On n-Polynomial convexity and some related inequalities

1.   Preliminaries

In this paper, a new class of functions, which is a special state of $n$ -polynomial convexity, has been defined and some algebraic properties of this class of function have been investigated. In addition, some Hermite-Hadamard type inequalities were obtained.

A function $f:I\rightarrow \mathbb{R}$ is said to be convex if the inequality

is valid for all $x, y\in I$ and $t\in \lbrack 0, 1]$. If this inequality reverses, then $f$ is said to be concave on interval $I\neq \emptyset$.

Convexity theory provides powerful principles and techniques to study a wide class of problems in both pure and applied mathematics. See articles ^[1,2,3,4,5] and the references therein.

Let $f:I\rightarrow \mathbb{R}$ be a convex function. Then the following inequalities hold

for all $a, b\in I$ with $a < b$. Both inequalities hold in the reversed direction if the function $f$ is concave. This double inequality is well known as the Hermite-Hadamard inequality ^[6]. Some refinements of the Hermite-Hadamard inequality for convex functions have been obtained ^[7,8]. Note that some of the classical inequalities for means can be derived from Hermite-Hadamard integral inequalities for appropriate particular selections of the mapping $f$.

Definition 1. ^[9] Let $h:J\rightarrow \mathbb{R}$ be a non-negative function, $h\neq 0.$ We say that $f:I\rightarrow \mathbb{R}$ is an $h$-convex function, or that $f$ belongs to the class $SX\left(h, I\right)$, if $f$ is non-negative and for all $x, y\in I,$ $\alpha \in \left(0, 1\right)$ we have

If this inequality is reversed, then $f$ is said to be $h$-concave, i.e. $f\in SV\left(h, I\right)$. It is clear that, if we choose $h(\alpha) = \alpha$ and $h(\alpha) = 1$, then the $h$-convexity reduces to convexity and definition of $P$-function, respectively.

Readers can look at ^[10,11] for studies on $h$-convexity.

In ^[12], İșcan gave a refinement of the Hölder integral inequality as follows:

Theorem 1 (Hölder-İșcan integral inequality ^[12]). Let $p > 1$ and $\frac{1}{p}+\frac{1}{q} = 1$. If $f$ and $g$ are real functions defined on interval $\left[a, b\right]$ and if $\left\vert f\right\vert ^{p}$, $\left\vert g\right\vert ^{q}$ are integrable functions on $\left[ a, b\right]$ then

An refinement of power-mean integral inequality as a different version of the Hölder-İșcan integral inequality can be given as follows:

Theorem 2 (Improved power-mean integral inequality ^[13]). Let $q\geq 1$. If $f$ and $g$ are real functions defined on interval $\left[ a, b\right]$ and if $\left\vert f\right\vert$, $\left\vert f\right\vert \left\vert g\right\vert ^{q}$ are integrable functions on $\left[a, b\right]$ then

The main purpose of this paper is to introduce the concept of $n$-polynomial convex functions and establish some results connected with the right-hand side of new inequalities similar to the Hermite-hadamard inequality for these classes of functions. Some applications to special means of positive real numbers are also given.

2.   The definition of $n$-polynomial convex functions

In this section, we introduce a new concept, which is called $n$-polynomial convexity and we give by setting some algebraic properties for the $n$ -polynomial convex functions, as follows:

Definition 2. Let $n\in \mathbb{N}$. A non-negative function $f:I\subset \mathbb{R}\rightarrow \mathbb{R}$ is called $n$-polynomial convex function if for every $x, y\in I$ and $t\in \left[0, 1\right]$,

We will denote by $POLC\left(I\right)$ the class of all $n$-polynomial convex functions on interval $I$.

We note that, every $n$-polynomial convex function is an $h$-convex function with the function $h(t) = \frac{1}{n}\sum_{s = 1}^{n}\left[1-(1-t)^{s}\right]$. Therefore, if $f, g\in POLC\left(I\right)$, then

ⅰ.) $f+g\in POLC\left(I\right)$ and for $c\in \mathbb{R} \;$($c\geq 0$) $cf\in POLC\left(I\right)$ (see ^[9], Proposition 9).

ⅱ.) if $f$ and $g$ be a similarly ordered functions on $I$, then $fg\in POLC\left(I\right).$(see ^[9], Proposition 10).

Also, if $f:I\rightarrow J$ is a convex and $g\in POLC\left(J\right)$ and nondecreasing, then $g\circ f\in POLC\left(I\right)$ (see ^[9], Theorem 15).

Remark 1. We especially note that; if we take $n = 1$ in the inequality (2.1), then the $1$-polynomial convexity reduces to the clasical convexity.

Remark 2. Let the function $f:I\subset \mathbb{R}\rightarrow \left[0, \infty \right)$ be a $2$-polynomial convex function if for every $x, y\in I$ and $t\in \left[ 0, 1\right]$,

It is easily seen that

for all $t\in \left[0, 1\right].$ This shows that every nonnegative convex function is also a $2$-polynomial convex function.

