An extension of the Hermite-Hadamard inequality for a power of a convex function

Yamin Sayyari; Mehdi Dehghanian; Choonkil Park; Siriluk Paokanta

doi:10.1515/math-2022-0542

Open Access Published by De Gruyter Open Access February 1, 2023

An extension of the Hermite-Hadamard inequality for a power of a convex function

Yamin Sayyari , Mehdi Dehghanian , Choonkil Park and Siriluk Paokanta

From the journal Open Mathematics

https://doi.org/10.1515/math-2022-0542

Abstract

In this article, we obtain an extension of the classical Hermite-Hadamard inequality for convex functions (concave functions) extending it to the power functions [ f ( x ) ] n . Some related inequalities are also introduced. By applying those results in analysis, we obtain new upper and lower bounds for the error function.

Keywords: Hermite-Hadamard inequality; Jensen’s inequality; convex function; error function; power function

MSC 2010: 52A40; 52A41; 26D15; 26D07

1 Introduction

Let f be a real-valued function on I ≔ [ a , b ] ⊂ R for a < b . The function f is said to be a convex function on I if the inequality

f ( λ x + ( 1 − λ ) y ) ≤ λ f ( x ) + ( 1 − λ ) f ( y )

holds for all x , y ∈ I and λ ∈ [ 0 , 1 ] . Many inequalities have been established for convex functions, but the most famous is the Hermite-Hadamard inequality [1], due to its rich geometrical significance and applications, which is stated as follows:

Theorem 1.1

[1] (Hermite-Hadamard Inequality) Let f be a real-valued and convex function on [ a , b ] . Then

(1.1) f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( b ) + f ( a ) 2 .

Both the inequalities hold in reversed direction if f is concave.

It is well known that the Hermite-Hadamard inequality plays an important role in the analysis and theory of convex functions. In recent years, there have been many generalizations of the Hermite-Hadamard inequality [2–9].

The well-known Jensen’s inequality for a convex function is given as follows.

Theorem 1.2

[10] (Jensen’s inequality) Let μ be a positive measure on a σ -algebra ℱ in a set Ω such that μ ( Ω ) = 1 . If f is a real function in L 1 ( μ ) , a < f ( x ) < b for all x ∈ Ω and φ is convex on ( a , b ) , then

φ ∫ Ω f d μ ≤ ∫ Ω φ ( f ) d μ .

2 Refinement of Hermite-Hadamard inequality

In this section, assume that f is a real-valued, nonnegative, and convex function on [ a , b ] with a < b and n is a positive integer and f n ≔ f ( x ) n ≔ f ( x ) × f ( x ) × ⋯ × f ( x ) ︸ n .

Lemma 2.1

Let f : [ a , b ] → R be a nonnegative convex function. Then f n is convex and

(2.1) f a + b 2 n ≤ 1 b − a ∫ a b f ( x ) n d x ≤ f ( b ) n + f ( a ) n 2 .

Proof

Let g ( x ) ≔ x n . Since g and f are two convex functions,

[ f ( t a + ( 1 − t ) b ] n ≤ [ t f ( a ) + ( 1 − t ) f ( b ) ] n ≤ t f ( a ) n + ( 1 − t ) f ( b ) n .

Hence, f n is a convex function. The inequality (2.1) follows from Theorem 1.1.□

The following example shows that the nonnegativity in Lemma 2.1 is necessary.

Example 2.2

Set f : [ − 1 , 1 ] → R : f ( x ) = ∣ x ∣ − 1 . Then f ( x ) is convex, but [ f ( x ) ] 3 is not convex (see Figure 1).

$Figure 1 y = ( ∣ x ∣ − 1 ) 3 y={\left(| x| -1)}^{3} .$

Figure 1

y = ( ∣ x ∣ − 1 ) 3 .

We now obtain an extension of Lemma 2.1.

