# Reverses of the Jensen-Type Inequalities for Signed Measures

Authors: Rozarija Jakšić; Josip Pečarić; Mirna Rodić Lipanović
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

In this paper we derive refinements of the Jensen type inequalities in the case of real Stieltjes measuredλ, not necessarily positive, which are generalizations of Jensen's inequality and its reverses for positive measures. Furthermore, we investigate the exponential and logarithmic convexity of the difference between the left-hand and the right-hand side of these inequalities and give several examples of the families of functions for which the obtained results can be applied. The outcome is a new class of Cauchy-type means.


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## Body

## 1. Introduction

The authors in [1] gave some conditions on the real Stieltjes measure dλ, not necessarily positive, under which the Jensen inequality and the converse of the Jensen inequality hold for continuous convex function φ. These results are derived by using the Green function G defined on [α,β]×[α,β] by
(1)G(t,s)={(t-β)(s-α)β-αforα≤s≤t,(s-β)(t-α)β-αfort≤s≤β.
The function G is convex and continuous with respect to both s and t.Several interesting results concerning the Jensen type inequalities have been derived by means of the functionG. The first one, which is stated in the following theorem, gives the conditions on the real Stieltjes measure dλ, not necessarily positive, under which the Jensen inequality holds for continuous convex function φ.Theorem 1 (see [1]).
Letg:[a,b]→R be continuous function and [α,β] interval such that the image of g is a subset of [α,β]. Let λ:[a,b]→R be continuous function or the function of bounded variation such that λ(a)≠λ(b) and ∫ab‍g(x)dλ(x)/∫ab‍dλ(x)∈[α,β]. Then the following two statements are equivalent.(1)
For every continuous convex functionφ:[α,β]→R(2)φ(∫ab‍g(x)dλ(x)∫ab‍dλ(x))≤∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)holds.(2)
For alls∈[α,β](3)G(∫ab‍g(x)dλ(x)∫ab‍dλ(x),s)≤∫ab‍G(g(x),s)dλ(x)∫ab‍dλ(x)holds, where the functionG:[α,β]×[α,β]→R is defined in (1).Furthermore, the statements (1) and (2) are also equivalent if we change the sign of inequality in both (2) and (3). Also note that for every continuous concave function φ:[α,β]→R the inequality (2) is reversed.Remark 2.
Note that in the case of positive measuredλ we get some well known results. If the function λ is increasing and bounded with λ(a)≠λ(b), then inequality (2) becomes Jensen’s integral inequality. On the other hand, if the function g is continuous and monotonic, and λ is either continuous or of bounded variation, satisfying
(4)λ(a)≤λ(x)≤λ(b),∀x∈[a,b],λ(a)<λ(b),
then inequality (2) becomes the Jensen-Steffensen inequality given by Boas in [2] (see also [3], page 59). Several other theorems concerning the inequality (2) or its reverse can be found in [3].Similar results have also been derived for the converse of the Jensen inequality. The following theorem from [1] gives the conditions on the real Stieltjes measure dλ, not necessarily positive such that λ(a)≠λ(b), under which the converse of the Jensen inequality holds for continuous convex function φ.Theorem 3 (see [1]).
Letg:[a,b]→R be continuous function and [α,β] be an interval such that the image of g is a subset of [α,β]. Let m,M∈[α,β]  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let λ:[a,b]→R be continuous function or the function of bounded variation, and λ(a)≠λ(b). Then the following two statements are equivalent.(1)
For every continuous convex functionφ:[α,β]→R(5)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)≤M-g¯M-mφ(m)+g¯-mM-mφ(M)holds, whereg¯=∫ab‍g(x)dλ(x)/∫ab‍dλ(x).(2)
For alls∈[α,β](6)∫ab‍G(g(x),s)dλ(x)∫ab‍dλ(x)≤M-g¯M-mG(m,s)+g¯-mM-mG(M,s)holds, where the functionG:[α,β]×[α,β]→R is defined in (1).Furthermore, the statements (1) and (2) are also equivalent if we change the sign of inequality in both (5) and (6).Remark 4.
If we set in Theorem3  m=α and M=β, the inequality (6) transforms into (see also [1])
(7)∫ab‍G(g(x),s)dλ(x)∫ab‍dλ(x)≤0.In his papers [4, 5], Dragomir gave some inequalities concerning reverses of the Jensen inequality for positive measure. In this paper we give a generalization of those results and derive similar Jensen-type inequalities in the case of real Stieltjes measure dλ which is not necessarily positive. Furthermore, we investigate the exponential and logarithmic convexity of the differences between the left-hand and the right-hand side of the obtained inequalities and give several examples of the families of functions for which those results can be applied.


