hermite-hadamard type inequalities for the functions … · many inequalities have been established...
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Applied Mathematical Sciences, Vol. 9, 2015, no. 1, 15 - 36HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2015.411944
Hermite-Hadamard Type Inequalities for
the Functions whose Second Derivatives
in Absolute Value are Convex and Concave
Jaekeun Park
Department of MathematicsHanseo University
Seosan, Choongnam, 356-706, Korea
Copyright c© 2014 Jaekeun Park. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
Abstract
In this article, the author obtain some generalization on Hermite-Hadamard-like type inequalities which gives an new estimate between1
b−a∫ ba f(x)dx and 1
2
[f(a+b2
)+ f(a)+f(b)
2
]for functions whose second
derivatives in absolute value at certain powers are, respectively, convexand concave.
Mathematics Subject Classification: 26D15, 26A51
Keywords: Hermite-Hadamard inequality; Convex functions; Concavefunction; Holder inequality; Power-mean inequality; Hypergeometric function;Betta function
1 Introduction
Recall that a function f : I ⊆ R → R is said to be convex on I if theinequality
f(tx+ (1− t)y) ≤ tf(x) + (1− t)f(y) (1)
holds for all x, y ∈ I and t ∈ [0, 1], and f is said to be concave on I if theinequality (1) holds in reversed direction.
16 Jaekeun Park
Many inequalities have been established for convex functions but the mostfamous is the Hermite-Hadamard inequality, due to its rich geometrical signif-icance and applications, which is stated as follow:
Let f : I ⊆ R → R be a convex function define on an interval I of realnumbers, and a, b ∈ I with a < b. Then the following double inequalities hold:
f(a+ b
2
)≤ 1
b− a
∫ b
a
f(t)dt ≤ f(a) + f(b)
2. (2)
Both inequalities hold in the reversed direction if f is concave.
It was first discovered by Hermite in 1881 in the Journal Mathesis. Thisinequality (2) was nowhere mentioned in the mathematical literature untill1893. In [1], Beckenbach, a leading expert on the theory of convex functions,wrote that the inequality (2) was proved by Hadamard in 1893. In 1974,Mitrinovic found Hermite and Hadamard’s note in Mathesis. That is why,the inequality (2) was known as Hermite-Hadamard inequality. We note thatHermite-Hadamard’s inequality may be regarded as a refinements of the con-cept of convexity and it follows easily from Jensen’s inequality. This inequality(2) has been received renewed attention in recent years and a remarkable va-riety of refinements and generalizations have been found in [2]-[17].
In recent paper[14], Tseng et. al established the following result whichgives a refinement of (2):
f(a+ b
2
)≤f(3a+b4
)+ f(a+3b4
)2
≤ 1
b− a
∫ b
a
f(x)dx
≤ 1
2
[f(a+ b
2
)+f(a) + f(b)
2
]≤ f(a) + f(b)
2, (3)
where f : [a, b]→ R is a convex function.
In[9], Latif established some new Hadamard-type inequalities for whosederivatives in absolute values are convex:
Theorem 1.1. Let f : I ⊆ R → R be a differentiable function define on theinterior I0 of an interval I in R such that f ′ ∈ L([a, b]), where a, b ∈ I0 witha < b. If | f ′ |q is convex on [a, b] for some fixed q ≥ 1, then the following
Hermite-Hadamard type inequalities 17
inequality holds:∣∣∣f(x) +(b− x)f(b) + (x− a)f(a)
b− a− 2
b− a
∫ b
a
f(u)du∣∣∣
≤(1
4
)[(x− a)2
b− a
{(5 | f ′(x) |q + | f ′(a) |q
6
) 1q
+( | f ′(x) |q +5 | f ′(a) |q
6
) 1q
}+
(b− x)2
b− a
{(5 | f ′(x) |q + | f ′(b) |q
6
) 1q
+( | f ′(x) |q +5 | f ′(b) |q
6
) 1q
}](4)
for all x ∈ [a, b].
Here, if we choose x = a+b2
in (4), we have some Hermite-Hadamard inequal-
ities which gives an estimate between 1b−a
∫ b
af(x)dx and 1
2
[f(a+b2
)+ f(a)+f(b)
2
]for functions whose derivatives in absolute value are convex.
Here we recall the definitions of the Gamma function
Γ(x) =
∫ ∞0
e−xtx−1dt
and the Betta function
β(x, y) =
∫ 1
0
tx−1(1− t)y−1dt.
Note that
β(x, y) =Γ(x)Γ(y)
Γ(x+ y).
The integral form of the hypergeometric function is defined by
2F1[x, y, c, z] =1
β(y, c− y)
∫ 1
0
ty−1(1− t)c−y−1(1− zt)−xdt
for | z |< 1, c > y > 0.