Theorem 2.3

Let f be a real-valued, nonnegative, and convex function on [ a , b ] and n be a positive integer. Then

(2.2) ∫ a b f ( x ) d x n ( b − a ) n ≤ 1 b − a ∫ a b f ( x ) n d x ≤ f ( b ) n + 1 − f ( a ) n + 1 ( n + 1 ) [ f ( b ) − f ( a ) ] ,

where we assume that if f ( a ) = f ( b ) , then f ( b ) n + 1 − f ( a ) n + 1 f ( b ) − f ( a ) ≔ ( n + 1 ) f ( a ) n .

Proof

Since φ ( t ) = t n is a convex function on ( 0 , ∞ ) , by the use of Theorem 1.2 with φ ( t ) = t n , we have

∫ 0 1 f ( t a + ( 1 − t ) b ) d t n ≤ ∫ 0 1 f ( t a + ( 1 − t ) b ) n d t .

By substituting x = t a + ( 1 − t ) b , it is easy to observe that

1 b − a ∫ a b f ( x ) d x n ≤ 1 b − a ∫ a b f ( x ) n d x .

On the other hand,

1 b − a ∫ a b f ( x ) n d x = ∫ 0 1 f ( t a + ( 1 − t ) b ) n d t ≤ ∫ 0 1 [ t f ( a ) + ( 1 − t ) f ( b ) ] n d t = ∫ 0 1 ∑ k = 0 n n k t k ( 1 − t ) n − k f ( a ) k f ( b ) n − k d t = ∑ k = 0 n n k f ( a ) k f ( b ) n − k ∫ 0 1 t k ( 1 − t ) n − k d t = ∑ k = 0 n n k f ( a ) k f ( b ) n − k B ( k + 1 , n − k + 1 ) ,

where B is the beta function

B ( m , n ) = ∫ 0 1 t m − 1 ( 1 − t ) n − 1 d t = ( m − 1 ) ! ( n − 1 ) ! ( m + n − 1 ) ! ,

for all m , n ∈ N . Since

n k ∫ 0 1 t k ( 1 − t ) n − k d t = n k B ( k + 1 , n − k + 1 ) = 1 n + 1

for all n ∈ N 0 and k ≤ n ,

1 b − a ∫ a b f ( x ) n d x ≤ ∑ k = 0 n 1 n + 1 f ( a ) k f ( b ) n − k = f ( b ) n + 1 − f ( a ) n + 1 ( n + 1 ) [ f ( b ) − f ( a ) ]

for f ( a ) ≠ f ( b ) , since A n + 1 − B n + 1 = ( A − B ) ( A n + A n − 1 B + ⋯ + A B n − 1 + B n ) . For f ( a ) = f ( b ) , the right-hand side of the aforementioned inequality is f ( a ) n . This completes the proof.□

Lemma 2.4

Let a and b be positive real numbers with a ≠ b . Then

b n + 1 − a n + 1 ( n + 1 ) [ b − a ] ≤ b n + a n 2 .

Proof

Without the loss of generality, suppose that a < b . By applying Theorem 1.1 with f ( x ) = x n , we obtain

a + b 2 n ≤ b n + 1 − a n + 1 ( n + 1 ) [ b − a ] ≤ b n + a n 2 .

This completes the proof.□

Theorem 2.3 and Lemma 2.4 together yield the following theorem.

Theorem 2.5

Let f be a real-valued, nonnegative, and convex function on [ a , b ] and n be a positive integer. Then

(2.3) f a + b 2 n ≤ 1 b − a ∫ a b f ( x ) d x n ≤ 1 b − a ∫ a b f ( x ) n d x ≤ f ( b ) n + 1 − f ( a ) n + 1 ( n + 1 ) [ f ( b ) − f ( a ) ] ≤ f ( b ) n + f ( a ) n 2 .