## 2. Main Results

Throughout this paper we will use the notation(8)g¯=∫ab‍g(x)dλ(x)∫ab‍dλ(x).The following result holds.Theorem 5.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉 be such that m≠M and m≤g(t)≤M for all t∈[a,b]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[m,M]. If (6) holds, for all s∈[α,β], then
(9)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤(M-g¯)(g¯-m)M-m(φ-′(M)-φ+′(m))≤14(M-m)(φ-′(M)-φ+′(m))
holds for every continuous convex function φ:[α,β]→R.Proof.
Let (6) hold for all s∈[α,β]. Then from Theorem 3 it follows that for every continuous convex function φ:[α,β]→R we have
(10)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤M-g¯M-mφ(m)+g¯-mM-mφ(M)-φ(g¯).
By the convexity of φ (as in [4]), we have the gradient inequality
(11)φ(t)-φ(M)≥φ-′(M)(t-M),
for any t∈[m,M]. If we multiply this inequality with t-m≥0, we get
(12)(t-m)φ(t)-(t-m)φ(M)≥φ-′(M)(t-M)(t-m),kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkit∈[m,M].
Analogously, we get
(13)(M-t)φ(t)-(M-t)φ(m)≥φ+′(m)(t-m)(M-t),kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkit∈[m,M].
Adding (12) to (13) and dividing by M-m, we get that
(14)(t-m)φ(M)+(M-t)φ(m)M-m-φ(t)≤(M-t)(t-m)M-m(φ-′(M)-φ+′(m))
holds for any t∈[m,M]. Substituting t with g¯∈[m,M], we obtain the first inequality in (9).
To prove the second inequality in (9), it is enough to notice that the function h(t)=(M-t)(t-m)/(M-m) is concave on [m,M], so for every t∈[m,M] inequality h(t)≤(1/4)(M-m) is valid. That completes the proof of (9).If we set thatm=α and M=β, we get the following result.Corollary 6.
Letg:[a,b]→R be continuous function and [α,β] (where α≠β) an interval such that the image of g is a subset of [α,β]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[α,β]. If (7) holds for all s∈[α,β], then for every continuous convex function φ:[α,β]→R,
(15)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤(β-g¯)(g¯-α)β-α(φ-′(β)-φ+′(α))≤14(β-α)(φ-′(β)-φ+′(α)),
provided that φ+′(α) and φ-′(β) are finite.Theorem 7.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[m,M]. If (6) holds for all s∈[α,β], then for every continuous convex function φ:[α,β]→R we have
(16)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤(M-g¯)(g¯-m)M-m·sup⁡t∈(m,M)Ψφ(t;m,M)≤(M-g¯)(g¯-m)M-m(φ-′(M)-φ+′(m))≤14(M-m)(φ-′(M)-φ+′(m)),
where Ψφ(·;m,M):〈m,M〉→R is defined by
(17)Ψφ(t;m,M)=φ(M)-φ(t)M-t-φ(t)-φ(m)t-m.
If  g¯∈〈m,M〉, then we also have
(18)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤14(M-m)Ψφ(g¯;m,M)≤14(M-m)(φ-′(M)-φ+′(m)).Proof.
Let (6) hold for all s∈[α,β]. Then from Theorem 3 it follows that
(19)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤M-g¯M-mφ(m)+g¯-mM-mφ(M)-φ(g¯)
holds for every continuous convex function φ:[α,β]→R. If g¯∈〈m,M〉, it is easy to verify that the term on the right side of (19) is equal to
(20)(M-g¯)(g¯-m)M-m[φ(M)-φ(g¯)M-g¯-φ(g¯)-φ(m)g¯-m],
and the term in the square brackets in (20) is equal to Ψφ(g¯;m,M), where the function Ψφ(·;m,M) is defined in (24). As in [5], it follows that
(21)Ψφ(g¯;m,M)≤sup⁡t∈(m,M)Ψφ(t;m,M)=sup⁡t∈(m,M)[φ(M)-φ(t)M-t-φ(t)-φ(m)t-m]≤sup⁡t∈(m,M)[φ(M)-φ(t)M-t]+sup⁡t∈(m,M)[-φ(t)-φ(m)t-m]=sup⁡t∈(m,M)[φ(M)-φ(t)M-t]-inf⁡t∈(m,M)[φ(t)-φ(m)t-m]=φ-′(M)-φ+′(m),
and since
(22)(M-g¯)(g¯-m)M-m≤14(M-m),
the inequality (16) follows. For g¯=m or g¯=M, the inequality (16) is obvious. The inequality in (18) follows directly from the proof of inequality (16).If we set thatm=α and M=β, we get the following result.Corollary 8.
