# Some New Dynamic Inequalities Involving Monotonic Functions on Time Scales

Authors: S. H. Saker; E. Awwad; A. Saied
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we prove some new dynamic inequalities involvingC- monotonic functions on time scales. The main results will be proved by employing Hölder’s inequality, integration by parts, and a chain rule on time scales. As a special case when T=R, our results contain the continuous inequalities proved by Heinig, Maligranda, Pečarić, Perić, and Persson and when T=N, the results to the best of the authors’ knowledge are essentially new.


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## Body

## 1. Introduction

In 1995 Heinig and Maligranda [1] proved that if -∞≤a<b≤∞, f, g≥0, f is decreasing on (a,b) and g is increasing on (a,b) with g(a)=0, then for any γ∈(0,1](1)∫abfxdgx≤∫abfγxdgγx1/γ.Inequality (1) is reversed when 1≤γ<∞. Also in [1] they proved that if f is increasing on (a,b) and g is decreasing on (a,b) with g(b)=0, then, for any γ∈(0,1],(2)∫abfxd-gx≤∫abfγxd-gγx1/γ.In [1] the authors generalized (1) and proved that if 0<p≤q<∞, and u, v are positive functions, then there exists a constant D>0 such that the inequality,(3)∫0∞uxfqxdx1/q≤D∫0∞vxfpxdx1/p,holds for all nonnegative decreasing function f if and only if (4)∫0tuxdx1/q≤D∫0tvxdx1/p,∀t>0.In [1] it is also proved that inequality (3) holds for all nonnegative increasing functions f and 0<p≤q<∞ if and only if (5)∫t∞uxdx1/q≤D∫t∞vxdx1/p,∀t>0.In 1997 Pečarić et al. [2] generalized (1) and proved that if f is C-decreasing with C≥1 and g is an increasing function on [a,b] such that g(a)=0, and φ:[0,∞)→R is a concave, nonnegative, and differentiable function such that φ(0)=0, then(6)φC∫abfxdgx≤C∫abφ′fxgxfxdgx.The function f is said to be C-decreasing if s≤t implies that f(t)≤Cf(s) for C>1. Pečarić et al. [2] also proved that if f is C-increasing with C≥1 and g is increasing on [a,b], such that g(a)=0, then(7)φ1C∫abfxdgx≥1C∫abφ′fxgxfxdgx.The function f is said to be C-increasing if s≤t implies f(s)≤Cf(t). Furthermore they also considered the case when f is C- increasing with C≥1 and g is decreasing on [a,b], such that g(b)=0, and proved that(8)φC∫abfxd-gx≤C∫abφ′fxgxfxd-gx.Finally they proved that if f is C-decreasing with C≥1 and g is decreasing on [a,b], such that g(b)=0, then(9)φ1C∫abfxd-gx≥1C∫abφ′fxgxfxd-gx.In the last decades some authors have been interested in finding some discrete results on lp(N) analogues to Lp(R)- bounds in different fields in analysis and as a result this subject becomes a topic of ongoing research. One reason for this upsurge of interest in discrete case is also due to the fact that discrete operators may even behave differently from their continuous counterparts. In this paper, we obtain the discrete inequalities as special cases of the results with a general domain called the time scale T, which is an arbitrary nonempty closed subset of the real numbers R. These new results on the time scale T contain the classical continuous and discrete inequalities as special cases when T=R and T=N and can be extended to different inequalities on different time scales such as T=hN, h>0, T=qN for q>1.In recent years the study of dynamic inequalities on time scales has received a lot of attention and has become a major field in pure and applied mathematics. For more details about the dynamic inequalities on time scales, we refer the reader to the books [3–5] and the papers [6–11].The natural question now is the following: Is it possible to prove some new inequalities withC-monotonic rd-continuous functions defined on a time scale T and as special cases contain the above results?Our aim in this paper is to give the answer to this question and find the relation between the weighted functions which ensure that the inequality(10)∫a∞wxfqxΔx1/q≤D∫a∞hxfpxΔx1/p,holds for all nonnegative decreasing function f such that limx→∞f(x)=0, D>0, and 0<p≤q<∞. Also, we establish some new dynamic inequalities involving C-monotonic functions in the form (11)φC∫abfxgΔxΔx≤C∫abfxgΔxφ′fxgxΔx,where φ:[0,∞)→R is a concave, nonnegative, and differentiable function such that φ(0)=0, and f is C- decreasing with C≥1, and g is an increasing function on [a,b]T, such that g(a)=0.The paper is organized as follows. In Section2, we present some preliminaries concerning the theory of time scales and prove the basic lemmas that will be needed in the proofs. In Section 3, we prove the main results by using Hölder’s inequality, integration by parts, and a chain rule on time scales. Our results when T=R give inequalities (1), (2), (3), (6), (7), (8), and (9) proved by Heinig, Maligranda, Pečarić, Perić, and Persson. When T=N, our results are essentially new.