Proof

By using Lemma 2.4, we obtain

(2.4) f ( b ) n + 1 − f ( a ) n + 1 ( n + 1 ) [ f ( b ) − f ( a ) ] ≤ f ( b ) n + f ( a ) n 2

and by using Theorem 1.1, we have

(2.5) f a + b 2 n ≤ 1 b − a ∫ a b f ( x ) d x n .

The desired result follows from (2.4), (2.5), and Theorem 2.3.□

Corollary 2.6

Let n be a positive integer and f n be a real-valued, nonnegative, and convex function on [ a , b ] . Then

(2.6) f a + b 2 ≤ 1 b − a ∫ a b f ( x ) n d x n ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( b ) f ( b ) n − f ( a ) f ( a ) n ( n + 1 ) [ f ( b ) n − f ( a ) n ] ≤ f ( b ) + f ( a ) 2 ,

where we assume that if f ( b ) = f ( a ) , then

f ( b ) f ( b ) n − f ( a ) f ( a ) n f ( b ) n − f ( a ) n ≔ ( n + 1 ) f ( a ) .

Proof

If f ( x ) is replaced by f ( x ) n , (2.6) follows from (2.3).□

Remark 2.7

With the notations in Corollary 2.6, if n = 1 , then

f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( b ) + f ( a ) 2 ,

which is analogous to inequality (1.1).

Now, we provide an example, which is the importance and superiority of our result compared to the existing results.

Example 2.8

Let n = 3 and f : [ 0 , 2 ] → R be defined by f ( x ) = e 3 x 2 . Then by Corollary 2.6,

e 3 ≤ 1 2 ∫ 0 2 e 3 x 2 3 d x 3 ≤ 1 2 ∫ 0 2 e 3 x 2 d x ≤ e 16 − 1 4 ( e 4 − 1 ) ≤ e 12 + 1 2 .

We have the following:

e 3 ≃ 20.1 is a lower bound of the Hermite-Hadamard inequality , 1 2 ∫ 0 2 e 3 x 2 3 d x 3 ≃ 556.7 is a lower bound of the inequality in Corollary 2.6 , e 16 − 1 4 ( e 4 − 1 ) ≃ 41447.8 is an upper bound of the inequality in Corollary 2.6 , e 12 + 1 2 ≃ 81377.9 is an upper bound of Hermite-Hadamard inequality .

20.1 ≤ 556.7 ≤ 1 2 ∫ 0 2 e 3 x 2 d x ≤ 41447.8 ≤ 81377.9 .

Corollary 2.9

Let f ≥ 0 and f be a convex function on [ a , b ] . Then

f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( a ) + f ( b ) + f ( a ) f ( b ) 3 ≤ f ( b ) + f ( a ) 2 .

Example 2.10

Let f ( x ) = e − x 2 . Note that f ( x ) is not convex on [ 0 , 1 ) . But f ( x ) is convex on [ 1 , b ] for all b > 1 . Thus,

e − 1 + b 2 ≤ 1 b − 1 ∫ 1 b e − x 2 2 d x 2 ≤ 1 b − 1 ∫ 1 b e − x 2 d x ≤ e − 1 + e − b 2 + e − 1 + b 2 2 3 ≤ e − 1 + e − b 2 2

for all b > 1 .

Proposition 2.11

Let a < b . Then

( b − a ) e a + b 2 ≤ ∫ a b e x 2 d x ≤ e b 2 − e a 2 b + a ≤ ( b − a ) ( e b 2 + e a 2 ) 2 .

Proof

Let a < b . Define f ( x ) = e x 2 on [ a , b ] . Since e x 2 n = e x 2 n is convex on [ a , b ] for all n ∈ N , and by Corollary 2.6, we have

e a + b 2 ≤ 1 b − a ∫ a b e x 2 d x ≤ e 1 + 1 n b 2 − e 1 + 1 n a 2 ( n + 1 ) ( e 1 n b 2 − e 1 n a 2 ) ≤ e b 2 + e a 2 2

for all n ∈ N . By taking the limit as n → ∞ , we have that

e a + b 2 ≤ 1 b − a ∫ a b e x 2 d x ≤ lim n → ∞ e 1 + 1 n b 2 − e 1 + 1 n a 2 ( n + 1 ) ( e 1 n b 2 − e 1 n a 2 ) ≤ e b 2 + e a 2 2 .

