ON A MEAN VALUE THEOREM CONNECTED WITH HERMITE-HADAMARD’S INEQUALITY

. In this paper we prove a mean-value theorem for integral calculus, then we demonstrate properties of the mean point. In the end we give an extension of Hermite-Hadamard’s inequality.


INTRODUCTION
In this article, we will give two interpolations of the Hermite-Hadamard's inequality.We start from: .
For the proofs we refer to [3] or [8].
x 2 ≤ b, and considering Lemma 2, we find that there exists t ∈ (0, 1  2 ) so that a+b Considering these relations and by use of the fact that function f is convex, we obtain Adding the above relations, we obtain that is g(x 1 ) ≤ g(x 2 ).So, the function g is increasing on 0, b−a 2 ., such that Applying Rolle's theorem to function F on the intervals 0, 1 2 and 1 2 , 1 , we obtain that there exists We have , and so we have that As an observation, it is possible that c 2 = 1 − c 1 .Consequently, from (4) we have that there exists c ∈ 0, 1  2 such that ( 5) Then, considering (5), we have that there exists α ∈ 0, b−a 2 such that (2) holds.
Since the function g is continuous on interval , such that Replacing the function g, we obtain that and there from comes (6).
has one and only one solution on the interval 0, 1 2 .
Proof.Let f : 0, 1 2 → R be the function defined by Then and from the variation of function f , it results that the function f has only one zero on the interval 0, 1 2 .
Proof.We apply now Theorem 5 to the restriction of the function f on the interval [a, x].
Theorem 8. We have the function f : [a, b] → R which verifies the conditions: (i) there exists a neighborhood V of the point a so that the function given by Theorem 7 has the property that there exists lim x a c(x) x−a = l, l ∈ 0, 1  2 , and l is the unique solution of the equation (8) Proof.We consider the function We calculate the limit According to Taylor's formula with the rest of Lagrange, for each and from where, considering (ii) it results that , from where we obtain that x−a are bounded on V ∩ (a, b] and considering the condition (i), it results that the functions From these observation and from (11), it results that (12) lim Considering ( 7) and ( 10), we have and using ( 12) and ( 13), we obtain x−a n .

Theorem 1 (f
Hermite-Hadamard).Let f : [a, b] → R be a convex function.Then (x)dx ≤ f (a)+f (b) 2 x 1 , x 2 ∈ [a, b], x 1 ≤ x 2 are symmetrical with respect to the middle point of the interval [a, b] iff there exists c ∈ 0, b−a 2 such that x 1 = a+b 2 − c and x 2 = a+b 2 + c.Lemma 2. Let [a, b], a < b, be an interval.The points x 1 , x 2 ∈ [a, b], x 1 ≤ x 2 are symmetrical with respect to the middle point of the interval [a, b]

Lemma 4 .
Let f : [a, b] → R be a continuous function on [a, b] with the property that b a f (x)dx = 0. Then there exists α ∈ 0, b−a 2

Corollary 9 . 2 √ 3 , 2 whichTheorem 10 .
In the conditions of Theorem 8, for n = 2, we obtain that l = lim x a c(x)x−a = 1 and for n = 4, we obtain that l = lim x Let 0 < a and f :[a, b] → R be the function f (x) = 1 x .According to Theorem 7, ∀x ∈ (a, b], ∃ c(x) ∈ 0, x−a Let f : [a, b] → R be a convexfunction, f is continuous on the right at a and continuous on the left at b. Then there exists α ∈ 0, b−a 2 so that ∀x ∈ [0, α], ∀y ∈ α, b−a 2 , we have

1 1
b), and since f is continuous on the right at a and continuous on the left at b, we know that f is continuous on interval [a, b].Next Theorem 3 and Theorem 5 are to be applied.Remarks.Theorem 10 is an extension and refinement of Hermite-Hadamard's inequality.Next, we will show that the maximal value of α with the property from Theorem 5, isα max = b−a 2 .Example.Let f : [−1, 1] → R be the function defined by f c (x) = x − c 1−c , x ∈ [−1, −c) 0 , x ∈ [−c, c ] −c x − c 1−c ,x ∈ (c, 1],