On Hermite–Hadamard-Type Inequalities for Functions Satisfying Second-Order Differential Inequalities

Abstract

Hermite–Hadamard inequality is a double inequality that provides an upper and lower bounds of the mean (integral) of a convex function over a certain interval. Moreover, the convexity of a function can be characterized by each of the two sides of this inequality. On the other hand, it is well known that a twice differentiable function is convex, if and only if it admits a nonnegative second-order derivative. In this paper, we obtain a characterization of a class of twice differentiable functions (including the class of convex functions) satisfying second-order differential inequalities. Some special cases are also discussed.

1. Introduction

Inequalities involving convex functions are very useful in many branches of mathematics. The Hermite–Hadamard inequality is the one of the most important inequality for convex functions. This inequality provides an upper and lower bounds of the mean of a convex function over a certain interval. It is mostly used in mathematics to study the properties of convex functions and their applications in optimization and approximation theory, see, e.g., [1,2,3].

A real-valued function f defined in an interval I is convex if:

$$f(\iota y+(1-\iota \left)z\right)\le \iota f\left(y\right)+(1-\iota )f\left(z\right)$$

for every $\iota \in [0,1]$ and $y,z\in I$. If f is twice differentiable, then f is convex, if and only if its second derivative is nonnegative. The Hermite–Hadamard inequality can be stated as follows: Let f be a real-valued convex function in an interval I. Then, for all $x,y\in I$ with $x<y$, we have:

$$f\left(\frac{x+y}{2}\right)\le \frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \le \frac{f\left(x\right)+f\left(y\right)}{2}.$$

(1)

Many generalizations and extensions of (1) can be found in the literature. For instance, Dragomir and Agarwal [2] studied the following class of functions:

$$\mathcal{F}=\left\{f:[a,b]\to \mathbb{R}:f \mathrm{is} \mathrm{differentiable},|f^{\prime }| \mathrm{is} \mathrm{convex}\right\}.$$

They proved that, if $f\in \mathcal{F}$, then:

$$\left|\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx-\frac{f\left(a\right)+f\left(b\right)}{2}\right|\le \frac{b-a}{8}\left(|f^{\prime }\left(a\right)|+|f^{\prime }\left(b\right)|\right).$$

Some improvements and extensions of the above result have been obtained by some authors, see, e.g., [4,5,6,7]. Other extensions of (1) to various classes of functions have been obtained: s-convex functions [8,9,10,11], log-convex functions [12,13,14], h-convex functions [15,16], and m-convex functions [17,18,19,20]. For other classes of functions, we refer to [21,22,23,24] and the references therein. Some extensions of Hermite–Hadamard inequality to a higher dimension can be found in [25,26,27,28,29].

It is interesting to notice that each of the two sides of (1) provides a characterization of convex functions. Namely, if f is a real valued continuous function in an interval I, then the following statements are equivalent:

(i)

f is convex;

(ii)

For all $x,y\in I$ with $x<y$:

$$\frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \ge f\left(\frac{x+y}{2}\right);$$

(iii)

For all $x,y\in I$ with $x<y$:

$$\frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \le \frac{f\left(x\right)+f\left(y\right)}{2}.$$

(2)

The proof of the implication (ii)⇒(i) can be found in ([30], p. 98). For the proof of the implication (iii)⇒(i), we refer to (Problem Q, [31], p. 15). On the other hand, one can check easily that (iii) is equivalent to:

$${\int }_x^{y}f\left(\tau \right)d\tau \le H^{\prime }\left(x\right)f\left(x\right)-H^{\prime }\left(y\right)f\left(y\right)$$

(3)

for all $x,y\in I$ with $x<y$, where:

$$H\left(z\right)=\frac{1}{2}(z-x)(y-z),x\le z\le y.$$

Observe that H is the unique (nonnegative) solution to the boundary value problem:

$$\left\{\begin{array}{c}{H}^{″}\left(z\right)=-1,x<z<y,\hfill \\ H\left(x\right)=H\left(y\right)=0.\hfill \end{array}\right.$$

From the above remarks, we deduce that, if f is twice differentiable in I, then $f^{″}\ge 0$ (i.e., f is convex), if and only if (3) holds for all $x,y\in I$ with $x<y$. Thus, (3) provides a characterization of twice differentiable functions in I, having a nonnegative second derivative.

