New Jensen's bounds for HA-convex mappings with applications to Shannon entropy

Yamin Sayyari; Saad Ihsan Butt; Mehdi Dehghanian; Praveen Agarwal; Juan J. Nieto

doi:10.1515/dema-2025-0122

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New Jensen's bounds for HA-convex mappings with applications to Shannon entropy

Yamin Sayyari , Saad Ihsan Butt , Mehdi Dehghanian , Praveen Agarwal and Juan J. Nieto

Published/Copyright: July 31, 2025

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From the journal Demonstratio Mathematica Volume 58 Issue 1

Abstract

The aim of this article is to establish some new extensions and variants of Jensen’s discrete and Simic-type inequalities for HA-convex and uniformly HA-convex functions. We introduce uniformly HA-convex functions, which are the generalized class of HA-convex (harmonic-arithmetic)-convex functions and provide some examples too. As an application point of view for the new Jensen’s bounds, we present some new improved bounds for Shannon’s entropy. Finally, some analysis is given by graphical illustration to prove the improvements of bounds.

Keywords: Shannon’s entropy; Jensen’s inequality; HA-convex function; uniformly HA-convex function

MSC 2010: 26A51; 94A17; 26B25; 52A41

1 Introduction and preliminary results

One of the key ideas that students studying physics and chemistry need to grasp with clarity is entropy. More importantly, entropy has several definitions and may be used in a variety of contexts, including the thermodynamic stage, cosmology, and even economics. Entropy is a notion that mostly discusses the spontaneous changes that take place in commonplace phenomena or the universe’s propensity toward chaos. Entropy is sometimes stated as a quantifiable physical quality that is most frequently connected with uncertainty, in addition to being merely a scientific notion. However, the term is also used in a variety of disciplines, including information theory, statistical physics, and classical thermodynamics. In addition to its uses in physics and chemistry, it has also applications in sociology, weather science, and the study of climate change, as well as the quantification of information transfer in telecommunication.

Thus, complex dynamical systems, thermodynamics, coding theory, information sciences, physics, and various other branches of mathematics all actively pursue the study of entropy problems [1–4]. Many scholars have worked hard to compute this criteria in recent years and have significantly developed the idea of entropies. The numerical range of entropy is not always easy to find. So optimized bounds are important to explore. As a result, several academics worked on the idea of convexity to introduce a new bound on entropy with distinct characteristics in the literature [5,6].

In the literature, the well-known Jensen’s discrete inequality [7] states that if f is a convex function on I ≔ [ b 1 , b 2 ] and { ϰ j } j = 1 n ⊆ I , then

∑ i = 1 n p j f ( ϰ j ) − f ∑ j = 1 n p j ϰ j ≥ 0 ,

where

(1.1) p j ≥ 0 , j = 1 , … , n and ∑ j = 1 n p j = 1 .

Also, when convexity is applied, Jensen’s inequality is quite magical to improvise new entropic bounds [8,9]. For the applications of convex functions and entropy, interested readers can refer to [10–12]. Furthermore, for the entropy estimation, see [13–15].

Definition 1.1

[11] Let an increasing function Θ : R ≥ 0 ⟶ [ 0 , + ∞ ) vanish only at 0, a mapping f : I ⟶ R is called uniformly convex with modulus Θ if

f ( λ ϰ 1 + ( 1 − λ ) ϰ 2 ) + λ ( 1 − λ ) Θ ( ∣ ϰ 1 − ϰ 2 ∣ ) ≤ λ f ( ϰ 1 ) + ( 1 − λ ) f ( ϰ 2 ) ,

for every λ ∈ [ 0 , 1 ] and ϰ 1 , ϰ 2 ∈ I .

Proposition 1.2

[16] If p = { p j } j = 1 n , η ≔ min 1 ≤ j ≤ n { p j } and ϑ ≔ max 1 ≤ j ≤ n { p j } , then

(1.2) m ( η , ϑ ) ≔ η log 2 η η + ϑ + ϑ log 2 ϑ η + ϑ ≤ log n − H ( p ) ≤ log ( η + ϑ ) 2 4 η ϑ ≔ M ( η , ϑ ) .

Lemma 1.3

[17] If f is a differentiable convex mapping on I, then we have

(1.3) 0 ≤ ∑ j = 1 n p j f ( ϰ j ) − f ∑ j = 1 n p j ϰ j ≤ 1 4 ( b 2 − b 1 ) ( f ′ ( b 2 ) − f ′ ( b 1 ) ) ≔ D f ( b 1 , b 2 ) .

So (1.3) leads to conclude, 0 ≤ log n − H ( p ) ≤ ( ϑ − η ) 2 4 η ϑ ≔ D ( η , ϑ ) .

Definition 1.4

[18] Let b 1 b 2 > 0 and f : [ b 1 , b 2 ] ⟶ R be a function. Then, f is harmonically convex (HA-convex) if

f ϰ 1 ϰ 2 λ ϰ 1 + ( 1 − λ ) ϰ 2 ≤ ( 1 − λ ) f ( ϰ 1 ) + λ f ( ϰ 2 ) ,

for every λ ∈ [ 0 , 1 ] and ϰ 1 , ϰ 2 ∈ I .

Now, for the rest of this article, we consider I is the interval [ b 1 , b 2 ] not containing 0.

Theorem 1.5

[19] (Jensen-type inequality) If f is a HA-convex function on an interval I containing ϰ 1 , ϰ 2 , … , ϰ n , then the following inequality is true:

f 1 ∑ j = 1 n p j ϰ j ≤ ∑ j = 1 n p j f ( ϰ j ) ,

where p j is as in (1.1).

The motive of this work is to introduce new perspectives of Jensen’s discrete and Simic’s inequalities for HA-convex and uniformly HA-convex mappings and as an application give new improved entropy bounds.

2 Refinement of Jensen’s inequality for HA-convex functions

In this section, we give an extension of Jensen’s inequality for HA-convex functions on an interval not containing zero.

Theorem 2.1

Let f be an HA-convex function on an interval I containing ϰ 1 , ϰ 2 , … , ϰ n . Then, the following inequality is true:

o ≤ max 1 ≤ η < ϑ ≤ n p η f ( ϰ η ) + p ϑ f ( ϰ ϑ ) − ( p η + p ϑ ) f ( p η + p ϑ ) ϰ η ϰ ϑ p η ϰ ϑ + p ϑ ϰ η ≤ ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ 1 n p j ϰ j .

where p j is defined in (1.1).