Letg:[a,b]→R be a continuous function and [α,β] (where α≠β) an interval such that the image of g is a subset of  [α,β]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[α,β]. If (7) holds for all s∈[α,β], then
(23)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤(β-g¯)(g¯-α)β-α·sup⁡t∈(α,β)Ψφ(t;α,β)≤(β-g¯)(g¯-α)β-α(φ-′(β)-φ+′(α))≤14(β-α)(φ-′(β)-φ+′(α)),
where Ψφ(·;α,β):〈α,β〉→R is defined by
(24)Ψφ(t;α,β)=φ(β)-φ(t)β-t-φ(t)-φ(α)t-α,
holds for every continuous convex function φ:[α,β]→R, provided that φ+′(α) and φ-′(β) are finite. If  g¯∈〈α,β〉, then we also have
(25)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤14(β-α)Ψφ(g¯;α,β)≤14(β-α)(φ-′(β)-φ+′(α)).Theorem 9.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[m,M]. If (6) holds for all s∈[α,β], then for every continuous convex function φ:[α,β]→R we have
(26)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤max⁡{M-g¯M-m,g¯-mM-m}·[φ(m)+φ(M)-2φ(m+M2)]={12+1M-m|m+M2-g¯|}·[φ(m)+φ(M)-2φ(m+M2)]≤12max⁡{M-g¯,g¯-m}{φ-′(M)-φ+′(m)}.In order to prove the theorem, we need the following lemma, which is a special case of [6, page 717, Theorem 1] for n=2.Lemma 10.
Letφ be a convex function on Dφ, x,y∈Dφ and p,q∈[0,1] such that p+q=1. Then
(27)min⁡{p,q}[φ(x)+φ(y)-2φ(x+y2)]≤pφ(x)+qφ(y)-φ(px+qy)≤max⁡{p,q}[φ(x)+φ(y)-2φ(x+y2)].Proof of Theorem9.
Let (6) hold for all s∈[α,β]. Then from Theorem 3 it follows that
(28)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)≤M-g¯M-mφ(m)+g¯-mM-mφ(M)
holds for every continuous convex function φ:[α,β]→R. Denote p=(M-g¯)/(M-m); so we have p∈[0,1] and g¯=p·m+(1-p)M. As in [7], by Lemma 10 we get the following:
(29)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤M-g¯M-mφ(m)+g¯-mM-mφ(M)-φ(g¯)=pφ(m)+(1-p)φ(M)-φ(p·m+(1-p)M)≤max⁡{p,1-p}·[φ(m)+φ(M)-2φ(m+M2)]=max⁡{M-g¯M-m,g¯-mM-m}·[φ(m)+φ(M)-2φ(m+M2)]={12+1M-m|m+M2-g¯|}·[φ(m)+φ(M)-2φ(m+M2)].
In order to prove the second inequality, it is easy to verify that(30)φ(m)+φ(M)-2φ((m+M)/2)M-m=12[φ(M)-φ((m+M)/2)M-((m+M)/2)-φ((m+M)/2)-φ(m)((m+M)/2)-m].
Since φ is a convex function on [α,β], by the gradient inequality we have (as in [5])
(31)φ(M)-φ((m+M)/2)M-((m+M)/2)≤φ-′(M),φ((m+M)/2)-φ(m)((m+M)/2)-m≥φ+′(m),
and we finally obtain
(32)φ(m)+φ(M)-2φ((m+M)/2)M-m≤12[φ-′(M)-φ+′(m)],
so the second inequality in (26) follows.If we set thatm=α and M=β, we get the following result.Corollary 11.
Letg:[a,b]→R be a continuous function and [α,β] (where α≠β) an interval such that the image of g is a subset of [α,β]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[α,β]. If (7) holds for all s∈[α,β], then for every continuous convex function φ:[α,β]→R we have
(33)∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)≤max⁡{β-g¯β-α,g¯-αβ-α}·[φ(α)+φ(β)-2φ(α+β2)]={12+1β-α|α+β2-g¯|}·[φ(α)+φ(β)-2φ(α+β2)]≤12max⁡{β-g¯,g¯-α}{φ-′(β)-φ+′(α)}
provided that φ+′(α) and φ-′(β) are finite.Remark 12.
If in Theorem5, Corollary 6, Theorem 7, Corollary 8, Theorem 9, and Corollary 11 we also request that (3) holds for all s∈[α,β], then by applying Theorem 1 we will get that
(34)0≤∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)-φ(g¯)
also holds for every continuous convex function φ:[α,β]→R.Remark 13.
Result given in Theorem5 together with Remark 12 represents a generalization of [4, Theorem 2], result given in Theorem 7 together with Remark 12 represents a generalization of [5, Theorem 2], and result given in Theorem 9 with Remark 12 represents a generalization of [5, Theorem 3].