## 2. Preliminaries and Basic Lemmas

In this section, we recall the following concepts related to the notion of time scales. For more details of time scale analysis we refer the reader to the two books by Bohner and Peterson [5, 12] which summarize and organize much of the time scale calculus. A time scale T is an arbitrary nonempty closed subset of the real numbers R. The forward jump operator and the backward jump operator are defined by σ(t)≔inf⁡s∈T:s>t, and ρ(t)≔sup⁡s∈T:s<t, where sup⁡∅=inf⁡T. A point, t∈T, is said to be left–dense if ρ(t)=t and t>inf⁡T, is right-dense if σ(t)=t, is left–scattered if ρ(t)<t, and is right–scattered if σ(t)>t. A function f:T→R is said to be right-dense continuous (rd-continuous) provided f is continuous at right dense points and, at left-dense points in T, left hand limits exist and are finite. The set of all such rd-continuous functions is denoted by Crd(T)=Crd(T,R). The product and quotient rules for the derivative of the product fg and the quotient f/g (where ggσ≠0, here gσ=g∘σ) of two differentiable functions f and g are given by(12)fgΔ=fgΔ+fΔgσ=fΔg+fσgΔ,fgΔ=fΔg-fgΔggσ.Let f:R→R be continuously differentiable and suppose that g:T→R is delta differentiable. Then f∘g:T→R is delta differentiable and there exists d in the real interval [t,σ(t)] for t∈T, such that(13)f∘gΔt=f′gdgΔt.Another shape of the chain rule is the formula(14)f∘gΔt=∫01f′gt+hμtgΔtdhgΔt.A special case of (14) is(15)uλtΔ=λ∫01huσ+1-huλ-1uΔtdh.We define the time scale interval a,bT by a,bT≔[a,b]∩T. In this paper, we will use Cauchy (delta) integral which we can define as follows. If GΔ(t)=g(t), then the Cauchy (delta) integral of g is defined by ∫atg(x)Δx≔G(t)-G(a). It can be shown (see [5]) that if g∈Crd(T), then the Cauchy integral G(t)≔∫t0tg(x)Δx exists, t0∈T, and satisfies GΔ(t)=g(t),t∈T. In case T=R, we have (16)σt=ρt=t,μt=0,fΔ=f′,∫abftΔt=∫abftdt,The integration on discrete time scales is defined by ∫abg(t)Δt=∑t∈[a,b)μ(t)g(t) and then, in case T=Z, we have (17)σt=t+1,ρt=t-1,μt=1,fΔ=Δf,∫abftΔt=∑t=ab-1ft.The integration by parts formula on time scales is given by (18)∫abuΔtvσtΔt=utvtab-∫abutvΔtΔt.Also, we have, for f∈Crd and t∈T, that (19)∫tσtfτΔτ=μtft.The Hölder inequality on time scales is given by(20)∫abftgtΔt≤∫abfγtΔt1/γ∫abgνtΔt1/ν,where >1, 1/γ+1/ν=1, and f, g∈Crd[a,b]T,R+. Inequality (20) is reversed if 0<γ<1 or γ<0.Definition 1.
A setK⊂R is convex if, for all x, y∈K and λ∈[0,1], we have (21)λx+1-λy∈K.A function F:K⊂R→R is concave if K is convex and, for all x, y∈K and λ∈[0,1], we have (22)Fλx+1-λy≥λFx+1-λFy.A function F:K⊂R→R is convex if -F is concave.Definition 2.
Assume thatT is a time scale, f:T→R and C≥1. If s≤t implies f(t)≤Cf(s), then f is C- decreasing. If s≤t implies f(s)≤Cf(t), then f is C-increasing. As a special case when C=1 we get the classical definitions.Now, we prove the basic lemmas that will be used to prove our main results. Throughout the paper, we assume that the functions (without mentioning) are rd-continuous nonnegative andΔ-differentiable functions, locally Δ-integrable on a,∞T and the integrals considered are assumed to exist and finite.Lemma 3.