Since

lim n → ∞ e 1 + 1 n b 2 − e 1 + 1 n a 2 ( n + 1 ) ( e 1 n b 2 − e 1 n a 2 ) = lim x → 0 x ( e ( 1 + x ) b 2 − e ( 1 + x ) a 2 ) ( x + 1 ) ( e x b 2 − e x a 2 ) = e b 2 − e a 2 b 2 − a 2 ,

(2.7) e a + b 2 ≤ 1 b − a ∫ a b e x 2 d x ≤ e b 2 − e a 2 b 2 − a 2 ≤ e b 2 + e a 2 2 .

Let b 2 = a 2 + y . Then y > 0 , and the last inequality in (2.7) becomes e a 2 e y − 1 y ≤ e a 2 e y + 1 2 . Let g ( y ) = y e y + y − 2 e y + 2 . Then g ′ ( y ) = e y ( y − 1 ) + 1 , g ″ ( y ) = y e y ≥ 0 , and g ′ ( 0 ) = g ( 0 ) = 0 and so g ′ ( y ) ≥ 0 and g ( y ) ≥ 0 for all y ≥ 0 . Since e a 2 > 0 , the last inequality in (2.7) holds true. Thus,

( b − a ) e a + b 2 ≤ ∫ a b e x 2 d x ≤ e b 2 − e a 2 b + a ≤ ( b − a ) ( e b 2 + e a 2 ) 2 .

This completes the proof.□

Corollary 2.12

Let a ≠ b be two positive real numbers. Then

a + b 2 k n ≤ b k n + 1 − a k n + 1 ( b − a ) ( k n + 1 ) ≤ b k n + k − a k n + k ( b − a ) ( n + 1 ) ≤ b k n + a k n 2 ,

for all k , n ∈ N .

Proof

It follows from Corollary 2.6 by choosing f ( x ) = x k n and appropriate elementary calculations.□

Remark 2.13

Under the notation of Corollary 2.12, if k = 1 , then

a + b 2 n ≤ b n + 1 − a n + 1 ( b − a ) ( n + 1 ) ≤ b n + a n 2 ,

which is a generalization of the following Haber inequality [11]

a + b 2 n ≤ 1 n + 1 ∑ i = 0 n a i b n − i .

3 Extension of Hermite-Hadamard inequality for the concave functions

In this section, we extend the Hermite-Hadamard inequality for the class of concave functions.

Theorem 3.1

Let f be a real-valued, nonnegative, and concave function on [ a , b ] , and let n be a positive integer. Then

(3.1) [ f ( b ) ] n + 1 − [ f ( a ) ] n + 1 ( n + 1 ) [ f ( b ) − f ( a ) ] ≤ 1 b − a ∫ a b [ f ( x ) ] n d x ≤ 2 n − 1 [ f a + b 2 ] n ,

where we assume that if f ( a ) = f ( b ) , then [ f ( b ) ] n + 1 − [ f ( a ) ] n + 1 f ( b ) − f ( a ) ≔ ( n + 1 ) [ f ( a ) ] n .

Theorem 3.2

Let f n be a real-valued, nonnegative, and concave function on [ a , b ] , and let n be a positive integer. Then

(3.2) f ( b ) f ( b ) n − f ( a ) f ( a ) n ( n + 1 ) [ f ( b ) n − f ( a ) n ] ≤ 1 b − a ∫ a b f ( x ) d x ≤ 2 n − 1 f a + b 2 ,

where we assume that if f ( b ) = f ( a ) , then

f ( b ) f ( b ) n − f ( a ) f ( a ) n f ( b ) n − f ( a ) n ≔ ( n + 1 ) f ( a ) .

