In 1974, B. Nagy in [1] developed a stochastic processes characterization to solving a generalization of the Cauchy functional equation. Then, in 1980, K. Nikodem in [2] obtained some properties of convex stochastic processes and gave generalizations of several results proved in [1] . Both research works began a line of research on convex stochastic processes; as a result of this line of research, the following papers have been obtained [3] - [9] .

The concept of m-convex function was introduced by G. H. Toader in [10] . We can find this type of functions in the articles [11] - [29] , in which some algebraic properties for this type of functions were demonstrated, classical integral inequalities of the Hermite-Hadamard type and some stability results and sandwich theorems.

Interesting and important inequalities for m-convex functions were developed by M. K. Bakula in [11] and M. E. Özdemir in [30] . On the other hand, some Venezuelan researchers have developed numerous works on this topic; some examples can be found in the papers [3] [4] [16] - [24] . Additionally, in 1994, K. Baron, J. Matkowski and K. Nikodem in [12] developed a characterization of real functions that can be separated by a convex function and in 1995, K. Nikodem and S. Wasowicz proved a sandwich theorem for affine functions in [28] .

For their part, in 2007, K. Nikodem and Z. Páles in [29] studied the classic Kakutani theorem and extended it to the convexity in the sense of Beckenbach, getting as consequences stability results of the Hyers-Ulam type. Subsequently, in 2016, N. Merentes and K. Nikodem in [27] proved that a pair of functions can be separated by functions strongly convex, approximately concave, or c-quadratic-affine functions, obtaining as a consequence, stability results of the Hyers-Ulam type. In [27] it has been shown an analogue result of the sandwich theorem for convex functions that is not true in the class of m-convex functions with $m\in \left(0,1\right)$. However, T. Lara in 2017 proved a useful sandwich result for the function m-convex in [20] .

The main objective of this paper is to perform a sandwich-type theorem and a Hyer-Ulam’s stability theorem for m-convex stochastic processes as a counterpart to those performed for m-convex functions.

Definition 2.1. A function $f:I\to ℝ$, where $I\subseteq ℝ$ is an interval, is convex if

$f\left(tx+\left(1-t\right)y\right)\le tf\left(x\right)+\left(1-t\right)f\left(y\right),$

para todo $x,y\in I$ y $t\in \left(0,1\right)$.

If the inequality is strict (<) for $x,y\in I$, $t\in \left(0,1\right)$, then we say that the function $f$ is strictly convex. If the inequality holds in the opposite direction (≥) we say that $f$ is concave and if it is verified in the strict sense (>) we say that f is strictly concave.

Definition 2.2. Let $\left(\Omega ,\mathbb{A},\mathbb{P}\right)$ be a probability space. A function $X:\Omega \to ℝ$ is a random variable if it is $\mathbb{A}$-measurable. A function $X:I\times \Omega \to ℝ$, where $I\subseteq ℝ$ is an interval, is a stochastic process if for each $t\in I$ the function $X\left(t,\cdot \right)$, is a random variable.

A stochastic process $X:I\times \Omega \to ℝ$ is:

Definition 2.3. Jensen-Convex if, for each $a,b\in I$, the following inequality is satisfied:

$X\left(\frac{a+b}{2},\cdot \right)\le \frac{X\left(a,\cdot \right)+X\left(b,\cdot \right)}{2}.\left(a.e.\right)$

Definition 2.4. Convex if, for each $a,b\in I$, $t\in \left(0,1\right)$ the following inequality is satisfied:

$X\left(ta+\left(1-t\right)b,\cdot \right)\le tX\left(a,\cdot \right)+\left(1-t\right)X\left(b,\cdot \right).\left(a.e.\right)$

Definition 2.5. Quasi-Convex if, for each $a,b\in I$, $t\in \left(0,1\right)$ the following inequality is satisfied:

$X\left(ta+\left(1-t\right)b,\cdot \right)\le \max \left[X\left(a,\cdot \right),X\left(b,\cdot \right)\right].\left(a.e.\right)$

Also, we say that a stochastic process $X:I\times \Omega \to ℝ$ is:

Definition 2.6. Continuous in probability in the interval $I$, if for all $t_0\in I$ we have

$P-\underset{t\to t_0}{\text{lim}}X\left(t,\cdot \right)=X\left(t_0,\cdot \right)\left(a.e.\right)$

where $P-\text{lim}$ denotes the limit in probability.