In this article, a new general identity for continuously twice differentiablefunctions is established. By making use of this equality, author has obtainednew estimates on generalization of Hermite-Hadamard-like type inequalitiesfor functions whose second derivatives in absolute value at certain powers areconvex and concave.
18 Jaekeun Park
2 Main results
In this section, for the simplicity of the notation, let
If (x) ≡ 1
b− a
∫ b
a
f(u)du− 1
2
{f(x) +
f(a) + f(b)
2
}+
1
2
(x− a+ b
2
)f ′(x)
for any x ∈ [a, b].
In order to prove our main results, we need the following lemma:
Lemma 1. Let f : I ⊆ R→ R be a twice differentiable function on the interiorI0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I with a < b. Then,for any x ∈ [a, b] we have the following identity:
If (x) =1
2(b− a)
[ ∫ x
a
(t− a)(t− a+ b
2
)f ′′(t)dt
+
∫ b
x
(t− b)(t− a+ b
2
)f ′′(t)dt]
Proof. By integration by parts, we can state
(i)I1 ≡∫ x
a
(t− a)(t− a+ b
2
)f ′′(t)dt
=(x− a
)(x− a+ b
2
)f ′(x)− b− a
2f(a)
− 2(x− 3a+ b
4
)f(x) + 2
∫ x
a
f(u)du. (5)
By similar way we get
(ii)I2 ≡∫ b
x
(t− b)(t− a+ b
2
)f ′′(t)dt
=(b− x
)(x− a+ b
2
)f ′(x)− b− a
2f(b)
+ 2(x− a+ 3b
4
)f(x) + 2
∫ b
x
f(u)du. (6)
By the equalities (5) and (6), we get the desired result.
Hermite-Hadamard type inequalities 19
Theorem 2.1. Let f : I ⊂ [0,∞)→ R be a twice differentiable function on theinterior I0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I0 with a < b.If | f ′′ | is convex on [a, b], then for any x ∈ [a, b] the following inequalitieshold:
(a) For x ≤ a+b2
, we have:∣∣∣If (x)∣∣∣ ≤ 1
192(b− a)
[λ11∣∣f ′′(a)∣∣+ λ12
∣∣f ′′(b)∣∣+ λ13
∣∣f ′′(x)∣∣+ λ14∣∣f ′′(a+ b
2
)∣∣], (7)
where
λ11 = 8(x− a)2(b− x),
λ12(x, q) = (b− a)3,
λ13 = 8(x− a)2(a+ 2b− 3x) + (a+ b− 2x)(7b− a− 6x),
λ14 = (a+ b− 2x)2(3b− a− 2x) + (b− a)3.
(b) For x > a+b2
, we have:∣∣∣If (x)∣∣∣ ≤ 1
192(b− a)
[λ15∣∣f ′′(a)∣∣+ λ16
∣∣f ′′(b)∣∣+ λ17
∣∣f ′′(x)∣∣+ λ18∣∣f ′′(a+ b
2
)∣∣], (8)
where
λ15 = (b− a)3,
λ16 = 8(x− a)(b− x)2,
λ17 = 8(b− x)2(3x− 2a− b) + (2x− a− b)2(6x− 7a+ b),
λ18 = (a+ b− 2x)2(2x− 3a+ b) + (b− x)3.
