# Integral Majorization Type Inequalities for the Functions in the Sense of Strong Convexity

Authors: Syed Zaheer Ullah; Muhammad Adil Khan; Zareen Abdulhameed Khan; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this article, we establish several integral majorization type and generalized Favard’s inequalities for the class of strongly convex functions. Our results generalize and improve the previous known results.


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## Body

## 1. Introduction

It is well known that convex functions are a class of important functions in the fields of mathematics and other natural sciences; they have been studied for more than one hundred years. In recent years there is a growing interest in generalized convex functions (such as quasi-convex function [1], strongly convex function [2–4], s-convex function [5], approximately convex function [6], logarithmically convex function [7, 8], midconvex function [9], pseudo-convex function [10], φ-convex function [11], λ-convex function [12], h-convex function [13], delta-convex function [14], Schur convex function [15–21], and other convex functions [22–29]) among the researchers of applied mathematics due to the fact that mathematical models with these functions are more suitable to describe problems of the real world than models using conventional convex functions. Recently, a large number of remarkable results and applications for the generalized convex functions can be found in the literature [30–49].In the article, our focus is on the integral majorization type inequalities for the strongly convex functions.Definition 1.
Letϕ be a real-valued function defined on the interval [λ1,ξ1] and c a positive real number. Then ϕ is said to be strongly convex with modulus c if the inequality(1)ϕηu1+1-ηv1≤ηϕu1+1-ηϕv1-cη1-ηu1-v12holds for all u1,v1∈[λ1,ξ1] and η∈[0,1]. From (1) we clearly see that(2)ϕu1-ϕv1≥ϕ+′v1u1-v1+cu1-v12.The following Lemma2 for strongly convex function is given in [2] (see also [50, Proposition 1.1.2]).Lemma 2.
A real-valued functionϕ:[λ1,ξ1]→R is a strongly convex function with modulus c if and only if the function φ:[λ1,ξ1]→R defined by φ(r)=ϕ(r)-cr2 is a convex function.Every strongly convex function is convex, but the converse is not true in general. Strongly convex functions have been utilized for showing the convergence of a gradient type algorithm for minimizing a function. They play a significant role in mathematical economics, approximation theory, and optimization theory; many applications and properties for strongly functions can be found in [2–4, 13, 30].Next we are going to present some basic theories of majorization.There is a natural description of the indefinite notion that the entries ofn-tuple δ are more nearly equal, or less spread out than, to the entries of n-tuple β. The applicable assertion is that δ majorizes β; it means that the sum of l largest entries of β does not exceed the sum of l largest entries of δ for all l=1,2,…,n-1 with equality for l=n. That is, let δ=(δ1,δ2,…,δn) and β=(β1,β2,…,βn) be two real n-tuples and let(3)δ1↓≥δ2↓≥⋯≥δn↓,β1↓≥β2↓≥⋯≥βn↓be their ordered entries. Then the n-tuple δ is said to majorize β (or β is said to be majorized by δ), in symbol δ≻β, if (4)∑j=1lδj↓≥∑j=1lβj↓holds for l=1,2,…,n-1 and (5)∑j=1nδj=∑j=1nβj.The theory of majorization is a very significant topic in mathematics; a remarkable and complete reference on the majorization subject is the book by Olkin and Marshall [51]. For example, the theory of majorization is an essential tool that permits us to transform nonconvex complicated constrained optimization problems that involve matrix valued variables into simple problems with scalar variables that can be easily solved [52–55].The definition of majorization for integrable functions can be stated as follows (see [7]).Definition 3.
Letf and g be two decreasing real-valued integrable functions on the interval [λ1,ξ1]. Then f is said to majorize g (or g is said to be majorized by f), in symbol, f≻g, if the inequality(6)∫λ1xgrdr≤∫λ1xfrdrholds for all x∈[λ1,ξ1) and (7)∫λ1ξ1grdr=∫λ1ξ1frdr.Theorem 4 (See [56]).