Motivated by the above discussion, our aim in this paper is to obtain a characterization of the class of twice continuously differentiable functions f in I, satisfying second-order differential inequalities of the form:

$${\left(\alpha f^{\prime }\right)}^{\prime }+\beta f+\gamma \ge 0,$$

(4)

where $\alpha$ is twice continuously differentiable in I and $\beta ,\gamma$ are continuous in I. We shall assume that for all $x,y\in I$ with $x<y$, there exists a unique nonnegative solution H to the boundary value problem:

$$\left\{\begin{array}{c}{\left(\alpha \left(z\right)H^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)H\left(z\right)=-1,x<z<y,\hfill \\ H\left(x\right)=H\left(y\right)=0.\hfill \end{array}\right.$$

The rest of the paper is organized as follows. Section 2 is devoted to the main results and their proofs. Namely, we establish a characterization of the class of functions f satisfying differential inequalities of the form (4). In Section 3, we discuss some special cases of (4).

2. Main Results

For any interval J of $\mathbb{R}$, by $C^n\left(J\right)$, where $n\ge 0$ is a natural number, we mean the space of n-continuously differentiable functions in J.

Let I be an open interval of $\mathbb{R}$. Let $\alpha \in C^1\left(I\right)$ and $\beta ,\gamma \in C\left(I\right)$. Throughout this section, it is assumed that for all $x,y\in I$ with $x<y$, there exists a unique nonnegative solution $H\in C^2\left(\right[x,y\left]\right)\cap C\left([x,y]\right)$ to the Dirichlet boundary value problem:

$$\left\{\begin{array}{c}{\left(\alpha \left(z\right)H^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)H\left(z\right)=-1,x<z<y,\hfill \\ H\left(x\right)=H\left(y\right)=0.\hfill \end{array}\right.$$

(5)

We are concerned with the class of functions $f\in C^2\left(I\right)$ satisfying the second-order differential inequality:

$${\left(\alpha \left(z\right)f^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)f\left(z\right)+\gamma \left(z\right)\ge 0,z\in I.$$

(6)

Our main result, which is stated below, provides a characterization of this class of functions.

Theorem 1. 

Let $\alpha \in C^1\left(I\right)$, $\beta ,\gamma \in C\left(I\right)$ and $f\in C^2\left(I\right)$. The following statements are equivalent:

(i)

(6) holds;

(ii)

For all $x,y\in I$ with $x<y$, it holds that:

$${\int }_x^{y}f\left(\tau \right)d\tau \le H^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-H^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+{\int }_x^y\gamma \left(\tau \right)H\left(\tau \right)d\tau .$$

(7)

Proof. 

Assume that (6) holds. Let $x,y\in I$ with $x<y$. Multiplying (6) by H (notice that $H\ge 0$) and integrating over $[x,y]$, we obtain:

$${\int }_x^y{\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }H\left(\tau \right)d\tau +{\int }_x^y\beta \left(\tau \right)f\left(\tau \right)H\left(\tau \right)d\tau \ge -{\int }_x^y\gamma \left(\tau \right)H\left(\tau \right)d\tau .$$

(8)

An integration by parts gives us that:

$${\int }_x^y{\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }H\left(\tau \right)d\tau ={\left[\alpha \left(\tau \right)f^{\prime }\left(\tau \right)H\left(\tau \right)\right]}_{\tau =x}^y-{\int }_x^y{f}^{\prime }\left(\tau \right)\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)d\tau .$$

On the other hand, by (5), we have $H\left(x\right)=H\left(y\right)=0$, which yields:

$${\left[\alpha \left(\tau \right)f^{\prime }\left(\tau \right)H\left(\tau \right)\right]}_{\tau =x}^y=0.$$

Then, it holds that:

$${\int }_x^y{\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }H\left(\tau \right)d\tau =-{\int }_x^y{f}^{\prime }\left(\tau \right)\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)d\tau .$$