Proof. From Lemma 1, we have∣∣∣If (a, b;x)∣∣∣ ≤ 1
2(b− a)
[ ∫ x
a
∣∣(t− a)(t− a+ b
2
)∣∣∣∣f ′′(t)∣∣dt+
∫ b
x
∣∣(t− b)(t− a+ b
2
)∣∣∣∣f ′′(t)∣∣dt]=
1
2(b− a)
{∣∣∣I1∣∣∣+∣∣∣I2∣∣∣} (Say) (9)
Since | f ′′ | is convex on [a, b], we have:(a) For x ≤ a+b
2, we have:
20 Jaekeun Park
(i)∣∣∣I1∣∣∣ =
∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣dt=
∫ x
a
(t− a)(a+ b
2− t)∣∣∣f ′′(t)∣∣∣dt
=
∫ x
a
(t− a)(a+ b
2− t)∣∣∣f ′′( t− a
x− ax+
x− tx− a
a)∣∣∣dt
≤∫ x
a
(t− a)(a+ b
2− t){ t− ax− a
∣∣∣f ′′(x)∣∣∣+x− tx− a
∣∣∣f ′′(a)∣∣∣}dt
=(x− a)2
12
{(b− x)
∣∣∣f ′′(a)∣∣∣+ (a+ 2b− 3x)∣∣∣f ′′(x)∣∣∣}. (10)
By similar way we get
(ii)∣∣∣I2∣∣∣ =
∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣dt=
∫ a+b2
x
(b− t)(a+ b
2− t)∣∣∣f ′′(t)∣∣∣dt
+
∫ b
a+b2
(b− t)(t− a+ b
2
)∣∣∣f ′′(t)∣∣∣dt. (11)
Here, note that∫ a+b2
x
(b− t)(a+ b
2− t)∣∣∣f ′′(t)∣∣∣dt
=
∫ a+b2
x
(b− t)(a+ b
2− t)∣∣∣f ′′( t− x
a+b2− x
(a+ b
2
)+
a+b2− t
a+b2− x
x)∣∣∣dt
≤ (a+ b− 2x)2
96
×{(
3b− a− 2x)∣∣∣f ′′(a+ b
2
)∣∣∣+(7b− a− 6x
)∣∣∣f ′′(x)∣∣∣} (12)
and ∫ b
a+b2
(b− t)(t− a+ b
2
)∣∣∣f ′′(t)∣∣∣dt=
∫ b
a+b2
(b− t)(t− a+ b
2
)∣∣∣f ′′( t− a+b2
b− a+b2
b+b− tb− a+b
2
(a+ b
2
))∣∣∣dt≤ (b− a)3
96
{∣∣∣f ′′(a+ b
2
)∣∣∣+∣∣∣f ′′(b)∣∣∣}. (13)
Hermite-Hadamard type inequalities 21
By substituting (12) and (13) in (11), we have:∣∣∣I2∣∣∣ ≤ 1
96
[(a+ b− 2x)2(7b− a− 6x)
∣∣∣f ′′(x)∣∣∣+ (b− a)3∣∣∣f ′′(b)∣∣∣
+{
(a+ b− 2x)2(3b− a− 2x) + (b− a)3}∣∣∣f ′′(a+ b
2
)∣∣∣]. (14)
By substituting (10) and (14) in (9), we have the desired result (8).(b) For x > a+b
2, we have:
(i)∣∣∣I1∣∣∣ =
∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣dt=
∫ a+b2
a
(t− a)(a+ b
2− t)∣∣∣f ′′(t)∣∣∣dt
+
∫ x
a+b2
(t− a)(t− a+ b
2
)∣∣∣f ′′(t)∣∣∣dt. (15)
Note that∫ a+b2
a
(t− a)(a+ b
2− t)∣∣∣f ′′(t)∣∣∣dt
=
∫ a+b2
a
(t− a)(a+ b
2− t)∣∣∣f ′′( t− a
a+b2− a
(a+ b
2
)+
a+b2− t
a+b2− a
a)∣∣∣dt
≤ (b− a)3
96
{∣∣∣f ′′(a)∣∣∣+∣∣∣f ′′(a+ b
2
)∣∣∣} (16)
and ∫ x
a+b2
(t− a)(t− a+ b
2
)∣∣∣f ′′(t)∣∣∣dt=
∫ x
a+b2
(t− a)(t− a+ b
2
)∣∣∣f ′′( t− a+b2
x− a+b2
x+x− tx− a+b
2
(a+ b
2
))∣∣∣dt≤ (2x− a− b)2
96
×{(
6x− 7a+ b)∣∣∣f ′′(x)∣∣∣+
(2x− 3a+ b
)∣∣∣f ′′(a+ b
2
)∣∣∣}. (17)
By substituting (16) and (17) in (15), we get∣∣∣I1∣∣∣ ≤ 1
96
[(b− a
)3∣∣∣f ′′(a)∣∣∣+(2x− a− b
)2(6x− 7a+ b
)∣∣∣f ′′(x)∣∣∣+{(b− x
)3+(2x− a− b
)2(2x− 3a+ b
)}∣∣∣f ′′(a+ b
2
)∣∣∣]. (18)
22 Jaekeun Park
By similar way we get
(ii)∣∣∣I2∣∣∣ =
∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣dt=
∫ b
x
(b− t)(t− a+ b
2
)∣∣∣f ′′( t− xb− x
b+b− tb− x
x)∣∣∣dt
≤ (b− x)2
12
{(x− a
)∣∣∣f ′′(b)∣∣∣+(3x− 2a− b
)∣∣∣f ′′(x)∣∣∣}. (19)
By substituting (18) and (19) in (9), we have the desired result (8).