Letf and g be two continuous and increasing real-valued functions defined on [λ1,ξ1], and let Ω:[λ1,ξ1]→R be a bounded variation function. Then the following statements are true. (a)
If(8)∫λ1xfrdΩr≤∫λ1xgrdΩrfor allx∈[λ1,ξ1] and(9)∫λ1ξ1frdΩr=∫λ1ξ1grdΩr,then(10)∫λ1ξ1ΨfrdΩr≤∫λ1ξ1ΨgrdΩrholds for every continuous convex functionΨ.(b)
If (8) and (9) hold, then (10) holds for every continuous increasing convex function Ψ.Theorem 5 (See [57]).
LetΨ:[0,∞)→R be a convex function, f, g and Ω be three positive and integrable functions defined on [λ1,ξ1] such that (11)∫λ1xfrΩrdr≤∫λ1xgrΩrdrfor all x∈[λ1,ξ1] and (12)∫λ1ξ1frΩrdr=∫λ1ξ1grΩrdr.Then the following statements are true: (a)
Iff is decreasing on [λ1,ξ1], then(13)∫λ1ξ1ΨfrΩrdr≤∫λ1ξ1ΨgrΩrdr.(b)
Ifg is increasing on [λ1,ξ1], then(14)∫λ1ξ1ΨgrΩrdr≤∫λ1ξ1ΨfrΩrdr.Letp>1, f be a positive and continuous concave function defined on [λ1,ξ1], and let Ψ be a convex function defined on [0,2f¯1] with (15)f¯1=1ξ1-λ1∫λ1ξ1frdr.Then Favard [58] proved that the inequalities (16)∫01Ψ2sf¯1=12f¯1∫02f¯1Ψydy≥1ξ1-λ1∫λ1ξ1Ψfrdrand (17)1ξ1-λ1∫λ1ξ1fprdr≤2pp+11ξ1-λ1∫λ1ξ1frdrphold.The main purpose of the article is to establish several integral majorization type and generalized Favard’s inequalities for strongly convex functions.


## 2. Main Results

Theorem 6.
Letc>0, Ψ:[0,∞)→R be a continuous strongly convex function with modulus c, and let f, g, and Ω be three positive and integrable functions defined on [λ1,ξ1] such that(18)∫λ1xfrΩrdr≤∫λ1xgrΩrdrfor all x∈[λ1,ξ1] and(19)∫λ1ξ1frΩrdr=∫λ1ξ1grΩrdr.Then the following statements are true. (a)
Iff is decreasing on [λ1,ξ1], then we have (20)∫λ1ξ1ΨgrΩrdr≥∫λ1ξ1ΨfrΩrdr+c∫λ1ξ1fr-gr2Ωrdr.(b)
Ifg is increasing on [λ1,ξ1], then one has(21)∫λ1ξ1ΨfrΩrdr≥∫λ1ξ1ΨgrΩrdr+c∫λ1ξ1fr-gr2Ωrdr.Proof.
(a) Letv1=f and u1=g. Then it follows from (2) and the proof of Lemma 2 given in [57] that (22)ΨfrΩr+Ψ+′frgr-frΩr+cgr-fr2Ωr≤ΨgrΩr.Let F(x)=∫λ1xf(r)-g(r)Ω(r)dr. Then (18) and (19) lead to F(x)≤0 for all x∈[λ1,ξ1], F(λ1)=F(ξ1)=0, and (23)∫λ1ξ1Ψfr-ΨgrΩrdr+c∫λ1ξ1fr-gr2Ωrdr≤∫λ1ξ1Ψ+′frfr-grΩrdr=∫λ1ξ1Ψ+′frdFr=Ψ+′frFrλ1ξ1-∫λ1ξ1FrdΨ+′fr=-∫λ1ξ1FrdΨ+′fr≤0.Since f is decreasing on [λ1,ξ1], therefore inequality (20) can be deduced easily from the above inequality. Similarly, we can prove part (b) for increasing function g defined on [λ1,ξ1].Theorem 7.