Definition 2.7. Mean-Square contiuous in $I$, if for all $t_0\in I$ we have

$\underset{t\to t_0}{\text{lim}}\mathbb{E}\left[{\left(X\left(t,\cdot \right)-X\left(t_0,\cdot \right)\right)}^2\right]=0.\left(a.e.\right)$

where $\mathbb{E}\left[X\left(t,\cdot \right)\right]$ denotes the expectation value of the random variable $X\left(t,\cdot \right)$.

Definition 2.8. Differentiable at a point $t\in I$, if there is a random variable $X^{\prime }\left(t,\cdot \right):I\times \Omega \to ℝ$ defined as follows:

$X^{\prime }\left(t,\cdot \right)=P-\underset{t\to t_0}{\text{lim}}\frac{X\left(t,\cdot \right)-X\left(t_0,\cdot \right)}{t-t_0}.\left(a.e.\right)$

Remark 2.9. Every mean-square continuous stochastic process is a continuous in probability stochastic process; however, the converse is not true.

Definition 2.10. Let $X:I\times \Omega \to ℝ$ be a stochastic process such that $\mathbb{E}\left[{\left(X\left(t,\cdot \right)\right)}^2\right]<+\infty$ for all $t\in I.$ The random variable $Y:\Omega \to ℝ$ is called the mean-square integral of the stochastic process $X$ en $\left[a,b\right]\subseteq I$, if for any partition $a=t_0<t_1<\cdots <t_n=b$ of the interval $\left[a,b\right]$ y ${\Theta }_k\in \left[t_{k-1},t_k\right]$ ( $k=1,\cdots ,n$), we have

$\underset{n\to +\infty }{\text{lim}}\mathbb{E}\left[{\left(\underset{k=1}{\overset{+\infty }{{\displaystyle \sum }}}X\left({\text{Θ}}_k,\cdot \right)\left(t_k-t_{k-1}\right)-Y(\cdot )\right)}^2\right]=0.\left(a.e.\right)$

In this case, the following notation is used:

$Y(\cdot )={\displaystyle {\int }_a^{b}X\left(s,\cdot \right)\text{d}s}.\left(a.e.\right)$

Remark 2.11. For the existence of the mean-square integral of the stochastic process $X$, it is sufficient that $X$ be mean-square continuous. Basic properties of the mean-square integral can be read in [31] .

Definition 2.12. Let $m\in \left[0,1\right]$ and $I=\left[0,c\right]$. A mean-square continuous stochastic process $X:I\times \Omega \to ℝ$, is m-convex, if for all $a,b\in I$ y $t\in \left[0,1\right]$, the following inequality is satisfied:

$X\left(ta+m\left(1-t\right)b,\cdot \right)\le tX\left(a,\cdot \right)+m\left(1-t\right)X\left(b,\cdot \right).\left(a.e.\right)$

We denote by $S_m\left(c,\cdot \right)$, the class of stochastic processes m-convex in $I$, such that $X\left(0,\cdot \right)\le 0$.

Remark 2.13. If in the previous definition, we take $t=0$, then

$X\left(mb,\cdot \right)\le mX\left(b,\cdot \right),\left(a.e.\right)$

for all $b\in I$.

First, let’s establish some algebraic properties for m-convex stochastic processes.

Lemma 3.1. Let $Y:\left[a,b\right]\times \Omega \to ℝ$ be a mean-square stochastic process, $m_1,m_2\in \left[0,1\right]$, such that $m_1\le m_2$ and $m_2\ne 0.$ If $Y$ is $m_2$-convex almost everywhere, then $Y$ is $m_1$-convex almost everywhere.

Proof. Since $Y$ is $m_2$-convexo and $m_1\le m_2$, we have

$\begin{array}{c}Y\left(ta+m_1\left(1-t\right)b,\cdot \right)=Y\left(ta+m_2\left(1-t\right)\left(\frac{m_1}{m_2}b,\cdot \right)\right)\\ \le tY\left(a,\cdot \right)+m_2\left(1-t\right)Y\left(\frac{m_1}{m_2}b,\cdot \right)\\ \le tY\left(a,\cdot \right)+m_2\left(1-t\right)\left(\frac{m_1}{m_2}\right)Y\left(b,\cdot \right)\\ =tY\left(a,\cdot \right)+m_1\left(1-t\right)Y\left(b,\cdot \right).\left(a.e.\right)\end{array}$

Therefore, $Y$ is $m_1$-convex almost everywhere. □

Proposition 3.2. Let $a\ge 0$ and $X,Y:\left[a,b\right]\times \Omega \to ℝ$ be mean-square stochastic processes, If $X$ is $m_1$-convex and $Y$ is $m_2$-convex almost everywhere, with $m_1\le m_2$ and $m_2\ne 0$, then $X+Y$ y $\alpha X$, $\alpha \ge 0$ are $m_1$-convex almost everywhere.

