SOME INEQUALITIES OF HERMITE-HADAMARD TYPE FOR TRIGONOMETRICALLY ρ-CONVEX FUNCTIONS

In this paper we establish several Hermite-Hadamard type integral inequalities for trigonometrically ρ-convex functions. Applications for special means are provided as well.


Introduction
The following integral inequality which holds for any convex function f : [a, b] → R, is well known in the literature as the Hermite-Hadamard inequality.
There is an extensive amount of literature devoted to this simple and nice result which has many applications in the Theory of Special Means and in Information Theory for divergence measures, for which we would like to refer the reader to the monograph [8], the recent survey paper [7] and the references therein.
Let I be a finite or infinite open interval of real numbers and ρ > 0.
Following [1], we say that a function f :

If we take
We have the following properties of trigonometrically ρ-convex on I, [1] (i) A trigonometrically ρ-convex function f : I → R has finite right and left derivatives f + (x) and f − (x) at every point x ∈ I and f − (x) ≤ f + (x) .The function f is differentiable on I with the exception of an at most countable set.(ii) A necessary and sufficient condition for the function f : I → R to be trigonometrically ρ-convex function on I is that it satisfies the gradient inequality for any x, y ∈ I where (iii) A necessary and sufficient condition for the function f to be a trigonometrically ρ-convex in I, is that the function is nondecreasing on I, where a ∈ I. (iv) Let f : I → R be a two times continuously differentiable function on I.
Then f is trigonometrically ρ-convex on I if and only if for all x ∈ I we have For other properties of trigonometrically ρ-convex functions, see [1].
Using the condition (1.5) one can also observe that any nonnegative twice differentiable and convex function on I is also trigonometrically ρ-convex on I for any ρ > 0.
There exists also concave functions on an interval that are trigonometrically ρ-convex on that interval for some ρ > 0.
In this paper we establish several Hermite-Hadamard type integral inequalities for trigonometrically ρ-convex functions.Applications for special means are provided as well.

Some Hermite-Hadamard Type Inequalities
We start with the following lemma of interest in itself: Proof.From (1.2) we have by replacing x with a From (1.3) for t = 1 2 and a = u, b = v we get which implies that for any u, v ∈ I. Now, if we in (2.5) take v = x and u = a + b − x, then we get the first inequality in (2.1).
Remark 1.By taking x = (1 − t) a + tb in (2.1) we get the equivalent double inequality for any a, b ∈ I with 0 < |b − a| < π ρ and t ∈ [0, 1] .We have the following Hermite-Hadamard type inequality that was obtained in a different and slightly more difficult manner in [2] Theorem 1. Assume that the function f : I → R is trigonometrically ρ-convex on I. Then for any a, b ∈ I with 0 < b − a < π ρ we have Proof.Integrating the inequality (2.1) over x on [a, b] we get and by the change of variable y = a We use the notation sec t = 1 cos t for t = (2k + 1) π 2 , k is an integer, to state the following weighted integral inequality of Hermite-Hadamard type: Proof.From (2.1) we have Using the change of variable y = a and by (2.11) we obtain the desired result (2.9).
The following weighted inequality also holds: Proof.If we multiply the inequality (2.1) by cos ρ x − a+b 2 ≥ 0 we get and by (2.13) we get (2.12).

Related Results
Theorem 4. Assume that the function f : I → R is trigonometrically ρ-convex on I. Then for any a, b ∈ I with 0 < b − a < π ρ we have Proof.If we take the integral over y ∈ [a, b] in the gradient inequality (1.4) we get By utilising the inequality (3.3) we then get (3.1).
Remark 3. We observe that, if we take x = a+b 2 in (3.1), then we recapture the first inequality in (2.7).

Preprints
Proof.We have by the gradient inequality (1.4) that for any x ∈ [a, b] , hence by (3.7) we get (3.4).
Remark 4. We observe that, if we take x = a+b 2 in (3.4), then we recapture the first inequality in (2.9).
From a different perspective, we have: Theorem 6. Assume that the function f : I → R is trigonometrically ρ-convex on I. Then for any a, b ∈ I with 0 < b − a < π ρ we have In particular, for y = a+b 2 , we get Proof.The function f : I → R is differentiable almost everywhere and we have
It is well known that L p is monotonic nondecreasing over p ∈ R with L −1 := L and L 0 := I.
In particular, we have the inequalities the function is convex and therefore trigonometrically ρ-convex for any ρ > 0. If p ∈ (0, 1) then the function is concave and This shows that for p ∈ (0, 1) and ρ > 0 the function f (x) = x p is trigonometrically ρ-convex on 1 showing that the function g is strictly increasing on (0, ∞) and the equation g (x) = 0 has a unique solution.Therefore g (x) < 0 for x ∈ (0, x ρ ) and g (x) > 0 for x ∈ (x ρ , ∞) , where x ρ is the unique solution of the equation ln x = 1 ρ 2 x 2 .In conclusion, if ρ > 0, then the function f (x) = ln x is trigonometrically ρconcave on (0, x ρ ) and trigonometrically ρ-convex on (x ρ , ∞) .
is trigonometrically ρ-convex on I if for any closed subinterval [a, b] of I with 0 < b − a < π ρ we have (1.2) f (x) ≤ sin [ρ (b − x)] sin [ρ (b − a)] f (a) + sin [ρ (x − a)] sin [ρ (b − a)] f (b) for all x ∈ [a, b] .If the inequality (1.2) holds with " ≥ ", then the function will be called trigonometrically ρ-concave on I.Geometrically speaking, this means that the graph of f on [a, b] lies nowhere above the ρ-trigonometric function determined by the equationH (x) = H (x; a, b, f ) := A cos (ρx) + B sin (ρx)where A and B are chosen such that H (a) = f (a) and H (b) = f (b) .

Lemma 1 .
Assume that the function f : I → R is trigonometrically ρ-convex on I. Then for any a, b ∈ I with 0 < b − a < π ρ and x ∈ [a, b] we have

Theorem 3 .
Assume that the function f : I → R is trigonometrically ρ-convex on I. Then for any a, b ∈ I with 0 < b − a < π ρ we have the change of variable y = a + b − x we have b a f (a + b − x) cos ρ x − a

)
The logarithmic mean:L = L (a, b)