# Application of Functionals in Creating Inequalities

Authors: Zlatko Pavić; Shanhe Wu; Vedran Novoselac
Journal: Journal of Function Spaces
 (2016)

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## Abstract

The paper deals with the fundamental inequalities for convex functions in the bounded closed interval. The main inequality includes convex functions and positive linear functionals extending and refining the functional form of Jensen’s inequality. This inequality implies the Jensen, Fejér, and, thus, Hermite-Hadamard inequality, as well as their refinements.


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## Body

## 1. Introduction

In our research, we apply the theory of positive linear functionals to convex analysis. Let us remember the initial notions related to positive linear functionals on the space of real functions.LetS be a nonempty set, and let F=F(S) be a subspace of the linear space of all real functions on domain S. We assume that space F contains unit function u defined by u(s)=1 for every s∈S. Such space F contains every real constant κ within the meaning of κ=κu and every composite function f(g) of a function g∈F and an affine function f:R→R. Actually, if f(x)=κ1x+κ2, then the composition(1)fg=κ1g+κ2ubelongs to F.LetL=L(F) be the space of all linear functionals on space F. Functional L∈L is said to be unital (normalized) if L(u)=1. Such functional has property L(κu)=κ for every real constant κ. If g∈F is a function and if L∈L is a unital functional, then affine function f:R→R satisfies equality(2)fLg=Lfg.Functional L∈L is said to be positive (nonnegative) if inequality L(g)≥0 holds for every nonnegative function g∈F. If a pair of functions g1,g2∈F satisfies inequality g1(s)≤g2(s) for every s∈S, then it follows that(3)Lg1≤Lg2.If g∈F is a function with the image in interval [a,b] (i.e., au≤g≤bu), then every positive unital functional L∈L meets inclusion L(g)∈[a,b] (i.e., a≤L(g)≤b). The same is true for each closed interval I⊆R.Introducing a continuous convex function, we can expose the functional form of Jensen’s inequality.Theorem A.
Letg∈F be a function with the image in closed interval I⊆R, and let L∈L be a positive unital functional.
Then each continuous convex functionf:I→R such that f(g)∈F satisfies inequality(4)fLg≤Lfg.We will consider convex functions in bounded closed interval[a,b] with endpoints a<b. Each point x∈[a,b] can be represented by the unique binomial convex combination(5)x=αxa+βxb,where(6)αx=b-xb-a,βx=x-ab-a.Convex functionf:[a,b]→R is bounded by two lines. The secant line of function f passes through graph points A(a,f(a)) and B(b,f(b)), and its equation is(7)fabsecx=b-xb-afa+x-ab-afb.Let c∈(a,b) be an interior point. The support lines of function f pass through graph point C(c,f(c)). Each support line is specified by slope coefficient κ∈[f′(c-),f′(c+)], and its equation is(8)fcsupx=κx-c+fc.The support-secant line inequality(9)fcsupx≤fx≤fabsecxholds for every x∈[a,b].In 1931, Jessen (see [1, 2]) stated the functional form of Jensen’s inequality for convex functions in interval I⊆R. In 1988, I. Rasa and I. Raşa (see [3]) pointed out that I must be closed otherwise it could happen that L(g)∉I and that f must be continuous otherwise it could happen that the inequality in formula (4) does not apply. In Theorem A, we have taken into account I. Rasa and I. Raşa’s remarks. Some generalizations of the functional form of Jensen’s inequality can be found in [4].A concise book on functional analysis, which contains an essential overview of operator theory and indicates the importance of positive linear functionals, is certainly the book in [5].