Integrating again by parts, we obtain:

$$\begin{array}{cc}& {\int }_x^y{\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }H\left(\tau \right)d\tau \hfill \\ & =-{\left[f\left(\tau \right)\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right]}_{\tau =x}^y+{\int }_x^{y}f\left(\tau \right){\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)}^{\prime }d\tau \hfill \\ & =-H^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+H^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)+{\int }_x^{y}f\left(\tau \right){\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)}^{\prime }d\tau .\hfill \end{array}$$

However, due to (5), we have ${\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)}^{\prime }=-1-\beta \left(\tau \right)H\left(\tau \right)$, which yields:

$$\begin{array}{cc}\hfill & {\int }_x^y{\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }H\left(\tau \right)d\tau \hfill \\ \hfill & =-H^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+H^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-{\int }_x^{y}f\left(\tau \right)d\tau -{\int }_x^y\beta \left(\tau \right)f\left(\tau \right)H\left(\tau \right)d\tau .\hfill \end{array}$$

(9)

Thus, (7) follows from (8) and (9). This shows that (i)⇒(ii). Assume now that (ii) holds. Let $x\in I$ be fixed. Then, for all $\epsilon >0$ (sufficiently small), we have:

$${\int }_{x-\epsilon }^{x+\epsilon }f\left(\tau \right)d\tau \le H^{\prime }(x-\epsilon )\alpha (x-\epsilon )f(x-\epsilon )-H^{\prime }(x+\epsilon )\alpha (x+\epsilon )f(x+\epsilon )+{\int }_{x-\epsilon }^{x+\epsilon }\gamma \left(\tau \right)H\left(\tau \right)d\tau ,$$

(10)

where H is the unique positive solution to the boundary value problem:

$$\left\{\begin{array}{c}{\left(\alpha \left(z\right)H^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)H\left(z\right)=-1,x-\epsilon <z<x+\epsilon ,\hfill \\ H(x-\epsilon )=H(x+\epsilon )=0.\hfill \end{array}\right.$$

(11)

Moreover, by (11), we have:

$${\int }_{x-\epsilon }^{x+\epsilon }f\left(\tau \right)d\tau =-{\int }_{x-\epsilon }^{x+\epsilon }\left({\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)}^{\prime }+\beta \left(\tau \right)H\left(\tau \right)\right)f\left(\tau \right)d\tau .$$

Integrating by parts, we obtain:

$$\begin{array}{cc}& {\int }_{x-\epsilon }^{x+\epsilon }f\left(\tau \right)d\tau \hfill \\ & =-{\int }_{x-\epsilon }^{x+\epsilon }{\left(\alpha \left(\tau \right)H^{\prime }\left(\tau \right)\right)}^{\prime }f\left(\tau \right)d\tau -{\int }_{x-\epsilon }^{x+\epsilon }\beta \left(\tau \right)H\left(\tau \right)f\left(\tau \right)d\tau \hfill \\ & =-{\left[\alpha \left(\tau \right)H^{\prime }\left(\tau \right)f\left(\tau \right)\right]}_{\tau =x-\epsilon }^{x+\epsilon }+{\int }_{x-\epsilon }^{x+\epsilon }{H}^{\prime }\left(\tau \right)\alpha \left(\tau \right)f^{\prime }\left(\tau \right)d\tau -{\int }_{x-\epsilon }^{x+\epsilon }\beta \left(\tau \right)H\left(\tau \right)f\left(\tau \right)d\tau \hfill \\ & =\alpha (x-\epsilon )H^{\prime }(x-\epsilon )f(x-\epsilon )-\alpha (x+\epsilon )H^{\prime }(x+\epsilon )f(x+\epsilon )+{\left[H\left(\tau \right)\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right]}_{\tau =x-\epsilon }^{x+\epsilon }\hfill \\ & -{\int }_{x-\epsilon }^{x+\epsilon }H\left(\tau \right){\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }d\tau -{\int }_{x-\epsilon }^{x+\epsilon }\beta \left(\tau \right)H\left(\tau \right)f\left(\tau \right)d\tau .\hfill \end{array}$$

