On discrete inequalities for some classes of sequences

Mohamed Jleli; Bessem Samet

doi:10.1515/math-2024-0021

Article Open Access

On discrete inequalities for some classes of sequences

Mohamed Jleli and Bessem Samet

Published/Copyright: June 28, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

For a given sequence a = ( a 1 , … , a n ) ∈ R n , our aim is to obtain an estimate of E n ≔ a 1 + a n 2 − 1 n ∑ i = 1 n a i . Several classes of sequences are studied. For each class, an estimate of E n is obtained. We also introduce the class of convex matrices, which is a discrete version of the class of convex functions on the coordinates. For this set of matrices, new discrete Hermite-Hadamard-type inequalities are proved. Our obtained results are extensions of known results from the continuous case to the discrete case.

Keywords: discrete inequalities; convex sequences; convex matrices; Fejér inequality; Hermite-Hadamard inequality

MSC 2010: 26D15; 39A12; 40B05

1 Introduction

A great attention has been paid on the extensions of known inequalities from the continuous case to the discrete case, e.g., [1–11] and the references therein. In particular, the study of inequalities involving convex sequences has been considered by many authors, e.g., [12–16] and the references therein. The notion of convex sequences is a discrete version of the concept of convex functions. Namely, a sequence a = ( a 1 , … , a n ) ∈ R n , where n ≥ 3 , is said to be convex, if (e.g., [17])

a i ≤ a i − 1 + a i + 1 2 , i = 2 , … , n − 1 .

Latreuch and Belaïdi [12] established a discrete version of the Fejér double inequality for the class of convex sequences. We recall that the Fejér double inequality [18] states that, if f : [ a , b ] → R is convex and p : [ a , b ] → R is integrable, nonnegative, and symmetric with respect to the midpoint a + b 2 , then

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

In particular, if p ≡ 1 , (1.1) reduces the Hermite-Hadamard double inequality [19,20]

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

For some results related to inequalities (1.1) and (1.2), see e.g., [21–37] and the references therein. Latreuch and Belaïdi [12] proved that, if a = ( a 1 , … , a n ) ∈ R n is a convex sequence and p = ( p 1 , … , p n ) ∈ R n is a positive sequence, which is symmetric with respect to n + 1 2 , then

(1.3) a N + a n + 1 − N 2 ∑ i = 1 n p i ≤ ∑ i = 1 n p i a i ≤ a 1 + a n 2 ∑ i = 1 n p i ,

where N = n + 1 2 is the integer part of n + 1 2 . The double inequality (1.3) is a discrete version of the Fejér double inequality (1.1). If p i = 1 for all i , then (1.3) reduces to the double inequality

(1.4) a N + a n + 1 − N 2 ≤ 1 n ∑ i = 1 n a i ≤ a 1 + a n 2 ,

which is a discrete version of the Hermite-Hadamard double inequality (1.2). We also refer to [14], where some generalizations of the above results have been obtained.

Dragomir and Agarwal [26] established two interesting inequalities for the class of differentiable mappings. They first proved that, if f : [ a , b ] → R is differentiable and ∣ f ′ ∣ is convex on [ a , b ] , then

(1.5) f ( a ) + f ( b ) 2 − 1 b − a ∫ a b f ( x ) d x ≤ ( b − a ) ( ∣ f ′ ( a ) ∣ + ∣ f ′ ( b ) ∣ ) 8 .

Next they proved that, if f : [ a , b ] → R is differentiable and ∣ f ′ ∣ p p − 1 is convex on [ a , b ] for some p > 1 , then

(1.6) f ( a ) + f ( b ) 2 − 1 b − a ∫ a b f ( x ) d x ≤ b − a 2 ( p + 1 ) 1 p ∣ f ′ ( a ) ∣ p p − 1 + ∣ f ′ ( b ) ∣ p p − 1 2 p − 1 p .

Dragomir et al. [27] considered the class of L -Lipschitzian functions f : [ a , b ] → R , namely, the class of functions f satisfying

∣ f ( x ) − f ( y ) ∣ ≤ L ∣ x − y ∣ , x , y ∈ [ a , b ] ,

where L > 0 is a constant. They proved (among many other results) that

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

Dragomir [25] considered the class of convex functions on the coordinates. Namely, the class of functions f : [ a , b ] × [ c , d ] → R satisfying

for all x ∈ [ a , b ] , the function f ( x , ⋅ ) : [ c , d ] ∋ y ↦ f ( x , y ) ∈ R is convex;
for all y ∈ [ c , d ] , the function f ( ⋅ , y ) : [ a , b ] ∋ x ↦ f ( x , y ) ∈ R is convex.