Assume thatT is a time scale with a∈T and , β∈Crd[a,∞)T,R+. If γ≥1, then(23)∫a∞αx∫aσxβτΔτγΔx1/γ≤∫a∞βx∫x∞αsΔs1/γΔx.Proof.
LetF(x)=∫axβ(τ)Δτ. Then, the left hand side of (23) can be written in the form(24)∫a∞αx∫aσxβτΔτγΔx=∫a∞αxFσxγ-1FσxΔx.Integrating the right hand side of (24) by parts with (25)uΔx=αxFσxγ-1,νσx=Fσx,we have that(26)∫a∞αxFσxγ-1FσxΔx=uxFxa∞-∫a∞uxβxΔx,where u(x)=-∫x∞α(s)Fσ(s)γ-1Δs. Using the facts that limx→∞⁡u(x)=0 and F(a)=0, we see that(27)∫a∞αxFσxγ-1FσxΔx=∫a∞βx∫x∞αsFσsγ-1ΔsΔx.Substituting (27) into (24), we get that(28)∫a∞αx∫aσxβτΔτγΔx=∫a∞βx∫x∞αsFσsγ-1ΔsΔx=∫a∞βx∫x∞α1/γsαγ-1/γsFσsγ-1ΔsΔx.Applying the Hölder inequality with indices γ>1, and γ/(γ-1) on the term (29)∫x∞α1/γsαγ-1/γsFσsγ-1Δswe see that(30)∫x∞α1/γsαγ-1/γsFσsγ-1Δs≤∫x∞αsΔs1/γ∫x∞αsFσsγΔsγ-1/γ.Substituting (30) into (28), where β is a positive function, we obtain(31)∫a∞αx∫aσxβτΔτγΔx≤∫a∞βx∫x∞αsΔs1/γ∫x∞αsFσsγΔsγ-1/γΔx.Since x≥a and α, β are positive functions, we see that (32)∫x∞αsFσsγΔs≤∫a∞αsFσsγΔs,and then we have for γ>1 (note that (γ-1)/γ>0) that(33)∫x∞αsFσsγΔsγ-1/γ≤∫a∞αsFσsγΔsγ-1/γ.From (31) and (33), we have (34)∫a∞αx∫aσxβτΔτγΔx≤∫a∞αsFσsγΔsγ-1/γ∫a∞βx∫x∞αsΔs1/γΔx=∫a∞αs∫aσsβτΔτγΔsγ-1/γ∫a∞βx∫x∞αsΔs1/γΔx,and then(35)∫a∞αx∫aσxβτΔτγΔx1/γ≤∫a∞βx∫x∞αsΔs1/γΔx,which is the desired inequality (23). The proof is complete.As in the proof of Lemma3, we can easily prove the following dual lemma.Lemma 4.
Assume thatT is a time scale with a∈T and , β∈Crd[a,∞)T,R+. If γ≥1, then(36)∫a∞αx∫x∞βτΔτγΔx1/γ≤∫a∞βx∫aσxαsΔs1/γΔx.


## 3. Main Results

In this section, we state and prove our main results and for simplicity, we will assume thatw, h are positive rd-continuous functions on [a,∞)T and φ:[0,∞)→[0,∞) is a concave and differentiable function such that φ(0)=0. We begin with the time scale version of (3).Theorem 5.
Assume thatT is a time scale with a∈T, 0<p≤q<∞. Furthermore assume that f is nonnegative and decreasing function such that lims→∞⁡f(s)=0 and (37)∫a∞hxfpxΔx<∞.If there exists a constant D>0 such that(38)∫aσxwτΔτ1/q≤D∫aσxhτΔτ1/p,∀x∈a,∞T,then(39)∫a∞wxfqxΔx1/q≤D∫a∞hxfpxΔx1/p.Proof.