## 3. Mean-Value Theorems

As in the previous section, letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), and let g¯∈[m,M].Motivated by the inequalities (9) and (26) for continuous convex function φ:[α,β]→R, we define functionals Φ1(g,λ,φ) and Φ2(g,λ,φ) by
(35)Φ1(g,λ,φ)=(M-g¯)(g¯-m)M-m(φ-′(M)-φ+′(m))-∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)+φ(g¯),(36)Φ2(g,λ,φ)={12+1M-m|m+M2-g¯|}·[φ(m)+φ(M)-2φ(m+M2)]-∫ab‍φ(g(x))dλ(x)∫ab‍dλ(x)+φ(g¯).If (6) holds for all s∈[α,β], then from Theorems 5 and 9 it follows that Φi(g,λ,φ)≥0  (i=1,2). If the function φ is concave, that is, if (6) holds with reversed inequality sign for all s∈[α,β], then we have Φi(g,λ,φ)≤0  (i=1,2).Now we give two mean-value theorems for the functionalsΦi(g,λ,φ)  (i=1,2). In the following, φ0 will denote the function defined by φ0(t)=t2 on whatever domain we need.Theorem 14.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let φ:[α,β]→R, φ∈C2([α,β]). Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b) and g¯∈[m,M], and let Φ1 and Φ2 be functionals defined in (35) and (36), respectively. Then there exist ξ1,ξ2∈[α,β] such that the following equalities hold:
(37)Φi(g,λ,φ)=φ′′(ξi)2Φi(g,λ,φ0),i=1,2.Proof.
We only give the proof for the functionalΦ1, since both proofs have the same concept. Since φ∈C2([α,β]), there exist m1=min⁡t∈[α,β]φ′′(t) and M1=max⁡t∈[α,β]φ′′(t). It is easy to verify that the functions φ1,φ2 defined by
(38)φ1(t)=M12t2-φ(t),φ2(t)=φ(t)-m12t2
are continuous and convex, and therefore Φ1(g,λ,φ1)≥0, Φ1(g,λ,φ2)≥0. This implies that
(39)m12Φ1(g,λ,φ0)≤Φ1(g,λ,φ)≤M12Φ1(g,λ,φ0).
Hence, as the function φ′′ is continuous, there exists ξ1∈[α,β] such that
(40)Φ1(g,λ,φ)=φ′′(ξ1)2Φ1(g,λ,φ0).
The proof for the functional Φ2 is analogous.Theorem 15.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let φ,ψ:[α,β]→R, φ,ψ∈C2([α,β]). Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b) and g¯∈[m,M], and let Φ1 and Φ2 be functionals defined in (35) and (36), respectively. Then there exist ξ1,ξ2∈[α,β] such that the following equalities hold
(41)Φi(g,λ,φ)Φi(g,λ,ψ)=φ′′(ξi)ψ′′(ξi),i=1,2,
provided that the denominators are nonzero.Proof.
We give the proof for the functionalΦ1. Define the function χ as a linear combination of functions φ and ψ(42)χ=c1·φ-c2·ψ,wherec1=Φ1(g,λ,ψ),kkkkkic2=Φ1(g,λ,φ).
Now, applying previous theorem on our function χ, we obtain that there exists ξ1∈[α,β] such that
(43)(c1φ′′(ξ1)2-c2ψ′′(ξ1)2)Φ1(g,λ,φ0)=0.
Since Φ1(g,λ,φ0)≠0 (otherwise we would have a contradiction with Φ1(g,λ,ψ)≠0), we get
(44)Φ1(g,λ,φ)Φ1(g,λ,ψ)=φ′′(ξ1)ψ′′(ξ1).
The proof for the functional Φ2 is analogous.Remark 16.
If the inverse of the functionφ′′/ψ′′ exists, then (41) gives
(45)ξi=(φ′′ψ′′)-1(Φi(g,λ,φ)Φi(g,λ,ψ))∈[α,β],i=1,2.Remark 17.
Theorems14 and 15 also hold if we set m=α and M=β; that is, if we extend the image of the function g to the entire interval [α,β], provided that φ+′(α), φ-′(β), ψ+′(α), and ψ+′(α) are finite.


## 4.n-Exponential Convexity

At the beginning of this section, let us recall some definitions and facts about exponentially convex functions (for instance, see [8] or [9]).Definition 18.
A functionf:I→R is n-exponentially convex in the Jensen sense on I if
(46)∑i,j=1n‍pipjf(xi+xj2)≥0
holds for all pi∈R and xi∈I, i=1,…,n.
A functionf:I→R is n-exponentially convex if it is n-exponentially convex in the Jensen sense and continuous on I.Remark 19.
We can see from the definition that1-exponentially convex functions in the Jensen sense are in fact nonnegative functions. Also, n-exponentially convex functions in the Jensen sense are k-exponentially convex in the Jensen sense for every k∈N, k≤n.By definition of the positive semidefinite matrices and some basic linear algebra, we have the following result.Lemma 20.