Suppose that all the assumptions of Theorem6 hold. Then the following statements are true. (a)
Iff is decreasing on [λ1,ξ1], then(24)∫λ1ξ1ΨgrΩrdr≥∫λ1ξ1ΨfrΩrdr+c∫λ1ξ1g2r-f2rΩrdr.(b)
Ifg is increasing on [λ1,ξ1], then(25)∫λ1ξ1ΨgrΩrdr≤∫λ1ξ1ΨfrΩrdr+c∫λ1ξ1g2r-f2rΩrdr.Proof.
SinceΨ is a strongly convex function with modulus c, therefore Ψ(r)-cr2 is a convex function, and inequalities (24) and (25) follow easily from the convexity of the function Ψ(r)-cr2 and Lemma 2 given in [57].Remark 8.
Inequalities (13) and (14) can be obtained by (24) and (25) immediately.Remark 9.
Generally, the assumptions of both the functionsf and g are monotonic in majorization theorem, but in Theorems 6 and 7 we only need one of the functions f and g to be monotonic.Theorem 10.
Suppose thatΨ:[λ1,ξ1]→R is a continuous strongly convex function with modulus c, and f, g, and Ω are three integrable functions on [λ1,ξ1]. If g and f-g are nondecreasing (nonincreasing) functions on [λ1,ξ1] and ∫λ1ξ1f(r)Ω(r)dr=∫λ1ξ1g(r)Ω(r)dr, then (26)∫λ1ξ1ΨfrΩrdr≥∫λ1ξ1ΨgrΩrdr+c∫λ1ξ1fr-gr2Ωrdr.Proof.
SinceΨ is a strongly convex function, therefore using (2) for u1=f and v1=g, we have(27)∫λ1ξ1Ψfr-ΨgrΩrdr-c∫λ1ξ1fr-gr2Ωrdr≥∫λ1ξ1Ψ+′grfr-grΩrdr.It follows from the Čebyšev inequality [59] that(28)1∫λ1ξ1Ωrdr∫λ1ξ1Ψ+′grfr-grΩrdr≥1∫λ1ξ1Ωrdr∫λ1ξ1Ψ+′grΩrdr1∫λ1ξ1Ωrdr∫λ1ξ1fr-grΩrdr≥0.Therefore, inequality (26) follows from (27) and (28).Making use of the similar idea as in the proof of Theorem10, we can obtain the following Theorem 11 immediately.Theorem 11.
Suppose thatΨ:[λ1,ξ1]→R is a continuous strongly convex function with modulus c, and f, g, and Ω are three integrable functions on [λ1,ξ1]. If g and f-g are nondecreasing (nonincreasing) functions on [λ1,ξ1], and ∫λ1ξ1f(r)Ω(r)dr≥∫λ1ξ1g(r)Ω(r)dr, then inequality (26) holds.Theorem 12.
The inequality(29)∫λ1ξ1ΨfrΩrdr≥∫λ1ξ1ΨgrΩrdr+c∫λ1ξ1f2r-g2rΩrdrholds if all the assumptions of Theorem 10 are satisfied.Proof.
SinceΨ is a strongly convex function with modulus c, therefore Ψ(r)-cr2 is a convex function and inequality (29) can be deduced by applying this convex function in Theorem 6 of [59].Using strongly convex function we can give an extension of [60, Theorem 2] in the following form.Theorem 13.
Letφ,Ψ:[0,∞)→R be two functions such that φ is a strictly increasing and Ψ∘φ-1 is strongly convex with modulus c, f, g, and Ω being three positive and integrable functions on [λ1,ξ1] such that (30)∫λ1xφfrΩrdr≤∫λ1xφgrΩrdrfor all x∈[λ1,ξ1] and (31)∫λ1ξ1φfrΩrdr=∫λ1ξ1φgrΩrdr.Then the following statements are true. (a)
Iff is decreasing on [λ1,ξ1], then inequality (20) holds.(b)
Ifg is increasing on [λ1,ξ1], then inequality (21) holds.Proof.