Since $H(x-\epsilon )=H(x+\epsilon )=0$, we obtain:

$$\begin{array}{cc}& {\int }_{x-\epsilon }^{x+\epsilon }f\left(\tau \right)d\tau \hfill \\ & =\alpha (x-\epsilon )H^{\prime }(x-\epsilon )f(x-\epsilon )-\alpha (x+\epsilon )H^{\prime }(x+\epsilon )f(x+\epsilon )\hfill \\ & -{\int }_{x-\epsilon }^{x+\epsilon }H\left(\tau \right)\left({\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }+\beta \left(\tau \right)f\left(\tau \right)\right)d\tau .\hfill \end{array}$$

Hence, by (10), it holds that:

$${\int }_{x-\epsilon }^{x+\epsilon }H\left(\tau \right)\left({\left(\alpha \left(\tau \right)f^{\prime }\left(\tau \right)\right)}^{\prime }+\beta \left(\tau \right)f\left(\tau \right)+\gamma \left(\tau \right)\right)d\tau \ge 0.$$

Since $H\ge 0$ and ${\left(\alpha f^{\prime }\right)}^{\prime }+\beta f+\gamma \in C\left([x-\epsilon ,x+\epsilon ]\right)$, then there exists $z_{\epsilon }\in [x-\epsilon ,x+\epsilon ]$ such that:

$${\left(\alpha \left(z_{\epsilon }\right)f^{\prime }\left(z_{\epsilon }\right)\right)}^{\prime }+\beta \left(z_{\epsilon }\right)f\left(z_{\epsilon }\right)+\gamma \left(z_{\epsilon }\right)\ge 0.$$

Passing to the limit as $\epsilon \to 0^{+}$ in the above inequality, we obtain:

$${\left(\alpha \left(x\right)f^{\prime }\left(x\right)\right)}^{\prime }+\beta \left(x\right)f\left(x\right)+\gamma \left(x\right)\ge 0,$$

which proves that (6) holds. This shows that (ii)⇒(i). □

Replacing f by $-f$ and $\gamma$ by $-\gamma$ in Theorem 1, we obtain the following result.

Theorem 2. 

Let $\alpha \in C^1\left(I\right)$, $\beta ,\gamma \in C\left(I\right)$ and $f\in C^2\left(I\right)$. The following statements are equivalent:

(i)

${\left(\alpha \left(z\right)f^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)f\left(z\right)+\gamma \left(z\right)\le 0,z\in I$;

(ii)

For all $x,y\in I$ with $x<y$, it holds that:

$${\int }_x^{y}f\left(\tau \right)d\tau \ge H^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-H^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+{\int }_x^y\gamma \left(\tau \right)H\left(\tau \right)d\tau .$$

From Theorem 1, we deduce the following result.

Corollary 1. 

Let $\alpha \in C^1\left(I\right)$, $\beta ,\gamma \in C\left(I\right)$ and $f\in C^2\left(I\right)$. If (6) holds, then for all $x,y\in I$ with $x<y$, we have:

$$\begin{array}{ccc}\hfill {\int }_x^{y}f\left(\tau \right)d\tau & \le & H_1^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-H_2^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+{\int }_x^y\gamma \left(\tau \right)H\left(\tau \right)d\tau \hfill \\ & & -\left[H_1^{\prime }\left(\frac{x+y}{2}\right)-H_2^{\prime }\left(\frac{x+y}{2}\right)\right]\alpha \left(\frac{x+y}{2}\right)f\left(\frac{x+y}{2}\right),\hfill \end{array}$$

(12)

where $H_1$ and $H_2$ are the unique nonnegative solutions to the boundary value problems:

$$\left\{\begin{array}{c}{\left(\alpha \left(z\right)H_1^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)H_1\left(z\right)=-1,x<z<\frac{x+y}{2}\hfill \\ H_1\left(x\right)=H_1\left(\frac{x+y}{2}\right)=0\hfill \end{array}\right.,$$

(13)

$$\left\{\begin{array}{c}{\left(\alpha \left(z\right)H_2^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)H_2\left(z\right)=-1,\frac{x+y}{2}<z<y\hfill \\ H_2\left(\frac{x+y}{2}\right)=H_2\left(y\right)=0\hfill \end{array}\right.,$$

(14)

and

$$H\left(z\right)=\left\{\begin{array}{ccc}{H}_1\left(z\right)\hfill & if\hfill & x\le z\le \frac{x+y}{2},\hfill \\ H_2\left(z\right)\hfill & if\hfill & \frac{x+y}{2}<z\le y.\hfill \end{array}\right.$$

(15)

Proof. 