Note that, if f is convex on [ a , b ] × [ c , d ] , then f is convex on the coordinates. However, the converse is not true in general [25, Lemma 1]. Dragomir [25] proved that, if f : [ a , b ] × [ c , d ] → R is convex on the coordinates, then

(1.8) f a + b 2 , c + d 2 ≤ 1 2 1 b − a ∫ a b f x , c + d 2 d x + 1 d − c ∫ c d f a + b 2 , y d y ≤ 1 ( b − a ) ( d − c ) ∫ a b ∫ c d f ( x , y ) d y d x ≤ 1 4 ( b − a ) ∫ a b ( f ( x , c ) + f ( x , d ) ) d x + 1 4 ( d − c ) ∫ c d ( f ( a , y ) + f ( b , y ) ) d y ≤ f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 .

The aim of this study is to establish discrete versions of inequalities (1.5), (1.6), (1.7), and (1.8).

2 Main results and proofs

We first fix some notations. For n ≥ 4 and a = ( a 1 , … , a n ) ∈ R n , we denote by a ′ = ( a 1 ′ , … , a n − 1 ′ ) ∈ R n − 1 the sequence defined by

a i ′ = ∣ a i + 1 − a i ∣ , i = 1 , … , n − 1 .

For a real number q > 1 , we denote by a ′ q ∈ R n − 1 the sequence defined by

a ′ q = ( a 1 ′ q , … , a n − 1 ′ q ) .

2.1 Discrete version of inequality (1.5)

Let us consider the set of sequences

A n = { a ∈ R n : a ′ is convex } ,

i.e., a ∈ A n , if and only if

a i ′ ≤ a i + 1 ′ + a i − 1 ′ 2 , i = 2 , … , n − 2 .

Our first main result is a discrete version of inequality (1.5), which was established in [26].

Theorem 2.1

Let n ≥ 4 . If a = ( a 1 , … , a n ) ∈ A n , then

(2.1) a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n n 2 n − 1 − n 2 a 1 ′ + a n − 1 ′ 2 ,

where n 2 denotes the integer part of n 2 .

Proof

We first claim that

(2.2) a 1 + a n 2 − 1 n ∑ i = 1 n a i = 1 n ∑ i = 1 n − 1 i − n 2 ( a i + 1 − a i ) .

Indeed, we have

∑ i = 1 n − 1 i − n 2 ( a i + 1 − a i ) + ∑ i = 1 n a i = ∑ i = 1 n − 1 i a i + 1 − i a i − n 2 a i + 1 + n 2 a i + ∑ i = 1 n a i = ∑ i = 1 n − 1 i a i + 1 − ∑ i = 1 n − 1 i a i + − n 2 ∑ i = 1 n − 1 a i + 1 + n 2 ∑ i = 1 n − 1 a i + ∑ i = 1 n a i = ∑ i = 1 n − 1 ( i + 1 ) a i + 1 − ∑ i = 1 n − 1 i a i + − n 2 ∑ i = 1 n − 1 a i + 1 + n 2 ∑ i = 1 n − 1 a i + ∑ i = 1 n a i − ∑ i = 1 n − 1 a i + 1 = ∑ i = 2 n i a i − ∑ i = 1 n − 1 i a i + − n 2 ∑ i = 2 n a i + n 2 ∑ i = 1 n − 1 a i + ∑ i = 1 n a i − ∑ i = 2 n a i = n a n − a 1 + − n 2 a n + n 2 a 1 + a 1 = n a 1 + a n 2 .

Then, multiplying the above identity by 1 n , we obtain (2.2).

We next use (2.2) to obtain

a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n ∑ i = 1 n − 1 i − n 2 ∣ a i + 1 − a i ∣ ,

i.e.,

(2.3) a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n ∑ i = 1 n − 1 p i a ′ i ,

where p = ( p 1 , … , p n − 1 ) ∈ R n − 1 is the sequence defined by

p i = i − n 2 , i = 1 , … , n − 1 .