Iff is an n-exponentially convex function in the Jensen sense, then the matrix [f((xi+xj)/2)]i,j=1k is positive semidefinite for all k∈N, k≤n. Particularly, det⁡[f((xi+xj)/2)]i,j=1k≥0 for all k∈N, k≤n.Definition 21.
A functionf:I→R is exponentially convex in the Jensen sense on I, if it is n-exponentially convex in the Jensen sense for all n∈N.
A functionf:I→R is exponentially convex if it is exponentially convex in the Jensen sense and continuous.Remark 22.
Some examples of exponentially convex functions are as follows (see [10]):(i)
f:I→R defined by f(x)=cekx, where c≥0 and k∈R,(ii)
f:R+→R defined by f(x)=x-k, where k>0,(iii)
f:R+→R+ defined by f(x)=e-kx, where k>0.Remark 23.
It is known that a functionf:I→R+ is log-convex in the Jensen sense on I if and only if the relation
(47)α2f(x)+2αβf(x+y2)+β2f(y)≥0
holds for every α,β∈R and x,y∈I. It follows that a positive function is log-convex in the Jensen sense if and only if it is 2-exponentially convex in the Jensen sense. Also, using basic theory of convex functions, it follows that a positive function is log-convex if and only if it is 2-exponentially convex.The following lemma is equivalent to the definition of convex function (see [3], page 2).Lemma 24.
Ifx1,x2,x3∈I are such that x1<x2<x3, then the function f:I→R is convex if and only if the following inequality holds:
(48)(x3-x2)f(x1)+(x1-x3)f(x2)+(x2-x1)f(x3)≥0.We will also need the following result (see [3], page 2).Lemma 25.
Iff:I→R is a convex function and x1,x2,y1,y2∈I are such that x1≤y1, x2≤y2, x1≠x2, and y1≠y2, then the following inequality is valid:
(49)f(x2)-f(x1)x2-x1≤f(y2)-f(y1)y2-y1.
If the function f is concave, then the reverse inequality in (49) holds.When dealing with functions with different degrees of smoothness, divided differences are found to be very useful.Definition 26.
The second order divided difference of a functionf:I→R at mutually different points y0,y1,y2∈I is defined recursively by
(50)[yi]f=f(yi),i=0,1,2,[yi,yi+1]f=f(yi+1)-f(yi)yi+1-yi,i=0,1,[y0,y1,y2]f=[y1,y2]f-[y0,y1]fy2-y0.Remark 27.
The value[y0,y1,y2]f is independent of the order of the points y0,y1, and y2. This definition may be extended to include the case in which some or all of the points coincide (see [3], page 16). Taking the limit y1→y0 in (50), we get
(51)lim⁡y1→y0[y0,y1,y2]f=[y0,y0,y2]f=f(y2)-f(y0)-f′(y0)(y2-y0)(y2-y0)2,y2≠y0,
provided that f′ exists. Furthermore, taking the limits yi→y0,  i=1,2 in (50), we get
(52)lim⁡y2→y0lim⁡y1→y0[y0,y1,y2]f=[y0,y0,y0]f=f′′(y0)2,
provided that f′′ exists.
A functionf:I→R is convex if and only if for every choice of three mutually different points y0,y1,y2∈I[y0,y1,y2]f≥0 holds.Now, we use an idea from [10] to give an elegant method of producing n-exponentially convex functions and exponentially convex functions by applying functionals Φi  (i=1,2) to a given family of functions with the same property.For the rest of this section we assume that (6) holds, so from Theorems 5 and 9 it follows that Φi(g,λ,φ)≥0  (i=1,2) under the appropriate assumptions on functions g, λ, and φ.Theorem 28.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let Ω={φp:p∈J} (where J is an interval in R) be a family of functions φp:[α,β]→R, φp∈C([α,β]), such that the function p↦[y0,y1,y2]φp is n-exponentially convex in the Jensen sense on J for every three mutually different points y0,y1,y2∈[α,β]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), let g¯∈[m,M], and let Φ1 and Φ2 be linear functionals defined in (35) and (36), respectively. Then the function p↦Φi(g,λ,φp)  (i=1,2) is n-exponentially convex in the Jensen sense on J. If the function p↦Φi(g,λ,φp) is continuous on J, then it is n-exponentially convex on J.Proof.
Forqj∈R  (j=1,…,n) we define the function
(53)h(x)=∑j,k=1n‍qjqkφ(pj+pk)/2(x),
where pj,pk∈J, 1≤j, k≤n, and φ(pj+pk)/2∈Ω. Since p→[y0,y1,y2]φp is n-exponentially convex in the Jensen sense by assumption, for every three mutually different points y0,y1,y2∈[α,β], we have
(54)[y0,y1,y2]h=∑j,k=1n‍qjqk[y0,y1,y2]φ(pj+pk)/2≥0.
