Refinements on the discrete Hermite–Hadamard inequality
Refinements on the discrete HermiteHadamard inequality
Ferhan M. Atıcı 0 1
Hatice Yaldız 0 1
Mathematics Subject Classification 0 1
0 R is a convex function , then the following
1 F. M. Atıcı Department of Mathematics, Western Kentucky University , Bowling Green, KY 421013576 , USA
In this paper, we use techniques and tools from time scale calculus to state and prove many refinements on the discrete HermiteHadamard inequality. 26D10 · 26D15 · 26A33 · 39A12 · 39A70

1 Introduction
The Hermite–Hadamard inequality [
9,10
] states that if f : I →
inequality is satisfied:
f
a + b
2
1
≤ b − a
≤
f (a) + f (b)
2
(1)
where a, b ∈ I and I is an interval in R. Hermite–Hadamard’s inequality for convex functions has received
renewed attention in recent years and a remarkable variety of refinements and generalizations have been found
(see, for example, [
3–6,12,13
]).
Regarding the definition of a convex function on a time scale, a pioneering work has been done by Mozyrska
and Torres [
11
]. They introduced the convexity of a function defined on a time scale (a nonempty closed subset
of R) and proved a theorem which characterizes the convex function in terms of its second derivative.
Since the midpoint condition plays an important role in the proofs related to the Hermite–Hadamard
inequality, the authors first defined the midpoint condition for the functions on the set of integers, Z, in
[
1
]. Then, some equivalent conditions for convexity have been given, and the discrete Hermite–Hadamard
inequality has been proven.
Let a, b ∈ Z with a < b. [a, b]Z means [a, b] ∩ Z. We define
T[a,b] =
t − b
u u = a − b
for t ∈ [a, b]Z .
We note that T[a,b] is a subset of the real interval [
0, 1
]. Since the set T[a,b] is an isolated time scale, from now
on we use the notations of the time scale calculus.
Definition 1.1 [
1
] f : Z → R is called convex on Z if for every x , y ∈ Z with x < y the following inequality
f (λx + (1 − λ) y) ≤ λ f (x ) + (1 − λ) f (y)
is satisfied for all λ ∈ T[x,y].
Theorem 1.2 [
1
] Suppose f : Z → R is a convex function on [a, b]Z with a, b ∈ Z and a < b, and a + b an
even number. Then
f
a + b
2
1
≤ 2 (b − a) ⎣
⎡
a
b
f (t ) t +
a
b
⎤
f (t )∇t ⎦ ≤
(2)
(3)
(4)
b
b
a
a
f (σ (t ))g (t ) t = ( f g) (b) − ( f g) (a) −
f (t ) g (t ) t
f (ρ (t ))g∇ (t ) ∇t = ( f g) (b) − ( f g) (a) −
f ∇ (t ) g (t ) ∇t.
a
a
b
b
Theorem 2.2 (Substitution rule) [
2,8
] Let ν : T → R be monotone.
(i) Assume ν is strictly increasing and T : ν (T) is a time scale. If f : T → R is a an rdcontinuous function
and ν is differentiable with rdcontinuous derivative, then if a, b ∈ T,
or
f (t )ν (t ) t =
f ◦ ν−1 (s) s
or
(ii) Assume ν is strictly decreasing and T : −ν (T) is a time scale. If f : T → R is an rdcontinuous
function and ν is differentiable with rdcontinuous derivative, then if a, b ∈ T,
− b − a
a
b
f (x )dx =
(1 − 2t ) f (t a + (1 − t ) b) dt.
(5)
Proof First, we claim that
=
(1 − 2ρ (t )) f (t a + (1 − t ) b) ∇t =
f (u) u.
(7)
− 2 (b − a)
⎡ f (a) + f (b)
2
1
(1 − 2ρ (t )) f (t a + (1 − t ) b) ∇t −
(1 − 2ρ (t )) f ∇ ((1 − t )a + t b))∇t ⎦ . (6)
to get that
by parts to the last integral, we have
where g(t ) = (1 − 2( b −b−σ a(t) )), (g. f )(ν−1(t )) = (1 − 2( b−σ (ν−1(t)) )) f (ν−1(t )). Now, we apply integration
b−a
0
1
If we multiply the last quantity by b−4 a , we have the desired result.
Next, we claim that
b − a
− 4
0
1
(1 − 2ρ (t )) f ∇ ((1 − t )a + t b)∇t =
f (a) + f (b)
4
1
− 2 (b − a) a
b
f (t )∇t.