We clearly see that it is sufficient to prove the case ofφ(r)=r, but this case is already proved in Theorem 6.Similarly, we have Theorem14 as follows.Theorem 14.
Suppose that all the assumptions of Theorem13 are satisfied. Then the following statements are true. (a)
Iff is decreasing on [λ1,ξ1], then inequality (24) holds.(b)
Ifg is increasing on [λ1,ξ1], then inequality (25) holds.The following Lemma15 was given in [57].Lemma 15.
Letχ be a positive integrable function and let ħ be an increasing function on (λ1,ξ1); then(32)∫λ1xħrχrdr∫λ1ξ1χrdr≤∫λ1ξ1ħrχrdr∫λ1xχrdr.If ħ is a decreasing function on (λ1,ξ1), then inequality (32) holds in reverse directions.Lemma 16.
Letf be a real-valued function defined on [λ1,ξ1]. Then the following statements are true. (i)
Iff is a strongly concave function with modulus c, then
(a)
the functionħ1(r)=f(r)/r-λ1-cr is decreasing on (λ1,ξ1];(b)
the functionħ2(r)=f(r)/ξ1-r+cr is increasing on [λ1,ξ1).(ii)
Iff is a strongly convex function with modulus c, then
(c)
the functionħ1(r)=f(r)/r-λ1-cr is increasing on (λ1,ξ1] if f(λ1)=0;(d)
the functionħ2(r)=f(r)/ξ1-r+cr is decreasing on [λ1,ξ1) if f(ξ1)=0.Proof.
(i) Suppose thatf is a strongly concave function with modulus c.
(a) To show that the functionħ1(r)=f(r)/r-λ1-cr is decreasing on (λ1,ξ1], in fact, for λ1<r1≤r2≤ξ1 we have (33)fr1=fr1-λ1r2-λ1r2+r2-λ1-r1-λ1r2-λ1λ1≥r1-λ1r2-λ1fr2+1-r1-λ1r2-λ1fλ1-cr1-λ1r2-λ11-r1-λ1r2-λ1r2-λ12≥r1-λ1r2-λ1fr2-cr1-λ1r2-r1,which shows that the function ħ1(r)=f(r)/r-λ1-cr is decreasing on (λ1,ξ1].
(b) To show that the functionħ2(r)=f(r)/ξ1-r+cr is increasing on [λ1,ξ1), in fact, for λ1≤r1≤r2<ξ1 we have (34)fr2=fξ1-r2ξ1-r1r1+ξ1-r1-ξ1-r2ξ1-r1ξ1≥ξ1-r2ξ1-r1fr1+1-ξ1-r2ξ1-r1fξ1-c1ξ1-r2ξ1-r11-ξ1-r2ξ1-r1r1-ξ12≥ξ1-r2ξ1-r1fr1-c1ξ1-r2r2-r1,which shows that the function ħ2(r)=f(r)/ξ1-r+cr is increasing on [λ1,ξ1).
(ii) Suppose thatf is a strongly convex function with modulus c.
(c) Sincef(λ1)=0, by similar method of (a) we can easily prove that the function ħ1(r)=f(r)/r-λ1-cr is increasing on (λ1,ξ1].
(d) Sincef(ξ1)=0, by similar method of (b) we can prove that the function ħ2(r)=f(r)/ξ1-r+cr is decreasing on (λ1,ξ1].Next, we establish several Favard type inequalities for strongly convex functions.Theorem 17.