Proof. By the previous lemma, we have that $Y$ es $m_1$-convex almost everywhere.

For ${\lambda }_1,{\lambda }_2\in \left[a,b\right]$ and $t\in \left[0,1\right]$, we obtain

$\begin{array}{l}\left(X+Y\right)\left(t{\lambda }_1+m_1\left(1-t\right){\lambda }_2,\cdot \right)\\ =X\left(t{\lambda }_1+m_1\left(1-t\right){\lambda }_2,\cdot \right)+Y\left(t{\lambda }_1+m_1\left(1-t\right){\lambda }_2,\cdot \right)\\ \le tX\left({\lambda }_1,\cdot \right)+m_1\left(1-t\right)X\left({\lambda }_2,\cdot \right)+tY\left({\lambda }_1,\cdot \right)+m_1\left(1-t\right)Y\left({\lambda }_2,\cdot \right)\\ =tX\left({\lambda }_1,\cdot \right)+tY\left({\lambda }_1,\cdot \right)+m_1\left(1-t\right)X\left({\lambda }_2,\cdot \right)+m_1\left(1-t\right)Y\left({\lambda }_2,\cdot \right)\\ =t\left(X+Y\right)\left({\lambda }_1,\cdot \right)+m_1\left(1-t\right)\left(X+Y\right)\left({\lambda }_2,\cdot \right).\left(a.e.\right)\end{array}$

From where, $X+Y$ is $m_1$-convex almost everywhere.

Besides,

$\begin{array}{c}\left(\alpha X\right)\left(t{\lambda }_1+m_1\left(1-t\right){\lambda }_2,\cdot \right)=\alpha \left(X\left(t{\lambda }_1+m_1\left(1-t\right){\lambda }_2,\cdot \right)\right)\\ \le \alpha \left(tX\left({\lambda }_1,\cdot \right)+m_1\left(1-t\right)X\left({\lambda }_2,\cdot \right)\right)\\ =\alpha tX\left({\lambda }_1,\cdot \right)+\alpha m_1\left(1-t\right)X\left({\lambda }_2,\cdot \right)\\ =t\left(\alpha X\right)\left({\lambda }_1,\cdot \right)+m_1\left(1-t\right)\left(\alpha X\right)\left({\lambda }_2,\cdot \right).\left(a.e.\right)\end{array}$

From where, $\alpha X$ is $m_1$-convex almost everywhere. □

Proposition 3.3. Let $X,Y:\left[0,b\right]\times \Omega \to ℝ$ be nonnegative stochastic processes such that

$\left(X\left({\lambda }_1,\cdot \right)-X\left({\lambda }_2,\cdot \right)\right)\left(Y\left({\lambda }_1,\cdot \right)-Y\left({\lambda }_2,\cdot \right)\right)\ge 0\left(a.e.\right)$

for all ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$. If $X,Y$ are m-convex stochastic processes, then $XY$ is m-convex almost everywhere.

Proof. Let ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$ and $t\in \left(0,1\right)$.

Since $X$ and $Y$ are m-convex stochastic processes, we have

$\begin{array}{l}XY\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\\ =X\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)Y\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\\ \le \left(tX\left({\lambda }_1,\cdot \right)+m\left(1-t\right)X\left({\lambda }_2,\cdot \right)\right)\left(tY\left({\lambda }_1,\cdot \right)+m\left(1-t\right)Y\left({\lambda }_2,\cdot \right)\right)\\ =t^{2}X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+tX\left({\lambda }_1,\cdot \right)m\left(1-t\right)Y\left({\lambda }_2,\cdot \right)\\ +m\left(1-t\right)X\left({\lambda }_2,\cdot \right)tY\left({\lambda }_1,\cdot \right)+m^2{\left(1-t\right)}^{2}X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\\ =t^{2}X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+mt\left(1-t\right)\left[X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_2,\cdot \right)+X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_1,\cdot \right)\right]\\ +m^2{\left(1-t\right)}^{2}X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right).\end{array}$

Furthermore, the inequality:

$\left(X\left({\lambda }_1,\cdot \right)-X\left({\lambda }_2,\cdot \right)\right)\left(Y\left({\lambda }_1,\cdot \right)-Y\left({\lambda }_2,\cdot \right)\right)\ge 0.$

Which implies

$X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\ge X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_2,\cdot \right)+X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_1,\cdot \right).$

Hence,

$\begin{array}{l}XY\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\\ \le t^{2}X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+mt\left(1-t\right)\left[X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_2,\cdot \right)+X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_1,\cdot \right)\right]\\ +m^2{\left(1-t\right)}^{2}X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\\ \le t^{2}X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+mt\left(1-t\right)\left[X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\right]\\ +m^2{\left(1-t\right)}^{2}X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\\ =\left[t^2+mt\left(1-t\right)\right]X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+\left[mt\left(1-t\right)+m^2{\left(1-t\right)}^2\right]X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\\ =t\left[t+m\left(1-t\right)\right]X\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+m\left(1-t\right)\left[t+m\left(1-t\right)\right]X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\\ =\left[t+m\left(1-t\right)\right]\left[tX\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+m\left(1-t\right)X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\right].\end{array}$

On the other hand,

$t+m\left(1-t\right)\le t+\left(1-t\right)=1,$

since, $0\le m\le 1$ and $t\in \left(0,1\right)$.

Therefore

$\begin{array}{l}XY\left(t{\lambda }_1+m_1\left(1-t\right){\lambda }_2,\cdot \right)\\ \le \left[t+m\left(1-t\right)\right]\left[tX\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+m\left(1-t\right)X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\right]\\ \le tX\left({\lambda }_1,\cdot \right)Y\left({\lambda }_1,\cdot \right)+m\left(1-t\right)X\left({\lambda }_2,\cdot \right)Y\left({\lambda }_2,\cdot \right)\\ =tXY\left({\lambda }_1,\cdot \right)+m\left(1-t\right)XY\left({\lambda }_2,\cdot \right)\left(a.e.\right).\end{array}$

From this last inequality, we conclude that $XY$ is m-convex almost everywhere. □

Proposition 3.4. Let $X:\left[0,b\right]\times \Omega \to ℝ$ and $Y:\left[0,r\right]\times \Omega \to ℝ$ be m-convex stochastic processes such that $rang\left[X\left(\lambda ,\cdot \right)\right]\subset \left[0,r\right]$ for all $\lambda \in \left[0,b\right]$. If $Y$ is increasing, then the composition function $Y\circ X$ is m-convex in $\left[0,b\right]$ almost everywhere.

Proof. Let ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$ y $t\in \left[0,1\right]$, then we have

$\begin{array}{c}\left(Y\circ X\right)\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)=Y\left(X\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right),\cdot \right)\\ \le Y\left(tX\left({\lambda }_1,\cdot \right)+m\left(1-t\right)X\left({\lambda }_2,\cdot \right),\cdot \right)\\ \le tY\left(X\left({\lambda }_1,\cdot \right),\cdot \right)+m\left(1-t\right)Y\left(X\left({\lambda }_2,\cdot \right),\cdot \right)\\ =t\left(Y\circ X\right)\left({\lambda }_1,\cdot \right)+m\left(1-t\right)\left(Y\circ X\right)\left({\lambda }_2,\cdot \right)\left(a.e.\right).\end{array}$

Therefore, the function $Y\circ X$ is m-convex in $\left[0,b\right]$ almost everywhere. □

The following result gives necessary conditions under which a pair of stochastic processes may be separated by a stochastic process m-convex. We shall prove a sandwich type theorem inspired in [20] .

Note that:

Remark 3.5. If $I=\left(0,+\infty \right)$ or $I=\left[0,+\infty \right)$ and $X:I\times \Omega \to ℝ$ is a m-convex stochastic process, then

$X\left(ma,\cdot \right)\le mX\left(a,\cdot \right),a\in I.\left(a.e.\right)$(1)

Theorem 3.6. Let $I=\left(0,+\infty \right)$ or $I=\left[0,+\infty \right)$ and $X:I\times \Omega \to ℝ$ be a mean-square integrable stochastic process, no negative and m-convex, then there exist a convex stochastic process $Z:I\times \Omega \to ℝ$ such that

$X\left(a,\cdot \right)\le Z\left(a,\cdot \right)\le mX\left(\frac{a}{m},\cdot \right),foralla\in I.\left(a.e.\right)$

or equivalent

$\frac{1}{m}Z\left(ma,\cdot \right)\le X\left(a,\cdot \right)\le Z\left(a,\cdot \right),foralla\in I.\left(a.e.\right)$

Proof. Let $a,b\in I$. Since $X:I\times \Omega \to ℝ$ is a m-convex stochastic process, we have

$X\left(ta+m\left(1-t\right)b,\cdot \right)\le tX\left(a,\cdot \right)+m\left(1-t\right)X\left(b,\cdot \right),$

for all $t\in \left[0,1\right]$.