Writing (7) with $\frac{x+y}{2}$ instead of y, we obtain:

$$\begin{array}{cc}\hfill & {\int }_x^{\frac{x+y}{2}}f\left(\tau \right)d\tau \hfill \\ \hfill & \le H_1^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-H_1^{\prime }\left(\frac{x+y}{2}\right)\alpha \left(\frac{x+y}{2}\right)f\left(\frac{x+y}{2}\right)+{\int }_x^{\frac{x+y}{2}}\gamma \left(\tau \right)H_1\left(\tau \right)d\tau .\hfill \end{array}$$

(16)

Similarly, writing (7) with $\frac{x+y}{2}$ instead of x, we obtain:

$$\begin{array}{cc}\hfill & {\int }_{\frac{x+y}{2}}^{y}f\left(\tau \right)d\tau \hfill \\ \hfill & \le H_2^{\prime }\left(\frac{x+y}{2}\right)\alpha \left(\frac{x+y}{2}\right)f\left(\frac{x+y}{2}\right)-H_2^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+{\int }_{\frac{x+y}{2}}^y\gamma \left(\tau \right)H_2\left(\tau \right)d\tau .\hfill \end{array}$$

(17)

Adding (16) to (17), we obtain (12). □

Similarly, from Theorem 2, we deduce the following result.

Corollary 2. 

Let $\alpha \in C^1\left(I\right)$, $\beta ,\gamma \in C\left(I\right)$ and $f\in C^2\left(I\right)$. If:

$${\left(\alpha \left(z\right)f^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)f\left(z\right)+\gamma \left(z\right)\le 0,z\in I,$$

then for all $x,y\in I$ with $x<y$, we have:

$$\begin{array}{ccc}\hfill {\int }_x^{y}f\left(\tau \right)d\tau & \ge & H_1^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-H_2^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+{\int }_x^y\gamma \left(\tau \right))H\left(\tau \right)d\tau \hfill \\ & & -\left[H_1^{\prime }\left(\frac{x+y}{2}\right)-H_2^{\prime }\left(\frac{x+y}{2}\right)\right]\alpha \left(\frac{x+y}{2}\right)f\left(\frac{x+y}{2}\right),\hfill \end{array}$$

where $H_1$ (resp. $H_2$) is the unique nonnegative solution to (13) (resp. (14)) and H is defined by (15).

From Corollary 1, we deduce the following refinement of Hermite–Hadamard inequality (see [29]).

Corollary 3. 

Let $f\in C^2\left(I\right)$ be a convex function. Then, for all $x,y\in I$ with $x<y$, we have:

$$\frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \le \frac{1}{2}\left(\frac{f\left(x\right)+f\left(y\right)}{2}+f\left(\frac{x+y}{2}\right)\right).$$

(18)

Proof. 

Taking:

$$\alpha =1,\beta =\gamma =0$$

in Corollary 1, we obtain:

$$H_1\left(z\right)=\frac{1}{4}(z-x)(x+y-2z),x\le z\le \frac{x+y}{2}$$

and

$$H_2\left(z\right)=\frac{1}{4}(y-z)(2z-x-y),\frac{x+y}{2}\le z\le y.$$

Then, by (12), we obtain (18). □

Similarly, from Corollary 2, we deduce the following result.

Corollary 4. 