It follows that h is convex (and continuous) function on [α,β], so
(55)Φi(g,λ,h)≥0,
and hence
(56)∑j,k=1n‍qjqkΦi(g,λ,φ(pj+pk)/2)≥0.
We conclude that the function p↦Φi(g,λ,φp) is n-exponentially convex on J in the Jensen sense.
If the functionp↦Φi(g,λ,φp) is also continuous on J, then it is n-exponentially convex by definition.The following corollary is an immediate consequence of the above theorem.Corollary 29.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let Ω={φp:p∈J} (where J is an interval in R) be a family of functions φp:[α,β]→R, φp∈C([α,β]), such that the function p↦[y0,y1,y2]φp is exponentially convex in the Jensen sense on J for every three mutually different points y0,y1,y2∈[α,β]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), let g¯∈[m,M], and let Φ1 and Φ2 be linear functionals defined in (35) and (36), respectively. Then the function p↦Φi(g,λ,φp)  (i=1,2) is exponentially convex in the Jensen sense on J. If the function p↦Φi(g,λ,φp) is continuous on J, then it is exponentially convex on J.Corollary 30.
Letg:[a,b]→R be a continuous function and [α,β] an interval. Let m,M∈〈α,β〉  (m≠M) be such that m≤g(t)≤M for all t∈[a,b]. Let Ω={φp:p∈J} (where J is an interval in R) be a family of functions φp:[α,β]→R, φp∈C([α,β]), such that the function p↦[y0,y1,y2]φp is 2-exponentially convex in the Jensen sense on J for every three mutually different points y0,y1,y2∈[α,β]. Let λ:[a,b]→R be a continuous function or a function of bounded variation such that λ(a)≠λ(b), let g¯∈[m,M], and let Φ1 and Φ2 be linear functionals defined in (35) and (36), respectively.
Then the following statements hold.(i)
If the functionp↦Φi(g,λ,φp) is continuous on J, then it is 2-exponentially convex on J. If p↦Φi(g,λ,φp) is additionally strictly positive, then it is also log-convex on J, and for r,s,t∈J such that r<s<t, we have
(57)(Φi(g,λ,φs))t-r≤(Φi(g,λ,φr))t-s(Φi(g,λ,φt))s-r.(ii)
If the functionp↦Φi(g,λ,φp) is strictly positive and differentiable on J, then for every p,q,u,v∈J such that p≤u and q≤v, we have
(58)μp,q(g,Φi,Ω)≤μu,v(g,Φi,Ω),(i=1,2),where(59)μp,q(g,Φi,Ω)={(Φi(g,λ,φp)Φi(g,λ,φq))1/(p-q),p≠q,exp⁡((d/dp)Φi(g,λ,φp)Φi(g,λ,φp)),p=q,forφp, φq∈Ω.Proof.
(i) The first part is an immediate consequence of Theorem28, and in the second part, log-convexity is an immediate consequence of Remark 23. Since p↦Φi(g,λ,φp) is strictly positive, then by applying Lemma 24 on the function f(x)=log⁡Φi(g,λ,φx) and r,s,t∈J  (r<s<t) we get
(60)(t-s)Φi(g,λ,φr)+(r-t)Φi(g,λ,φs)+(s-r)Φi(g,λ,φt)≥0,
which is equivalent to inequality (57).
(ii) Since by (i) the functionp↦Φi(g,λ,φp) is log-convex on J, that is, the function p↦log⁡Φi(g,λ,φp) is convex on J, by applying Lemma 25 for p≤u, q≤v, p≠q, and u≠v, we get
(61)log⁡Φi(g,λ,φp)-log⁡Φi(g,λ,φq)p-q≤log⁡Φi(g,λ,φu)-log⁡Φi(g,λ,φv)u-v,
and therefore we conclude that
(62)μp,q(g,Φi,Ω)≤μu,v(g,Φi,Ω).
The cases p=q and u=v follow from (61) as limit cases.Remark 31.
Note that the results from Theorem28 and Corollaries 29 and 30 still hold when two of the points y0,y1,y2∈[α,β] coincide (say y1=y0), for a family of differentiable functions φp such that the function p↦[y0,y1,y2]φp is n-exponentially convex in the Jensen sense (exponentially convex in the Jensen sense, log-convex in the Jensen sense). Furthermore, these results still hold when all three points coincide for a family of twice differentiable functions with the above mentioned properties. The proofs are obtained by recalling Remark 27 and suitable characterization of convexity.Remark 32.
The results from this section also hold if we setm=α and M=β, that is, if we extend the image of the function g to the entire interval [α,β], provided that φ+′(α) and φ-′(β) are finite.