Replacing $b$ with $\frac{b}{m}$ in the previous inequality, we obtain:

$X\left(ta+m\left(1-t\right)\frac{b}{m},\cdot \right)\le tX\left(a,\cdot \right)+m\left(1-t\right)X\left(\frac{b}{m},\cdot \right).$

where from,

$X\left(ta+\left(1-t\right)b,\cdot \right)\le tX\left(a,\cdot \right)+m\left(1-t\right)X\left(\frac{b}{m},\cdot \right).$

On the other hand, from inequality (1), it follows that:

$\begin{array}{c}X\left(ta+\left(1-t\right)b,\cdot \right)\le tX\left(a,\cdot \right)+m\left(1-t\right)X\left(\frac{b}{m},\cdot \right)\\ \le tmX\left(\frac{a}{m},\cdot \right)+\left(1-t\right)mX\left(\frac{b}{m},\cdot \right).\end{array}$

Let $Y:I\times \Omega \to ℝ$ be the convex stochastic process, defined as follows:

$Y\left(a,\cdot \right):=mX\left(\frac{a}{m},\cdot \right),$

we get

$X\left(ta+\left(1-t\right)b,\cdot \right)\le tY\left(a,\cdot \right)+m\left(1-t\right)Y\left(b,\cdot \right).$

Applying the sandwich theorem for convex functions, we conclude that there exists a convex stochastic process $Z:I\times \Omega \to ℝ$, such that

$X\left(a,\cdot \right)\le Z\left(a,\cdot \right)\le Y\left(a,\cdot \right),a\in I.$

Hence,

$X\left(a,\cdot \right)\le Z\left(a,\cdot \right)\le mX\left(\frac{a}{m},\cdot \right),a\in I,\left(a.e.\right)$

or equivalent

$\frac{1}{m}Z\left(ma,\cdot \right)\le X\left(a,\cdot \right)\le Z\left(a,\cdot \right),a\in I.\left(a.e.\right)$

□

Theorem 3.7. Let $b>0$, $m\in \left[0,1\right]$ and $X,Y:\left[0,b\right]\times \Omega \to ℝ$ be stochastic processes, where $Y$ is non-negative. There exists a m-convex stochastic process $Z:\left[0,b\right]\times \Omega \to ℝ$, such that $X\le Z$ in $\left[0,mb\right]$ and $Z\le Y$ in $\left[0,b\right]$, if and only if, for any ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$ and $t\in \left[0,1\right]$, the following inequality holds almost everywhere

$X\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\le tY\left({\lambda }_1,\cdot \right)+m\left(1-t\right)Y\left({\lambda }_2,\cdot \right).$

Proof. ( $\Rightarrow$) Suppose there exists a m-convex stochastic process $Z:\left[0,b\right]\times \text{Ω}\to ℝ$, such that $X\le Z$ in $\left[0,mb\right]$ and $Z\le Y$ in $\left[0,b\right]$.

If ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$ and $t\in \left[0,1\right]$, then

$\begin{array}{c}X\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\le Z\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\\ \le tZ\left({\lambda }_1,\cdot \right)+m\left(1-t\right)Z\left({\lambda }_2,\cdot \right)\\ \le tY\left({\lambda }_1,\cdot \right)+m\left(1-t\right)Y\left({\lambda }_2,\cdot \right).\left(a.e.\right)\end{array}$

( $\Leftarrow$) Suppose for all ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$ y $t\in \left[0,1\right]$, the following inequality holds

$X\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\le tY\left({\lambda }_1,\cdot \right)+m\left(1-t\right)Y\left({\lambda }_2,\cdot \right).$

Let us consider the following set

$\mathbb{A}=EnvCon{v}_Y\left\{\left(p,q\right)\in \left[0,b\right]\times ℝ:Y\left(p,\cdot \right)\le q\right\}.$

That is, $\mathbb{A}$ is the set of the convex hull of the epigraphs of $Y$.