## 2. Main Results

We firstly present the extension of the inequality in formula (4) concerning interval [a,b].Lemma 1.
Letg∈F be a function with the image in [a,b], and let L∈L be a positive unital functional.
Then each continuous convex functionf:[a,b]→R such that f(g)∈F satisfies double inequality(10)fLg≤Lfg≤fabsecLg.Proof.
The pointl=L(g) is in interval [a,b]. We sketch the proof in two steps depending on the position of l.
Ifl∈(a,b), we take support line flsup of f at l. By applying positive functional L to the support-secant inequality in formula (9) with x=g(s), where s∈S, we get (11)Lflsupg≤Lfg≤Lfabsecg.By utilizing the affinity of functions flsup and fabsec via formula (2), the above inequality takes the form(12)flsupLg≤Lfg≤fabsecLg,where the first term (13)flsupLg=fLg.
Ifl∈{a,b}, we rely on the continuity of f using a support line at a point of open interval (a,b) that is close enough to l. Given ε>0, we can find c∈(a,b) so that (14)fl-ε<fcsupl.By combining the above inequality and the inequality in formula (12) with the support line at c, we obtain (15)fl-ε<fcsupl≤Lfg≤fabsecl=fl.Letting ε approach zero, we reach the conclusion L(f(g))=f(l). In this case, trivial inequality f(l)≤f(l)≤f(l) represents formula (10).Formula (10) can be expressed in the form which includes the convex combination of interval endpoints a and b. The respective form of Lemma 1 is as follows.Corollary 2.
Letg∈F be a function with the image in [a,b], and let L∈L be a positive unital functional. Let(16)l=Lg=αla+βlb.
Then each continuous convex functionf:[a,b]→R such that f(g)∈F satisfies double inequality(17)fαla+βlb≤Lfg≤αlfa+βlfb.Proof.
As regards to the last terms of formulae (10) and (17), we have (18)fabsecLg=αlfabseca+βlfabsecb=αlfa+βlfbbecause of the affinity of fabsec and its coincidence with f at endpoints.In order to refine the inequality in formula (10), we will combine the secant lines of convex function f with positive unital functionals.Lemma 3.
Letc∈(a,b) be a point.
Then each convex functionf:[a,b]→R satisfies the secant lines inequality(19)min⁡facsecx,fcbsecx≤fx≤max⁡facsecx,fcbsecxfor every x∈[a,b].Proof.
Casesx∈[a,c] and x∈[c,b] should be considered.Suppose that functiong∈F has the image in [a,b] and is not identically equal to a or b. Such function satisfies inequality g(s1)≤c≤g(s2) for some number c∈(a,b) and some pair of points s1,s2∈S. In that case, we can find a pair of functionals L1,L2∈L meeting related inequality(20)L1g≤c≤L2g.For example, we can take the point evaluations at s1 and s2, that is, the functionals defined by L1(q)=q(s1) and L2(q)=q(s2) for every function q∈F.In the main theorem, we use functionalsL1 and L2 satisfying the inequality in formula (20).Theorem 4.
Letc∈(a,b) be a point. Let g∈F be a function with the image in [a,b], and let L1,L2∈L be positive unital functionals such that L1(g)∈[a,c] and L2(g)∈[c,b]. Let L=λ1L1+λ2L2 be a convex combination of L1 and L2.
Then each continuous convex functionf:[a,b]→R such that f(g)∈F satisfies the series of inequalities(21)fLg≤λ1fL1g+λ2fL2g≤Lfg≤λ1facsecL1g+λ2fcbsecL2g≤fabsecLg.Proof.
By applying the convexity off to convex combination L(g)=λ1L1(g)+λ2L2(g), we get(22)fLg≤λ1fL1g+λ2fL2g.
By applying the left-hand side of formula (10) to L1 and L2, we obtain(23)λ1fL1g+λ2fL2g≤λ1L1fg+λ2L2fg=Lfg.
As the right-hand side of formula (19) with x=g(s), the inequality (24)fgs≤max⁡facsecgs,fcbsecgsholds for every s∈S. By acting with L1 in the above inequality and using assumption L1(g)∈[a,c], we find (25)L1fg≤L1facsecg=facsecL1g,and similarly, by acting with L2 and using the assumption L2(g)∈[c,b], we find (26)L2fg≤L2fcbsecg=fcbsecL2g.Multiplication by λ1 and λ2 and then summation yield(27)λ1L1fg+λ2L2fg≤λ1facsecL1g+λ2fcbsecL2g.
Using main secantfabsec, we reach conclusion(28)λ1facsecL1g+λ2fcbsecL2g≤λ1fabsecL1g+λ2fabsecL2g=fabsecLg.
Putting together the inequalities in formulae (22), (23), (27), and (28) into a series, we achieve the inequality in formula (21).The geometric presentation of the series of inequalities in formula (21) is created in Figure 1. The inequality terms are represented by five black dots above point L(g).Figure 1
Geometric presentation of the inequality in formula (21).To emphasize interval endpointsa and b, we present the following version of Theorem 4.Corollary 5.
Letc∈(a,b) be a point. Let g∈F be a function with the image in [a,b], and let L1,L2∈L be positive unital functionals such that L1(g)∈[a,c] and L2(g)∈[c,b]. Let L=λ1L1+λ2L2 be a convex combination of L1 and L2, and let l=L(g)=αla+βlb.
Then each continuous convex functionf:[a,b]→R such that f(g)∈F satisfies the series of inequalities(29)fαla+βlb≤λ1fL1g+λ2fL2g≤Lfg≤αfa+βfb+γfc≤αlfa+βlfb,where(30)α=λ1c-L1gc-a,β=λ2L2g-cb-c,γ=λ1L1g-ac-a+λ2b-L2gb-c.Proof.
To calculate coefficientsα, β, and γ, we include convex combinations L1(g)=α1a+γ1c and L2(g)=γ2c+β2b. Then the fourth term of formula (21) takes the form (31)λ1facsecL1g+λ2fcbsecL2g=λ1α1fa+λ2β2fb+λ1γ1+λ2γ2fc.Taking the coefficient of f(a) and using formula (6), we calculate (32)α=λ1α1=λ1c-L1gc-a.Similarly we determine β and γ.Let us finish the section by presenting the generalization of Theorem4 that uses several secant lines.Corollary 6.
Leta=c0<c1<⋯<cn-1<cn=b be points. Let g∈F be a function with the image in [a,b], and let Li∈L be positive unital functionals such that Li(g)∈[ci-1,ci] for i=1,…,n. Let L=∑i=1n‍λiLi be a convex combination of functionals Li.
Then each continuous convex functionf:[a,b]→R such that f(g)∈F satisfies the series of inequalities(33)fLg≤∑i=1nλifLig≤Lfg≤∑i=1nλifci-1cisecLig≤fabsecLg.


