Reverses of the Jensen-Type Inequalities for Signed Measures

and Applied Analysis 3 Adding (12) to (13) and dividing byM − m, we get that (t − m) φ (M) + (M − t) φ (m) M − m − φ (t) ≤ (M − t) (t − m) M − m (φ 󸀠 − (M) − φ 󸀠 + (m)) (14) 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 that m = α and M = β, we get the following result. Corollary 6. Let g : [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, ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (x) − φ (g) ≤ (β − g) (g − α) β − α (φ 󸀠 − (β) − φ 󸀠 + (α)) ≤ 1 4 (β − α) (φ 󸀠 − (β) − φ 󸀠 + (α)) , (15) provided that φ + (α) and φ − (β) are finite. Theorem 7. Let g : [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 ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (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)) ≤ 1 4 (M − m) (φ 󸀠 − (M) − φ 󸀠 + (m)) , (16) where Ψφ(⋅; m,M) : ⟨m,M⟩ → R is defined by Ψφ (t; m,M) = φ (M) − φ (t) M − t − φ (t) − φ (m) t − m . (17) If g ∈ ⟨m,M⟩, then we also have ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (x) − φ (g) ≤ 1 4 (M − m)Ψφ (g;m,M) ≤ 1 4 (M − m) (φ 󸀠 − (M) − φ 󸀠 + (m)) . (18) Proof. Let (6) hold for all s ∈ [α, β]. Then fromTheorem 3 it follows that ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (x) − φ (g) ≤ M − g M − m φ (m) + g − m M − m φ (M) − φ (g) (19) 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 (M − g) (g − m) M − m [ φ (M) − φ (g) M − g − φ (g) − φ (m) g − m ] , (20) 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 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) , (21) and since (M − g) (g − m) M − m ≤ 1 4 (M − m) , (22) 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). 4 Abstract and Applied Analysis If we set that m = 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 ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (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)) ≤ 1 4 (β − α) (φ 󸀠 − (β) − φ 󸀠 + (α)) , (23) where Ψφ(⋅; α, β) : ⟨α, β⟩ → R is defined by Ψφ (t; α, β) = φ (β) − φ (t) β − t − φ (t) − φ (α) t − α , (24) 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 ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (x) − φ (g) ≤ 1 4 (β − α)Ψφ (g; α, β) ≤ 1 4 (β − α) (φ 󸀠 − (β) − φ 󸀠 + (α)) . (25) Theorem 9. Let g : [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 ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (x) − φ (g) ≤ max{ M − g M − m , g − m M − m } ⋅ [φ (m) + φ (M) − 2φ ( m + M 2 )] = { 1 2 + 1 M − m 󵄨󵄨󵄨󵄨󵄨󵄨 m + M 2 − g 󵄨󵄨󵄨󵄨󵄨󵄨 } ⋅ [φ (m) + φ (M) − 2φ ( m + M 2 )] ≤ 1 2 max {M − g, g − m} {φ − (M) − φ 󸀠 + (m)} . (26) 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. Let φ be a convex function on Dφ, x, y ∈ Dφ and p, q ∈ [0, 1] such that p + q = 1. Then min {p, q} [φ (x) + φ (y) − 2φ ( x + y 2 )] ≤ pφ (x) + qφ (y) − φ (px + qy) ≤ max {p, q} [φ (x) + φ (y) − 2φ ( x + y 2 )] . (27) Proof of Theorem 9. Let (6) hold for all s ∈ [α, β]. Then from Theorem 3 it follows that ∫ b a φ (g (x)) dλ (x) ∫ b a dλ (x) ≤ M − g M − m φ (m) + g − m M − m φ (M) (28) 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: ∫ b a φ (g (x)) dλ (x)


Introduction
The authors in [1] gave some conditions on the real Stieltjes measure , not necessarily positive, under which the Jensen inequality and the converse of the Jensen inequality hold for continuous convex function .These results are derived by using the Green function  defined on [ The function  is convex and continuous with respect to both  and .
Several interesting results concerning the Jensen type inequalities have been derived by means of the function .The first one, which is stated in the following theorem, gives the conditions on the real Stieltjes measure , not necessarily positive, under which the Jensen inequality holds for continuous convex function .
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  : [, ] → R the inequality (2) is reversed.
Remark 2. Note that in the case of positive measure  we get some well known results.If the function  is increasing and bounded with () ̸ = (), then inequality (2) becomes Jensen's integral inequality.On the other hand, if the function  is continuous and monotonic, and  is either continuous or of bounded variation, satisfying 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 , not necessarily positive such that () ̸ = (), under which the converse of the Jensen inequality holds for continuous convex function .
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 Theorem 3  =  and  = , the inequality (6) transforms into (see also [1]) 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  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.

Main Results
Throughout this paper we will use the notation The following result holds.
If we set that  =  and  = , we get the following result.
If  ∈ ⟨, ⟩, it is easy to verify that the term on the right side of ( 19) is equal to and the term in the square brackets in (20) is equal to Ψ  (; , ), where the function Ψ  (⋅; , ) is defined in (24).As in [5], it follows that and since the inequality (16) follows.For  =  or  = , the inequality ( 16) is obvious.The inequality in (18) follows directly from the proof of inequality ( 16).
If we set that  =  and  = , we get the following result.
where Ψ  (⋅; , ) : ⟨, ⟩ → R is defined by In order to prove the theorem, we need the following lemma, which is a special case of [6, page 717, Theorem 1] for  = 2.As in [7], by Lemma 10 we get the following: In order to prove the second inequality, it is easy to verify that ( Since  is a convex function on [, ], by the gradient inequality we have (as in [5]) and we finally obtain so the second inequality in (26) follows.
If we set that  =  and  = , we get the following result.
provided that the denominators are nonzero.
Proof.We give the proof for the functional Φ 1 .Define the function  as a linear combination of functions  and Now, applying previous theorem on our function , we obtain that there exists  1 ∈ [, ] such that Since Φ 1 (, ,  0 ) ̸ = 0 (otherwise we would have a contradiction with Φ 1 (, , ) ̸ = 0), we get The proof for the functional Φ 2 is analogous.
Remark 16.If the inverse of the function   /  exists, then (41) gives Remark 17. Theorems 14 and 15 also hold if we set  =  and  = ; that is, if we extend the image of the function  to the entire interval [, ], provided that   + (),   − (),   + (), and   + () are finite.

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]).
We will also need the following result (see [3], page 2).
If the function  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 function  :  → R at mutually different points  0 ,  1 ,  2 ∈  is defined recursively by Remark 27.The value [ 0 ,  1 ,  2 ] is independent of the order of the points  0 ,  1 , and  2 .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  1 →  0 in (50), we get lim provided that   exists.Furthermore, taking the limits   →  0 ,  = 1, 2 in (50), we get lim provided that   exists.A function  :  → R is convex if and only if for every choice of three mutually different points  0 ,  1 ,  2 ∈  [ 0 ,  1 ,  2 ] ≥ 0 holds.Now, we use an idea from [10] to give an elegant method of producing -exponentially convex functions and exponentially convex functions by applying functionals Φ  ( = 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 Φ  (, , ) ≥ 0 ( = 1, 2) under the appropriate assumptions on functions , , and .