If $\left(p,q\right)\in \mathbb{A}$, then by Caratheodory’s theorem, $\left(p,q\right)$ belongs to the interior of $S\subset \mathbb{A}$, where $S$ is the affine convex set of the form $t\left(p_1,q_1\right)+\left(1-t\right)\left(p_2,q_2\right)$ for $t\in \left[0,1\right]$ and $\left(p_1,q_1\right),\left(p_2,q_2\right)$ vertices of $S$.

Let $q_0=inf\left\{r\in ℝ:\left(p,r\right)\in S\right\}$.

Since $Y$ is non-negative, we have $S$ is bounded set. Therefore, $q\ge q_0$ and $\left(p,q_0\right)$ is a limit point of $S$, where from

$\left(p,q_0\right)=t\left(p_1,q_1\right)+\left(1-t\right)\left(p_2,q_2\right),$

for some, $t\in \left[0,1\right]$ and $\left(p_1,q_1\right),\left(p_2,q_2\right)$ vertices of $S$.

Hence

$\begin{array}{c}{q}_0\ge tY\left(p_1,\cdot \right)+\left(1-t\right)Y\left(p_2,\cdot \right)\\ \ge tY\left(p_1,\cdot \right)+\left(1-t\right)mY\left(p_2,\cdot \right)\\ \ge X\left(t{p}_1+m\left(1-t\right)p_2,\cdot \right)\\ =X\left(p,\cdot \right).\end{array}$

Let’s define

$Z\left(p,\cdot \right):=inf\left\{q\in ℝ:\left(p,q\right)\in \mathbb{A}\right\},$

this infimum exists because $Y$ is non-negative.

It is clear that, $X\le Z$ in $\left[0,mb\right]$, given that $p\in \left[p_1,m{p}_2\right]$.

Besides, $\left(p,Y\left(p,\cdot \right)\right)\in \mathbb{A}$ for any $p\in \left[0,b\right]$ and $Z\le Y$ by the definition of infimum.

It remains to be shown that, $Z$ es m-convex.

Let $p_1,p_2\in \left[0,b\right]$ and $t\in \left[0,1\right]$. If $q_1,q_2$ are such that

$\left(p_1,q_1\right),\left(m{p}_2,m{q}_2\right)\in \mathbb{A},$

then

$\left(t{p}_1+\left(1-t\right)m{p}_2,t{q}_1+\left(1-t\right)m{q}_2\right)\in \mathbb{A}.$

Consequently,

$Z\left(t{p}_1+\left(1-t\right)m{p}_2,\cdot \right)\le t{q}_1+\left(1-t\right)m{q}_2,$

for any $q_1,q_2$, in particular for the infimum.

Therefore

$Z\left(t{p}_1+\left(1-t\right)m{p}_2,\cdot \right)\le tZ\left(p_1,\cdot \right)+\left(1-t\right)mZ\left(p_2,\cdot \right).\left(a.e.\right)$

□

Corollary 3.8. Let $b>0$, $m\in \left[0,1\right)$ and $X,Y:\left[0,b\right]\times \Omega \to ℝ$ be stochastic processes, with $Y$ non-negative, such that

$X\left(t{\lambda }_1+\left(1-t\right)m{\lambda }_2,\cdot \right)\le tY\left({\lambda }_1,\cdot \right)+\left(1-t\right)mY\left({\lambda }_2,\cdot \right),\left(a.e.\right)$

for all $t\in \left[0,1\right)$, then $X\left(0,\cdot \right)\le 0$.

Proof. By the previous theorem, there exists a m-convex stochastic process, $Z:\left[0,b\right]\times \Omega \to ℝ$, such that $X\le Z$ in $\left[0,mb\right]$ y $Z\le Y$ in $\left[0,b\right]$.

Therefore

$X\left(t{\lambda }_1+\left(1-t\right)m{\lambda }_2,\cdot \right)\le Z\left(t{\lambda }_1+\left(1-t\right)m{\lambda }_2,\cdot \right)\le tZ\left({\lambda }_1,\cdot \right)+\left(1-t\right)mZ\left({\lambda }_2,\cdot \right).$

If ${\lambda }_1={\lambda }_2=0$, then

$\begin{array}{c}X\left(0,\cdot \right)=X\left(t0+\left(1-t\right)m0,\cdot \right)\\ \le Z\left(t0+\left(1-t\right)m0,\cdot \right)\\ =Z\left(0,\cdot \right)\\ \le tZ\left(0,\cdot \right)+\left(1-t\right)mZ\left(0,\cdot \right)\\ \le \left(t+m\left(1-t\right)\right)Z\left(0,\cdot \right)\\ \le 0,\left(a.e.\right)\end{array}$

given that, $t+m\left(1-t\right)<1$. □

Next definition is the counterpart to the given for m-convex functions in [19] .