Theorem 2.2. Let f : I ⊂ [0,∞) → R be a twice differentiable function onthe interior I0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I0 witha < b. If | f ′′ |q is convex on [a, b] for some fixed q > 1 with 1
p+ 1
q= 1, then
for any x ∈ [a, b] the following inequalities hold:
(a) For x ≤ a+b2
, we have:
∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[µ
1p
21
{x− a2
(∣∣∣f ′′(a)∣∣∣q +∣∣∣f ′′(x)∣∣∣q)} 1
q
+ µ1p
22
{a+ b− 2x
4
(∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(x)∣∣∣q)
+b− a
4
(∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(b)∣∣∣q)} 1
q], (20)
where
µ21 =(b− a)p(x− a)1+p
2p(1 + p)2F1[1 + p,−p, 2 + p,
2(x− a)
b− a],
µ22 =(b− a)p
2p(1 + p)
{(1 + p)(b− a)1+p
22+3pβ(
1
2, 1 + p)(1 + (−1)1+p)
+ (x− b)p 2F1[1 + p,−p, 2 + p,2(b− x)
b− a]}.
(b) For x > a+b2
, we have:
∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[µ
1p
23
{2x− a− b4
(∣∣∣f ′′(x)∣∣∣q +∣∣∣f ′′(a+ b
2
)∣∣∣q)+b− a
4
(∣∣∣f ′′(a)∣∣∣q +∣∣∣f ′′(a+ b
2
)∣∣∣q)} 1q
+ µ1p
24
{b− x2
(∣∣∣f ′′(x)∣∣∣q +∣∣∣f ′′(b)∣∣∣q)} 1
q], (21)
Hermite-Hadamard type inequalities 23
where
µ23 =(b− a)p
21+p
{(b− a)1+p
21+pβ(
1
2, 1 + p)
+(2x− a− b)1+p
1 + p2F1[1 + p,−p, 2 + p,
a+ b− 2x
b− a]},
µ24 =(b− a)p
21+p
{(b− a)1+p
21+pβ(
1
2, 1 + p)
− (2x− a− b)1+p
1 + p2F1[1 + p,−p, 2 + p,
2x− a− bb− a
]}.
Proof. From Lemma 1 and using the well-known Holder integral inequality,we have
∣∣∣If (x)∣∣∣
≤ 1
2(b− a)
[{∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣pdt} 1p{∫ x
a
∣∣∣f ′′(t)∣∣∣qdt} 1q
+{∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣pdt} 1p{∫ b
x
∣∣∣f ′′(t)∣∣∣qdt} 1q]. (22)
Since | f ′′ | is convex on [a, b], we have:
(a) For x ≤ a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣pdt = µ21,
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣pdt = µ22,
(iii)
∫ x
a
∣∣∣f ′′(t)∣∣∣qdt ≤ x− a2
{∣∣∣f ′′(a)∣∣∣q +∣∣∣f ′′(x)∣∣∣q},
(iv)
∫ b
x
∣∣∣f ′′(t)∣∣∣qdt ≤ a+ b− 2x
4
{∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(x)∣∣∣q}
+b− a
4
{∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(b)∣∣∣q}. (23)
By substituting the above equalities (a):(i)-(ii) and the above inequalities(a):(iii)-(iv) in (22), we have the desired result (20).
24 Jaekeun Park
(b) For x > a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣pdt = µ23,
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣pdt = µ24,
(iii)
∫ x
a
∣∣∣f ′′(t)∣∣∣qdt ≤ b− a4
{∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(a)∣∣∣q}
+2x− a− b
4
{∣∣∣f ′′(x)∣∣∣q +∣∣∣f ′′(a+ b
2
)∣∣∣q},(iv)
∫ b
x
∣∣∣f ′′(t)∣∣∣qdt ≤ b− x2
{∣∣∣f ′′(x)∣∣∣q +∣∣∣f ′′(b)∣∣∣q}. (24)
By substituting the above equalities (b):(i)-(ii) and the above inequalities(b):(iii)-(iv) in (22), we have the desired result (21).
Theorem 2.3. Let f : I ⊂ [0,∞) → R be a twice differentiable function onthe interior I0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I0 witha < b. If | f ′′ |q is convex on [a, b] for some fixed q > 1 with 1
p+ 1
q= 1, then
for any x ∈ [a, b] the following inequalities hold:
(a) For x ≤ a+b2
, we have:
∣∣∣If (x)∣∣∣
≤ 1
2(b− a)
[µ
1p
31
{ 1
x− a
(λ31
∣∣∣f ′′(a)∣∣∣q + λ32
∣∣∣f ′′(x)∣∣∣q)} 1q
+ µ1p
32
{ 2
a+ b− 2x
(λ33
∣∣∣f ′′(x)∣∣∣q + λ34
∣∣∣f ′′(a+ b
2
)∣∣∣q)+
2
b− a
(λ35
∣∣∣f ′′(a+ b
2
)∣∣∣q + λ36
∣∣∣f ′′(b)∣∣∣q)} 1q], (25)
Hermite-Hadamard type inequalities 25
where
µ31 =(b− a)1+p − (a+ b− 2x)1+p
21+p(1 + p),
µ32 =(b− a)1+p + (a+ b− 2x)1+p
21+p(1 + p),
λ31 =(x− a)2+q
(1 + q)(2 + q),
λ32 =(x− a)2+q
2 + q,
λ33 =(b− a)2+q + 21+q(b− x)1+q
{(2 + q)a+ qb− 2(1 + q)x
}22+q(1 + q)(2 + q)
,
λ34 =22+q(b− x)2+q − (b− a)1+q
{(1 + q)a+ (3 + q)b− 2(2 + q)x
}22+q(1 + q)(2 + q)
,
λ35 =(b− a)2+q
22+q(2 + q),
λ36 =(b− a)2+q
22+q(1 + q)(2 + q).