( a ) Let f be a strongly concave function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)-c1r(r-λ1) is a positive increasing function, let Ψ be a strongly convex function with modulus c2 on [0,2f¯1], z¯1=λ1(1-s)+ξ1s, and(35)f¯1=ξ1-λ1∫λ1ξ1grΩrdr2∫λ1ξ1r-λ1Ωrdr.Then(36)1ξ1-λ1∫λ1ξ1ΨgrΩrdr≤∫01Ψ2sf¯1Ωz¯1ds-c2∫01fz¯1-c1z¯1sξ1-sλ1-2sf¯12Ωz¯1ds.
Iff is a strongly convex function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)-c1r(r-λ1) is a positive increasing function and f(λ1)=0, then the reverse inequality in (36) holds.
( b ) Let f be a strongly concave function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)+c1r(ξ1-r) is a positive decreasing function, let Ψ be a strongly convex function with modulus c2 on [0,2f¯2], z¯2=λ1s+ξ1(1-s), and(37)f¯2=ξ1-λ1∫λ1ξ1grΩrdr2∫λ1ξ1ξ1-rΩrdr.Then(38)1ξ1-λ1∫λ1ξ1ΨgrΩrdr≤∫01Ψ2sf¯2Ωz¯2ds-c2∫01fz¯2+c1z¯2sξ1-sλ1-2sf¯22Ωz¯2ds.
Iff is a strongly convex function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)+c1r(ξ1-r) is a positive decreasing function and f(ξ1)=0, then the reverse inequality in (38) holds.Proof.
(a) From Lemma16(a) we know that the function ħ1(r)=f(r)/r-λ1-c1r is decreasing; then using Lemma 15 to the functions χ(r)=(r-λ1)Ω(r) and ħ1(r), we obtain(39)∫λ1ξ1grΩrdr∫λ1xr-λ1Ωrdr≤∫λ1xgrΩrdr∫λ1ξ1r-λ1Ωrdr.It follows from (35) that inequality (39) can be rewritten as(40)∫λ1xr-λ1ξ1-λ12f¯1Ωrdr≤∫λ1xgrΩrdrfor all x∈[λ1,ξ1].
Asg is an increasing function, and by use of Theorem 6(b), we have(41)∫λ1ξ1ΨgrΩrdr≤∫λ1ξ1Ψr-λ1ξ1-λ12f¯1Ωrdr-c2∫λ1ξ1gr-r-λ1ξ1-λ12f¯12Ωrdr.Note that (42)1ξ1-λ1∫λ1ξ1Ψr-λ1ξ1-λ12f¯1Ωrdr-c2ξ1-λ1∫λ1ξ1gr-r-λ1ξ1-λ12f¯12Ωrdr=12f¯1∫02f¯1ΨyΩλ1+yξ1-λ12f¯1dy-c22f¯1∫02f¯1fλ1+yξ1-λ12f¯1-c1λ1+yξ1-λ12f¯1yξ1-λ12f¯1-y2×Ωλ1+yξ1-λ12f¯1dy=∫01Ψ2sf¯1Ωλ11-s+ξ1sds-c2∫01fλ11-s+ξ1s-c1λ11-s+ξ1ssξ1-sλ1-2sf¯12×Ωλ11-s+ξ1sds.
Therefore, we get(43)1ξ1-λ1∫λ1ξ1ΨgrΩrdr≤∫01Ψ2sf¯1Ωz¯1ds-c2∫01fz¯1-c1z¯1sξ1-sλ1-2sf¯12Ωz¯1ds.
Iff is a strongly convex function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)-c1r(r-λ1) is a positive increasing function and f(λ1)=0, then the reverse inequality in (36) can be proved by using a similar method as in the proof of part (a) and Lemma 16(c).
(b) From Lemma16(b) we know that the function ħ2(r)=f(r)/ξ1-r+c1r is increasing; then using Lemma 15 to the functions χ(r)=(ξ1-r)Ω(r) and ħ2(r), we obtain(44)∫λ1xgrΩrdr∫λ1ξ1ξ1-rΩrdr≤∫λ1ξ1grΩrdr∫λ1xξ1-rΩrdr.From (37) we clearly see that inequality (44) can be rewritten as (45)∫λ1xgrΩrdr≤∫λ1xξ1-rξ1-λ12f¯2Ωrdrfor all x∈[λ1,ξ1].