Proof

Since

( p η + p ϑ ) f ( p η + p ϑ ) ϰ η ϰ ϑ p η ϰ ϑ + p η ϰ ϑ = ( p η + p ϑ ) f 1 p η p η + p ϑ . 1 ϰ η + p η p η + p ϑ . 1 ϰ ϑ ≤ p η f ( ϰ η ) + p ϑ f ( ϰ ϑ ) ,

the first inequality holds. The second inequality is verified as follows. Let s , r ∈ { 1 , … , n } be arbitrary. Theorem 1.5 yields

(2.1) f 1 ∑ j = 1 n p j ϰ j = f 1 ∑ j ≠ r , s p j ϰ j + ( p r + p s ) p r ϰ s + p s ϰ r ( p r + p s ) ϰ r ϰ s ≤ ∑ j ≠ r , s p j f ( ϰ j ) + ( p r + p s ) f ( p r + p s ) ϰ r ϰ s p r ϰ s + p s ϰ r .

Therefore,

∑ 1 n p j f ( ϰ j ) − f 1 ∑ 1 n p j ϰ j ≥ p r f ( ϰ r ) + p s f ( ϰ s ) − ( p r + p s ) f ( p r + p s ) ϰ r ϰ s p r ϰ s + p s ϰ r .

Since ϰ r and ϰ s are arbitrary, the assertion of theorem follows.□

The following corollary follows from the combination of Theorems 1.5 and 2.1.

Corollary 2.2

Let f be a HA-convex function on an interval containing ϰ 1 , ϰ 2 , … , ϰ n . Then, the following inequality is true:

(2.2) 0 ≤ max 1 ≤ r ≤ s ≤ n p r f ( ϰ r ) + p s f ( ϰ s ) − ( p r + p s ) f ( p r + p s ) ϰ r ϰ s p r ϰ s + p s ϰ r ≤ ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ 1 n p j ϰ j .

Lemma 2.3

Let f be a HA-convex function on [ b 1 , b 2 ] and 0 ≤ μ , ν ≤ 1 with μ + ν = 1 . Then,

2 f 2 b 1 b 2 b 1 + b 2 ≤ f 1 ν b 1 + μ b 2 + f 1 μ b 1 + ν b 2 ≤ ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 ≤ f ( b 1 ) + f ( b 2 ) .

Proof

Let μ + ν = 1 . So,

2 f 2 b 1 b 2 b 1 + b 2 = 2 f 1 1 2 b 1 + 1 2 b 2 = 2 f 1 1 2 ν b 1 + μ b 2 + 1 2 μ b 1 + ν b 2 ≤ 2 1 2 f 1 ν b 1 + μ b 2 + 1 2 f 1 μ b 1 + ν b 2 ≤ ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 ≤ f ( b 1 ) + f ( b 2 ) .

Therefore, the result follows.□

Theorem 2.4

Let f be HA-convex on [ b 1 , b 2 ] such that ϰ j ∈ [ b 1 , b 2 ] for all j = 1 , … , n and p j be as defined in (1.1). Then, the following inequality is valid:

∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j ≤ f ( b 1 ) + f ( b 2 ) − 2 f 2 b 1 b 2 b 1 + b 2 .

Proof

Let { t j } ( 0 ≤ t j ≤ 1 ) be the sequence such that 1 ϰ j = t j b 1 + 1 − t j b 2 . Hence,

Γ ≔ ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j = ∑ j = 1 n p j f 1 1 ϰ j − f 1 ∑ j = 1 n p j ϰ j

= ∑ j = 1 n p j f 1 t j b 1 + 1 − t j b 2 − f 1 ∑ j = 1 n p j t j b 1 + 1 − t j b 2 ≤ ∑ j = 1 n p j ( t j f ( b 1 ) + ( 1 − t j ) f ( b 2 ) ) − f 1 ∑ j = 1 n p j t j b 1 + 1 − ∑ j = 1 n p j t j b 2 .

Denote μ ≔ ∑ j = 1 n p j t j and ν ≔ 1 − ∑ j = 1 n p j t j . Consequently,

(2.3) Γ ≤ μ f ( b 1 ) + ν f ( b 2 ) − f 1 μ b 1 + ν b 2 = f ( b 1 ) + f ( b 2 ) − ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 .

Thus, by employing Lemma 2.3, we obtain

(2.4) 0 ≤ f ( b 1 ) + f ( b 2 ) − ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 ≤ f ( b 1 ) + f ( b 2 ) − 2 f 2 b 1 b 2 b 1 + b 2 .

Together, (2.3) and (2.4) imply

□ ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j ≤ f ( b 1 ) + f ( b 2 ) − 2 f 2 b 1 b 2 b 1 + b 2 .

Theorem 2.5

Let f be an HA-convex function on I. If η = min 1 ≤ j ≤ n { ϰ j } and ϑ = max 1 ≤ j ≤ n { ϰ j } , then

1 n f ( η ) + f ( ϑ ) − 2 f 2 η ϑ η + ϑ ≤ 1 n ∑ j = 1 n f ( ϰ j ) − f n ∑ 1 n 1 ϰ j ≤ f ( η ) + f ( ϑ ) − 2 f 2 η ϑ η + ϑ .

Proof

Note that inequality (2.2) is true for every p r , ϰ r , p s , ϰ s . Thus, by substituting p j = 1 n , j = 1 , … , n , ϰ r ≔ η , and ϰ s ≔ ϑ , the first inequality follows from Corollary 2.2. However, the second inequality follows from Theorem 2.4 with b 1 ≔ η , b 2 ≔ ϑ and p j = 1 n , j = 1 , … , n .□

3 Jensen’s inequality via uniformly HA-convexity

We start this section by introducing the following new class of function.

Definition 3.1

[20, Definition 4-(i)] Consider an increasing function Θ : R ≥ 0 ⟶ [ 0 , + ∞ ) that vanishes only at 0. For b 1 b 2 > 0 , a mapping f : [ b 1 , b 2 ] ⟶ R is called uniform harmonic convex function (or uniformly HA-convex) with modulus Θ if

f ϰ 1 ϰ 2 λ ϰ 1 + ( 1 − λ ) ϰ 2 + λ ( 1 − λ ) Θ 1 ϰ 1 − 1 ϰ 2 ≤ ( 1 − λ ) f ( ϰ 1 ) + λ f ( ϰ 2 ) ,

for every λ ∈ [ 0 , 1 ] and ϰ 1 , ϰ 2 ∈ [ b 1 , b 2 ] .