(b) For x > a+b2
, we have:
∣∣∣If (a, b;x)∣∣∣
≤ 1
2(b− a)
[µ
1p
41
{ 2
b− a
(λ41
∣∣∣f ′′(a)∣∣∣q + λ42
∣∣∣f ′′(a+ b
2
)∣∣∣q)+
2
2x− a− b
(λ43
∣∣∣f ′′(a+ b
2
)∣∣∣q + λ44
∣∣∣f ′′(x)∣∣∣q)} 1q
+ µ1p
42
{ 1
b− x
(λ45
∣∣∣f ′′(x)∣∣∣q + λ46
∣∣∣f ′′(b)∣∣∣q)} 1q], (26)
26 Jaekeun Park
where
µ41 =(b− a)1+p + (2x− a− b)1+p
21+p(1 + p),
µ42 =(b− a)1+p − (2x− a− b)1+p
21+p(1 + p),
λ41 =(b− a)2+q
22+q(1 + q)(2 + q),
λ42 =(b− a)2+q
22+q(2 + q),
λ43 =22+q(x− a)2+q + (b− a)1+q
{(3 + q)a+ (1 + q)b− 2(2 + q)x
}22+q(1 + q)(2 + q)
,
λ44 =(b− a)2+q + 21+q(x− a)1+q
{qa+ (2 + q)b− 2(1 + q)x
}22+q(1 + q)(2 + q)
,
λ45 =(b− x)2+q
2 + q,
λ46 =(b− x)2+q
(1 + q)(2 + q).
Proof. From Lemma 1 and using the well-known Holder integral inequality,we have∣∣∣If (a, b;x)
∣∣∣≤ 1
2(b− a)
[{∫ x
a
∣∣∣(t− a+ b
2
)∣∣∣pdt} 1p{∫ x
a
∣∣∣(t− a)∣∣∣q∣∣∣f ′′(t)∣∣∣qdt} 1q
+{∫ b
x
∣∣∣(t− a+ b
2
)∣∣∣pdt} 1p{∫ b
x
∣∣∣(t− b)∣∣∣q∣∣∣f ′′(t)∣∣∣qdt} 1q]. (27)
Since | f ′′ | is convex on [a, b], we have:
(a) For x ≤ a+b2
, we have:
(i)
∫ x
a
∣∣∣t− a+ b
2
∣∣∣pdt = µ31,
(ii)
∫ b
x
∣∣∣t− a+ b
2
∣∣∣pdt = µ32,
(iii)
∫ x
a
∣∣∣(t− a)∣∣∣q∣∣∣f ′′(t)∣∣∣qdt ≤ 1
x− a
{λ31
∣∣∣f ′′(a)∣∣∣q + λ32
∣∣∣f ′′(x)∣∣∣q},
Hermite-Hadamard type inequalities 27
(iv)
∫ b
x
∣∣∣(t− b)∣∣∣q∣∣∣f ′′(t)∣∣∣qdt≤ 2
a+ b− 2x
{λ33
∣∣∣f ′′(x)∣∣∣q + λ34
∣∣∣f ′′(a+ b
2
)∣∣∣q}+
2
b− a
{λ35
∣∣∣f ′′(a+ b
2
)∣∣∣q + λ36
∣∣∣f ′′(b)∣∣∣q}.By substituting the above equalities (a):(i)-(ii) and the above inequalities
(a):(iii)-(iv) in (27), we have the desired result (25).(b) For x > a+b
2, we have:
(i)
∫ x
a
∣∣∣t− a+ b
2
∣∣∣pdt = µ41,
(ii)
∫ b
x
∣∣∣t− a+ b
2
∣∣∣pdt = µ42,
(iii)
∫ x
a
∣∣∣t− a∣∣∣q∣∣∣f ′′(t)∣∣∣qdt≤ 2
b− a
{λ41
∣∣∣f ′′(a)∣∣∣q + λ42
∣∣∣f ′′(a+ b
2
)∣∣∣q}+
2
2x− a− b
{λ43
∣∣∣f ′′(a+ b
2
)∣∣∣q + λ44
∣∣∣f ′′(x)∣∣∣q},(iv)
∫ b
x
∣∣∣t− b∣∣∣q∣∣∣f ′′(t)∣∣∣qdt=
∫ b
x
(b− t
)q∣∣∣f ′′( t− xb− x
b+b− tb− x
x)∣∣∣qdt
≤ 1
b− x
{λ45
∣∣∣f ′′(x)∣∣∣q + λ46
∣∣∣f ′′(b)∣∣∣q}.By substituting the above equalities (b):(i)-(ii) and the above inequalities
(b):(iii)-(iv) in (27), we have the desired result (26).