Asg is decreasing function, and by using Theorem 6(a) we have (46)∫λ1ξ1ΨgrΩrdr≤∫λ1ξ1Ψξ1-rξ1-λ12f¯2Ωrdr-c2∫λ1ξ1gr-ξ1-rξ1-λ12f¯22Ωrdr.Note that (47)1ξ1-λ1∫λ1ξ1Ψξ1-rξ1-λ12f¯2Ωrdr-c2ξ1-λ1∫λ1ξ1gr-ξ1-rξ1-λ12f¯22Ωrdr=12f¯2∫02f¯2ΨyΩξ1-yξ1-λ12f¯2dy-c22f¯2∫02f¯2fξ1-yξ1-λ12f¯2+c1ξ1-yξ1-λ12f¯2yξ1-λ12f¯2-y2×Ωξ1-yξ1-λ12f¯2dy=∫01Ψ2sf¯2Ωλ1s+ξ11-sds-c2∫01fλ1s+ξ11-s+c1λ1s+ξ11-ssξ1-sλ1-2sf¯22×Ωλ1s+ξ11-sds.Therefore, (48)1ξ1-λ1∫λ1ξ1ΨgrΩrdr≤∫01Ψ2sf¯2Ωz¯2ds-c2∫01fz¯2+c1z¯2sξ1-sλ1-2sf¯22Ωz¯2ds.
Iff is a strongly convex function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)+c1r(ξ1-r) is a positive decreasing function and f(ξ1)=0, then the reverse inequality in (38) can be proved by using a similar method as in the proof of part (b) and Lemma 16(d).Theorem 18.
The following statements are true under the assumptions of Theorem17. (a)
Ifg is a positive increasing function, then(49)1ξ1-λ1∫λ1ξ1ΨgrΩrdr≤∫01Ψ2sf¯1Ωz¯1ds+c2∫01fz¯1-c1z¯1sξ1-sλ12-2sf¯12Ωz¯1ds.(b)
Ifg is a positive decreasing function, then(50)1ξ1-λ1∫λ1ξ1ΨgrΩrdr≤∫01Ψ2sf¯2Ωz¯2ds+c2∫01fz¯2+c1z¯2sξ1-sλ12-2sf¯22Ωz¯2ds.Proof.
(a) We clearly see thatΨ(r)-c2r2 is convex function due to Ψ being a strongly convex function with modulus c2. Therefore, inequality (49) follows easily from [57, Theorem 1(i)] and the strong convexity of Ψ(r)-c2r2 together with the fact that f is a strongly concave function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)-c1r(r-λ1) is positive increasing function.
(b) Similarly, inequality (50) follows easily from [57, Theorem 1(ii)] and the strong convexity of Ψ(r)-c2r2 together with the fact that f is a strongly concave function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)+c1r(ξ1-r) is positive decreasing function.Theorem 19.
Letg(r)=f(r)-c1r(r-λ1) be an increasing function on (0,1), let g/h be a decreasing function on (0,1), let g, h, and Ω be three positive functions on (0,1), and let gΩ and hΩ be integrable on (0,1) such that(51)ϕ=∫01grΩrdr∫01hrΩrdr≥0.And let Ψ be a strongly convex function with modulus c2. Then the inequality holds(52)∫01ΨkϕhrΩrdr≥∫01ΨkgrΩrdr+c2k2∫01gr-ϕhr2Ωrdrfor all k>0.Proof.
Fromh>0 and (51), applying Lemma 15 to the function χ(r)=h(r)Ω(r) and the decreasing function ħ(r)=g(r)/h(r) we get (53)∫0xkϕhrΩrdr≤∫0xkgrΩrdr.Since g is increasing, therefore by using Theorem 6 we have (54)∫01ΨkϕhrΩrdr≥∫01ΨkgrΩrdr+c2k2∫01gr-ϕhr2Ωrdr.Theorem 20.
