Abstract
1. Introduction
2. Main Results
Author Contributions
Competing Interests
References

Open Journal of Mathematical Analysis (OMA)

Volume 3 (2019) Issue 2
Pages: 103
- 105

ISSN: 2616-8111 (Online) 2616-8103 (Print)

DOI: https://www.doi.org/10.30538/psrp-oma2019.0043

Research article

New inequalities based on harmonic log-convex functions

Author(s): ^¹,², ^³

¹Abdus Salam School of Mathematical Sciences, GC University, Lahore, Pakistan.

²Govt. College for Boys, Gulberg Higher Education Department, Punjab, Pakistan.

³Mathematics, College of Engineering and Science, Victoria University, Melbourne City, Australia.

Copyright © Imran Abbas Baloch, Silvestru Sever Dragomir. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Received: 02/09/2019
Accepted: 22/12/2019
Published: 31/12/2019

Abstract

Harmonic convexity is very important new class of non-convex functions, it gained prominence in the Theory of Inequalities and Applications as well as in the rest of Mathematics’s branches. The harmonic convexity of a function is the basis for many inequalities in mathematics. Furthermore, harmonic convexity provides an analytic tool to estimate several known definite integrals like $\int_{a}^{b} \frac{e^{x}}{x^{n}} d x$ , $\int_{a}^{b} e^{x^{2}} d x$ , $\int_{a}^{b} \frac{\sin x}{x^{n}} d x$ and $\int_{a}^{b} \frac{\cos x}{x^{n}} d x$ $\forall n \in N$ , where $a, b \in (0, \infty)$ . In this article, some un-weighted inequalities of Hermite-Hadamard type for harmonic log-convex functions defined on real intervals are given.

Keywords: Harmonic convex functions, Hermite-Hadamard type inequalities, integral inequalities, harmonic log-convex functions.

1. Introduction

During the investigation of convexity, many researchers founded new classes of functions which are not convex in general. Some of them are the so called harmonic convex functions [1], harmonic

(α, m)

-convex functions [2], harmonic

(s, m)

-convex functions [4, 5] and harmonic

(p, (s, m))

-convex functions [3]. For a quick glance on importance of these classes and applications, see [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19] and references therein.

Definition1. A function $f : I \subseteq R ∖ {0} \to R$ is said to be harmonic convex function on $I$ if

f (\frac{x y}{t x + (1 - t) y}) \leq t f (y) + (1 - t) f (x)

(1)

holds for all

x, y \in I

and

t \in [0, 1]

. If the inequality is reversed, then

f

is said to be harmonic concave.

In [5, 20], Baloch et al. and Noor et al. also gave the definition of harmonic log-convex functions as follow:

Definition 2. A function $f : I \subseteq R ∖ {0} \to (0, \infty)$ is said to be harmonic log-convex function on $I$ if

f (\frac{x y}{t x + (1 - t) y}) \leq [f (x)]^{1 - t} [f (y)]^{t}

(2)

% holds for all

x, y \in I

and

t \in [0, 1]

. If the inequality is reversed, then

f

is said to be harmonic log-concave.

In [20], Noor et al. proved the following result for harmonic log-convex functions:

Theorem 3. Let $I \subseteq R ∖ {0}$ be an interval. If $f : I \to (0, \infty)$ is harmonic convex function, then

f (\frac{2 a b}{a + b}) \leq \exp [\frac{a b}{b - a} \int_{a}^{b} \log (\frac{f (x)}{x^{2}}) d x] \leq \sqrt{f (a) f (b)}

(3)

for all

a, b \in I

and

a < b

Here, motivated by the above result we study the class of harmonic log-convex functions and present some new inequalities for this class of functions.

2. Main Results

The following result holds.