The notation f ∈ U ( Θ ; [ b 1 , b 2 ] ) means that the mapping f : [ b 1 , b 2 ] → R is a uniformly HA-convex mapping.

Theorem 3.2

Let f ∈ U ( Θ ; [ b 1 , b 2 ] ) , { ϰ j } j = 1 n ⊆ [ b 1 , b 2 ] and let S be a permutation on { 1 , … , n } satisfying the following relation:

(3.1) ϰ S ( n ) ≤ … ≤ ϰ S ( 2 ) ≤ ϰ S ( 1 ) .

Then, the inequality

(3.2) f 1 ∑ j = 1 n p j ϰ j ≤ ∑ j = 1 n p j f ( ϰ j ) − ∑ j = 1 n − 1 p S ( j ) p S ( j + 1 ) Θ 1 ϰ S ( j + 1 ) − 1 ϰ S ( j )

holds for every ϰ j ∈ [ b 1 , b 2 ] .

Proof

Without loss of generality, the sequence of { ϰ j } j = 1 n can be taken to be decreasing. Using mathematical induction, we can complete the proof. The base case for n = 2 is straightforward; thus, we proceed induction by assuming that the results hold up to n . So, we have

(3.3) f 1 ∑ j = 1 n + 1 p j ϰ j = f 1 ∑ j = 1 n − 1 p j ϰ j + ( p n + p n + 1 ) p n ϰ n + 1 + p n + 1 ϰ n ( p n + p n + 1 ) ϰ n ϰ n + 1 ≤ ∑ j = 1 n − 1 p j f ( ϰ j ) + ( p n + p n + 1 ) f ( p n + p n + 1 ) ϰ n ϰ n + 1 p n ϰ n + 1 + p n + 1 ϰ n − p 1 p 2 Θ 1 ϰ 2 − 1 ϰ 1 − … − p n − 2 p n − 1 Θ 1 ϰ n − 1 − 1 x n − 2 − p n − 1 ( p n + p n + 1 ) Θ 1 ϰ n − 1 − p n ϰ n + 1 + p n + 1 ϰ n ( p n + p n + 1 ) ϰ n ϰ n + 1 .

Since 1 ϰ n − 1 ≤ 1 ϰ n ≤ 1 ϰ n + 1 ,

p n ϰ n + 1 + p n + 1 ϰ n ( p n + p n + 1 ) ϰ n ϰ n + 1 − 1 ϰ n − 1 = p n p n + p n + 1 . 1 ϰ n + p n + 1 p n + p n + 1 . 1 ϰ n + 1 − 1 ϰ n − 1 ≥ 1 ϰ n − 1 ϰ n − 1 .

Therefore,

(3.4) Θ p n ϰ n + 1 + p n + 1 ϰ n ( p n + p n + 1 ) ϰ n ϰ n + 1 − 1 ϰ n − 1 ≥ Θ 1 ϰ n − 1 ϰ n − 1 .

Also, we have

(3.5) f ( p n + p n + 1 ) ϰ n ϰ n + 1 p n ϰ n + 1 + p n + 1 ϰ n = f 1 p n p n + p n + 1 . 1 ϰ n + p n + 1 p n + p n + 1 . 1 ϰ n + 1 ≤ p n p n + p n + 1 f ( ϰ n ) + p n + 1 p n + p n + 1 f ( ϰ n + 1 ) − p n p n + 1 ( p n + p n + 1 ) 2 . Θ 1 ϰ n + 1 − 1 ϰ n .

Using a combination of (3.3), (3.4), and (3.5), we obtain

f 1 ∑ j = 1 n + 1 p j ϰ j ≤ ∑ j = 1 n − 1 p j f ( ϰ j ) + ( p n + p n + 1 ) f ( p n + p n + 1 ) ϰ n ϰ n + 1 p n ϰ n + 1 + p n + 1 ϰ n − p 1 p 2 Θ 1 ϰ 2 − 1 ϰ 1 − … − p n − 2 p n − 1 Θ 1 ϰ n − 1 − 1 x n − 2 − p n − 1 ( p n + p n + 1 ) Θ 1 ϰ n − 1 − p n ϰ n + 1 + p n + 1 ϰ n ( p n + p n + 1 ) ϰ n ϰ n + 1 ≤ ∑ j = 1 n + 1 p j f ( ϰ j ) − p n p n + 1 p n + p n + 1 . Θ 1 ϰ n + 1 − 1 ϰ n − p 1 p 2 Θ 1 ϰ 2 − 1 ϰ 1 − … − p n − 2 p n − 1 Θ 1 ϰ n − 1 − 1 x n − 2 − p n − 1 ( p n + p n + 1 ) Θ 1 ϰ n − 1 ϰ n − 1 ≤ ∑ j = 1 n + 1 p j f ( ϰ j ) − p n p n + 1 Θ 1 ϰ n + 1 − 1 ϰ n − ∑ j = 1 n − 2 p j p j + 1 Θ 1 ϰ j + 1 − 1 ϰ j − p n − 1 p n Θ 1 ϰ n − 1 ϰ n − 1 = ∑ j = 1 n + 1 p j f ( ϰ j ) − ∑ j = 1 n p j p j + 1 Θ 1 ϰ j + 1 − 1 ϰ j ,

achieving the inequality in formula (3.2).□

Theorem 3.2 yields the following result.

Corollary 3.3

Under the assumptions of Theorem 3.2, the following inequality

f n ∑ j = 1 n 1 ϰ j ≤ 1 n ∑ j = 1 n f ( ϰ j ) − 1 n 2 ∑ j = 1 n − 1 Θ 1 ϰ S ( k + 1 ) − 1 ϰ S ( k )

holds.

Proof

A direct consequence of Theorem 3.2 with p j = 1 n for all j = 1 , … , n .□

4 Refined Simic’s inequality for uniformly HA-convex functions

Now, we give the refinement of Jensen bounds obtained in [16] by employing uniformly HA-convex functions.