LetΨ be a strongly convex function with modulus c2. Then the inequality(55)∫01ΨkϕhrΩrdr≥∫01ΨkgrΩrdr+c2k2∫01ϕhr2-gr2Ωrdr.holds for all k>0 if all the assumptions of Theorem 19 are satisfied.Proof.
We clearly see thatΨ(r)-c2r2 is a convex function due to Ψ being a strongly convex function with modulus c2. Therefore, inequality (55) follows easily from [60, Theorem 3] and the convexity of the function Ψ(r)-c2r2.Remark 21.
Clearly, [60, Theorem 3] can be deduced from (52) due to (56)∫01gr-ϕhr2Ωrdr≥0or from (55) due to (57)∫01ϕhr2-gr2Ωrdr≥0for convex function Ψ(r)=r2.The following Theorem22 is an extension of Theorem 19.Theorem 22.
LetΨ:[0,∞)→R be a continuous strongly convex function with modulus c2, g(r)=f(r)-c1r(r-λ1) (c1 is a nonnegative real number), h and Ω two positive integrable functions on [λ1,ξ1], z1(r)=g(r)/∫λ1ξ1g(r)Ω(r)dr, and z2(r)=h(r)/∫λ1ξ1h(r)Ω(r)dr. Then the following statements are true. (a)
Ifg is increasing on [λ1,ξ1] and g/h is decreasing on [λ1,ξ1], then(58)∫λ1ξ1Ψz2rΩrdr≥∫λ1ξ1Ψz1rΩrdr+c2∫λ1ξ1z1r-z2r2Ωrdr.(b)
Ifh is increasing on [λ1,ξ1] and g/h is increasing on [λ1,ξ1], then(59)∫λ1ξ1Ψz1rΩrdr≥∫λ1ξ1Ψz2rΩrdr+c2∫λ1ξ1z1r-z2r2Ωrdr.Proof.
(a) Leth>0; then applying Lemma 15 to the function χ(r)=h(r)Ω(r) and the decreasing function ħ(r)=g(r)/h(r), we have (60)∫λ1xz2rΩrdr≤∫λ1xz1rΩrdr.Therefore, inequality (58) follows from Theorem 6 and the fact that g is an increasing function on [λ1,ξ1].
(b) Leth>0; then applying Lemma 15 to the function χ(r)=h(r)Ω(r) and the increasing function ħ(r)=g(r)/h(r), we get (61)∫λ1xz1rΩrdr≤∫λ1xz2rΩrdr.Therefore, inequality (59) follows from Theorem 6 and the fact that h is an increasing function on [λ1,ξ1].Theorem 23.
The following statements are true under the assumptions of Theorem22. (a)
Ifg is an increasing function on [λ1,ξ1] and g/h is a decreasing function on [λ1,ξ1], then(62)∫λ1ξ1Ψz1rΩrdr≤∫λ1ξ1Ψz2rΩrdr+c2∫λ1ξ1z12r-z22rΩrdr.(b)
Ifh is increasing on [λ1,ξ1] and g/h is an increasing function on [λ1,ξ1], then(63)∫λ1ξ1Ψz2rΩrdr≤∫λ1ξ1Ψz1rΩrdr+c2∫λ1ξ1z22r-z12rΩrdr.Proof.
We clearly see thatΨ(r)-c2r2 is a convex function due to Ψ being a strongly convex function with modulus c2. Therefore, inequalities (62) and (63) follow from [61, Theorem 2.3] and the convexity of the function Ψ(r)-c2r2.Remark 24.
Clearly, Theorem 2.3(1) and Theorem 2.3(2) given in [57] can be deduced by (62) and (63), respectively.Remark 25.