Observe that p is symmetric with respect to n − 1 2 . Moreover, the sequence a ′ is convex. Hence, by the right discrete Fejér inequality (1.3), we have

∑ i = 1 n − 1 p i a ′ i ≤ ∑ i = 1 n − 1 p i a ′ 1 + a ′ n − 1 2 ,

i.e.,

(2.4) ∑ i = 1 n − 1 p i a ′ i ≤ ∑ i = 1 n − 1 i − n 2 a ′ 1 + a ′ n − 1 2 .

On the other hand, we have

∑ i = 1 n − 1 i − n 2 = ∑ i = 1 n 2 n 2 − i + ∑ i = n 2 + 1 n − 1 i − n 2 = n 2 n 2 − ∑ i = 1 n 2 i + ∑ i = n 2 + 1 n − 1 i − n 2 n − 1 − n 2 = n 2 n 2 − 1 2 n 2 n 2 + 1 + 1 2 n − 1 − n 2 n + n 2 − n 2 n − 1 − n 2 = n 2 n 2 − 1 2 n 2 n 2 + 1 + 1 2 n − 1 − n 2 n 2 = 1 2 n 2 n − 1 − n 2 + 1 2 n − 1 − n 2 n 2 = 1 2 n − 1 − n 2 n 2 + n 2 = n 2 n − 1 − n 2 .

Finally, (2.1) follows from (2.3), (2.4), and the above identity.□

If n is even, then n 2 = n 2 . Hence, from Theorem 2.1, we deduce the following result.

Corollary 2.2

Let n ≥ 4 be even. If a = ( a 1 , … , a n ) ∈ A n , then

a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ n − 2 8 ( a 1 ′ + a n − 1 ′ ) .

If n is odd, then n 2 = n − 1 2 . Hence, from Theorem 2.1, we deduce the following result.

Corollary 2.3

Let n ≥ 4 be odd. If a = ( a 1 , … , a n ) ∈ A n , then

a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ ( n − 1 ) 2 8 n ( a 1 ′ + a n − 1 ′ ) .

2.2 Discrete version of inequality (1.6)

For a real number q > 1 and n ≥ 4 , let us consider the set of sequences

A n , q = { a ∈ R n : a ′ q is convex } ,

i.e., a ∈ A n , q if and only if

a i ′ q ≤ a i + 1 ′ q + a i − 1 ′ q 2 , i = 2 , … , n − 2 .

Our second main result is a discrete version of inequality (1.6), which was established in [26].

Theorem 2.4

Let n ≥ 4 , p > 1 , and 1 p + 1 q = 1 . If a = ( a 1 , … , a n ) ∈ A n , q , then

(2.5) a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n n − 1 2 1 q ∑ i = 1 n − 1 i − n 2 p 1 p ( a 1 ′ q + a n − 1 ′ q ) 1 q .

Proof

Making use of (2.3) and Hölder’s inequality, we obtain

(2.6) a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n ∑ i = 1 n − 1 i − n 2 a ′ i ≤ 1 n ∑ i = 1 n − 1 i − n 2 p 1 p ∑ i = 1 n − 1 a ′ i q 1 q .

On the other hand, since a ∈ A n , q , making use of the right discrete Hermite-Hadamard inequality (1.4), we obtain

(2.7) ∑ i = 1 n − 1 a ′ i q ≤ ( n − 1 ) a ′ 1 q + a ′ n − 1 q 2 .

Then, combining (2.6) with (2.7), we obtain (2.5).□

Remark that

∑ i = 1 n − 1 i − n 2 p = ∑ i = 1 n 2 n 2 − i p + ∑ i = n 2 + 1 n − 1 i − n 2 p = ∑ i = 1 n 2 n 2 − i p + ∑ j = 1 n − n 2 − 1 n 2 − j p .

Hence, we obtain

∑ i = 1 n − 1 i − n 2 p = 2 ∑ i = 1 n 2 n 2 − i p if n is even , 2 ∑ i = 1 n − 1 2 n 2 − i p if n is odd .

Thus, from Theorem 2.4, we deduce the following results.

Corollary 2.5

Let n ≥ 4 be even, p > 1 and 1 p + 1 q = 1 . If a = ( a 1 , … , a n ) ∈ A n , q , then

a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n n − 1 2 1 q 2 1 p ∑ i = 1 n 2 n 2 − i p 1 p ( a 1 ′ q + a n − 1 ′ q ) 1 q .

Corollary 2.6

Let n ≥ 4 be odd, p > 1 and 1 p + 1 q = 1 . If a = ( a 1 , … , a n ) ∈ A n , q , then

a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ 1 n n − 1 2 1 q 2 1 p ∑ i = 1 n − 1 2 n 2 − i p 1 p ( a 1 ′ q + a n − 1 ′ q ) 1 q .