Theorem 2.4. Let f : I ⊂ [0,∞) → R be a twice differentiable function onthe interior I0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I0 witha < b. If | f ′′ |q is convex on [a, b] for some fixed q ≥ 1, then for any x ∈ [a, b]the following inequalities hold:
(a) For x ≤ a+b2
, we have:∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[µ1− 1
q
51
{λ51
∣∣∣f ′′(a)∣∣∣q + λ52
∣∣∣f ′′(x)∣∣∣q)} 1q
+ µ1− 1
q
52
{λ53
∣∣∣f ′′(x)∣∣∣q + λ54
∣∣∣f ′′(a+ b
2
)∣∣∣q)+ λ55
(∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(b)∣∣∣q)} 1
q], (28)
28 Jaekeun Park
where
µ51 =(x− a)2(a+ 3b− 4x)
12,
µ52 =(b− a)3 + (a+ b− 2x)2(5b− a− 4x)
48,
λ51 =(x− a)2(b− x)
12,
λ52 =(x− a)2(a+ 2b− 3x)
12,
λ53 =(a+ b− 2x)2(7b− a− 6x)
96,
λ54 =(a+ b− 2x)2(3b− a− 2x)
96,
λ55 =(b− a)3
96.
(b) For x > a+b2
, we have:
∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[µ1− 1
q
61
{λ61
(∣∣∣f ′′(a)∣∣∣q +∣∣∣f ′′(a+ b
2
)∣∣∣q)+ λ62
∣∣∣f ′′(x)∣∣∣q + λ63
∣∣∣f ′′(a+ b
2
)∣∣∣q)} 1q
+ µ1− 1
q
62
{λ64
∣∣∣f ′′(x)∣∣∣q + λ65
∣∣∣f ′′(b)∣∣∣q)} 1q], (29)
where
µ61 =(b− a)3 + (4x− 5a+ b)(a+ b− 2x)2
48,
µ61 =(b− x)2(4x− 3a− b)
12,
λ61 =(b− a)3
96,
λ62 =(2x− a− b)2(6x− 7a+ b)
96,
λ63 =(2x− a− b)2(2x− 3a+ b)
96,
λ64 =(b− x)2(3x− 2a− b)
12,
λ65 =(b− x)2(x− a)
12.
Hermite-Hadamard type inequalities 29
Proof. Suppose that q ≥ 1. From Lemma 1 and using the well-knownpower-mean integral inequality, we have
∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[{∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣dt}1− 1q
×{∫ x
a
∣∣∣(t− a)(t− a+ b
2
∣∣∣∣∣∣f ′′(t)∣∣∣qdt} 1q
+{∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣dt}1− 1q
×{∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣qdt} 1q]. (30)
Since | f ′′ | is convex on [a, b], we have:
(a) For x ≤ a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣dt = µ51,
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣dt = µ52,
(iii)
∫ x
a
∣∣∣(t− a)((t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣qdt≤ λ51(x)
∣∣∣f ′′(a)∣∣∣q + λ52(x)∣∣∣f ′′(x)∣∣∣q,
(iv)
∫ b
x
∣∣∣(t− b)((t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣qdt≤ λ53(x)
∣∣∣f ′′(x)∣∣∣q + λ54(x)∣∣∣f ′′(a+ b
2
)∣∣∣q+ λ55(x)
(∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(b)∣∣∣q).
By substituting the above equalities (a):(i)-(ii) and the above inequalities(a):(iii)-(iv) in (30), we have the desired result (28).