Integrating the term(40)∫a∞wxfqxΔx,by parts formula with u(x)=fq(x) and vΔ(x)=w(x), we have that (41)∫a∞wxfqxΔx=fqxvxa∞-∫a∞fqxΔvσxΔx,where v(x)=∫axw(τ)Δτ. Using the facts that v(a)=0 and limx→∞⁡f(x)=0, we obtain (42)∫a∞wxfqxΔx=∫a∞-fqxΔvσxΔx=∫a∞-fqxΔ∫aσxwτΔτΔx,and then(43)∫a∞wxfqxΔxp/q=∫a∞-fqxΔ∫aσxwτΔτΔxp/q.Substituting (38) into (43), we see (note f is decreasing) that(44)∫a∞wxfqxΔxp/q≤Dp∫a∞-fqxΔ∫aσxhτΔτq/pΔxp/q.Applying (23) with γ=q/p, α(x)=-fq(x)Δ and β(τ)=h(τ) on the right side of (44), we see that (45)∫a∞-fqxΔ∫aσxhτΔτq/pΔxp/q≤∫a∞hx∫x∞-fqsΔΔsp/qΔx.Using the assumption that limx→∞⁡f(x)=0, we get(46)∫a∞-fqxΔ∫aσxhτΔτq/pΔxp/q≤∫a∞hxfpxΔx.Substituting (46) into (44), we have (47)∫a∞wxfqxΔxp/q≤Dp∫a∞hxfpxΔx,and then(48)∫a∞wxfqxΔx1/q≤D∫a∞hxfpxΔx1/p,which is the desired inequality (39). The proof is complete.Remark 6.
Suppose that the inequality(49)∫a∞wxfqxΔx1/q≤D∫a∞hxfpxΔx1/p,holds for all nonnegative decreasing functions f. Then it holds when (50)fx=1,ifx∈a,σtT,0,ifx∉a,σtT,for any fixed number t∈[a,∞)T and becomes (51)∫aσtwxΔx1/q≤D∫aσthxΔx1/p.This proves the necessary condition of Theorem 5.From Theorem5 and Remark 6, we have the following corollary.Corollary 7.
Assume thatT is a time scale with a∈T, 0<p≤q<∞. If f is nonnegative and decreasing function such that lims→∞⁡f(s)=0, and (52)∫a∞hxfpxΔx<∞,then there exists a constant D>0 such that the inequality (53)∫a∞wxfqxΔx1/q≤D∫a∞hxfpxΔx1/p,holds if and only if (54)∫aσxwτΔτ1/q≤D∫aσxhτΔτ1/p,∀x∈a,∞T.Remark 8.
As a special case of Corollary7 when T=R, we get the integral inequality (3) proved by Heinig and Maligranda [1].Remark 9.
As a special case of Corollary7 when T=N and a=1, we see that the inequality (55)∑n=1∞wnfnq1/q≤D∑n=1∞hnfnp1/p,holds if and only if (56)∑k=1nwk1/q≤D∑k=1nhk1/p,∀n∈N,and 0<p≤q<∞, for nonnegative and decreasing sequences f when limn→∞⁡f(n)=0.Theorem 10.
Assume thatT is a time scale with a∈T, 0<p≤q<∞. Furthermore assume that f is a nonnegative bounded and increasing function such that f(a)=0 and (57)∫a∞hxfσxpΔx<∞.If there exists a constant D>0 such that(58)∫x∞wτΔτ1/q≤D∫x∞hτΔτ1/p,∀x∈a,∞T,then(59)∫a∞wxfqxΔx1/q≤D∫a∞hxfσxpΔx1/p.Proof.
Applying the integration by parts on the term∫a∞w(x)fq(x)Δx with u(x)=fq(x) and vΔ(x)=w(x), we have (60)∫a∞wxfqxΔx=fqxvxa∞-∫a∞fqxΔvσxΔx,where v(x)=-∫x∞w(τ)Δτ. Using the facts that limx→∞⁡v(x)=0, f is bounded, and f(a)=0, we get (61)∫a∞wxfqxΔx=∫a∞-fqxΔvσxΔx=∫a∞fqxΔ∫σx∞wτΔτΔx.Since f is increasing, we obtain(62)∫a∞wxfqxΔx≤∫a∞fqxΔ∫x∞wτΔτΔx.From (58) and (62), we see (note that f is increasing) that (63)∫a∞wxfqxΔx≤Dq∫a∞fqxΔ∫x∞hτΔτq/pΔx,and then(64)∫a∞wxfqxΔxp/q≤Dp∫a∞fqxΔ∫x∞hτΔτq/pΔxp/q.Applying (36) with γ=q/p, α(x)=fq(x)Δ and β(τ)=h(τ) on the term (65)∫a∞fqxΔ∫x∞hτΔτq/pΔxp/q,we have(66)∫a∞fqxΔ∫x∞hτΔτq/pΔxp/q≤∫a∞hx∫aσxfqsΔΔsp/qΔx.Substituting (66) into (64), we see (note that f(a)=0) that (67)∫a∞wxfqxΔxp/q≤Dp∫a∞hx∫aσxfqsΔΔsp/qΔx=Dp∫a∞hxfσxpΔx,and then (68)∫a∞wxfqxΔx1/q≤D∫a∞hxfσxpΔx1/p,which is the desired inequality (59). The proof is complete.Theorem 11.