Proof

If f ( x ) is replaced by f ( x ) n , (3.2) follows from (3.1).□

Proposition 3.3

Let 0 ≤ a < b . Then

e − a 2 − e − b 2 a + b ≤ ∫ a b e − x 2 d x ≤ ( b − a ) 2 m − 1 e − a + b 2 2 ,

where m is the smallest integer greater than or equal to 2 b 2 .

Proof

Let 0 ≤ a < b . Define f ( x ) = e − x 2 on [ a , b ] . Since e − x 2 n = e − x 2 n is concave on 0 , n 2 for all n ∈ N , e − x 2 n = e − x 2 n is concave on [ a , b ] ⊆ 0 , n 2 for all n ≥ 2 b 2 . By Theorem 3.2, we have

(3.3) e − b 2 1 + 1 n − e − a 2 1 + 1 n ( n + 1 ) e − b 2 n − e − a 2 n ≤ 1 b − a ∫ a b e − x 2 d x ≤ 2 n − 1 e − a + b 2 2 ,

for all n ≥ 2 b 2 . Hence,

1 b − a ∫ a b e − x 2 d x ≤ 2 m − 1 e − a + b 2 2 ,

where m is the smallest integer greater than or equal to 2 b 2 .

By taking the limit as n → ∞ in (3.3), we deduce that

lim n → ∞ e − b 2 1 + 1 n − e − a 2 1 + 1 n ( n + 1 ) e − b 2 n − e − a 2 n ≤ 1 b − a ∫ a b e − x 2 d x .

Since

lim n → ∞ e − b 2 1 + 1 n − e − a 2 1 + 1 n ( n + 1 ) e − b 2 n − e − a 2 n = lim x → 0 x ( e − b 2 ( 1 + x ) − e − a 2 ( 1 + x ) ) ( 1 + x ) [ e − x b 2 − e − x a 2 ] = e − a 2 − e − b 2 b 2 − a 2 ,

e − a 2 − e − b 2 b 2 − a 2 ≤ 1 b − a ∫ a b e − x 2 d x .

Therefore,

e − a 2 − e − b 2 a + b ≤ ∫ a b e − x 2 d x ≤ ( b − a ) 2 m − 1 e − a + b 2 2 .

This completes the proof.□

Remark 3.4

Note that m ≤ 2 b 2 + 1 . Thus,

e − a 2 − e − b 2 a + b ≤ ∫ a b e − x 2 d x ≤ ( b − a ) 4 b 2 e − a + b 2 2 ,

for all b > a ≥ 0 .

The error function erf ( x ) and the complementary error function erfc ( x ) are defined as follows:

erf ( x ) = ∫ 0 x 2 e − t 2 π d t , erfc ( x ) = ∫ x ∞ 2 e − t 2 π d t .

Proposition 3.5

Let x > 0 and m be the smallest integer greater than or equal to 2 x 2 . Then

2 ( 1 − e − x 2 ) π x ≤ erf ( x ) ≤ 2 m π x e − 1 4 x 2 ≤ 2 x 4 x 2 π e − 1 4 x 2 , for all x > 0 .
1 − 2 x 4 x 2 π e − 1 4 x 2 ≤ erfc ( x ) ≤ π x − 2 ( 1 − e − x 2 ) π x .

Proof

Applying Proposition 3.3 and Remark 3.4 with a = 0 , b = x , after some calculations the desired assertion follows.
Since erfc ( x ) = 1 − erf ( x ) , the result follows from (1).

The proof is completed.□

4 Hermite-Hadamard inequality for the power of a convex function

Let f be a convex function on [ a , b ] . We define

f 0 ≡ min { min { f ( x ) : a ≤ x ≤ b } , 0 } ,

and the convex positive part of a real-valued function f is defined by the formula:

f c ( x ) ≔ f ( x ) − f 0 .