(b) For x > a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣dt = µ61,
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣dt = µ62,
30 Jaekeun Park
(iii)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣∣∣∣f ′′(t)∣∣∣qdt
≤ λ61
{∣∣∣f ′′(a+ b
2
)∣∣∣q +∣∣∣f ′′(a)∣∣∣q}
+ λ62(x)∣∣∣f ′′(x)∣∣∣q + λ63(x)
∣∣∣f ′′(a+ b
2
)∣∣∣q},(iv)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣q∣∣∣f ′′(t)∣∣∣qdt
≤ λ64
∣∣∣f ′′(x)∣∣∣q + λ65
∣∣∣f ′′(b)∣∣∣q}.By substituting the above equalities (b):(i)-(ii) and the above inequalities
(b):(iii)-(iv) in (30), we have the desired result (29).
Theorem 2.5. Let f : I ⊂ [0,∞) → R be a twice differentiable function onthe interior I0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I0 witha < b. If | f ′′ |q is concave on [a, b] for some fixed q > 1 with 1
p+ 1
q= 1, then
for any x ∈ [a, b] the following inequalities hold:
(a) For x ≤ a+b2
, we have:
∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[µ
1p
71
{(x− a)
∣∣∣f ′′(x+ a
2
)∣∣∣q} 1q
+ µ1p
72
{(b− a2
)(∣∣∣f ′′(a+ b+ 6x
4
)∣∣∣q +∣∣∣f ′′(a+ 3b
4
)∣∣∣q} 1q], (31)
where
µ71 =(b− a)p(x− a)1+p
2p(1 + p)2F1[1 + p,−p, 2 + p,
2(x− a)
b− x],
µ72 =(b− a)p
2p
{(b− a)1+p
22+3p(1 + (−1)1+p)β(
1
2, 1 + p)
+(x− b)p
1 + p2F1[1 + p,−p, 2 + p,
2(b− x)
b− a]}. (32)
(b) For x > a+b2
, we have:
∣∣∣If (x)∣∣∣ ≤ 1
2(b− a)
[µ
1p
74
{(b− x)
∣∣∣f ′′(x+ b
2
)∣∣∣q} 1q
+ µ1p
73
{(b− a2
)(∣∣∣f ′′(3a+ b
4
)∣∣∣q+
2
2x− a− b
∣∣∣f ′′(a+ b+ 2x
4
)∣∣∣q} 1q], (33)
Hermite-Hadamard type inequalities 31
where
µ73 =(b− a)p
21+2p
{(b− a)1+p
21+2pβ(
1
2, 1 + p)
+(2x− a− b)1+p
1 + p2F1[1 + p,−p, 2 + p,
a+ b− 2x
b− a]},
µ74 =(b− a)p
21+2p
{(b− a)1+p
21+2pβ(
1
2, 1 + p)
− (2x− a− b)1+p
1 + p2F1[1 + p,−p, 2 + p,
2x− a− bb− a
]}. (34)
Proof. From Lemma 1 and using the well-known Holder inequality fot q > 1with 1
p+ 1
q= 1, we have
∣∣∣If (a, b;x)∣∣∣
≤ 1
2(b− a)
[{∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣pdt} 1p{∫ x
a
∣∣∣f ′′(t)∣∣∣qdt} 1q
+{∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣pdt} 1p{∫ b
x
∣∣∣f ′′(t)∣∣∣qdt} 1q]. (35)
Since | f ′′ | is concave on [a, b], we have:
(a) For x ≤ a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣pdt = µ71,
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣dt = µ72,
(iii)
∫ x
a
∣∣∣f ′′(t)∣∣∣qdt ≤ (x− a)∣∣∣f ′′(x+ a
2
)∣∣∣q,(iv)
∫ b
x
∣∣∣f ′′(t)∣∣∣qdt ≤ (b− a2
){∣∣∣f ′′(a+ b+ 6x
4
)∣∣∣q +∣∣∣f ′′(a+ 3b
4
)∣∣∣q}.By substituting the above equalities (a):(i)-(ii) and the above inequalities
(a):(iii)-(iv) in (35), we have the desired result (31).
32 Jaekeun Park
(b) For x > a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣dt = µ73,
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣dt = µ74,
(iii)
∫ x
a
∣∣∣f ′′(t)∣∣∣qdt≤(b− a
2
)∣∣∣f ′′(3a+ b
4
)∣∣∣q +( 2
2x− a− b)∣∣∣f ′′(a+ b+ 2x
4
)∣∣∣q,(iv)
∫ b
x
∣∣∣f ′′(t)∣∣∣qdt ≤ (b− x)∣∣∣f ′′(x+ b
2
)∣∣∣q.By substituting the above equalities (b):(i)-(ii) and the above inequalities
(b):(iii)-(iv) in (35), we have the desired result (33).