## 3. Applications to Integral and Discrete Inequalities

We firstly utilize Lemma1 to obtain a very general integral inequality.Corollary 7.
Letg:[a,b]→R be an integrable function with the image in [a,b], and let h:[a,b]→R be a positive integrable function.
Then each convex functionf:[a,b]→R satisfies double inequality(34)f∫abghdx∫abhdx≤∫abfghdx∫abhdx≤∫abb-ghdx∫abb-ahdxfa+∫abg-ahdx∫abb-ahdxfb.Proof.
LetF be the space of all integrable functions over domain S=[a,b]. Composition f(g) is bounded in [a,b] and continuous almost everywhere in [a,b]. Therefore f(g) is integrable over [a,b], that is, f(g)∈F.
The integrating linear functionalL defined by(35)Lq=Lq;h=∫abqhdx∫abhdxfor every q∈F is positive and unital. The first term of formula (34) is equal to f(L(g)), the second term is equal to L(f(g)), and the third term is equal to fabsec(L(g)). Thus, formula (34) fits into the frame of formula (10), and it is valid for a continuous convex function f.
Let us verify that the inequality in formula (34) applies to a convex function which is not continuous at endpoints. We observe the position of point(36)l=Lg=∫abghdx∫abhdx.
Ifl∈(a,b), then we may utilize continuous extension f~ of f/(a,b) to [a,b] in formula (34). The first two terms are the same as we use f, and the last terms satisfy inequality(37)αlf~a+βlf~b=f~absecl<fabsecl=αlfa+βlfb.So, formula (34) applies to f in this case.
Ifl∈{a,b}, then either g(x)-l≥0 or g(x)-l≤0 for every x∈[a,b]. We rearrange formula (36) to be integral equation(38)∫abg-lhdx=0,from which it follows that g(x)=l for almost every x∈[a,b]. Thus we have that f(g(x))=f(l) for almost every x∈[a,b], and inequality f(l)≤f(l)≤f(l) represents formula (34).
Respecting all considerations, we may conclude that the inequality in formula (34) applies to any convex function f.The inequality in formula (34) is the extended version of Jensen’s inequality for the ratio of integrals in interval [a,b], as well as the generalized form of the Fejér and Hermite-Hadamard inequality.Let us demonstrate the simplifications of the inequality in formula (34) relating to the identity, unit, and symmetric function.Usingh(x)=1, we get the extension of the classical integral form of Jensen’s inequality (see [6])(39)f∫abgdxb-a≤∫abfgdxb-a≤∫abb-gdxb-a2fa+∫abg-adxb-a2fb.Using identity functiong(x)=x and function h(x), satisfying equation h(x)=h(a+b-x), which represents the symmetry with a center at midpoint (a+b)/2, we have the classical form of Fejér inequality (see [7])(40)fa+b2≤∫abfhdx∫abhdx≤fa+fb2.Namely, as a consequence of the symmetry we have(41)∫abxhdx∫abhdx=∫abx-a+b/2hdx∫abhdx+∫aba+b/2hdx∫abhdx=a+b2because (42)∫abx-a+b2hdx=0.Following formula (41), we can conclude that the Fejér inequality is valid for function h satisfying weaker condition ∫ab‍xhdx/∫ab‍hdx=(a+b)/2.Usingh(x)=1 in Fejér’s inequality in formula (40), we obtain the classical form of Hermite-Hadamard inequality (see [8, 9])(43)fa+b2≤∫abfdxb-a≤fa+fb2.To obtain refinements of the inequality in formula (34), we use point c∈(a,b) and apply Theorem 4.Corollary 8.