The function f can be expressed in terms of f c and f 0 as f ( x ) = f c ( x ) + f 0 .

Example 4.1

Set f : [ 3 2 , 4 ] ⟶ R : f ( x ) = x − ln ( x − 1 ) − 3 . Then f 0 = − 1 and f c ( x ) = x − ln ( x − 1 ) − 2 (see Figure 2).

$Figure 2 f , f 0 f,{f}_{0} , and f c {f}_{c} .$

Figure 2

f , f 0 , and f c .

Lemma 4.2

Let f be a real-valued and convex function on [ a , b ] . Then f c = f c ( x ) is a nonnegative convex function on [ a , b ] and f 0 is nonpositive.

Proof

It is easy.□

Denote E n ≔ { 2 k : 2 k ≤ n , k = 0 , 1 , 2 , … } and O n ≔ { 2 k + 1 : 2 k + 1 ≤ n , k = 0 , 1 , 2 , … } .

Theorem 4.3

Let f be a real-valued and convex function on [ a , b ] , and let n be a positive integer. Then

∑ k ∈ O n n k ( f 0 ) k f c ( b ) n − k + f c ( a ) n − k 2 + ∑ k ∈ E n n k ( f 0 ) k f c a + b 2 n − k ≤ ∑ k ∈ O n n k ( f 0 ) k f c ( b ) n − k + 1 − f c ( a ) n − k + 1 ( n − k + 1 ) ( f c ( b ) − f c ( a ) ) + ∑ k ∈ E n n k ( f 0 ) k f c a + b 2 n − k ≤ 1 b − a ∫ a b f ( x ) n d x ≤ ∑ k ∈ E n n k ( f 0 ) k f c ( b ) n − k + 1 − f c ( a ) n − k + 1 ( n − k + 1 ) ( f c ( b ) − f c ( a ) ) + ∑ k ∈ O n n k ( f 0 ) k f c a + b 2 n − k ≤ ∑ k ∈ E n n k ( f 0 ) k f c ( b ) n − k + f c ( a ) n − k 2 + ∑ k ∈ O n n k ( f 0 ) k f c a + b 2 n − k ,

where f c ( b ) n + 1 − f c ( a ) n + 1 f c ( b ) − f c ( a ) ≔ ( n + 1 ) f c ( a ) n for f c ( b ) = f c ( a ) .

Proof

Let f be a convex function. By the use of Lemma 4.2, the function f c is a nonnegative function and f 0 is nonpositive. Since

f ( x ) n = [ f c ( x ) + f 0 ] n = ∑ k = 0 n n k f 0 k f c ( x ) n − k ,

by Theorem 2.3, we have

1 b − a ∫ a b f ( x ) n d x = 1 b − a ∫ a b [ f c ( x ) + f 0 ] n d x = 1 b − a ∑ k ∈ E n n k f 0 k ∫ a b f c ( x ) n − k d x + ∑ k ∈ O n n k f 0 k ∫ a b f c ( x ) n − k d x ≤ ∑ k ∈ E n n k f 0 k f c ( b ) n − k + 1 − f c ( a ) n − k + 1 ( n − k + 1 ) ( f c ( b ) − f c ( a ) ) + ∑ k ∈ O n n k f 0 k f c a + b 2 n − k .

Also, by Theorem 2.3, we have

1 b − a ∫ a b f ( x ) n d x = 1 b − a ∫ a b [ f c ( x ) + f 0 ] n d x = 1 b − a ∑ k ∈ O n n k f 0 k ∫ a b f c ( x ) n − k d x + ∑ k ∈ E n n k f 0 k ∫ a b f c ( x ) n − k d x ≥ ∑ k ∈ O n n k f 0 k f c ( b ) n − k + 1 − f c ( a ) n − k + 1 ( n − k + 1 ) ( f c ( b ) − f c ( a ) ) + ∑ k ∈ E n n k f 0 k f c a + b 2 n − k ,

which completes the proof.□

Theorem 4.4

With the notations in Theorem 4.3, if n = 1 , then

f a + b 2 ≤ 1 b − a ∫ a b f ( x ) d x ≤ f ( b ) + f ( a ) 2 ,

which is analogous to inequality (1.1).