Definition 3.9. Let $\epsilon \ge 0$ and $m\in \left[0,1\right]$. A stochastic process $X:I\times \Omega \to ℝ$ is $\epsilon$-m-convex, if for any ${\lambda }_1,{\lambda }_2\in \left[0,b\right]$ and $t\in \left[0,1\right]$, we have

$X\left(t{\lambda }_1+m\left(1-t\right){\lambda }_2,\cdot \right)\le tX\left({\lambda }_1,\cdot \right)+m\left(1-t\right)X\left({\lambda }_2,\cdot \right)+\epsilon .\left(a.e.\right)$

An important consequence of the previous theorem, is the following Hyers-Ulam-type stability result for m-convex stochastic processes. More in detail.

Corollary 3.10. Let $\epsilon >0$ and $m\in \left[0,1\right]$. If $X:\left[0,b\right]\times \Omega \to \left[-\frac{\epsilon }{m},+\infty \right)$ is a $\epsilon$-m-convexo stochastic process, ther exixts a function $Z:\left[0,b\right]\times \Omega \to ℝ$ m-convex , such almost everywhere that

$\left|X\left(\lambda ,\cdot \right)-Z\left(\lambda ,\cdot \right)\right|\le \frac{\epsilon }{2m},\lambda \in \left[0,mb\right].$

Proof. Let $Y:=X+\frac{\epsilon }{m}$.

We have, $Y$ is a non-negative stochastic process.

On the other hand,

$\begin{array}{c}X\left(t{\lambda }_1+\left(1-t\right)m{\lambda }_2,\cdot \right)\le tX\left({\lambda }_1,\cdot \right)+\left(1-t\right)mX\left({\lambda }_2,\cdot \right)+\epsilon \\ =tX\left({\lambda }_1,\cdot \right)+t\frac{\epsilon }{m}+\left(1-t\right)mX\left({\lambda }_2,\cdot \right)+\left(1-t\right)\epsilon +\epsilon \\ -t\frac{\epsilon }{m}-\left(1-t\right)\epsilon \\ =t\left(X\left({\lambda }_1,\cdot \right)+\frac{\epsilon }{m}\right)+\left(1-t\right)m\left(X\left({\lambda }_2,\cdot \right)+\frac{\epsilon }{m}\right)+t\epsilon -t\frac{\epsilon }{m}\\ =tY\left({\lambda }_1,\cdot \right)+\left(1-t\right)mY\left({\lambda }_2,\cdot \right)+\epsilon t\left(1-\frac{1}{m}\right)\\ \le tY\left({\lambda }_1,\cdot \right)+\left(1-t\right)mY\left({\lambda }_2,\cdot \right),\text{since}\epsilon t\left(1-\frac{1}{m}\right)\le 0.\end{array}$

By the previous theorem, there exists a m-convex stochastic process $H:\left[0,b\right]\times \Omega \to ℝ$, such that $X\le H$ in $\left[0,mb\right]$ and $H\le Y$ in $\left[0,b\right]$.

where from,

$X\le H\le Y=X+\frac{\epsilon }{m}\text{in}\left[0,mb\right]$

Defining, $Z:=H-\frac{\epsilon }{2m}$, we have to $Z$ is m-convex.

Besides,

$\left|X\left(\lambda ,\cdot \right)-Z\left(\lambda ,\cdot \right)\right|\le \frac{\epsilon }{2m},\lambda \in \left[0,mb\right].\left(a.e.\right)$

□

Theorem 3.11. Let $X:\left[0,b\right]\times \Omega \to ℝ$ be a twice differentiable mean square stochastic process and $k_1,k_2\in ℝ$, such that

$k_1\le X^{″}\le k_2.$

Then, for $m\in \left[0,1\right]$ fixed, $a\in \left[0,b\right]$ and $t\in \left[0,1\right]$ arbitrary, we have almost everywhere

$\begin{array}{c}{k}_1\frac{{\left(mb-a\right)}^{2}t\left(1-t\right)}{2}\le tX\left(a,\cdot \right)+\left(1-t\right)X\left(mb,\cdot \right)-X\left(ta+m\left(1-t\right)b,\cdot \right)\\ \le k_2\frac{{(mb-a)}^{2}t\left(1-t\right)}{2}.\end{array}$

Proof. We define,

$Y\left(t,\cdot \right):=tX\left(a,\cdot \right)+\left(1-t\right)X\left(mb,\cdot \right)-X\left(ta+m\left(1-t\right)b,\cdot \right)-k_1\frac{{\left(mb-a\right)}^{2}t\left(1-t\right)}{2},$

with $t\in \left[0,1\right]$.