Letc∈(a,b) be a point. Let g:[a,b]→R be an integrable function such that g(x)∈[a,c] for x∈[a,c] and g(x)∈[c,b] for x∈[c,b], and let h:[a,b]→R be a positive integrable function.
Then each convex functionf:[a,b]→R satisfies the series of inequalities(44)f∫abghdx∫abhdx≤∫achdx∫abhdxf∫acghdx∫achdx+∫cbhdx∫abhdxf∫cbghdx∫cbhdx≤∫abfghdx∫abhdx≤∫acc-ghdx∫abc-ahdxfa+∫cbg-chdx∫abb-chdxfb+∫acg-ahdx∫abc-ahdx+∫cbb-ghdx∫abb-chdxfc≤∫abb-ghdx∫abb-ahdxfa+∫abg-ahdx∫abb-ahdxfb.Proof.
Just as in Corollary7, we use F as the space of all integrable functions in interval [a,b]. In order to apply Theorem 4, we define integrating linear functionals(45)L1q=L1q;h=∫acqhdx∫achdx,L2q=L2q;h=∫cbqhdx∫cbhdxfor every q∈F. Functionals L1 and L2 are positive and unital.
Sinceg(x)∈[a,c] for x∈[a,c], point L1(g) falls into [a,c], and similarly point L2(g) falls into [c,b]. Using coefficients(46)λ1=∫achdx∫abhdx,λ2=∫cbhdx∫abhdxand functional convex combination L=λ1L1+λ2L2, we get(47)Lg=λ1L1g+λ2L2g=∫abghdx∫abhdx,Lfg=∫abfghdx∫abhdx.By further calculating the functional terms according to formula (21), we obtain the integral terms of formula (44).The series of inequalities in formula (44) with g(x)=x and h(x)=1 gives the refinement of the Hermite-Hadamard inequality in formula (43) as follows:(48)fa+b2≤c-ab-afa+c2+b-cb-afb+c2≤∫abfdxb-a≤c-a2b-afa+b-c2b-afb+12fc≤fa+fb2.The above refinement holds for each c∈(a,b). The version of the above refinement was obtained in [10] by using the Jensen type inequality for convex combinations with the common center. That inequality was used to refine some important means.The inequality in formula (44) with identity function g(x)=x and a symmetric function satisfying equation h(x)=h(a+b-x) provides the refinements of the Fejér inequality in formula (40).At the end, let us present the discrete version of Corollary8. Point evaluations g(xi) and h(xi) will be shortened by gi and hi, respectively.Corollary 9.
Letc∈(a,b) be a point. Let g:[a,b]→R be a function such that g(x)∈[a,c] for x∈[a,c] and g(x)∈[c,b] for x∈[c,b], and let h:[a,b]→R be a positive function. Let x1,…,xk∈[a,c] and xk+1,…,xn∈[c,b] be points.
Then each convex functionf:[a,b]→R satisfies the series of inequalities(49)f∑i=1ngihi∑i=1nhi≤∑i=1khi∑i=1nhif∑i=1kgihi∑i=1khi+∑i=k+1nhi∑i=1nhif∑i=k+1ngihi∑i=k+1nhi≤∑i=1nfgihi∑i=1nhi≤∑i=1kc-gihi∑i=1nc-ahifa+∑i=k+1ngi-chi∑i=1nb-chifb+∑i=1kgi-ahi∑i=1nc-ahi+∑i=k+1nb-gihi∑i=1nb-chifc≤∑i=1nb-gihi∑i=1nb-ahifa+∑i=1ngi-ahi∑i=1nb-ahifb.Proof.
LetF be the space of all real functions on domain S=[a,b]. We can apply the proof of Corollary 8 to summarizing linear functionals(50)L1q=L1q;h=∑i=1kqihi∑i=1khi,L2q=L2q;h=∑i=k+1nqihi∑i=k+1nhiacting to every q∈F, and coefficients(51)λ1=∑i=1khi∑i=1nhi,λ2=∑i=k+1nhi∑i=1nhi.Functionals L1 and L2 are certainly positive and unital.The inequality in formula (49) is the extension and refinement of the famous discrete form of Jensen’s inequality (see [11]).



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