Theorem 4.1

Let b 1 > 0 , f ∈ U ( Θ ; [ b 1 , b 2 ] ) , { ϰ j } j = 1 n ⊆ [ b 1 , b 2 ] and S be a permutation satisfying (3.1). Then, for any convex combination ∑ j = 0 n p j ϰ j of points ϰ j ∈ [ b 1 , b 2 ] , the following inequality

(4.1) ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j ≥ p S ( 1 ) f ( ϰ S ( 1 ) ) + p S ( n ) f ( ϰ S ( n ) ) − ( p S ( 1 ) + p S ( n ) ) f ( p S ( 1 ) + p S ( n ) ) ϰ S ( 1 ) ϰ S ( n ) p S ( 1 ) ϰ S ( n ) + p S ( n ) ϰ S ( 1 )

(4.1) + ( p S ( 1 ) + p S ( n ) ) ∑ j = 2 n − 1 p S ( j ) ∑ j = 2 n − 2 p S ( j ) p S ( j + 1 ) Θ 1 ϰ S ( j + 1 ) − 1 ϰ S ( j )

holds.

Proof

Without loss of generality, we assume that ϰ n ≤ … ≤ ϰ 2 ≤ ϰ 1 . Let σ = ∑ j ≠ 2 n − 1 p j . We conclude from Theorem 3.2 that

f 1 ∑ j = 1 n p j ϰ j = f 1 σ ∑ j = 2 n − 1 p j σ ϰ j + ( p 1 + p n ) p 1 ϰ n + p n ϰ 1 ( p 1 + p n ) ϰ 1 ϰ n ≤ σ f σ ∑ j = 2 n − 1 p j ϰ j + ( p 1 + p n ) f ( p 1 + p n ) ϰ 1 ϰ n p 1 ϰ n + p n ϰ 1 − σ ( p 1 + p n ) Θ ∑ j = 2 n − 1 p j σ ϰ j − p 1 ϰ n + p n ϰ 1 ( p 1 + p n ) ϰ 1 ϰ n ≤ ∑ j = 2 n − 1 p j f ( ϰ j ) − ∑ j = 2 n − 2 p j p j + 1 Θ 1 ϰ j + 1 − 1 ϰ j + ( p 1 + p n ) f ( p 1 + p n ) ϰ 1 ϰ n p 1 ϰ n + p n ϰ 1 − σ ( p 1 + p n ) Θ ∑ j = 2 n − 1 p j σ ϰ j − p 1 ϰ n + p n ϰ 1 ( p 1 + p n ) ϰ 1 ϰ n .

Therefore,

∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j ≥ p 1 f ( ϰ 1 ) + p n f ( ϰ 1 ) − ( p 1 + p n ) f ( p 1 + p n ) ϰ 1 ϰ n p 1 ϰ n + p n ϰ 1 + ( p 1 + p n ) ∑ j = 2 n − 1 p j ∑ j = 2 n − 2 p j p j + 1 Θ 1 ϰ j + 1 − 1 ϰ j ,

which completes the proof.□

Lemma 4.2

Let f ∈ U ( Θ ; [ b 1 , b 2 ] ) and 0 ≤ μ , ν ≤ 1 ; μ + ν = 1 . Then,

2 f 2 b 1 b 2 b 1 + b 2 + 1 2 Θ ∣ μ − ν ∣ 1 b 1 − 1 b 2 ≤ f 1 ν b 1 + μ b 2 + f 1 μ b 1 + ν b 2 ≤ ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 − μ ν Θ 1 b 1 − 1 b 2 ≤ f ( b 1 ) + f ( b 2 ) − 2 μ ν Θ 1 b 1 − 1 b 2 .

Proof

Let μ + ν = 1 . So,

2 f 2 b 1 b 2 b 1 + b 2 = 2 f 1 1 2 b 1 + 1 2 b 2 = 2 f 1 1 2 ν b 1 + μ b 2 + 1 2 μ b 1 + ν b 2 ≤ 2 1 2 f 1 ν b 1 + μ b 2 + 1 2 f 1 μ b 1 + ν b 2 − 1 4 Θ ν b + μ a b 1 b 2 − μ b + ν a b 1 b 2 ≤ ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 − 1 2 Θ ∣ μ − ν ∣ 1 b 1 − 1 b 2 − μ ν Θ 1 b 1 − 1 b 2 ≤ f ( b 1 ) + f ( b 2 ) − 1 2 Θ ∣ μ − ν ∣ 1 b 1 − 1 b 2 − 2 μ ν Θ 1 b 1 − 1 b 2 .

Therefore, the result follows.□

Theorem 4.3

Let b 1 b 2 > 0 and f ∈ U ( Θ ; [ b 1 , b 2 ] ) , such that ϰ ¯ ≔ { ϰ j } ⊆ [ b 1 , b 2 ] . Then,

∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j ≤ f ( b 1 ) + f ( b 2 ) − 2 f 2 b 1 b 2 b 1 + b 2 − J Θ , ϰ ¯ ( b 1 , b 2 ) ,

where p j is given in (1.1) and

J Θ , ϰ ¯ ( b 1 , b 2 ) ≔ 1 2 Θ b 1 + b 2 − 2 b 1 b 2 ∑ j = 1 n p j ϰ j b 1 b 2 + 3 b 1 b 2 b 2 − b 1 ( b 1 + b 2 ) ∑ j = 1 n p j ϰ j − b 1 b 2 ∑ j = 1 n p j ϰ j 2 − 1 Θ 1 b 1 − 1 b 2 .

Proof

As b 1 b 2 > 0 , consider a sequence { t j } ( 0 ≤ t j ≤ 1 ) in such way that 1 ϰ j = t j b 1 + 1 − t j b 2 . Hence,

I ≔ ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j = ∑ j = 1 n p j f 1 1 ϰ j − f 1 ∑ j = 1 n p j ϰ j = ∑ j = 1 n p j f 1 t j b 1 + 1 − t j b 2 − f 1 ∑ j = 1 n p j t j b 1 + 1 − t j b 2 = ∑ j = 1 n p j ( t j f ( b 1 ) + ( 1 − t j ) f ( b 2 ) ) − f 1 ∑ j = 1 n p j t j b 1 + 1 − ∑ j = 1 n p j t j b 2 .

Denote μ ≔ ∑ j = 1 n p j t j and ν ≔ 1 − ∑ j = 1 n p j t j . Consequently,

(4.2) I ≤ μ f ( b 1 ) + ν f ( b 2 ) − f 1 μ b 1 + ν b 2 − μ ν Θ 1 b 1 − 1 b 2 = f ( b 1 ) + f ( b 2 ) − ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 − μ ν Θ 1 b 1 − 1 b 2 .