Theorem 2.6. Let f : I ⊂ [0,∞) → R be a twice differentiable function onthe interior I0 of an interval I such that f
′′ ∈ L1[a, b], where a, b ∈ I0 witha < b. If | f ′′ |q is concave on [a, b] for some fixed q ≥ 1, then for any x ∈ [a, b]the following inequality holds:
(a) For x ≤ a+b2
, we have:∣∣∣If (x, a, b)∣∣∣
≤ 1
2(b− a)
[µ1− 1
q
81
{λ81
∣∣∣f ′′((a+ 3b− 4x)b− 3(b− x)2
a+ 3b− 4x
)∣∣∣q} 1q
+ µ1− 1
q
82
{λ82
∣∣∣f ′′((2x− a+ 3b)(2x− a− b)− 8x(b− x)
4(4x+ a− 5b)
)∣∣∣q+ λ83
∣∣∣f ′′(a+ 3b
4
)∣∣∣q} 1q], (36)
where
µ81 =(x− a)2(a+ 3b− 4x)
12,
µ82 =(a+ b− 2x)2(5b− a− 4x)
48,
λ81 =(x− a)2(a+ 3b− 4x)
12,
λ82 =(a+ b− 2x)2(5b− a− 4x)
48,
λ83 =(b− a)3
48.
Hermite-Hadamard type inequalities 33
(b) For x > a+b2
, we have:
∣∣∣If (x, a, b)∣∣∣
≤ 1
2(b− a)
[µ1− 1
q
91 (x){λ91(x)
∣∣∣f ′′(3a+ b
4
)∣∣∣q+ λ92(x)
∣∣∣f ′′(3a− b+ 2x)(a+ b− 2x) + 8x(2a− b− x)
4(5a− b− 4x)
)∣∣∣q} 1q
+ µ1− 1
q
92 (x){λ93(x)
∣∣∣f ′′((2x− a− b)x+ (x− a)(x+ a)
4x− 3b− b)∣∣∣q} 1
q], (37)
where
µ91(x) =(b− a)3 + (a+ b− 2x)2(4x− 5a+ b)
48,
µ92(x) =(b− x)2(4x− 3a− b)
12,
λ91(x) =(b− a)3
48,
λ92(x) =(a+ b− 2x)2(4x− 5a+ b)
48,
λ93(x) =(b− x)2(4x− 3a− b)
12.
Proof. From Lemma 1 and using the well-known power-mean integral in-equality fot q ≥ 1, we have
∣∣∣If (a, b;x)∣∣∣
≤ 1
2(b− a)
[{∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣dt} 1p
×{∫ x
a
∣∣∣(t− a)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣qdt} 1q
+{∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣dt} 1p
×{∫ b
x
∣∣∣(t− b)(t− a+ b
2
)∣∣∣∣∣∣f ′′(t)∣∣∣qdt} 1q]. (38)
Since | f ′′ | is concave on [a, b], we have:
34 Jaekeun Park
(a) For x ≤ a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣dt = µ81(x, p),
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣dt = µ82(x, p),
(iii)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣∣∣∣f ′′(t)∣∣∣dt
≤ λ81(x)∣∣∣f ′′((a+ 3b− 4x)b− 3(b− x)2
a+ 3b− 4x
)∣∣∣q,(iv)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣∣∣∣f ′′(t)∣∣∣dt
≤ λ82(x)∣∣∣f ′′((2x− a+ 3b)(2x− a− b)− 8x(b− x)
4(4x+ a− 5b)
)∣∣∣q+ λ83(x)
∣∣∣f ′′(a+ 3b
4
)∣∣∣q.By substituting the above equalities (a):(i)-(ii) and the above inequalities
(a):(iii)-(iv) in (38), we have the desired result (36).
(b) For x > a+b2
, we have:
(i)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣dt = µ91(x),
(ii)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣dt = µ92(x)
(iii)
∫ x
a
∣∣∣(t− a)(t− a+ b
2)∣∣∣∣∣∣f ′′(t)∣∣∣qdt
≤ λ92(x)∣∣∣f ′′((3a− b+ 2x)(a+ b− 2x) + 8x(2a− b− x)
4(5a− b− 4x)
)∣∣∣q+ λ91(x)
∣∣∣f ′′(3a+ b
4
)∣∣∣q,(iv)
∫ b
x
∣∣∣(t− b)(t− a+ b
2)∣∣∣∣∣∣f ′′(t)∣∣∣qdt
≤ λ93(x)∣∣∣f ′′(x(2x− a− b) + (x− a)(x+ a)
4x− 3a− b)∣∣∣q.
By substituting the above equalities (b):(i)-(ii) and the above inequalities(b):(iii)-(iv) in (38), we have the desired result (37).
Hermite-Hadamard type inequalities 35
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Received: November 25, 2014; Published: December 22, 2014