More generally, we can give the following remark together with proof:

Remark 3. Every nonnegative convex function is also an $n$-polynomial convex function. Indeed, this case is clear from the following inequalities

for all $t\in \left[0, 1\right]$ and $n\in \mathbb{N}.$ Now, we will prove that the inequality

holds for all $t\in \left[0, 1\right]$ and $n\in \mathbb{N}$: The following inequality is well known as Bernoulli's inequality in mathematical analysis $\left(1-t\right) ^{n}\geq 1-nt$ for all $t\in \left[ 0, 1\right]$ and $n\in \mathbb{N}.$ From the above inequality, we get

and thus

then we have

The cases of $t = 0$ and $t = 1$ are clear.

Example 1. In case of $n = 2$, the function $f:\left[0, \infty \right) \rightarrow \mathbb{R}$, $f(x) = x$ is a $2$-polynomial convex.

Theorem 3. Let $b > 0$ and $f_{\alpha }:\left[a, b\right] \rightarrow \mathbb{R}$ be an arbitrary family of $n$-polynomial convex functions and let $f(x) = \sup_{\alpha }f_{\alpha }(x)$. If $J = \left\{ u\in \left[a, b\right] :f(u) < \infty \right\}$ is nonempty, then $J$ is an interval and $f$ is an $n$ -polynomial convex function on $J$.

Proof. Let $t\in \left[0, 1\right]$ and $x, y\in J$ be arbitrary. Then

This shows simultaneously that $J$ is an interval, since it contains every point between any two of its points, and that $f$ is an $n$-polynomial convex function on $J$. This completes the proof of theorem.

3.   Hermite-Hadamard inequality for $n$-polynomial convex functions

The goal of this paper is to establish some inequalities of Hermite-Hadamard type for $n$-polynomial convex functions. In this section, we will denote by $L\left[a, b\right]$ the space of (Lebesgue) integrable functions on $\left[a, b\right].$

Theorem 4. Let $f:\left[a, b\right] \rightarrow \mathbb{R}$ be an $n$-polynomial convex function. If $\ a < b$ and $f\in L\left[a, b \right]$, then the following Hermite-Hadaamrd type inequalities hold:

Proof. From the propery of the $n$-polynomial convex function of $f$, we get

By taking integral in the last inequality with respect to $t\in \left[0, 1 \right]$, we deduce that

By using the property of the $n$-polynomial convex function $f$, if the variable is changed as $x = ta+(1-t)b,$ then

where

This completes the proof of theorem.

Remark 4. In case of $n = 1$, the inequality (3.1) coincides with the the inequality (1.1)

4.   New inequalities for $n$-polynomial convex functions

The main purpose of this section is to establish new estimates that refine Hermite-Hadamard inequality for functions whose first derivative in absolute value, raised to a certain power which is greater than one, respectively at least one, is $n$-polynomial convex function. Dragomir and Agarwal ^[14] used the following lemma:

Lemma 1 (^[14]). Let $f:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable mapping on $I^{\circ }$, $a, b\in I^{\circ }$ with $a < b.$ If $f^{\prime }\in L\left[a, b\right]$, then the following identity holds:

Theorem 5. Let $f:I\rightarrow \mathbb{R}$ be a differentiable function on $I^{\circ }$, $a, b\in I^{\circ }$ with $a < b$ and assume that $f^{\prime }\in L\left[a, b\right]$. If $\left\vert f^{\prime }\right\vert$ is an $n$-polynomial convex function on interval $\left[a, b \right]$, then the following inequality holds for $t\in \left[0, 1\right]$.

Proof. Using Lemma 1 and the inequality

we get

where

and $A$ is the arithmetic mean. This completes the proof of theorem.

Corollary 1. If we take $n = 1$ in the inequality (4.1), we get the following inequality:

This inequality coincides with the inequality in ^[14].

Theorem 6. Let $f:I\rightarrow \mathbb{R}$ be a differentiable function on $I^{\circ }$, $a, b\in I^{\circ }$ with $a < b,$ $q > 1, \frac{1}{p}+\frac{1}{q} = 1$ and assume that $f^{\prime }\in L\left[a, b\right]$. If $\left\vert f^{\prime }\right\vert ^{q}$ is an $n$-polynomial convex function on interval $\left[a, b\right]$, then the following inequality holds for $t\in \left[0, 1\right]$.

Proof. Using Lemma 1, Hölder's integral inequality and the following inequality

which is the $n$-polynomial convex function of $\left\vert f^{\prime }\right\vert ^{q}$, we get

where

and $A$ is the arithmetic mean. This completes the proof of theorem.

Corollary 2. If we take $n = 1$ in the inequality (4.2), we get the following inequality:

This inequality coincides with the inequality in ^[14].

Theorem 7. Let $f:I\rightarrow \mathbb{R}$ be a differentiable function on $I^{\circ }$, $a, b\in I^{\circ }$ with $a < b,$ $q\geq1$, and assume that $f^{\prime }\in L\left[a, b\right]$. If $\left\vert f^{\prime }\right\vert ^{q}$ is an $n$ -polynomial convex function on the interval $\left[a, b\right]$, then the following inequality holds for $t\in \left[0, 1\right]$.