Thus, taking into account Lemma 4.2, we obtain

(4.3) 0 ≤ f ( b 1 ) + f ( b 2 ) − ν f ( b 1 ) + μ f ( b 2 ) + f 1 μ b 1 + ν b 2 ≤ f ( b 1 ) + f ( b 2 ) − 2 f 2 b 1 b 2 b 1 + b 2 − 1 2 Θ ∣ μ − ν ∣ 1 b 1 − 1 b 2 − 2 μ ν Θ 1 b 1 − 1 b 2 .

Together, (4.2) and (4.3) imply

(4.4) ∑ j = 1 n p j f ( ϰ j ) − f 1 ∑ j = 1 n p j ϰ j ≤ f ( b 1 ) + f ( b 2 ) − 2 f 2 b 1 b 2 b 1 + b 2 − 1 2 Θ ∣ μ − ν ∣ 1 b 1 − 1 b 2 − 3 μ ν Θ 1 b 1 − 1 b 2 .

On the other hand, we have

t j = b 1 ( b 2 − ϰ j ) ϰ j ( b 2 − b 1 ) .

Therefore,

(4.5) μ − ν = 1 b 2 − b 1 b 1 + b 2 − 2 b 1 b 2 ∑ j = 1 n p j ϰ j

and

(4.6) μ ν = 1 b 2 − b 1 ( b 1 + b 2 ) b 1 b 2 ∑ j = 1 n p j ϰ j − b 1 2 b 2 2 ∑ j = 1 n p j ϰ j 2 − b 1 b 2 .

Hence,

(4.7) 1 2 Θ ∣ μ − ν ∣ 1 b 1 − 1 b 2 + 3 μ ν Θ 1 b 1 − 1 b 2 = 1 2 Θ b 1 + b 2 − 2 b 1 b 2 ∑ j = 1 n p j ϰ j b 1 b 2 + 3 b 2 − b 1 ( b 1 + b 2 ) b 1 b 2 ∑ j = 1 n p j ϰ j − b 1 2 b 2 2 ∑ j = 1 n p j ϰ j 2 − b 1 b 2 Θ 1 b 1 − 1 b 2 = J Θ , ϰ ¯ ( b 1 , b 2 ) .

The assertion of theorem follows from the combination of (4.4) and (4.7).□

Theorem 4.4

Let b 1 b 2 > 0 and f ∈ U ( Θ ; [ b 1 , b 2 ] ) such that { ϰ j } j = 1 n ⊆ [ b 1 , b 2 ] is a decreasing sequence. Then,

f ( ϰ n ) + f ( ϰ 1 ) n − 2 n f 2 ϰ n ϰ 1 ϰ n + ϰ 1 + 2 ( n − 2 ) n 4 ∑ j = 2 n − 2 Θ 1 x j + 1 − 1 x j ≤ 1 n ∑ j = 1 n f ( ϰ j ) − f n ∑ j = 1 n 1 ϰ j ≤ f ( ϰ n ) + f ( ϰ 1 ) − 2 f 2 ϰ n ϰ 1 ϰ n + ϰ 1 − J Θ , ϰ ¯ ( ϰ n , ϰ 1 ) ,

where

J Θ , ϰ ¯ ( ϰ n , ϰ 1 ) = 1 2 Θ ϰ n + ϰ 1 − 2 ϰ n ϰ 1 n ∑ j = 1 n 1 ϰ j ϰ n ϰ 1 + 3 ϰ n ϰ 1 ϰ 1 − ϰ n ( ϰ n + ϰ 1 ) n ∑ j = 1 n 1 ϰ j − ϰ n ϰ 1 n 2 ∑ j = 1 n 1 ϰ j 2 − 1 Θ 1 ϰ n − 1 ϰ 1 .

Proof

The first inequality follows from Theorem 4.1 with p j = 1 n , j = 1 , … , n , b 1 ≔ ϰ n , and b 2 ≔ ϰ 1 . The second inequality follows from Theorem 4.3 with b 1 = ϰ n , b 2 = ϰ 1 , and p j = 1 n , j = 1 , … , n and

□ J Θ , x ¯ ( ϰ n , ϰ 1 ) = 1 2 Θ ϰ n + ϰ 1 − 2 ϰ n ϰ 1 n ∑ j = 1 n 1 ϰ j ϰ n ϰ 1 + 3 ϰ n ϰ 1 ϰ 1 − ϰ n ( ϰ n + ϰ 1 ) n ∑ j = 1 n 1 ϰ j − ϰ n ϰ 1 n 2 ∑ j = 1 n 1 ϰ j 2 − 1 Θ 1 ϰ n − 1 ϰ 1 .

5 Inequalities in information theory

Information theory is the study of data that manage the capacity, measurement, and correspondence of data. The subject studies all the theoretical problems related to information transformation over the communication channels. Being a theoretical substance, data cannot be measured without any problem. Claude Shannon, who is nowadays regarded as the “Father of Information Theory,” presented a theory in order to quantify the communication of information [21]. The issue of how to transmit information through a specific channel most effectively is addressed by Shannon’s theory. The topic of communication security is also covered. According to Shannon’s formula, when working with systems where individual alternative outcomes are equally likely to occur, we will learn more from Shannon’s information. The potential reduction in knowledge uncertainty is quantified by Shannon’s entropy. Information theory, dynamical systems, statistical physics, molecular ecology, and population genetics all make growing use of Shannon’s entropy and related measurements. Now, we use the obtained results of Theorems 4.1, 4.3, and 4.4 in information theory to improve Shannon’s entropy bounds.

Definition 5.1

Considering a positive distribution p = ( p 1 , … , p n ) , Shannon’s entropy is defined by H ( p ) ≔ ∑ j = 1 n p j log 1 p j .

First, we construct an example of uniformly HA-convex function.

Lemma 5.2

If b 1 > 0 and f : [ b 1 , b 2 ] ⟶ R defined by f ( ϰ ) = log ( ϰ ) , then f is uniformly HA-convex with modulus Θ ( r ) ≔ b 1 2 2 r 2 .