(8)
To prove this claim, we first use the substitution rule (Theorem 2.2) and then use integration by parts formula
(Theorem 2.1). Here, we have
−
1=ν(b)
0=ν(a)
1
= − b − a a
(1 − 2ρ (t )) f ∇ ((1 − t )a + t b) ∇t
b
1 − 2
ρ (u) − a
b − a
f ∇ (u) ∇u,
t−a .
wheNreexνt(,t w)=eubs−eathe integration by parts formula to the last integral above. We have
1
− b − a a
b
⎧
1 ⎨
= − b − a
⎩
1
= b − a
Proof Using Lemma (3.1) and the convexity on [a, b]Z of f
and f ∇ , we obtain
⎛ b
.
Theorem 3.2 Suppose f : Z → R is a function. Let a, b ∈ Z with a < b, and a + b be an even number. If
f and f ∇ are convex on [a, b]Z, then the following inequality holds:
(1 − 2ρ (t)) f (ta + (1 − t) b) ∇t +
(2ρ (t) − 1) f ∇ ((1 − t)a + tb) ∇t
1 − 2ρ (t)  f (ta + (1 − t) b) ∇t +
2ρ (t) − 1 f ∇ ((1 − t)a + tb) ∇t
⎛ b
a
a
b
⎛ b
a
a
b
0
1
1
0
1
(9)
(10)
Corollary 3.3 Suppose f : Z → R is a function on [a, b]Z with a, b ∈ Z and a < b, and a + b an even
number. If f and f ∇ are convex on [a, b]Z, and  f (a) +  f ∇ (b) =  f ∇ (a) +  f (b), then
Corollary 3.4 Suppose f : Z → R is a function on [a, b]Z with a, b ∈ Z and a < b, and a + b an even
number. If f and f ∇ are convex on [a, b]Z, then
Corollary 3.5 Suppose f : Z → R is a function on [a, b]Z with a, b ∈ Z and a < b, and a + b an even
number. If f and f ∇ are convex on [a, b]Z, then
⎛ b
Theorem 3.6 Suppose f : Z → R is a convex function on [a, b]Z with a, b ∈ Z, a < b and a + b an even
number. Then, the refinement on the right side of the discrete Hermite–Hadamard inequality follows:
⎛ b
1
a
for all a, b ∈ Z and t ∈ T[a,b]\ {0, 1}. Integrating each side of the inequality on T[a,b] , we obtain
f (a) + f (b) −
f (ta + (1 − t) b) t −
f ((1 − t) a + tb) t
Using the substitution method on time scales (Theorem 2.2), we have
This completes the proof of the inequality (11).
Corollary 3.7 With the assumptions in Theorem 3.6 and the condition that f (a + b − x) = f (x) for all
x ∈ [a, b]Z, the following inequality is satisfied:
1
f (a) − 2 (b − a) ⎝
⎛ b
a
1
≥  f (a) − 2 (b − a) ⎣⎢
⎛ b
f (x) + f (2a + b − x) x − f a +2 b ≥ 0.
(12)
Theorem 3.8 With the assumptions in Theorem 3.6, the following inequality is satisfied:
a
2
f (ta + (1 − t)b) + f ((1 − t)a + tb) − f a + b
2 2
≥
f (ta + (1 − t)b) + f ((1 − t)a + tb) − f a + b
2 2
for all t ∈ T[a,b]\{0,1}.
Integrating this inequality each side over T[a,b] , we get that
x = ta + (1 − t)b
y = (1 − t)a + tb.
It is easy to see that x, y ∈ [a,b]Z and x + y is even. Hence, 21 ∈ T[x,y] or T[y,x] . Since f is convex on
[x, y]Z or [y, x]Z . We have that
f (x) + f (y) − f x + y
2 2 ≥
f (x) +2 f (y) − f x +2 y .
Next if we substitute x and y back in f , we have
! f (ta + (1 − t)b) t + ! f ((1 − t)a + tb) "t
T[a,b]⊂[
0,1
] T[a,b]⊂[
0,1
]
− f a + b
2
≥
f (ta + (1 − t)b) +2 f ((1 − t)a + tb) t − f a +2 b ,
Corollary 3.9 With the assumptions in Theorem 3.6, and the condition that f (a + b − x) = f (x) for all
x ∈ [a,b]Z, the following inequality is satisfied:
≥ b − a[a,b]Z
 f (x) x − f a +2 b ≥ 0.
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