2.3 Discrete version of inequality (1.7)

The following result provides a discrete version of inequality (1.7), which was established in [27].

Theorem 2.7

Let n ≥ 2 and a = ( a 1 , … , a n ) ∈ R n . We have

(2.8) a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ ( n − 2 ) ( n − 1 ) n L ,

where L = max ∣ a i − a j ∣ ∣ i − j ∣ : i , j ∈ { 1 , … , n } , i ≠ j .

Proof

For n = 2 , (2.8) is obvious. So, we may suppose that n ≥ 3 . For all natural number k ≥ 2 , we can write a k as

a k = ( a k − a k − 1 ) + ( a k − 1 − a k − 2 ) + … + ( a 2 − a 1 ) + a 1 ,

i.e.,

(2.9) a k = a 1 + ∑ j = 2 k ( a j − a j − 1 ) , k = 2 , 3 , …

Making use of (2.9), we obtain

∑ i = 1 n a i = a 1 + ∑ k = 2 n a k = a 1 + ∑ k = 2 n a 1 + ∑ j = 2 k ( a j − a j − 1 ) = a 1 + ( n − 1 ) a 1 + ∑ k = 2 n ∑ j = 2 k ( a j − a j − 1 ) = n a 1 + ∑ k = 2 n − 1 ∑ j = 2 k ( a j − a j − 1 ) + ∑ j = 2 n ( a j − a j − 1 ) = n a 1 + ∑ k = 2 n − 1 ∑ j = 2 k ( a j − a j − 1 ) + ( n − 1 ) ( a n − a 1 ) ,

which implies that

1 n ∑ i = 1 n a i − a 1 + a n 2 = a 1 + 1 n ∑ k = 2 n − 1 ∑ j = 2 k ( a j − a j − 1 ) + n − 1 n ( a n − a 1 ) − a 1 + a n 2 = a 1 1 − n − 1 n − 1 2 + a n n − 1 n − 1 2 + 1 n ∑ k = 2 n − 1 ∑ j = 2 k ( a j − a j − 1 ) = n − 2 2 n ( a n − a 1 ) + 1 n ∑ k = 2 n − 1 ∑ j = 2 k ( a j − a j − 1 ) .

Hence, it holds that

1 n ∑ i = 1 n a i − a 1 + a n 2 ≤ n − 2 2 n ∣ a n − a 1 ∣ + 1 n ∑ k = 2 n − 1 ∑ j = 2 k ∣ a j − a j − 1 ∣ ≤ ( n − 2 ) ( n − 1 ) 2 n L + L n ∑ k = 1 n − 1 ( k − 1 ) = ( n − 2 ) ( n − 1 ) n L ,

which proves (2.8).□

We also have the following result.

Theorem 2.8

Let n ≥ 1 and a = ( a 1 , … , a n ) ∈ R n . We have

(2.10) a 1 + a n 2 − 1 n ∑ i = 1 n a i ≤ n − 2 2 M ,

where M = max { ∣ a i − a j ∣ : i , j ∈ { 1 , … , n } } .

Proof

For n = 1 or n = 2 , (2.10) is obvious. For n ≥ 3 , from the proof of Theorem 2.7, we have

1 n ∑ i = 1 n a i − a 1 + a n 2 = n − 2 2 n ( a n − a 1 ) + 1 n ∑ k = 2 n − 1 ∑ j = 2 k ( a j − a j − 1 ) ,

which implies that

1 n ∑ i = 1 n a i − a 1 + a n 2 ≤ n − 2 2 n ∣ a n − a 1 ∣ + 1 n ∑ k = 2 n − 1 ∑ j = 2 k ∣ a j − a j − 1 ∣ ≤ n − 2 2 n M + M n ∑ k = 1 n − 1 ( k − 1 ) = n − 2 2 n M + M n ∑ i = 1 n − 2 j = n − 2 2 n M + ( n − 2 ) ( n − 1 ) 2 n M = n − 2 2 M ,

which proves (2.10).□

The following examples show that the upper bounds of a 1 + a n 2 − 1 n ∑ i = 1 n a i provided by Theorems 2.7 and 2.8 are not comparable in general.

Example 2.9

Let n ≥ 2 and a = ( a 1 , … , a n ) ∈ R n , where

a i = 1 i , i = 1 , … , n .