## 5. Examples

In this section we will vary on choice of a familyΩ={φp:p∈J}, presenting several families of functions which fulfil the conditions of Theorem 28 and Corollaries 29 and 30 (and Remark 31). This enables us to construct different examples of exponentially convex functions and to construct some Cauchy-type means.Example 1.
Let(63)Ω1={ψp:R→[0,∞):p∈R}
be a family of functions defined by
(64)ψp(x)={1p2epx,p≠0;12x2,p=0.
Since(d2/dx2)ψp(x)=epx>0 for x∈R, ψp is convex function on R for every p∈R. From Remark 22 it follows that the function p↦(d2/dx2)ψp(x) is exponentially convex, and from [10] we then also have that p↦[y0,y1,y2]ψp is exponentially convex (and so exponentially convex in the Jensen sense). So, our family Ω1 of functions ψp fulfills the condition given in Corollary 29, and we conclude that p↦Φi(g,λ,ψp) (for i=1,2) are exponentially convex in the Jensen sense. It is easy to verify that these mappings are continuous (although p↦ψp is not continuous at p=0), so they are exponentially convex.
Using Corollary30 for this family of functions, μp,q(g,Φi,Ω1)  (i=1,2) from (59) become
(65)μp,q(g,Φi,Ω1)={(Φi(g,λ,ψp)Φi(g,λ,ψq))1/(p-q),p≠q;exp⁡(Φi(g,λ,id·ψp)Φi(g,λ,ψp)-2p),p=q≠0;exp⁡(13Φi(g,λ,id·ψ0)Φi(g,λ,ψ0)),p=q=0;
and using (58) we have that they are monotonous in parameters p and q.
IfΦi  (i=1,2) are positive, then using Theorem 15 applied for φ=ψp∈Ω1 and ψ=ψq∈Ω1, it follows that
(66)Mp,q(g,Φi,Ω1)=log⁡μp,q(g,Φi,Ω1),fori=1,2,
satisfy
(67)α≤Mp,q(g,Φi,Ω1)≤β,fori=1,2.
If we set that the image of the function g is [α,β], then we have
(68)α=min⁡t∈[a,b]{g(t)}≤Mp,q(g,Φi,Ω1)≤max⁡t∈[a,b]{g(t)}=β,fori=1,2,
which shows that in this case Mp,q(g,Φi,Ω1) are means (of the function g). Notice that by (58) Mp,q(g,Φi,Ω1) are also monotonic.Example 2.
Let(69)Ω2={φp:R+⟶R:p∈R}
be a family of functions defined by
(70)φp(x)={xpp(p-1),p≠0,1;-log⁡x,p=0;xlog⁡x,p=1.
Since(d2/dx2)φp(x)=xp-2=e(p-2)log⁡x>0, φp is convex function for x>0. From Remark 22 it follows that p↦(d2/dx2)φp(x) is exponentially convex, and from [10] we then also have that p↦[y0,y1,y2]φp is exponentially convex (and so exponentially convex in the Jensen sense). So, our family Ω2 of functions φp fulfills the condition given in Corollary 29.
In this example we assume that our interval[α,β] from Corollaries 29 and 30 is a subset of R+, and for our family of functions we have the following:
(71)μp,q(g,Φi,Ω2)={(Φi(g,λ,φp)Φi(g,λ,φq))1/(p-q),p≠q;exp⁡(1-2pp(p-1)-Φi(g,λ,φ0φp)Φi(g,λ,φp)),p=q≠1,0;exp⁡(1-Φi(g,λ,φ02)2Φi(g,λ,φ0)),p=q=0.exp⁡(-1-Φi(g,λ,φ0φ1)2Φi(g,λ,φ1)),p=q=1.
As in the previous example, we conclude that functionsp↦Φi(g,λ,φp)  (i=1,2) are exponentially convex and that for μp,q(g,Φi,Ω2) the monotonicity property holds.
IfΦi  (i=1,2) are positive, then Theorem 15 applied for φ=φp∈Ω2 and ψ=φq∈Ω2 yields that there exist
(72)ξi∈[α,β],(i=1,2),
such that
(73)ξip-q=Φi(g,λ,φp)Φi(g,λ,φq).
Since the function ξ↦ξp-q is invertible for p≠q, we then have
(74)α≤(Φi(g,λ,φp)Φi(g,λ,φq))1/(p-q)≤β,fori=1,2.
As in the previous example, if we set that the image of the function g is [α,β], then we have
(75)α=min⁡t∈[a,b]{g(t)}≤(Φi(g,λ,φp)Φi(g,λ,φq))1/(p-q)≤max⁡t∈[a,b]{g(t)}=β,
which shows that in this case μp,q(g,Φi,Ω2) are means (of function g).