Corollary 4.5

Let f : [ a , b ] → R be a convex function. Then

f a + b 2 2 + 2 f 0 f ( a ) + f ( b ) 2 − f a + b 2 ≤ f a + b 2 2 + 2 f 0 f ( a ) + f ( b ) 2 − f a + b 2 + f 0 2 ≤ 1 b − a ∫ a b f ( x ) 2 d x ≤ f ( a ) 2 + f ( a ) f ( b ) + f ( b ) 2 3 + 2 f 0 f a + b 2 − f ( a ) + f ( b ) 2 ≤ f ( b ) 2 + f ( a ) 2 2 + 2 f 0 f a + b 2 − f ( a ) + f ( b ) 2 .

Proof

We apply Theorem 4.3 with n = 2 to obtain

f 0 ( f c ( b ) + f c ( a ) ) + f c a + b 2 2 + f 0 2 ≤ f 0 f c ( b ) 2 − f c ( a ) 2 f c ( b ) − f c ( a ) + f c a + b 2 2 + 2 f 0 2 ≤ 1 b − a ∫ a b f ( x ) n d x ≤ f c ( b ) 3 − f c ( a ) 3 3 ( f c ( b ) − f c ( a ) ) + f 0 2 + 2 f 0 f c a + b 2 ≤ f c ( b ) 2 + f c ( a ) 2 2 + f 0 2 + 2 f 0 f c a + b 2 .

Thus, by replacing f c ( a ) by f ( a ) − f 0 , f c ( b ) by f ( b ) − f 0 and f c a + b 2 by f a + b 2 − f 0 , we obtain

f a + b 2 2 + 2 f 0 f ( a ) + f ( b ) 2 − f a + b 2 ≤ f a + b 2 2 + 2 f 0 f ( a ) + f ( b ) 2 − f a + b 2 + f 0 2 ≤ 1 b − a ∫ a b f ( x ) 2 d x ≤ f ( a ) 2 + f ( a ) f ( b ) + f ( b ) 2 3 + 2 f 0 f a + b 2 − f ( a ) + f ( b ) 2 ≤ f ( b ) 2 + f ( a ) 2 2 + 2 f 0 a + b 2 − f ( a ) + f ( b ) 2 .

This completes the proof.□

Remark 4.6

The superiority of the results obtained in this article compared to the existing results for the Hermite-Hadamard’s inequality (1.1) is as follows:

In Theorem 2.5, we have generalized the Hermite-Hadamard bound for functions whose powers of n are convex.
In Corollary 2.6, we have improved the Hermite-Hadamard bound for functions, where f 1 n is convex, especially, if n = 2 , then Corollary 2.9 is obtained.
In Theorem 4.3, we have obtained the Hermite-Hadamard inequality for the power n of a convex function, especially, if n = 2 , then Corollary 4.5 is obtained.
Examples 2.8 and 2.10 and Corollary 2.12 show the superiority of the work compared to the existing results.

5 Conclusion

We obtained an extension of the classical Hermite-Hadamard inequality for convex functions (concave functions) extending it to the power functions [ f ( x ) ] n . Some related inequalities were also introduced. By applying those results in the analysis, we obtained new upper and lower bounds for the error function.

Acknowledgements

We would like to express our sincere gratitude to the anonymous referee for his/her helpful comments that will help to improve the quality of the manuscript.

Funding information: The authors declare that there is no funding available for this article.
Author contributions: The authors equally conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Not applicable.