Let $f\in C^2\left(I\right)$ be a concave function. Then, for all $x,y\in I$ with $x<y$, we have

$$\frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \ge \frac{1}{2}\left(\frac{f\left(x\right)+f\left(y\right)}{2}+f\left(\frac{x+y}{2}\right)\right).$$

3. Applications

In this section, some special cases of Theorems 1 and 2 are discussed. Namely, we provide characterizations of various classes of functions satisfying differential inequalities of type (6). We first consider the classes of functions:

$${\mathcal{F}}_{\ell }^{+}=\left\{f\in C^2\left(I\right):f^{″}\left(z\right)\ge \ell ,z\in I\right\}$$

(19)

and

$${\mathcal{F}}_{\ell }^{-}=\left\{f\in C^2\left(I\right):f^{″}\left(z\right)\le \ell ,z\in I\right\},$$

(20)

where $\ell \in \mathbb{R}$ is a constant. Observe that for $\ell =0$, ${\mathcal{F}}_0^{+}$ reduces to the class of twice continuously differentiable convex functions, while ${\mathcal{F}}_0^{-}$ reduces to the class of twice continuously differentiable concave functions. We recall that in [29], Niculescu and Persson proved that, if $f\in {\mathcal{F}}_{\ell }^{+}$, then for all $x,y\in I$ with $x<y$, it holds that:

$$\frac{f\left(x\right)+f\left(y\right)}{2}-\frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \ge \frac{\ell {(y-x)}^2}{12}.$$

(21)

Furthermore, if $f\in {\mathcal{F}}_{\ell }^{-}$, then for all $x,y\in I$ with $x<y$, it holds that:

$$\frac{f\left(x\right)+f\left(y\right)}{2}-\frac{1}{y-x}{\int }_x^{y}f\left(\tau \right)d\tau \le \frac{\ell {(y-x)}^2}{12}.$$

(22)

In this section, we show that (21) (resp. (22)) provides a characterization of the class of functions ${\mathcal{F}}_{\ell }^{+}$ (resp. ${\mathcal{F}}_{\ell }^{-}$). We next consider the classes of functions

$${\mathcal{G}}_{\lambda }^{+}=\left\{f\in C^2\left(I\right):f^{″}\left(z\right)-\lambda f\left(z\right)\ge 0,z\in I\right\}$$

(23)

and

$${\mathcal{G}}_{\lambda }^{-}=\left\{f\in C^2\left(I\right):f^{″}\left(z\right)-\lambda f\left(z\right)\le 0,z\in I\right\},$$

(24)

where $\lambda >0$. Observe that when $\lambda =0$, ${\mathcal{G}}_0^{+}$ reduces to the class of twice continuously differentiable convex functions, while ${\mathcal{G}}_0^{-}$ reduces to the class of twice continuously differentiable concave functions.

3.1. Characterizations of the Classes of Functions ${\mathcal{F}}_{\ell }^{\pm }$

Let I be an open interval of $\mathbb{R}$. Let $\ell \in \mathbb{R}$. The following result provides a characterization of the class of functions ${\mathcal{F}}_{\ell }^{+}$ defined by (19).

Corollary 5. 

Let $f\in C^2\left(I\right)$. The following statements are equivalent:

(i)

$f\in {\mathcal{F}}_{\ell }^{+}$;

(ii)

For all $x,y\in I$ with $x<y$, (21) holds.

Proof. 

Observe that:

$${\mathcal{F}}_{\ell }^{+}=\left\{f\in C^2\left(I\right):{\left(\alpha \left(z\right)f^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)f\left(z\right)+\gamma \left(z\right)\ge 0,z\in I\right\},$$

where

$$\alpha =1,\beta =0,\gamma =-\ell .$$

Hence, by Theorem 1, $f\in {\mathcal{F}}_{\ell }^{+}$, if and only if, for all $x,y\in I$ with $x<y$, it holds that:

$${\int }_x^{y}f\left(\tau \right)d\tau \le H^{\prime }\left(x\right)f\left(x\right)-H^{\prime }\left(y\right)f\left(y\right)-\ell {\int }_x^{y}H\left(\tau \right)d\tau ,$$

(25)

where

$$H\left(z\right)=\frac{1}{2}(z-x)(y-z),x\le z\le y$$

is the unique (nonnegative) solution to the boundary value problem:

$$\left\{\begin{array}{c}{H}^{″}\left(z\right)=-1,x<z<y,\hfill \\ H\left(x\right)=H\left(y\right)=0.\hfill \end{array}\right.$$