Proof. Assume first that $q > 1$. From Lemma 1, Hölder integral inequality and the property of the $n$-polynomial convex function of $\left\vert f^{\prime }\right\vert ^{q}$, we obtain

where

For $q = 1$ we use the estimates from the proof of Theorem 5, which also follow step by step the above estimates. This completes the proof of theorem.

Corollary 3. Under the assumption of Theorem 7 with $q = 1$, we get the conclusion of Theorem 5.

Corollary 4. If we take $n = 1$ in the inequality (4.3), we get the following inequality:

If we take $q = 1$ in the above inequality, then obtained inequality coincides with the inequality in ^[14].

Now, we will prove the Theorem 6 by using Hölder-İșcan integral inequality. Then we will show that the result we have obtained in this theorem gives a better approach than that obtained in the Theorem 6.

Theorem 8. Let $f:I\rightarrow \mathbb{R}$ be a differentiable function on $I^{\circ }$, $a, b\in I^{\circ }$ with $a < b,$ $q > 1, \frac{1}{p}+\frac{1}{q} = 1$ and assume that $f^{\prime }\in L\left[a, b\right]$. If $\left\vert f^{\prime }\right\vert ^{q}$ is an $n$-polynomial convex function on interval $\left[a, b\right]$, then the following inequality holds for $t\in \left[0, 1\right]$.

Proof. Using Lemma 1, Hölder-İșcan integral inequality and the following inequality

which is the $n$-polynomial convex function of $\left\vert f^{\prime }\right\vert ^{q}$, we get

where

This completes the proof of theorem.

Corollary 5. If we take $n = 1$ in the inequality (4.4), we get the following inequality:

This inequality coincides with the inequality of Theorem 3.2 in ^[12].

Remark 5. The inequality (4.4) gives better results than the inequality (4.2). Let us show that

Using concavity of the function $h:\left[0, \infty \right) \rightarrow \mathbb{R}$, $h(x) = x^{\lambda }, 0 < \lambda \leq 1$ by sample calculation we get

which is the required.

Theorem 9. Let $f:I\rightarrow \mathbb{R}$ be a differentiable function on $I^{\circ }$, $a, b\in I^{\circ }$ with $a < b,$ $q\geq1$, and assume that $f^{\prime }\in L\left[a, b\right]$. If $\left\vert f^{\prime }\right\vert ^{q}$ is an $n$ -polynomial convex function on the interval $\left[a, b\right]$, then the following inequality holds for $t\in \left[0, 1\right]$.

where

Proof. Assume first that $q > 1$. From Lemma 1, improved power-mean integral inequality and the property of the $n$-polynomial convex function of $\left\vert f^{\prime }\right\vert ^{q}$, we obtain

where

For $q = 1$ we use the estimates from the proof of Theorem 5, which also follow step by step the above estimates. This completes the proof of theorem.

Corollary 6. If we take $n = 1$ in the inequality (4.5), we get the following inequality:

Remark 6. The inequality (4.5) gives better result than the inequality (4.3). Let us show that

If we use the concavity of the function $h:\left[0, \infty \right) \rightarrow \mathbb{R}$, $h(x) = x^{\lambda }, 0 < \lambda \leq 1$, we get

where

which completes the proof of remark.

5.   Applications for special means

Throughout this section, for shortness, the following notations will be used for special means of two nonnegative numbers $a, b$ with $b > a$:

1. The arithmetic mean

2. The geometric mean

3. The harmonic mean

4. The logarithmic mean

5. The $p$-logaritmic mean

6.The identric mean

These means are often used in numerical approximation and in other areas. However, the following simple relationships are known in the literature: $H\leq G\leq L\leq I\leq A$. It is also known that $L_{p}$ is monotonically increasing over $p\in \mathbb{R},$ denoting $L_{0} = I$ and $L_{-1} = L.$

Proposition 1. Let $a, b\in \left[0, \infty \right)$ with $a < b$ and $n\in \left(-\infty, 0\right) \cup \left[1, \infty \right) \backslash \left\{ -1\right\}$. Then, the following inequalities are obtained:

Proof. The assertion follows from the inequalities (3.1) for the function $f(x) = x^{n}, \; \ x\in \left[0, \infty \right)$.

Proposition 2. Let $a, b\in \left(0, \infty \right)$ with $a < b$. Then, the following inequalities are obtained:

Proof. The assertion follows from (3.1) for the function $f(x) = x^{-1}, \; \ x\in \left(0, \infty \right)$.

Proposition 3. Let $a, b\in \left(0, 1\right]$ with $a < b$. Then, the following inequalities are obtained:

Proof. The assertion follows from the inequalities (3.1) for the function

6.   Conclusion

We established some refinements of the Hermite-Hadamard inequality for functions whose first derivative in absolute value, raised to a certain power which is greater than one, respectively at least one, is $n$-polynomial convexity. Similar method can be applied to the different type of convex functions.

Conflict of interest

The authors declare no conflict of interest in this paper.