Proof

Since Θ is increasing and vanishes only at 0, consider two fixed points ϰ 1 , ϰ 2 ∈ [ b 1 , b 2 ] and define

g ( λ ) ≔ log ϰ 1 ϰ 2 λ ϰ 1 + ( 1 − λ ) ϰ 2 − ( 1 − λ ) log ϰ 1 − λ log ϰ 2 + b 1 2 2 λ ( 1 − λ ) 1 ϰ 1 − 1 ϰ 2 2 ,

for all λ ∈ [ 0 , 1 ] and all ϰ 1 , ϰ 2 ∈ I . Now, we show that g ( λ ) ≤ 0 , for every λ ∈ [ 0 , 1 ] . Since g ( 0 ) = g ( 1 ) = 0 and d 2 g d λ 2 = ( ϰ 2 − ϰ 1 ) 2 ( λ ϰ 1 + ( 1 − λ ) ϰ 2 ) 2 − b 1 2 ( 1 ϰ 1 − 1 ϰ 2 ) 2 ≥ 0 , g ( λ ) ≤ 0 for every ϰ 1 , ϰ 2 ∈ [ b 1 , b 2 ] and λ ∈ [ 0 , 1 ] . Hence,

□ log ϰ 1 ϰ 2 λ ϰ 1 + ( 1 − λ ) ϰ 2 + b 1 2 λ ( 1 − λ ) 2 1 ϰ 1 − 1 ϰ 2 2 ≤ ( 1 − λ ) log ϰ 1 + λ log ϰ 2 .

Proposition 5.3

Let η ≔ p n ≤ p n − 1 ≤ … ≤ p 1 ≔ ϑ . Then,

m 1 ( η , ϑ ) ≔ η log 2 η η + ϑ + ϑ log 2 ϑ η + ϑ + η 2 ( ϑ + η ) ( 1 − ϑ − η ) 2 ∑ j = 2 n − 2 1 p j p j + 1 ( p j − p j + 1 ) 2 ≤ log n − H ( p ) .

Proof

Employing Theorem 4.1 and Lemma 5.2 for mapping f ( ϰ ) = log ( ϰ ) and modulus Θ ( r ) = p n 4 2 p 1 2 r 2 , ϰ j = p j , j = 1 , … , n and b 1 = p n , and b 2 = p 1 , we obtain

∑ j = 1 n p j log ( p j ) − log 1 n ≥ p 1 log ( p 1 ) + p n log ( p n ) − ( p 1 + p n ) log ( p 1 + p n ) p 1 p n p 1 p n + p n p 1 + ( p 1 + p n ) ( 1 − p 1 − p n ) ∑ j = 2 n − 2 p j p j + 1 2 p n 2 1 p j + 1 − 1 p j 2 .

Therefore,

log n − H ( p ) ≥ p 1 log ( p 1 ) + p n log ( p n ) − ( p 1 + p n ) log p 1 + p n 2 + p n 2 2 ( p 1 + p n ) ( 1 − p 1 − p n ) ∑ j = 2 n − 2 1 p j p j + 1 ( p j − p j + 1 ) 2 ,

which completes the proof.□

Proposition 5.4

Let p 1 ≤ p 2 ≤ … ≤ p n . Then,

(5.1) H ( p ) ≤ log n − p 1 2 2 ∑ j = 1 n − 1 ( p j + 1 − p j ) 2 p j p j + 1 .

Proof

Employing Theorem 3.2 and Lemma 5.2 for mapping f ( ϰ ) = log ( ϰ ) , b 1 ≔ p 1 , b 2 ≔ p n , modulus Θ ( u ) = p 1 2 u 2 2 , and ϰ j = p j , j = 1 , … , n ,

log 1 n ≤ ∑ j = 1 n p j log ( p j ) − p 1 2 2 ∑ j = 1 n − 1 p j p j + 1 1 p j − 1 p j + 1 2 ,

which completes the proof.□

Corollary 5.5

H ( p ) = log n if and only if p j = 1 n for all j = 1 , … , n .

Proof

It is clear that, for p j = 1 n , H ( p ) = log n . Conversely, let H ( p ) = log n . Then, using Proposition 5.4, we obtain

p 1 2 2 ∑ j = 1 n − 1 ( p j + 1 − p j ) 2 p j p j + 1 = 0 .

Therefore, p j + 1 = p j for all j = 1 , … , n − 1 . Since ∑ j = 1 n p j = 1 , p j = 1 n for all j = 1 , … , n .□

Proposition 5.6

Define η ≔ min { p j } and ϑ ≔ max { p j } . Then,

log n − H ( p ) ≤ log ( η + ϑ ) 2 4 η ϑ − 1 4 ϑ 2 ( η + ϑ − 2 n η ϑ ) 2 − 3 η ( ϑ − η ) ϑ ( n ( η + ϑ ) − n 2 η ϑ − 1 ) ≔ M 1 ( η , ϑ ) .

Proof

Taking into account Theorem 4.3 and Lemma 5.2 for f ( ϰ ) = log ( ϰ ) and modulus Θ ( r ) = η 4 2 ϑ 2 r 2 , ϰ j = p j , j = 1 , … , n and b 1 = η and b 2 = ϑ , we obtain

∑ j = 1 n p j log ( p j ) − log 1 n ≤ log ( η ) + log ( ϑ ) − 2 log 2 η ϑ η + ϑ − J Θ , ϰ ¯ ( η , ϑ ) .

Thus,

(5.2) log n − H ( p ) ≤ log ( η + ϑ ) 2 4 η ϑ − J Θ , ϰ ¯ ( η , ϑ ) .

Since

J Θ , ϰ ¯ ( η , ϑ ) = 1 4 ϑ 2 η + ϑ − 2 η ϑ ∑ j = 1 n p j ϰ j 2 + 3 η ( ϑ − η ) ϑ ( η + ϑ ) ∑ j = 1 n p j ϰ j − η ϑ ∑ j = 1 n p j ϰ j 2 − 1 = 1 4 ϑ 2 ( η + ϑ − 2 n η ϑ ) 2 + 3 η ( ϑ − η ) ϑ ( n ( η + ϑ ) − n 2 η ϑ − 1 ) ,

the result follows from (5.2).□

Now, we gave another HA-convex function.

Lemma 5.7

If b 1 > 0 and f : [ b 1 , b 2 ] ⟶ R defined by f ( ϰ ) = − log ( ϰ ) ϰ , then f is uniformly HA-convex with modulus Θ ( r ) ≔ b 1 2 b 2 r 2 .