In this case, we have

M = max 1 i − 1 j : i , j ∈ { 1 , … , n } = 1 − 1 n , L = max 1 i − 1 j ∣ i − j ∣ : i , j ∈ { 1 , … , n } , i ≠ j = max 1 i j : i , j ∈ { 1 , … , n } , i ≠ j = 1 2 , n − 2 2 M = ( n − 2 ) ( n − 1 ) n L = ( n − 2 ) ( n − 1 ) 2 n .

Hence, inequalities (2.8) and (2.10) are the same.

Example 2.10

Let n ≥ 2 and a = ( a 1 , … , a n ) ∈ R n , where

a i = i , i = 1 , … , n .

In this case, we have

M = max { ∣ i − j ∣ : i , j ∈ { 1 , … , n } } = n − 1 , L = max ∣ i − j ∣ ∣ i − j ∣ : i , j ∈ { 1 , … , n } , i ≠ j = 1 , n − 2 2 M = ( n − 2 ) ( n − 1 ) 2 , ( n − 2 ) ( n − 1 ) n L = ( n − 2 ) ( n − 1 ) n .

Observe that

( n − 2 ) ( n − 1 ) n L ≤ n − 2 2 M ,

which shows that (2.8) is more sharp than (2.10).

Example 2.11

Let n ≥ 3 and a = ( a 1 , … , a n ) ∈ R n , where

a 1 = 1 , a 2 = 2 , a i = 1 i + 1 , i = 3 , … , n .

In this case, we have

M = max { ∣ a i − a j ∣ : i , j ∈ { 1 , … , n } } = max 1 , 1 i , 1 − 1 i , 1 i − 1 j : i , j ∈ { 3 , … , n } = 1

and

L = max ∣ a i − a j ∣ ∣ i − j ∣ : i , j ∈ { 1 , … , n } , i ≠ j = max 1 , 1 i ( i − 1 ) , i − 1 i ( i − 2 ) , 1 i j : i , j ∈ { 3 , … , n } , i ≠ j = 1 .

Hence,

n − 2 2 M = n − 2 2 ≤ ( n − 2 ) ( n − 1 ) n = ( n − 2 ) ( n − 1 ) n L ,

which shows that (2.10) is more sharp than (2.8).

2.4 Discrete versions of inequalities (1.8)

Let A = ( a i , j ) 1 ≤ i ≤ n , 1 ≤ j ≤ m be a real matrix of size n × m , where n , m ≥ 3 . We first introduce the following definition.

Definition 2.12

We say that A is a convex matrix, if

for all i = 1 , … , n , the sequence A ( i , ⋅ ) = ( a i 1 , … , a i m ) ∈ R m is convex;
for all j = 1 , … , m , the sequence A ( ⋅ , j ) = ( a 1 j , … , a n j ) ∈ R n is convex.

The following result provides discrete versions of inequalities (1.8) established in [25].

Theorem 2.13

Let n , m ≥ 3 and A = ( a i , j ) 1 ≤ i ≤ n , 1 ≤ j ≤ m be a real convex matrix. We have

(2.11) n + m 4 ( a N n , N m + a N n , m + 1 − N m + a n + 1 − N n , N m + a n + 1 − N n , m + 1 − N m ) ≤ 1 2 ∑ i = 1 n ( a i , N m + a i , m + 1 − N m ) + ∑ j = 1 m ( a N n , j + a n + 1 − N n , j ) ≤ 1 m + 1 n ∑ i = 1 n ∑ j = 1 m a i j ≤ 1 2 ∑ i = 1 n ( a i 1 + a i m ) + ∑ j = 1 m ( a 1 j + a n j ) ≤ n + m 4 ( a 11 + a 1 m + a n 1 + a n m ) ,

where N m = m + 1 2 and N n = n + 1 2 .

Proof

Let i ∈ { 1 , … , n } . Since A ( i , ⋅ ) ∈ R m is a convex sequence, by the double discrete Hermite-Hadamard inequality (1.4), we have

a i , N m + a i , m + 1 − N m 2 ≤ 1 m ∑ j = 1 m a i j ≤ a i 1 + a i m 2 .

If we sum the above inequality over 1 ≤ i ≤ n , we obtain

(2.12) 1 2 ∑ i = 1 n ( a i , N m + a i , m + 1 − N m ) ≤ 1 m ∑ i = 1 n ∑ j = 1 m a i j ≤ 1 2 ∑ i = 1 n ( a i 1 + a i m ) .

