An Extension of the Hadamard-Type Inequality for a Convex Function Defined on Modulus of Complex Integral Functions

In this paper we extend the Hadamard’s type inequalities for convex functions defined on the modulus of integral functions in complex field. Firstly, by using the Principal of maximum modulus theorem we show that M(r) and lnM(r) are continuous and convex functions for any non-negative values of r. Finally we derive two inequalities analogous to well known Hadamard’s inequality by using elementary analysis.


Introduction
Let f : I ⊆ R → R is a convex mapping defined on the interval I ⊂ R. If a, b ∈ I, and a < b, then the following double inequality holds.This is called the Hermite-Hadamard inequality.Since its discovery in 1893, Hadamard's inequality (Hadamard, 1893) has been proved to be one of the most useful inequalities in mathematical analysis.A number of papers have been written on this inequality providing new proofs, noteworthy extensions, generalization and numerous applications (Hadamard, 1893;Heing & Maligranda, 1991/1992;Pachpatte, 2003;Mitrinovic, 1970;Tunc, 2012;Dragomir, 1990b) and reference cited therein.The main purpose of this paper is to establish some integral inequality involving the modulus of complex integral functions.Throughout this note, we write integral functions for complex integral functions.Here some necessary definitions and theorem are mentioned which are closely connected to our main result.
Definition 1.2 If the derivative f (z) exists at all points z of a region R, then f (z) is said to be analytic in R and referred to as an analytic function in R or a function analytic in R. The terms regular and holomorphic are sometimes used as synonyms for analytic.
Definition 1.3 If AB and BC are two rectifiable arcs of lengths l and l respectively, which have only the point B in common, the arc AC is evidently also rectifiable, its lengths being l + l .From this it follows that a Jordan arc which consists of a finite number of regular arcs is rectifiable, its length being the sum of the lengths of the regular arcs forming it.Such an arc we call a contour.Also a closed contour means a simple closed Jordan curve which consists of a finite number of regular arcs.Obviously a closed contour is rectifiable.
Definition 1.4An integral function is a function which is analytic for all finite values of z.For example e z , cosz, sinz, and all polynomials are integral functions.
Definition 1.5 The maximum and minimum modulus of an integral function usually denoted by M(r) and m(r) respectively, and defined by Where D is a region bounded by a closed contour C.
Theorem 1.6 (The Principal of Maximum Modulus Theorem) Let f (z) is an analytic function, regular in a region D and on its boundary C, where C is a simple closed contour.A more precise form of the theorem is as follows: Proof.See (Titchmarsh, 1939, pp. 173) and (Copson, 1935, pp. 164).
In Section 2, we are mentioned some necessary lemmas with proof, which are closely connected to our main results.
Mainly here we established that M(r) and lnM(r) are continuous and convex functions for any non-negative real values of r.Finally in Section 3, we established two integral inequalities analogous to the well-known Hadamard's inequality (1.1).