Then,

$\begin{array}{c}{Y}^{″}\left(t,\cdot \right)=-{\left(a-mb\right)}^2{X}^{″}\left(ta+m\left(1-t\right)b,\cdot \right)+k_1{\left(mb-a\right)}^2\\ ={\left(mb-a\right)}^2\left[k_1-X^{″}\left(ta+m\left(1-t\right)b,\cdot \right)\right]\\ \le 0.\end{array}$

where from, $Y$ ia a concave stochastic process on $\left[0,1\right]\times \Omega$, moreover $Y\left(0,\cdot \right)=Y\left(1,\cdot \right)=0$, therefore, $Y\left(t,\cdot \right)\ge 0$, for all $t\in \left[0,1\right]$, so left hand side of inequality holds.

On the other hand, we define

$\begin{array}{l}Z\left(0,\cdot \right):=tX\left(a,\cdot \right)+\left(1-t\right)X\left(mb,\cdot \right)-X\left(ta+m\left(1-t\right)b,\cdot \right)\\ -k_2\frac{{\left(mb-a\right)}^{2}t\left(1-t\right)}{2},\left(a.e.\right)\end{array}$

with $t\in \left[0,1\right]$.

Using a procedure analogous to the previous one, it is shown that $Z\left(0,\cdot \right)$ is a convex stochastic process on $\left[0,1\right]\times \Omega$, moreover $Z\left(0,\cdot \right)=Z\left(1,\cdot \right)=0$.

Therefore, $Y\left(t,\cdot \right)\le 0$, for all $t\in \left[0,1\right]$, so right hand side of inequality holds. □

As a consequence of the previous theorem, we obtain an integral inequality of Hermite-Hadamard type for m-convex stochastic processes.

In more detail, the following result is obtained.

Corollary 3.12. Let $X:\left[0,b\right]\times \Omega \to ℝ$ be a twice differentiable mean square stochastic process and $k_1,k_2\in ℝ$, such that

$k_1\le X^{″}\le k_2.$

Then, for $m\in \left[0,1\right]$ fixed, $a\in \left[0,b\right]$ and $t\in \left[0,1\right]$ arbitrary, we have almost everywhere

$\begin{array}{c}{k}_1\frac{{\left(mb-a\right)}^2}{12}\le \frac{X\left(a,\cdot \right)+X\left(mb,\cdot \right)}{2}-\frac{1}{mb-a}{\displaystyle {\int }_a^{mb}X\left(x,\cdot \right)\text{d}x}\\ \le k_2\frac{{\left(mb-a\right)}^2}{12}.\end{array}$

Moreover, if $X$ is a m-convexo stochastic process, then the following inequalities of Hermite-Hadamard type take place:

$\begin{array}{c}\frac{X\left(a,\cdot \right)+X\left(mb,\cdot \right)}{2}-k_2\frac{{\left(mb-a\right)}^2}{12}\le \frac{1}{mb-a}{\displaystyle {\int }_a^{mb}X\left(x,\cdot \right)\text{d}x}\\ \le \frac{X\left(a,\cdot \right)+mX\left(b,\cdot \right)}{2}-k_1\frac{{\left(mb-a\right)}^2}{12}.\end{array}$

Proof. By the previous Theorem, for $m\in \left[0,1\right]$ fixed, $a\in \left[0,b\right]$ and $t\in \left[0,1\right]$ arbitrary, we have

$\begin{array}{c}{k}_1\frac{{\left(mb-a\right)}^{2}t\left(1-t\right)}{2}\le tX\left(a,\cdot \right)+\left(1-t\right)X\left(mb,\cdot \right)-X\left(ta+m\left(1-t\right)b,\cdot \right)\\ \le k_2\frac{{\left(mb-a\right)}^{2}t\left(1-t\right)}{2}.\left(a.e.\right)\end{array}$

Integrating each term of the previous inequalities, with respect to $t\in \left[0,1\right]$, and by the change of variable $\alpha =ta+m\left(1-t\right)b$, we get the first inequalities.