Proof

We consider two fixed points ϰ 1 , ϰ 2 ∈ [ b 1 , b 2 ] and show that for every λ ∈ [ 0 , 1 ] g ( λ ) ≥ 0 , where

g ( λ ) ≔ λ ϰ 1 + ( 1 − λ ) ϰ 2 ϰ 1 ϰ 2 log ϰ 1 ϰ 2 λ ϰ 1 + ( 1 − λ ) ϰ 2 − b 1 λ ( 1 − λ ) 2 b 2 1 ϰ 1 − 1 ϰ 2 2 − ( 1 − λ ) log ( ϰ 1 ) ϰ 1 − λ log ( ϰ 2 ) ϰ 2 .

Since g ( 0 ) = g ( 1 ) = 0 and

d 2 g d λ 2 = − ( ϰ 1 − ϰ 2 ) 2 ϰ 1 ϰ 2 ( λ ϰ 1 + ( 1 − λ ) ϰ 2 ) 2 + b 1 b 2 1 ϰ 1 − 1 ϰ 2 2 < 0 ,

g ( λ ) ≥ 0 for every ϰ 1 , ϰ 2 ∈ [ b 1 , b 2 ] and λ ∈ [ 0 , 1 ] . Hence,

− λ ϰ 1 + ( 1 − λ ) ϰ 2 ϰ 1 ϰ 2 log ϰ 1 ϰ 2 λ ϰ 1 + ( 1 − λ ) ϰ 2 + b 1 λ ( 1 − λ ) 2 b 2 1 ϰ 1 − 1 ϰ 2 2 ≤ − ( 1 − λ ) log ( ϰ 1 ) ϰ 1 − λ log ( ϰ 2 ) ϰ 2 ,

which completes the proof.□

Proposition 5.8

Let p 1 ≤ p 2 ≤ … ≤ p n . Then,

H ( p ) ≤ log n − p 1 2 2 n ∑ j = 1 n − 1 ( p j + 1 − p j ) 2 .

Proof

Testing 3.3 and Lemma 5.7 for mapping f ( ϰ ) = − log ( ϰ ) ϰ , b 1 ≔ p 1 , b 2 ≔ p n , modulus Θ ( u ) = p 1 2 u 2 2 and ϰ j = 1 p j , j = 1 , … , n ,

− ∑ j = 1 n 1 ϰ j n log n ∑ j = 1 n 1 ϰ j ≤ − 1 n ∑ j = 1 n log ( ϰ j ) ϰ j − p 1 2 2 n 2 ∑ j = 1 n − 1 1 ϰ S ( j + 1 ) − 1 ϰ S ( j ) 2 .

Hence,

− log n n ≤ 1 n ∑ j = 1 n p j log ( p j ) − p 1 2 2 n 2 ∑ j = 1 n − 1 ( p j + 1 − p j ) 2 ,

which completes the proof.□

Proposition 5.9

Let η ≔ p 1 ≤ p 2 ≤ … ≤ p n ≔ ϑ . Then,

m 2 ( η , ϑ ) ≔ η log 2 η η + ϑ + ϑ log 2 ϑ η + ϑ + η ( n − 2 ) ϑ n 3 ∑ j = 2 n − 2 ( p j + 1 − p j ) 2 ≤ log n − H ( p ) ≤ n η log 2 η η + ϑ + n ϑ log 2 ϑ η + ϑ − η 2 n ϑ ( n η + n ϑ − 2 ) 2 − 3 ( n ϑ − 1 ) ( 1 − n η ) ( ϑ − η ) 2 n ϑ 2 ≔ M 2 ( η , ϑ ) .

Proof

Testing Theorem 4.4 and Lemma 5.7 for f ( x ) = − log ( ϰ ) ϰ and modulus Θ ( r ) = p 1 r 2 2 p n , ϰ j = 1 p j , j = 1 , … , n , and η = p 1 = 1 b 2 , ϑ = p n = 1 b 1 , ϰ n ≤ … ≤ ϰ 2 ≤ ϰ 1 . Therefore,

(5.3) 1 n ϑ log ϑ + η log η + ( η + ϑ ) log 2 η + ϑ + η ( n − 2 ) ϑ n 4 ∑ j = 2 n − 2 ( p j + 1 − p j ) 2 ≤ 1 n ( log n − H ( p ) ) ≤ ϑ log ϑ + η log η + ( η + ϑ ) log 2 η + ϑ − J Θ , ϰ ¯ 1 ϑ , 1 η .

Since

J Θ , ϰ ¯ 1 ϑ , 1 η = η 2 ϑ η + ϑ − 2 n 2 + 3 η ( ϑ − η ) 2 ϑ η + ϑ n η ϑ − 1 n 2 η ϑ − 1 = η 2 ϑ η + ϑ − 2 n 2 + 3 ( n ϑ − 1 ) ( 1 − n η ) ( ϑ − η ) 2 n 2 ϑ 2 ,

the result follows from (5.3).□

6 Examples with graphical representation

Now, we present some examples to compare the obtained bounds with the previous ones and confirm that bounds given in propositions of Section 5 are improved as compared to Proposition 1.2.

Example 6.1

Let n = k 3 , η = k − 4 , and ϑ = k − 2 , where k be a positive integer (Figure 1). Then,

n m ( η , ϑ ) = 1 k log 2 1 + k 2 + k log 2 k 2 1 + k 2 ,

and

M 2 ( η , ϑ ) = n m ( η , ϑ ) − ( k − 1 ) 2 2 k 7 − 3 ( k + 1 ) ( k − 1 ) 3 2 k 4 ,

(where k is an arbitrary positive integer and the notations n m ( ⋅ , ⋅ ) and M 2 ( ⋅ , ⋅ ) are in Propositions 1.2 and 5.9, respectively.)

$Figure 1 n m ( η , ϑ ) nm\left(\eta ,{\vartheta }) and M 2 ( η , ϑ ) {M}_{2}\left(\eta ,{\vartheta }) .$

Figure 1

n m ( η , ϑ ) and M 2 ( η , ϑ ) .

Example 6.2

Let n = 1 0 k , η = 1 0 − k − 1 , and ϑ = 1 0 − k + 1 , where k is a positive integer (Figure 2). Then, M ( η , ϑ ) ≃ 1.406 and

M 1 ( η , ϑ ) = M ( η , ϑ ) − 0.02025 − 24.057 1 0 k + 1 ,

where k is an arbitrary positive integer and the notations M ( ⋅ , ⋅ ) and M 1 ( ⋅ , ⋅ ) are in Propositions 1.2 and 5.6, respectively.