On the other hand, for all $x,y\in I$ with $x<y$, we have:

$$\begin{array}{cc}& H^{\prime }\left(x\right)f\left(x\right)-H^{\prime }\left(y\right)f\left(y\right)-\ell {\int }_x^{y}H\left(\tau \right)d\tau \hfill \\ & =\frac{y-x}{2}f\left(x\right)-\frac{1}{2}(x-y)f\left(y\right)-\frac{\ell }{2}{\int }_x^y(-{\tau }^2+(x+y)\tau -xy)d\tau \hfill \\ & =(y-x)\frac{f\left(x\right)+f\left(y\right)}{2}-\frac{\ell }{2}{\left[-\frac{{\tau }^3}{3}+\frac{x+y}{2}{\tau }^2-xy\tau \right]}_{\tau =x}^y\hfill \\ & =(y-x)\frac{f\left(x\right)+f\left(y\right)}{2}-\frac{\ell }{12}\left(y^3-x^3-3x{y}^2+3x^{2}y\right)\hfill \\ & =(y-x)\frac{f\left(x\right)+f\left(y\right)}{2}-\frac{\ell }{12}{(y-x)}^3,\hfill \end{array}$$

which shows that (25) is equivalent to (21). □

Similarly, using Theorem 2 (or replacing f by $-f$ and ℓ by $-\ell$ in Corollary 5), we obtain the following characterization of the class of functions ${\mathcal{F}}_{\ell }^{-}$ defined by (20).

Corollary 6. 

Let $f\in C^2\left(I\right)$. The following statements are equivalent:

(i)

$f\in {\mathcal{F}}_{\ell }^{-}$;

(ii)

For all $x,y\in I$ with $x<y$, (22) holds.

3.2. Characterizations of the Classes of Functions ${\mathcal{G}}_{\lambda }^{\pm }$

Let I be an open interval of $\mathbb{R}$ and $\lambda >0$. We first need the following lemma. Its proof is elementary; we omit the details.

Lemma 1. 

For all $x,y\in I$ with $x<y$, the following boundary value problem:

$$\left\{\begin{array}{c}{H}^{″}\left(z\right)-\lambda H\left(z\right)=-1,x<z<y,\hfill \\ H\left(x\right)=H\left(y\right)=0\hfill \end{array}\right.$$

admits a unique nonnegative solution given by:

$$H\left(z\right)=\frac{e^{-\sqrt{\lambda }z}\left(e^{\sqrt{\lambda }x}-e^{\sqrt{\lambda }z}\right)\left(e^{\sqrt{\lambda }z}-e^{\sqrt{\lambda }y}\right)}{\lambda \left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)},x\le z\le y.$$

(26)

The following result provides a characterization of the class of functions ${\mathcal{G}}_{\lambda }^{+}$ defined by (23).

Corollary 7. 

Let $f\in C^2\left(I\right)$. The following statements are equivalent:

(i)

$f\in {\mathcal{G}}_{\lambda }^{+}$;

(ii)

For all $x,y\in I$ with $x<y$, it holds that:

$${\int }_x^{y}f\left(\tau \right)d\tau \le \frac{e^{\sqrt{\lambda }y}-e^{\sqrt{\lambda }x}}{\sqrt{\lambda }\left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)}\left(f\left(x\right)+f\left(y\right)\right).$$

(27)

Proof. 

Observe that:

$${\mathcal{G}}_{\lambda }^{+}=\left\{f\in C^2\left(I\right):{\left(\alpha \left(z\right)f^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)f\left(z\right)+\gamma \left(z\right)\ge 0,z\in I\right\},$$

where

$$\alpha =1,\beta =-\lambda ,\gamma =0.$$

Hence, by Theorem 1, $f\in {\mathcal{G}}_{\lambda }^{+}$, if and only if, for all $x,y\in I$ with $x<y$, it holds that:

$${\int }_x^{y}f\left(\tau \right)d\tau \le H^{\prime }\left(x\right)f\left(x\right)-H^{\prime }\left(y\right)f\left(y\right),$$