References

[1] J. Hadamard, Etude sur les proprietes des fonctions entieres et en particulier dune fonction consideree par Riemann, J. Math. Pures Appl. 58 (1893), no. 9, 171–172. Search in Google Scholar

[2] H. Budak, H. Kara, M. A. Ali, S. Khan, and Y. Chu, Fractional Hermite-Hadamard-type inequailities for interval-valued coordinated convex functions, Open Math. 19 (2021), no. 1, 1081–1097, DOI: https://doi.org/10.1515/math-2021-0067. 10.1515/math-2021-0067Search in Google Scholar

[3] H. Budak, S. Khan, M. A. Ali, and Y. Chu, Refinements of quantum Hermite-Hadamard-type inequailities, Open Math. 19 (2021), no. 1, 724–734, DOI: https://doi.org/10.1515/math-2021-0029. 10.1515/math-2021-0029Search in Google Scholar

[4] F. Chen, Extensions of the Hermite-Hadamard Inequality for convex functions via fractional integrals, J. Math. Inequal. 10 (2016), no. 1, 75–81, DOI: https://doi.org/10.7153/jmi-10-07. 10.7153/jmi-10-07Search in Google Scholar

[5] M. Benbachir, M. Dahmane, and A. El Farissi, An extension of the Hermite-Hadamard inequality for convex symmetrized functions, Real Anal. Exchange 38 (2012/2013), no. 2, 467–474. 10.14321/realanalexch.38.2.0467Search in Google Scholar

[6] M. Z. Sarikaya and H. Budak, On generalized Hermite-Hadamard inequality for generalized convex function, Int. J. Nonlinear Anal. Appl. 8 (2017), no. 2, 209–222, DOI: https://doi.org/10.22075/IJNAA.2017.11313.1552. Search in Google Scholar

[7] R. Xiang, Refinements of Hermite-Hadamard type inequalities for convex function via fractional integrals, J. Appl. Math. Inform. 33 (2015), no. 1–2, 119–125, DOI: https://doi.org/10.14317/jami.2015.119. 10.14317/jami.2015.119Search in Google Scholar

[8] G. Zabandan, New inequalities of Hermite-Hadamard type for MN-convex functions, Adv. Inequal. Appl. 2016 (2016), 7. Search in Google Scholar

[9] G. Zabandan, A. Bodaghi, and A. Kılıçman, The Hemite-Hadamard inequality for r-convex function, J. Inequal. Appl. 2012 (2012), 215, DOI: https://doi.org/10.1186/1029-242X-2012-215. 10.1186/1029-242X-2012-215Search in Google Scholar

[10] W. Rudin, Real and Complex Analysis, McGraw-Hill, New York, 1986. Search in Google Scholar

[11] H. Haber, An elementary inequality, Int. J. Math. Math. Sci. 2 (1979), no. 3, 531–535. 10.1155/S0161171279000429Search in Google Scholar

Received: 2022-05-22

Revised: 2022-11-18

Accepted: 2022-12-05

Published Online: 2023-02-01

This work is licensed under the Creative Commons Attribution 4.0 International License.

An extension of the Hermite-Hadamard inequality for a power of a convex function

Abstract

1 Introduction

Theorem 1.1

Theorem 1.2

2 Refinement of Hermite-Hadamard inequality

Lemma 2.1

Proof

Example 2.2

Theorem 2.3

Proof

Lemma 2.4

Proof

Theorem 2.5

Proof

Corollary 2.6

Proof

Remark 2.7

Example 2.8

Corollary 2.9

Example 2.10

Proposition 2.11

Proof

Corollary 2.12

Proof

Remark 2.13

3 Extension of Hermite-Hadamard inequality for the concave functions

Theorem 3.1

Theorem 3.2

Proof

Proposition 3.3

Proof

Remark 3.4

Proposition 3.5

Proof

4 Hermite-Hadamard inequality for the power of a convex function

Example 4.1

Lemma 4.2

Proof

Theorem 4.3

Proof

Theorem 4.4

Corollary 4.5

Proof

Remark 4.6

5 Conclusion

Acknowledgements

References

Journal and Issue

Articles in the same Issue