (α, m)-convex mapping f : ∆ → R is (α, m)-convex on the coordinates, where ∆ = [0, b] × [0, d] and α, m ∈ [0, 1].
Proof. Suppose that f : ∆ → R is (α, m)-convex on ∆. Consider the function
fx : [0, d] → R, fx (v) = f (x, v) , (x ∈ [0, b]) .
226
M. E. Özdemir, E. Set, M. Z. Sarıkaya
Then for t ∈ [0, 1], (α, m) ∈ [0, 1]2 and v1 , v2 ∈ [0, d], one has
fx (tv1 + m (1 − t) v2 ) = f (x, tv1 + m (1 − t) v2 )
= f (tx + (1 − t) x, tv1 + m (1 − t) v2 )
≤ tα f (x, v1 ) + m (1 − tα ) f (x, v2 )
= tα fx (v1 ) + m (1 − tα ) fx (v2 ) .
Therefore, fx (v) = f (x, v) is (α, m)-convex on [0, d]. The fact that fy : [0, b] → R,
fy (u) = f (u, y) is also (α, m)-convex on [0, b] for all y ∈ [0, d] goes likewise, and we shall
omit the details.
3.3. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is an (α, m)-convex function on
the co-ordinates on ∆, where (α, m) ∈ (0, 1]2 . If 0 ≤ a < b < ∞, 0 ≤ c < d < ∞ and
fx ∈ L1 [0, d], fy ∈ L1 [0, b], then the following inequalities hold:
Z b Z d 1
c+d
1
a+b
f x,
dx +
f
, y dy
b−a a
2
d−c c
2
1
≤
(3.2)
(d − c) (b − a)
Z d Z b 2f (x, y) + m(2α − 1) f x, y + f x , y
m
m
dx dy,
×
2α
c
a
and
(3.3)
1
(d − c) (b − a)
≤
Z
d
c
Z
b
f (x, y) dx dy
a
1
1
min {w1 , w2 } +
min {w3 , w4 } ,
2 (α + 1) (b − a)
2 (α + 1) (d − c)
where
w1 =
Z
Z
b
f (x, c) dx + αm
a
b
Z
b
f
a
Z
b
x,
d
m
dx
c
dx
f x,
m
a
a
Z d
Z d b
, y dy
w3 =
f (a, y) dy + αm
f
m
c
c
Z d
Z d a w4 =
f (b, y) dy + αm
f
, y dy.
m
c
c
w2 =
f (x, d) dx + αm
Proof. Since f : ∆ → R is co-ordinated (α, m)-convex on ∆ it follows that the mapping
gx : [0, d] → R, gx (y) = f (x, y) is (α, m)-convex on [0, d] for all x ∈ [0, b]. Then by the
inequality (1.4) one has:
Z d
y
gx (y) + m(2α − 1)gx m
c+d
1
gx
≤
dy,
2
d−c c
2α
that is
f
c+d
x,
2
1
≤
d−c
Z
d
c
f (x, y) + m(2α − 1)f x,
2α
y
m
dy,
Co-ordinated m-Convex and (α, m)-Convex Functions
227
where 0 ≤ c < d < ∞ and (α, m) ∈ (0, 1]2 . Integrating this inequality on [a, b], we have
(3.4)
1
b−a
Z
b
f
a
x,
c+d
2
dx
1
≤
(d − c) (b − a)
Z
b
a
Z
d
c
f (x, y) + m(2α − 1)f x,
2α
y
m
dy dx,
where 0 ≤ a < b < ∞.
By a similar argument applied for the mapping gy : [0, b] → [0, ∞), gy (x) = f (x, y)
with 0 ≤ a < b < ∞, we get
Z d 1
a+b
f
, y dy
d−c c
2
(3.5)
Z bZ d
x
f (x, y) + m(2α − 1)f m
,y
1
≤
dy dx.
(d − c) (b − a) a c
2α
Summing the inequalities (3.4) and (3.5), we get the inequality (3.2).
The inequality (3.3) can be obtained in a similar way to the proof of Theorem 2.3 by
using (1.5).
3.4. Remark. If we take α = 1, (3.2) and (3.3) reduce to (2.6) and (2.1), respectively.
3.5. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is (α, m)-convex function on
the co-ordinates on ∆, where (α, m) ∈ (0, 1]2 . If 0 ≤ a < b < ∞, 0 ≤ c < d < ∞ and
fx ∈ L1 [0, d], fy ∈ L1 [0, b], then the following inequality holds:
1
(b − a) (d − c)
≤
(3.6)
Z
b
a
Z
d
f (x, y) dy dx
c
Z b
Z b
1
1
1
f (x, c) dx +
f (x, d) dx
4 (α + 1) b − a a
b−a a
Z b Z
b c
mα
d
mα
f x,
dx +
f x,
dx
+
b−a a
m
b−a a
m
Z d
Z d
1
1
+
f (a, y) dy +
f (b, y) dy
d−c c
d−c c
Z d Z d mα
a mα
b
+
f
, y dy +
f
, y dy .
d−c c
m
d−c c
m
Proof. Since f : ∆ → R is co-ordinated (α, m)-convex on ∆ it follows that the mapping
gx : [0, d] → R, gx (y) = f (x, y) is (α, m)-convex on [0, d] for all x ∈ [0, b]. Then by
inequality (1.6) one has:
"
#
Z d
c
d
1
1 gx (c) + gx (d) + mα gx m + gx m
gx (y) dy ≤
,
d−c c
2
α+1
that is
1
d−c
Z
d
c
1
f (x, y) dy ≤
2
"
f (x, c) + f (x, d) + mα f x,
α+1
c
m
d
+ f x, m
#
,
228
M. E. Özdemir, E. Set, M. Z. Sarıkaya
where 0 ≤ c < d < ∞ and (α, m) ∈ (0, 1]2 . Integrating this inequality on [a, b], we have
Z bZ d
1
f (x, y) dy dx
(b − a) (d − c) a c
Z b
Z b
1
1
1
(3.7)
≤
f (x, c) dx +
f (x, d) dx
2 (α + 1) b − a a
b−a a
Z b Z b mα
c
mα
d
+
f x,
dx +
f x,
dx
b−a a
m
b−a a
m
where 0 ≤ a < b < ∞.
By a similar argument applied to the mapping gy : [0, b] → [0, ∞), gy (x) = f (x, y)
with 0 ≤ a < b < ∞, we get
Z dZ b
1
f (x, y) dx dy
(d − c) (b − a) c a
Z d
Z d
1
1
1
(3.8)
≤
f (a, y) dy +
f (b, y) dy
2 (α + 1) d − c c
d−c c
Z d Z d mα
a mα
b
+
f
, y dy +
f
, y dy .
d−c c
m
d−c c
m
Summing the inequalities (3.7) and (3.8), we get the inequality (3.6).
3.6. Corollary. Choosing m = 1 in Theorem 3.5, we get the following inequality
Z bZ d
1
f (x, y) dy dx
(b − a) (d − c) a c
Z b
Z b
1
1
1
≤
f (x, c) dx +
f (x, d) dx
4 (α + 1) b − a a
b−a a
Z b
Z b
α
α
+
f (x, c) dx +
f (x, d) dx
b−a a
b−a a
Z d
Z d
1
1
+
f (a, y) dy +
f (b, y) dy
d−c c
d−c c
Z d
Z d
α
α
f (a, y) dy +
f (b, y) dy .
+
d−c c
d−c c
3.7. Remark. Choosing (α, m) = (1, 1) in (3.6), we get the third inequality of (1.7).
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