Similarly, let j ∈ { 1 , … , m } . Since A ( ⋅ , j ) ∈ R n is a convex sequence, by the double discrete Hermite-Hadamard inequality (1.4), we have

a N n , j + a n + 1 − N n , j 2 ≤ 1 n ∑ i = 1 n a i j ≤ a 1 j + a n j 2 .

If we sum the above inequality over 1 ≤ j ≤ m , we obtain

(2.13) 1 2 ∑ j = 1 m ( a N n , j + a n + 1 − N n , j ) ≤ 1 n ∑ i = 1 n ∑ j = 1 m a i j ≤ 1 2 ∑ j = 1 m ( a 1 j + a n j ) .

Summing inequalities (2.12) and (2.13), we obtain

(2.14) 1 2 ∑ i = 1 n ( a i , N m + a i , m + 1 − N m ) + ∑ j = 1 m ( a N n , j + a n + 1 − N n , j ) ≤ 1 m + 1 n ∑ i = 1 n ∑ j = 1 m a i j ≤ 1 2 ∑ i = 1 n ( a i 1 + a i m ) + ∑ j = 1 m ( a 1 j + a n j ) .

Using the left-hand side inequality in (1.4), we obtain

a N n , N m + a n + 1 − N n , N m 2 ≤ 1 n ∑ i = 1 n a i , N m

and

a N n , m + 1 − N m + a n + 1 − N n , m + 1 − N m 2 ≤ 1 n ∑ i = 1 n a i , m + 1 − N m .

Summing the above two inequalities, we obtain

(2.15) n 2 ( a N n , N m + a n + 1 − N n , N m + a N n , m + 1 − N m + a n + 1 − N n , m + 1 − N m ) ≤ ∑ i = 1 n ( a i , N m + a i , m + 1 − N m ) .

Similarly, we have

a N n , N m + a N n , m + 1 − N m 2 ≤ 1 m ∑ j = 1 m a N n , j

and

a n + 1 − N n , N m + a n + 1 − N n , m + 1 − N m 2 ≤ 1 m ∑ j = 1 m a n + 1 − N n , j .

Summing the above two inequalities, we obtain

(2.16) m 2 ( a N n , N m + a N n , m + 1 − N m + a n + 1 − N n , N m + a n + 1 − N n , m + 1 − N m ) ≤ ∑ j = 1 m ( a N n , j + a n + 1 − N n , j ) .

Summing (2.15) and (2.16), we obtain

(2.17) n + m 4 ( a N n , N m + a N n , m + 1 − N m + a n + 1 − N n , N m + a n + 1 − N n , m + 1 − N m ) ≤ 1 2 ∑ i = 1 n ( a i , N m + a i , m + 1 − N m ) + ∑ j = 1 m ( a N n , j + a n + 1 − N n , j ) .

Similarly, using the right-hand side inequality in (1.4), we obtain

1 n ∑ i = 1 n a i 1 ≤ a 11 + a n 1 2

and

1 n ∑ i = 1 n a i m ≤ a 1 m + a n m 2 ,

which implies that

(2.18) ∑ i = 1 n ( a i 1 + a i m ) ≤ n 2 ( a 11 + a n 1 + a 1 m + a n m ) .

Proceeding as above, we obtain

(2.19) ∑ j = 1 m ( a 1 j + a n j ) ≤ m 2 ( a 11 + a 1 m + a n 1 + a n m ) .

Summing (2.18) and (2.19), we obtain

(2.20) 1 2 ∑ i = 1 n ( a i 1 + a i m ) + ∑ j = 1 m ( a 1 j + a n j ) ≤ n + m 4 ( a 11 + a 1 m + a n 1 + a n m ) .

Finally, (2.11) follows from (2.14), (2.17), and (2.20).□

We provide below an example to check the validity of Theorem 2.13.