(28)

where H is given by (26). On the other hand, for all $x,y\in I$ with $x<y$, we have:

$$H^{\prime }\left(z\right)=\frac{e^{\sqrt{\lambda }(x-z+y)}-e^{\sqrt{\lambda }z}}{\sqrt{\lambda }\left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)},x\le z\le y,$$

which yields:

$$H^{\prime }\left(x\right)=\frac{e^{\sqrt{\lambda }y}-e^{\sqrt{\lambda }x}}{\sqrt{\lambda }\left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)}$$

and

$$H^{\prime }\left(y\right)=\frac{e^{\sqrt{\lambda }x}-e^{\sqrt{\lambda }y}}{\sqrt{\lambda }\left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)}=-H^{\prime }\left(x\right).$$

Hence, for all $x,y\in I$ with $x<y$, we have:

$$\begin{array}{cc}& H^{\prime }\left(x\right)f\left(x\right)-H^{\prime }\left(y\right)f\left(y\right)\hfill \\ & =H^{\prime }\left(x\right)(f\left(x\right)+f\left(y\right))\hfill \\ & =\frac{e^{\sqrt{\lambda }y}-e^{\sqrt{\lambda }x}}{\sqrt{\lambda }\left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)}\left(f\left(x\right)+f\left(y\right)\right),\hfill \end{array}$$

which shows that (28) is equivalent to (27). □

Remark 1. 

Passing to the limit as $\lambda \to 0^{+}$, (27) reduces to the standard Hermite–Hadamard inequality (2).

Similarly, using Theorem 2 (or replacing f by $-f$ in Corollary 7), we obtain the following characterization of the class of functions ${\mathcal{G}}_{\lambda }^{-}$ defined by (24).

Corollary 8. 

Let $f\in C^2\left(I\right)$. The following statements are equivalent:

(i)

$f\in {\mathcal{G}}_{\lambda }^{-}$;

(ii)

For all $x,y\in I$ with $x<y$, it holds that:

$${\int }_x^{y}f\left(\tau \right)d\tau \ge \frac{e^{\sqrt{\lambda }y}-e^{\sqrt{\lambda }x}}{\sqrt{\lambda }\left(e^{\sqrt{\lambda }x}+e^{\sqrt{\lambda }y}\right)}\left(f\left(x\right)+f\left(y\right)\right).$$

4. Conclusions

The Hermite–Hadamard inequality (Inequality (1)) provides an upper and lower bounds of the (integral) mean of a convex function over a certain interval. Moreover, each of the two sides of (1) provides a characterization of convex functions. In the special case when a function f is twice differentiable in a certain interval I, the convexity of f is equivalent to the differential inequality $f^{″}\ge 0$ in I. Thus, it is natural to ask whether it is possible to obtain a characterization of twice differentiable functions satisfying more general differential inequalities. In this paper, we gave a positive answer to this question for the class of functions f satisfying differential inequalities of the form ${\left(\alpha f^{\prime }\right)}^{\prime }+\beta f+\gamma \ge 0$ in I, where $\alpha \in C^1\left(I\right)$ and $\beta ,\gamma \in C\left(I\right)$. Namely, assuming that for every $x,y\in I$ with $x<y$, the Dirichlet boundary value problem:

$$\left\{\begin{array}{c}{\left(\alpha \left(z\right)H^{\prime }\left(z\right)\right)}^{\prime }+\beta \left(z\right)H\left(z\right)=-1,x<z<y,\hfill \\ H\left(x\right)=H\left(y\right)=0\hfill \end{array}\right.$$

admits a unique nonnegative solution H. We show that the considered differential inequality is equivalent to:

$${\int }_x^{y}f\left(\tau \right)d\tau \le H^{\prime }\left(x\right)\alpha \left(x\right)f\left(x\right)-H^{\prime }\left(y\right)\alpha \left(y\right)f\left(y\right)+{\int }_x^y\gamma \left(\tau \right)H\left(\tau \right)d\tau$$

for every $x,y\in I$ with $x<y$. The above inequality is a generalization of the right side of Hermite–Hadamard inequality (1), which can be obtained by taking $\alpha =1$ and $\beta =\gamma =0$. We also discussed some special cases of $\alpha$, $\beta$ and $\gamma$, and provided some characterizations in those cases.

In this work, only second-order differential inequalities are investigated. It would be interesting to show whether it is possible to obtain a characterization of functions f satisfying higher-order differential inequalities. For instance, the class of functions f satisfying $f^{⁗}\ge 0$ in I deserves to be studied.