Assume thatT is a time scale with , b∈T. If f is C-decreasing on [a,b]T for C≥1 and g is increasing on [a,b]T, such that g(a)=0, then(69)φC∫abfxgΔxΔx≤C∫abfxgΔxφ′fxgxΔx.Proof.
Denote(70)Ft≔φC∫atfxgΔxΔx-C∫atfxgΔxφ′fxgxΔx,and(71)Gt≔C∫atfxgΔxΔx.Therefore, we have from (70) and (71) that(72)Ft≔φGt-C∫atfxgΔxφ′fxgxΔx.Since f is C- decreasing, then we have, for t≥x, that f(t)≤Cf(x), and then we obtain (note g is increasing and g(a)=0) that(73)∫atCfxgΔxΔx≥∫atftgΔxΔx=ft∫atgΔxΔx=gt-gaft=ftgt.Substituting (71) into (73), we have (74)Gt≥ftgt.Applying the chain rule formula (13) on the term φG(t), we see that there exists ζ∈[t,σ(t)], such that(75)φΔGt=φ′GζGΔt.From (71), we obtain (note g is increasing) that(76)GΔt=CftgΔt≥0,and then G(t) is increasing on [a,b]T and then we have, for ζ≥t, that(77)Gζ≥Gt.Since φ is concave on [0,∞), then φ′′<0 (φ′ is decreasing on [0,∞)) and, then, we observe from (77) that(78)φ′Gζ≤φ′Gt.Substituting (76) and (78) into (75), we get(79)φΔGt≤CftgΔtφ′Gt.From (74), we have that φ′G(t)≤φ′f(t)g(t), and then we get (note f is positive and g is increasing) that (80)CftgΔtφ′Gt≤CftgΔtφ′ftgt,and thus we obtain from (79) that(81)φΔGt≤CftgΔtφ′ftgt.From (72), we have(82)FΔt=φΔGt-CftgΔtφ′ftgt.Substituting (81) into (82), we see that FΔ(t)≤0, and therefore F is decreasing on [a,b]T. Since b>a, we see that F(b)≤F(a). Since φ(0)=0, we have from (70) that (83)Fa=φ0=0,and then F(b)≤0, and then we have from (70), by sitting t=b, that (84)φC∫abfxgΔxΔx≤C∫abfxgΔxφ′fxgxΔx,which is the desired inequality (69). The proof is complete.Remark 12.
As a special case of Theorem11 when T=R, we get the integral inequality (6) proved by Pečarić et al. [2].Remark 13.
As a special case of Theorem11 when T=R, C=1, φ(t)=tp, and 0<p≤1, we obtain the integral inequality (1) proved by Heinig and Maligranda [1].Remark 14.
As a special case of Theorem11 when T=N and a=1, we see that the discrete inequality (85)φC∑k=1NfkΔgk≤C∑k=1NfkΔgkφ′fkgk,holds for the C-decreasing sequence f and the increasing sequence g with g(a)=0.Theorem 15.
Assume thatT is a time scale with , b∈T. If f is C-increasing on [a,b]T, C≥1 and g is increasing on [a,b]T, such that g(a)=0, then(86)φ1C∫abfxgΔxΔx≥1C∫abfxgΔxφ′fσxgσxΔx.Proof.