Example 2.14

Let n = m = 3 . In this case, we have N m = N n = 2 . Furthermore, we obtain

n + m 4 ( a N n , N m + a N n , m + 1 − N m + a n + 1 − N n , N m + a n + 1 − N n , m + 1 − N m ) = 6 4 ( a 2 , 2 + a 2 , 2 + a 2 , 2 + a 2 , 2 ) = 6 a 22 ,

1 2 ∑ i = 1 n ( a i , N m + a i , m + 1 − N m ) + ∑ j = 1 m ( a N n , j + a n + 1 − N n , j ) = 1 2 ∑ i = 1 3 ( a i , 2 + a i , 2 ) + ∑ j = 1 3 ( a 2 , j + a 2 , j ) = ∑ i = 1 3 a i , 2 + ∑ j = 1 3 a 2 , j ,

1 m + 1 n ∑ i = 1 n ∑ j = 1 m a i j = 2 3 ∑ i = 1 3 ∑ j = 1 3 a i j ,

1 2 ∑ i = 1 n ( a i 1 + a i m ) + ∑ j = 1 m ( a 1 j + a n j ) = 1 2 ∑ i = 1 3 ( a i 1 + a i 3 ) + ∑ j = 1 3 ( a 1 j + a 3 j ) ,

and

n + m 4 ( a 11 + a 1 m + a n 1 + a n m ) = 3 2 ( a 11 + a 13 + a 31 + a 33 ) .

Hence, (2.11) reduces to

(2.21) 6 a 22 ≤ ∑ i = 1 3 a i , 2 + ∑ j = 1 3 a 2 , j ≤ 2 3 ∑ i = 1 3 ∑ j = 1 3 a i j ≤ 1 2 ∑ i = 1 3 ( a i 1 + a i 3 ) + ∑ j = 1 3 ( a 1 j + a 3 j ) ≤ 3 2 ( a 11 + a 13 + a 31 + a 33 ) .

Consider now the matrix

A = 1 2 4 2 1 2 4 3 6 .

It can be easily seen that A is a convex matrix in the sense of Definition 2.12. On the other hand, we have

6 a 22 = 6 , ∑ i = 1 3 a i , 2 + ∑ j = 1 3 a 2 , j = 11 , 2 3 ∑ i = 1 3 ∑ j = 1 3 a i j = 50 3 , 1 2 ∑ i = 1 3 ( a i 1 + a i 3 ) + ∑ j = 1 3 ( a 1 j + a 3 j ) = 39 2 , 3 2 ( a 11 + a 13 + a 31 + a 33 ) = 45 2 ,

which confirms the validity of (2.21).

3 Conclusion

For a given sequence a = ( a 1 , … , a n ) ∈ R n , some upper bounds of the term E n ≔ a 1 + a n 2 − 1 n ∑ i = 1 n a i are obtained. Namely, we first considered the class of sequences a ∈ R n such that

a ′ = ( ∣ a 2 − a 1 ∣ , … , ∣ a n − a n − 1 ∣ ) ∈ R n − 1

is convex. For this class of sequences, a discrete version of inequality (1.5) [26] is established (Theorem 2.1). We next considered the class of sequences a ∈ R n such that a ′ q = ( ∣ a 2 − a 1 ∣ q , … , ∣ a n − a n − 1 ∣ q ) ∈ R n − 1 is convex for some real number q > 1 . For this class of sequences, a discrete version of inequality (1.6)[26] is proved (Theorem 2.4). We also derived two upper bounds of E n for an arbitrary sequence a ∈ R n . The first one (Theorem 2.7) involves the real number L = max ∣ a i − a j ∣ ∣ i − j ∣ : i , j ∈ { 1 , … , n } , i ≠ j . The obtained inequality is a discrete version of inequality (1.7) [27]. The second upper bound (Theorem 2.8) involves the real number M = max { ∣ a i − a j ∣ : i , j ∈ { 1 , … , n } } . We finally introduced the notion of convex matrices A = ( a i , j ) 1 ≤ i ≤ n , 1 ≤ j ≤ m , which is a discrete version of the notion of convex functions on the coordinates considered in [25]. For this set of matrices, discrete versions of inequalities (1.8) [25] are proved (Theorem 2.13).

Funding information: Bessem Samet was supported by Researchers Supporting Project number (RSP2024R4), King Saud University, Riyadh, Saudi Arabia.
Author contributions: All authors contributed equally to this work and have read and agreed to the published version of the manuscript.
Conflict of interest: The authors state no conflicts of interest. Bessem Samet was a Guest Editor of the Open Mathematics journal and was not involved in the review and decision-making process of this article.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2024-01-06

Revised: 2024-05-06

Accepted: 2024-05-14

Published Online: 2024-06-28

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

Articles in the same Issue

https://doi.org/10.1515/math-2024-0021

Keywords for this article

discrete inequalities; convex sequences; convex matrices; Fejér inequality; Hermite-Hadamard inequality

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