Theorem 4. Let $f : I \subseteq R ∖ {0} \to (0, \infty)$ be harmonic log-convex function. Then, for every $t \in [0, 1]$ , we have

\begin{array}{rcl} \int_{a}^{b} f (x) d x & \geq & \int_{a}^{b} [f (x)]^{1 - t} {[\frac{a^{2} b^{2}}{[(a + b) x - a b]^{2}} f (\frac{a b x}{(a + b) x - a b})]}^{t} d x \\ \geq & {\begin{cases} (1 - 2 t) a^{2} b^{2} \int_{\frac{a b}{t a + (1 - t) b}}^{\frac{a b}{(1 - t) a + t b}} \frac{[(a + b) t u - a b]^{2 (t - 1)}}{[a b - (1 - t) (a + b) u]^{2 t}} f (u) d u & if t \neq \frac{1}{2}; \\ \frac{2 a b}{a + b} \ln (\frac{b}{a}) f (\frac{2 a b}{a + b}) & if t = \frac{1}{2} . \end{cases} \end{array}

(4)

Proof. The cases $t = 0, \frac{1}{2}, 1$ are obvious. Assume that $t \in (0, 1) ∖ {\frac{1}{2}} .$ By the harmonic log-convexity of $f$ we have

\begin{array}{rcl} [f (x)]^{1 - t} {[f (\frac{a b x}{(a + b) x - a b})]}^{t} & \geq & f (\frac{\frac{a b x^{2}}{(a + b) x - a b}}{t x + (1 - t) \frac{a b x}{(a + b) x - a b}}) f (\frac{a b x}{(a + b) t x - (2 t - 1) a b}) \end{array}

(5)

for any

x \in [a, b]

. This allows that

[f (x)]^{1 - t} {[\frac{a^{2} b^{2}}{[(a + b) x - a b]^{2}} f (\frac{a b x}{(a + b) x - a b})]}^{t} \geq \frac{a^{2 t} b^{2 t}}{[(a + b) x - a b]^{2 t}} f (\frac{a b x}{(a + b) t x - (2 t - 1) a b}) .

(6)

Integrating the inequality (6) over

x

[a, b]

, we have

\int_{a}^{b} [f (x)]^{1 - t} {[\frac{a^{2} b^{2}}{[(a + b) x - a b]^{2}} f (\frac{a b x}{(a + b) x - a b})]}^{t} d x \geq \int_{a}^{b} \frac{a^{2 t} b^{2 t}}{[(a + b) x - a b]^{2 t}} f (\frac{a b x}{(a + b) t x - (2 t - 1) a b}) d x .

Since

t \neq \frac{1}{2},

then

u = \frac{a b x}{(a + b) t x - (2 t - 1) a b}

is the change of variable with

d x = \frac{(1 - 2 t) a^{2} b^{2}}{[(a + b) t u - a b]^{2}} d u

. For

x = a

, we get

u = \frac{a b}{t a + (1 - t) b}

and for

x = b

, we get

u = \frac{a b}{(1 - t) a + t b} .

Therefore,

\int_{a}^{b} \frac{a^{2 t} b^{2 t}}{[(a + b) x - a b]^{2 t}} f (\frac{a b x}{(a + b) t x - (2 t - 1) a b}) d x = (1 - 2 t) a^{2} b^{2} \int_{\frac{a b}{t a + (1 - t) b}}^{\frac{a b}{(1 - t) a + t b}} \frac{[(a + b) t u - a b]^{2 (t - 1)}}{[a b - (1 - t) (a + b) u]^{2 t}} f (u) d u,

and hence the second inequality (4) is proved. By the Holder integral inequality for

p = \frac{1}{1 - t}

q = \frac{1}{t}

, we have

\begin{array}{rcl} \int_{a}^{b} [f (x)]^{1 - t} {[\frac{a^{2} b^{2}}{[(a + b) x - a b]^{2}} f (\frac{a b x}{(a + b) x - a b})]}^{t} d x \\ \leq {(\int_{a}^{b} {([f (x)]^{1 - t})}^{\frac{1}{1 - t}} d x)}^{1 - t} {(\int_{a}^{b} {({[\frac{a^{2} b^{2}}{[(a + b) x - a b]^{2}} f (\frac{a b x}{(a + b) x - a b})]}^{t})}^{\frac{1}{t}} d x)}^{t} \\ = {(\int_{a}^{b} f (x) d x)}^{1 - t} {(\int_{a}^{b} \frac{a^{2} b^{2}}{[(a + b) x - a b]^{2}} f (\frac{a b x}{(a + b) x - a b}) d x)}^{t} \\ = {(\int_{a}^{b} f (x) d x)}^{1 - t} {(\int_{a}^{b} f (x) d x)}^{t} = \int_{a}^{b} f (x) d x . \end{array}

This proves the first part of inequality (4).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing Interests

The author(s) do not have any competing interests in the manuscript.

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