Theorem22 is an extension of Favard’s inequality given in Theorem 17. Indeed, let Ψ(r) be a strongly convex function with modulus c2, then Ψ(kr) is also a strongly convex function with modulus k2c2 for any k∈R. Substituting h(r)=r-λ1 in (58), one has(64)∫λ1ξ1Ψfr-c1rr-λ1Ωrdr≤∫λ1ξ1Ψ∫λ1ξ1fr-c1rr-λ1Ωrdr∫λ1ξ1r-λ1Ωrdrr-λ1Ωrdr-k2c2∫λ1ξ1fr-c1rr-λ1-∫λ1ξ1fr-c1rr-λ1Ωrdr∫λ1ξ1r-λ1Ωrdrr-λ12Ωrdr.Since f is a strongly concave function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)-c1r(r-λ1) is a positive increasing function, taking k=1 and using (35) and z¯1 in (64), we obtain the Favard’s inequalities given in Theorem 17.Remark 26.
From (64) we can easily obtain Remark 2.4 given in [61] due to (65)∫λ1ξ1fr-c1rr-λ1-∫λ1ξ1fr-c1rr-λ1Ωrdr∫λ1ξ1r-λ1Ωrdrr-λ12Ωrdr≥0.For an application of Theorem22, we get Corollary 27 as follows.Corollary 27.
Letr>1, p∈(-∞,0)∪(1,∞), Ψ(r)=rp, and Ω, g, h, z1, and z2 be stated as in Theorem 22. Then the following statements are true. (a)
Ifg is increasing on [λ1,ξ1] and g/h is decreasing on [λ1,ξ1], then(66)∫λ1ξ1grΩrdr∫λ1ξ1hrΩrdrp≥∫λ1ξ1gprΩrdr∫λ1ξ1hprΩrdr+c2∫λ1ξ1grΩrdrp∫λ1ξ1hprΩrdr∫λ1ξ1z1r-z2r2Ωrdr.(b)
Ifh is increasing on [λ1,ξ1] and g/h is an increasing function on [λ1,ξ1], then(67)∫λ1ξ1gprΩrdr∫λ1ξ1hprΩrdr≥∫λ1ξ1grΩrdr∫λ1ξ1hrΩrdrp+c2∫λ1ξ1grΩrdrp∫λ1ξ1hprΩrdr∫λ1ξ1z1r-z2r2Ωrdr.Proof.
(a) Sinceg is increasing on [λ1,ξ1] and g/h is a decreasing function on [λ1,ξ1], using (58) given in Theorem 22 and substituting Ψ(r)=rp, we get (68)∫λ1ξ1z2prΩrdr≥∫λ1ξ1z1prΩrdr+c2∫λ1ξ1z1r-z2r2Ωrdr,∫λ1ξ1hr∫λ1ξ1hrΩrdrpΩrdr≥∫λ1ξ1gr∫λ1ξ1grΩrdrpΩrdr+c2∫λ1ξ1z1r-z2r2Ωrdr.Therefore, inequality (66) follows from (68).
(b) Sinceh is increasing on [λ1,ξ1] and g/h is an increasing function on [λ1,ξ1], using (59) given in Theorem 22 and substituting Ψ(r)=rp we get inequality (67).Remark 28.
Leth(r)=r-λ1, Ω(r)=1, and f be a strongly concave function with modulus c1 on [λ1,ξ1] such that g(r)=f(r)-c1r(r-λ1) is a positive increasing function. Then inequality (66) leads to the classical Favard’s inequality for strongly convex functions with modulus c2:(69)2pp+11ξ1-λ1∫λ1ξ1grdrp≥1ξ1-λ1∫λ1ξ1gprdr+c2ξ1-λ1∫λ1ξ1grdrp∫λ1ξ1gr∫λ1ξ1grdr-2r-λ1ξ1-λ122dr.Remark 29.
From (69) we get the classical Favard’s inequality given in [58] due to (70)∫λ1ξ1grdrp∫λ1ξ1gr∫λ1ξ1grdr-2r-λ1ξ1-λ122dr≥0.



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