Some Useful Lemmas
In order to establish the result, we need the following lemmas, some of them are discussed in (Titchmarsh, 1939;Polya, 1926).M (r 1 ), hence M(r 1 ) obtained at z = r 1 e iα 1 .Now z = r 1 e iα 1 lies inside the circle |z| = r 2 .So, by using the principal of maximum modulus theorem, we get Therefore, M(r) is a steadily increasing function of r.
(2.9) Also, z = re iα 0 lies inside |z| = r 0 .Now by principal of maximum modulus theorem, we get From inequalities (2.9) and (2.10), we get From equations (2.7) and (2.11), we get lim Therefore, M (r) is continuous at r 0 .Now r 0 is arbitrary, so we say that M (r) is a continuous function for any non-negative real values of r.Proof.By using lemma, we say that M(r) is a increasing function for any non-negative real values of r.Choose, r 1 ≤ r ≤ r 2 , lies on line segment between the points r 1 and r 2 .And for 0 < λ < 1, we get r = λr 1 + (1 − λ) r 2 .Also the value M(r) lies on the curve which is downward of the line segment between the points M(r 1 ) and M(r 2 ), and λM(r 1 ) + (1 − λ) M(r 2 ) lies on this line segment.So we get M(r) ≤ λM(r 1 ) + (1 − λ) M(r 2 ).Therefore M(r) is a convex function for any non-negative real values of r.This completes the proof.Proof.Let r 1 < r 2 < r 3 ≤ R and M i (r i ) is the maximum modulus of | f (z)| on the region bounded by the circles |z| = r i , for i = 1, 2, 3, respectively.Then by using Hadamard's three circle theorem, we get (2.12) The sign of equality will occur only if the function f (z) is constant multiple of a power of z.Excluding this case, we get Taking logarithm on the both sides, we get ln r 3 r 1 lnM 2 < ln r 3 r 2 lnM 1 + ln r 2 r 1 lnM 3 , ⇒ lnM 2 < lnr 3 −lnr 2 lnr 3 −lnr 1 lnM 1 + lnr 2 −lnr 1 lnr 3 −lnr 1 lnM 3 . (2.13) Now lnM (r) is a continuous function of lnr, and so if we put x = lnr, then we get, lnM (r) = ϕ (lnr ) = ϕ(x) .Also consider, x i = lnr i , for i = 1, 2, 3, then So we obtained the following inequality from (2.13) where, λ = (x 3 − x 2 )/(x 3 − x 1 ) < 1, since x 1 < x 2 < x 3 .
Also here we choose x 2 arbitrarily.Hence we can say that ϕ (x) is a convex function of x, i.e. lnM (r) is a convex function of lnr.This completes the proof.

Main Results
Theorem 3.1 Let f (z) is a complex integral function defined on any finite region D: |z| = R, then for any a, b ∈ I ⊂ [0, ∞) with a < b, we get the following inequalities Proof.By using lemma, we get M: I ⊂ [0, ∞) → R is a convex mapping defined on the interval I of real numbers, and a, b ∈ I with a < b.Now by using the Hermite-Hadamard inequality on convex function, we derive the following double inequality Also, by using Lemma 2.2, we get The proof of this theorem is completed.(3.4) The proof of this theorem is completed.

Conclusion
Inequalities (3.1) to (3.4) indicated our final results.These are the extension of the Hadamard's type inequalities (1.1) for the functions M(r) and lnM(r).In the next paper we will give the applications on these results by obtaining some Hadamard's-type inequality for Meromorphic functions in complex field.Specially we will try to obtain Hadamard's-type inequalities for n(r, f ) and T (r, f ), defined on Meromorphic functions.
be an analytic function, regular within and on the closed curve C. Let M be the upper bound of | f (z)| on C. Then the inequality | f (z)| ≤ M holds everywhere within C.Moreover, | f (z)| = M, at a point within C if and only if f (z) is constant".Theorem 1.7 (The Principal of Minimum Modulus Theorem) If f (z) is a non-constant integral function without zeros within the region bounded by a closed contour C, then | f (z)| obtained its minimum value at a point on the boundary of C, i.e. if m is the minimum value of | f (z)| on C, then the inequality holds | f (z)| ≥ m, for any z lies inside C. Theorem 1.8 (Hadamard's Three Circle Theorem) Let f (z) is an analytic function, regular for any z lies inside the annulus r 1

Lemma 2. 2
Let f (z) is an integral function, and M(r) is the maximum modulus of | f (z)| on the region D: |z| ≤ r.Then | f (z)| ≤ M(r), and | f (r)| ≤ M(r).Proof.By using the principal of maximum modulus theorem, we get | f (z)| ≤ M(r), for any z lies inside and on the circle |z| = r.Also, if we take z = r + 0 × i = r, then this z lies on the circle |z| = r, hence for this z = r, we get | f (r)| ≤ M(r).This completes the proof.Lemma 2.3 Let f (z) is an integral function and M (r) be the maximum modulus of | f (z)| on the region D: |z| ≤ r.Then M(r) is a convex function for any non-negative real values of r.

Lemma 2. 4
Let f (z) is an analytic function in |z| ≤ R, f (0) 0, and M (r) the maximum modulus of | f (z)| on the region D: |z| ≤ r.Then lnM (r) is a convex function of lnr for any positive real values of r.