Denote(87)Ft≔φ1C∫atfxgΔxΔx-1C∫atfxgΔxφ′fσxgσxΔx,and(88)Gt≔1C∫atfxgΔxΔx.Therefore, we have from (87) and (88) that(89)Ft=φGt-1C∫atfxgΔxφ′fσxgσxΔx.Since f is C- increasing, then we have, for x≤σ(t), that f(x)≤Cfσ(t), and then we get (note g is increasing and g(a)=0) that (90)∫aσtfxgΔxΔx≤∫aσtCfσtgΔxΔx=Cfσt∫aσtgΔxΔx=Cfσtgσt-ga=Cfσtgσt,and thus(91)1C∫aσtfxgΔxΔx≤fσtgσt.From (88), inequality (91) becomes(92)Gσt≤fσtgσt.Applying the chain rule formula (13) on the term φG(t), we see that there exists ζ∈[t,σ(t)], such that(93)φΔGt=φ′GζGΔt.From (88), we obtain (note g is increasing) that(94)GΔt=1CftgΔt≥0;then G(t) is increasing on [a,b]T, and then we have, for ζ≤σ(t), that(95)Gζ≤Gσt.Since φ is concave on [0,∞), then φ′′<0 (φ′ is decreasing on [0,∞)) and then we observe from (95) that(96)φ′Gζ≥φ′Gσt.Substituting (94) and (96) into (93), we get(97)φΔGt≥1CftgΔtφ′Gσt.From (92), we have that φ′Gσ(t)≥φ′fσ(t)gσ(t), and then we get (note f is positive and g is increasing) that (98)1CftgΔtφ′Gσt≥1CftgΔtφ′fσtgσt;thus we obtain from (97) that(99)φΔGt≥1CftgΔtφ′fσtgσt.From (89), we have(100)FΔt=φΔGt-1CftgΔtφ′fσtgσt.Substituting (99) into (100), we see that FΔ(t)≥0, and therefore F is increasing on [a,b]T. Since b>a, we see that F(b)≥F(a). Since φ(0)=0, we have from (87) that (101)Fa=φ0=0;then F(b)≥0, and then we have, from (87) by sitting t=b, that (102)φ1C∫abfxgΔxΔx≥1C∫abfxgΔxφ′fσxgσxΔx,which is the desired inequality (86). The proof is complete.Remark 16.
As a special case of Theorem15 when T=R and σ(x)=x, we get the integral inequality (7) proved by Pečarić et al. [2].Remark 17.
As a special case of Theorem15 when T=N, σ(n)=n+1, and a=1, we see that the discrete inequality (103)φ1C∑k=1NfkΔgk≥1C∑k=1NfkΔgkφ′fk+1gk+1,holds for the C-increasing sequence f and the increasing sequence g with g(a)=0.Theorem 18.
Assume thatT is a time scale with a, b∈T. If f is C-increasing on [a,b]T , C≥1, and g is decreasing on [a,b]T, such that g(b)=0, then(104)φC∫abfx-gxΔΔx≤C∫abfx-gxΔφ′fσxgσxΔx.Proof.
Denote(105)Ft≔-φC∫tbfx-gxΔΔx-C∫tbfxgΔxφ′fσxgσxΔx,and(106)G∗t≔C∫tbfx-gxΔΔx.Therefore, we have from (105) and (106) that(107)Ft=-φG∗t-C∫tbfxgΔxφ′fσxgσxΔx.Since f is C- increasing, then we have, for σ(t)≤x, that fσ(t)≤Cf(x), and then we obtain (note g is decreasing and g(b)=0) that(108)C∫σtbfx-gxΔΔx≥∫σtbfσt-gxΔΔx=fσt∫σtb-gxΔΔx=fσtgσt-gb=fσtgσt.Substituting (106) into (108), we observe that(109)G∗σt≥fσtgσt.By applying the chain rule formula (13) on the term φG∗(t), we see that there exists ζ∈[t,σ(t)], such that(110)φΔG∗t=φ′G∗ζG∗tΔ.From (106), we obtain (note g is decreasing) that(111)G∗tΔ=CftgΔt≤0;then G∗(t) is decreasing on [a,b]T, and then we have for ζ≤σ(t) that(112)G∗ζ≥G∗σt.Since φ is concave on [0,∞), then φ′′<0 (φ′ is decreasing on [0,∞)) and then we observe from (112) that(113)φ′G∗ζ≤φ′G∗σt.Substituting (111) and (113) into (110), we get(114)φΔG∗t≥CftgΔtφ′G∗σt.From (109), we have that φ′G∗(σ(t))≤φ′fσ(t)gσ(t), and then we get (note f is positive and g is decreasing) that (115)CftgΔtφ′G∗σt≥CftgΔtφ′fσtgσt,and thus we obtain from (114) that (116)φΔG∗t≥CftgΔtφ′fσtgσt,and then(117)-φΔG∗t≤-CftgΔtφ′fσtgσt.From (107), we have(118)FΔt=-φΔG∗t+CftgΔtφ′fσtgσt.Substituting (117) into (118), we see that FΔ(t)≤0, and therefore F is decreasing on [a,b]T. Since b>a, we see that F(b)≤F(a). Since φ(0)=0, we have from (105) that (119)Fb=-φ0=0;then F(a)≥0, and then we have, from (105) by sitting t=a, that (120)φC∫abfx-gxΔΔx≤C∫abfx-gxΔφ′fσxgσxΔx,which is the desired inequality (104). The proof is complete.Remark 19.