Now, we impose one additional parameterr. For r≠0 by substituting g→gr, p→p/r, and q→q/r in (75), we get
(76)min⁡t∈[a,b]{(g(t))r}≤(Φi(gr,λ,φp)Φi(gr,λ,φq))r/(p-q)≤max⁡t∈[a,b]{(g(t))r},fori=1,2.
We define new generalized mean as follows:(77)μp,q;r(g,Φi,Ω2)={(μp/r,q/r(gr,Φi,Ω2))1/r,r≠0;μp,q(log⁡g,Φi,Ω1),r=0.
These new generalized means are also monotonic. If p,q,u,v∈R, r≠0 such that p≤u, q≤v, then we have
(78)μp,q;r(g,Φi,Ω2)≤μu,v;r(g,Φi,Ω2),fori=1,2.
This result follows from
(79)μp/r,q/r(gr,Φi,Ω2)=(Φi(gr,λ,φp/r)Φi(gr,λ,φq/r))r/(p-q)≤(Φi(gr,λ,φu/r)Φi(gr,λ,φv/r))r/(u-v)=μu/r,v/r(gr,Φi,Ω2),
for p,q,u,v∈R, r≠0, such that p/r≤u/r, q/r≤v/r, and the fact that μp,q(g,Φi,Ω2) for i=1,2 are monotonous in both parameters. For r=0, we obtain the required result by taking the limit r→0.Example 3.
Let(80)Ω3={θp:R+⟶R+:p∈R+}
be a family of functions defined by
(81)θp(x)=e-xpp.
Since (d2/dx2)θp(x)=e-xp>0, θp is convex function for x>0. From Remark 22 we have that p↦(d2/dx2)θp(x) is exponentially convex, and from [10] we then also have that p↦[y0,y1,y2]θp is exponentially convex function. Family Ω3 of functions θp fulfills the condition given in Corollary 29. Here in this example we again assume that interval [α,β] from our corollaries is a subset of R+, and so for our family of functions we have the following possible cases for μp,q:
(82)μp,q(g,Φi,Ω3)={(Φi(g,λ,θp)Φi(g,λ,θq))1/(p-q),p≠q;exp⁡(-Φi(g,λ,id·θp)2pΦi(g,λ,θp)-1p),p=q.
As before, we conclude that the functionsp↦Φi(g,λ,θp)  (i=1,2) are exponentially convex and that for μp,q(g,Φi,Ω3) the monotonicity property holds.
IfΦi  (i=1,2) are positive, then using Theorem 15 applied for φ=θp∈Ω3 and ψ=θq∈Ω3, it follows that
(83)Mp,q(g,Φi,Ω3)=-(p+q)log⁡μp,q(g,Φi,Ω3)
satisfy
(84)α≤Mp,q(g,Φi,Ω3)≤β,fori=1,2.
If we set that the image of the functiong is [α,β], then we have that in this case Mp,q(g,Φi,Ω3) are means (of the function g).Example 4.
Let(85)Ω4={ϕp:R+⟶R+:p∈R+}
be a family of functions defined by
(86)ϕp(x)={p-x(log⁡p)2,p≠1;x22,p=1.
Since (d2/dx2)ϕp(x)=p-x>0, ϕp is convex function for p>0. From Remark 22 it follows that p↦(d2/dx2)ϕp(x) is exponentially convex function, and from [10] we then also have that p↦[y0,y1,y2]ϕp is exponentially convex. Our family Ω4 of functions ϕp fulfills the condition given in Corollary 29. We assume again that interval [α,β] from our corollaries is a subset of  R+, and so for our family of functions we have the following:
(87)μp,q(g,Φi,Ω4)={(Φi(g,λ,ϕp)Φi(g,λ,ϕq))1/(p-q),p≠q;exp⁡(-Φi(g,λ,id·ϕp)pΦi(g,λ,ϕp)-2plog⁡p),p=q≠1;exp⁡(-23Φi(g,λ,id·ϕ1)Φi(g,λ,ϕ1)),p=q=1.
As before, we conclude that the functionsp↦Φi(g,λ,ϕp)  (i=1,2) are exponentially convex, and for μp,q(g,Φi,Ω4) the monotonicity property holds.
IfΦi  (i=1,2) are positive, then using Theorem 15 applied for φ=ϕp∈Ω4 and ψ=ϕq∈Ω4, it follows that
(88)Np,q(g,Φi,Ω4)=-L(p,q)log⁡μp,q(g,Φi,Ω4),
where L(p,q) denotes the logarithmic mean defined by L(p,q)=(p-q)/(log⁡p-log⁡q) for p≠q, and L(p,p)=psatisfy
(89)α≤Np,q(g,Φi,Ω4)≤β,fori=1,2.
If we set that the image of the functiong is [α,β], then we have that in this case Np,q(g,Φi,Ω4) are means (of the function g).



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