$Figure 2 M ( η , ϑ ) M\left(\eta ,{\vartheta }) and M 1 ( η , ϑ ) {M}_{1}\left(\eta ,{\vartheta }) .$

Figure 2

M ( η , ϑ ) and M 1 ( η , ϑ ) .

Example 6.3

Let n = k 3 , η = k − 4 , and ϑ = k − 2 , where k is a positive integer (Figure 3). Then, M ( η , ϑ ) = log ( k 2 + 1 ) 2 4 k 2 and

M 1 ( η , ϑ ) = M ( η , ϑ ) − ( k − 1 ) 2 4 k 4 − 3 ( k + 1 ) ( k − 1 ) 3 k 7 ,

where k is an arbitrary positive integer and the notations M ( ⋅ , ⋅ ) and M 1 ( ⋅ , ⋅ ) are in Propositions 1.2 and 5.6, respectively.

$Figure 3 M ( η , ϑ ) M\left(\eta ,{\vartheta }) and M 1 ( η , ϑ ) {M}_{1}\left(\eta ,{\vartheta }) .$

Figure 3

M ( η , ϑ ) and M 1 ( η , ϑ ) .

Example 6.4

Let k > 2 , n = k + 3 and let p 1 = … = p n − 4 = 1 4 k + 4 ≔ η , p n − 3 = 1 2 k + 6 , p n − 2 = k + 2 2 ( k + 1 ) ( k + 3 ) , p n − 1 = ( k + 2 ) 2 + 1 4 ( k + 1 ) ( k + 3 ) , and p n = k + 2 2 k + 6 ≔ ϑ (Figure 4). Then,

m ( η , ϑ ) = 1 4 k + 4 log 2 k + 6 2 ( k + 1 ) ( k + 2 ) + k + 3 + k + 2 2 k + 6 log 4 ( k + 1 ) ( k + 2 ) 2 ( k + 1 ) ( k + 2 ) + k + 3

and

m 1 ( η , ϑ ) = m ( η , ϑ ) + ( 2 k 2 + 7 k + 7 ) ( 2 k 2 + 9 k + 5 ) 2 9 ( k + 1 ) 4 ( k + 3 ) 3 × ( k − 1 ) 2 2 ( k + 1 ) ( k + 3 ) + 1 ( k + 1 ) ( k + 2 ) + ( k 2 + 2 k − 1 ) 2 2 ( k + 2 ) ( ( k + 2 ) 2 + 1 ) ,

where k is an arbitrary positive integer and the notations m ( ⋅ , ⋅ ) and m 1 ( ⋅ , ⋅ ) are in Propositions 1.2 and 5.3, respectively.

$Figure 4 m ( η , ϑ ) m\left(\eta ,{\vartheta }) and m 1 ( η , ϑ ) {m}_{1}\left(\eta ,{\vartheta }) .$

Figure 4

m ( η , ϑ ) and m 1 ( η , ϑ ) .

Example 6.5

Let k > 2 , n = k + 3 and let p 1 = … = p n − 4 = 1 4 k + 4 ≔ η , p n − 3 = 1 2 k + 6 , p n − 2 = k + 2 2 ( k + 1 ) ( k + 3 ) , p n − 1 = ( k + 2 ) 2 + 1 4 ( k + 1 ) ( k + 3 ) and p n = k + 2 2 k + 6 ≔ ϑ (Figure 5). Then,

m ( η , ϑ ) = 1 4 k + 4 log 2 k + 6 2 ( k + 1 ) ( k + 2 ) + k + 3 + k + 2 2 k + 6 log 4 ( k + 1 ) ( k + 2 ) 2 ( k + 1 ) ( k + 2 ) + k + 3

and

m 2 ( η , ϑ ) = m ( η , ϑ ) + k 2 ( k + 3 ) 2 + 4 + ( k + 2 ) 2 ( k + 3 ) 4 32 ( k + 1 ) 2 ( k + 2 ) 3 ( k + 3 ) 4 ,

where k is an arbitrary positive integer and the notations m ( ⋅ , ⋅ ) and m 2 ( ⋅ , ⋅ ) are in Propositions 1.2 and 5.9, respectively.

$Figure 5 m ( η , ϑ ) m\left(\eta ,{\vartheta }) and m 2 ( η , ϑ ) {m}_{2}\left(\eta ,{\vartheta }) .$

Figure 5

m ( η , ϑ ) and m 2 ( η , ϑ ) .

7 Conclusions

The main finding of our study is to introduce the generalized variant of Jensen’s inequality for newly introduced uniformly HA-convexity. We also first gave Jensen-type inequalities for simple HA-convex mappings. Furthermore, we introduced new Jensen-Simic-type inequality for HA-convex mappings, which is quite new in the literature. Since Jensen’s inequality is the most notable tool to formulate new information inequalities and entropy bounds, we gave applications of our obtained inequalities to formulate new bounds for Shannon’s entropy. Finally, a comparative analysis was given to prove the effectiveness of newly obtained entropic bounds. Our results can be further studied over fractal domains as lot of work is performed on fractal Jensen-type [22] and fractal Jensen-Mercer-type inequalities [23,24] for generalized convex functions on fractal sets.

Acknowledgments

This work has been partially supported by the Agencia Estatal de Investigacion (AEI) of Spain under Grant PID2020-113275GB-I00, cofinanced by the European Community fund FEDER, as well as Xunta de Galicia grant ED431C 2019/02 for Competitive Reference Research Groups (2019–2022).

Funding information: This work has been partially supported by the Agencia Estatal de Investigacion (AEI) of Spain under Grant PID2020-113275GB-I00, cofinanced by the European Community fund FEDER, as well as Xunta de Galicia grant ED431C 2019/02 for Competitive Reference Research Groups (2019–2022).
Author contributions: All authors have contributed equally to this article.
Conflict of interest: The authors state no conflict of interest.
Ethical approval: We would like to mention that this article does not contain any studies with animals and does not involve any studies over human being.
Data availability statement: Not applicable.

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Received: 2023-07-07

Revised: 2024-12-01

Accepted: 2025-03-07

Published Online: 2025-07-31

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

Articles in the same Issue

https://doi.org/10.1515/dema-2025-0122

Keywords for this article

Shannon’s entropy; Jensen’s inequality; HA-convex function; uniformly HA-convex function

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