As a special case of Theorem18 when T=R and σ(x)=x, we get the integral inequality (8) proved by Pečarić et al. [2].Remark 20.
As a special case of Theorem18 when T=R, σ(n)=n,C=1, φ(t)=tp, and 0<p≤1, we obtain the integral inequality (2) proved by Heinig and Maligranda [1].Remark 21.
As a special case of Theorem18 when T=N, σ(n)=n+1, and a=1, we see that the discrete inequality (121)φC∑k=1NfkΔ-gk≤C∑k=1NfkΔ-gkφ′fk+1gk+1,holds for the C-increasing sequence f and the decreasing sequence g with g(b)=0.Theorem 22.
Assume thatT is a time scale with a, b∈T. If f is C-decreasing on [a,b]T , C≥1, and g is decreasing on [a,b]T, such that g(b)=0, then(122)φ1C∫abfx-gxΔΔx≥1C∫abfx-gxΔφ′fxgxΔx.Proof.
Denote(123)Ft≔-φ1C∫tbfx-gxΔΔx+1C∫tbfx-gxΔφ′fxgxΔx,and(124)G∗t≔1C∫tbfx-gxΔΔx.Therefore, we have from (123) and (124) that(125)Ft=-φG∗t+1C∫tbfx-gxΔφ′fxgxΔx.Since f is C-decreasing, then we have, for x≥t, that f(x)≤Cf(t), and then we obtain (note g is decreasing and g(b)=0) that (126)∫tbfx-gxΔΔx≤∫tbCft-gxΔΔx=Cft∫tb-gxΔΔx=Cftgt-gb=Cftgt,and then(127)1C∫tbfx-gxΔΔx≤ftgt.Substituting (124) into (127), we get(128)G∗t≤ftgt.By applying the chain rule formula (13) on the term φG∗(t), we see that there exists d∈[t,σ(t)], such that(129)φΔG∗t=φ′G∗dG∗tΔ.From (124), we obtain (note g is decreasing) that(130)G∗tΔ=1CftgΔt≤0;then G∗(t) is decreasing on [a,b]T, and then we have, for d≥t, that(131)G∗d≤G∗t.Since φ is concave on [0,∞), then φ′′<0 (φ′ is decreasing on [0,∞)) and then, we observe from (131) that(132)φ′G∗d≥φ′(G∗t.Substituting (130) and (132) into (129), we get(133)φΔG∗t≤1CftgΔtφ′(G∗t.From (128), we have that φ′G∗(t)≥φ′f(t)g(t), and then we get (note f is positive and g is decreasing) that (134)1CftgΔtφ′G∗t≤1CftgΔtφ′ftgt;thus we obtain from (133) that (135)φΔG∗t≤1CftgΔtφ′ftgt,and then(136)-φΔG∗t≥-1CftgΔtφ′ftgt.From (125), we have(137)FΔt=-φΔG∗t+1CftgΔtφ′ftgt.Substituting (136) into (137), we see that FΔ(t)≥0, and therefore F is increasing on [a,b]T. Since b>a, we see that F(b)≥F(a). Since φ(0)=0, we have from (123) that(138)Fb=-φ0=0;then F(a)≤0, and then we have, from (123) by sitting t=a, that (139)-φ1C∫abfx-gxΔΔx+1C∫abfx-gxΔφ′fxgxΔx≤0,and thus(140)φ1C∫abfx-gxΔΔx≥1C∫abfx-gxΔφ′fxgxΔx,which is the desired inequality (122). The proof is complete.Remark 23.
As a special case of Theorem22 when T=R, we get the integral inequality (9) proved by Pečarić et al. [2].Remark 24.
As a special case of Theorem22 when T=N and a=1, we see that the discrete inequality (141)φ1C∑k=1NfkΔ-gk≥1C∑k=1NfkΔ-gkφ′fkgk,holds for the C-decreasing sequence f and the decreasing sequence g with g(b)=0.



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*Source: 7584836-2019-05-08.xml*