1 Introduction

In Ohlin [12] proved the following interesting and very useful result on convex functions in a probabilistic context (as usual, \({\mathbb {E}}[X]\) denotes the expectation of the random variable X):

Lemma 1

[12]. Let \(X_1\), \(X_2\) be two real valued random variables such that \(\mathbb {E}[X_1]=\mathbb {E}[X_2]\). If the distribution functions \(F_{X_1} , F_{X_2}\) cross one time, i.e. there exists \(t_0 \in {\mathbb R}\) such that

$$\begin{aligned} F_{X_1} (t)\le F_{X_2} (t) \quad if \; t<t_0 \;and\; F_{X_1} (t)\ge F_{X_2} (t)\quad if \; t>t_0, \end{aligned}$$
(1)

then

$$\begin{aligned} \mathbb {E}[f(X_1)]\le \mathbb {E}[f(X_2)] \end{aligned}$$

for every convex function \(f:{\mathbb R}\rightarrow {\mathbb R}.\)

For years the above Ohlin lemma was not well-known in the mathematical community. It has been rediscovered by Rajba [14], who found its various applications to the theory of functional inequalities. In [13, 15, 18], the Ohlin lemma is used, among others, to get a simple proof of the known Hermite–Hadamard inequalities, as well as to obtaining new Hermite–Hadamard type inequalities.

In this note we prove counterparts of the Ohlin theorem for convex set-valued maps. We present also applications of these results to obtain some inclusions connected with convex set-valued maps.

2 Preliminaries

Let \((Y, \Vert \cdot \Vert )\) be a separable Banach space, B be the closed unit ball in Y, \((\Omega ,\mathcal {A}, P)\) be a probability space with a non-atomic measure P and \(I\subset {\mathbb R}\) be an open interval. Denote by n(Y) the family of all nonempty subsets of Y and by cl(Y) the family of all closed nonempty subsets of Y. For a given set-valued map \(G:\Omega \rightarrow n(Y)\) the integral \(\int _{\Omega } G(\omega ) dP\) is understood in the sense of Aumann, i.e. it is the set of integrals of all integrable (in the sense of Bochner) selections of the map G (cf. [1, 2]):

$$\begin{aligned} \int _{\Omega } G(\omega ) dP = \left\{ \int _{\Omega } g(\omega ) dP: g:\Omega \rightarrow Y\quad \;\text {is integrable and} ~ g(\omega )\in G(\omega ), \omega \in \Omega \right\} . \end{aligned}$$

A set-valued map \(G:\Omega \rightarrow n(Y)\) is called integrable bounded if there exists a non-negative integrable function \(k:\Omega \rightarrow {\mathbb R}\) such that \(G(\omega ) \subset k(\omega ) B\), for all \(\omega \in \Omega .\) In this case every measurable selection of G is integrable and, consequently, the Aumann integral of G is nonempty whenever G is measurable.

The following properties of the Aumann integral will be needed in our investigations:

Lemma 2

[1], Theorems 8.6.3, 8.6.4 ]. Let \(G:\Omega \rightarrow cl(Y)\) be a measurable set-valued map. a) The closure of the integral of G is convex and

$$\begin{aligned} \overline{\int _{\Omega } G(\omega ) dP} = \overline{conv}\Big (\int _{\Omega } G(\omega ) dP\Big ). \end{aligned}$$

b) If Y is finite dimensional, then the integral of G is convex. In particular, if \(Y={\mathbb R}\) and \(G(\omega )=[g_1(\omega ), g_2(\omega )], \omega \in \Omega \), then

$$\begin{aligned} \int _{\Omega } G(\omega ) dP =\Big [ \int _{\Omega } g_1(\omega ) dP, \int _{\Omega } g_2(\omega ) dP\Big ]. \end{aligned}$$

c) If G is integrable bounded, then

$$\begin{aligned} \overline{\int _{\Omega } G(\omega ) dP} = \int _{\Omega }\overline{conv} G(\omega ) dP. \end{aligned}$$

Recall that a set-valued map \(G:I\rightarrow n(Y)\) is called convex if

$$\begin{aligned} t G(x_1)+ (1-t)G(x_2) \subset G(tx_1+(1-t)x_2) \end{aligned}$$
(2)

for all \(x_1,x_2\in I\) and \(t\in [0,1]\) (see e.g. [1, 3, 4, 8] and the references therein). Note that by (2), all values of G are convex subsets of Y.

The following lemma characterizes convex set-valued maps with values in \(cl({\mathbb R})\).

Lemma 3

[8] . A set-valued map \(G:I\rightarrow cl({\mathbb R})\) is convex if and only if it has one of the following forms:

  1. a)

    \(G(x)= [g_1(x),g_2(x)], \quad x\in I,\)

  2. b)

    \(G(x)= [g_1(x),+ \infty ), \quad x\in I,\)

  3. c)

    \(G(x)= (- \infty ,g_2(x)], \quad x\in I,\)

  4. d)

    \(G(x)= (-\infty , + \infty ), \quad x\in I,\)

where \(g_1:I\rightarrow {\mathbb R}\) is convex and \(g_2:I\rightarrow {\mathbb R}\) is concave.

Clearly, if \(G:I\rightarrow cl({\mathbb R})\) is convex and integrable bounded, then it is of the form a).

3 Ohlin-Type Result for Convex Set-Valued Maps

The following result is a counterpart the Ohlin lemma for convex set-valued maps.

Theorem 4

Let \((Y, \Vert \cdot \Vert )\) be a separable Banach space, \((\Omega ,\mathcal {A}, P)\) be a probability space with a non-atomic measure P and \(I\subset {\mathbb R}\) be an open interval. Assume that \(X_1, X_2:\Omega \rightarrow I\) are integrable random variables such that \(\mathbb {E}[X_1]=\mathbb {E}[X_2]\). If there exists \(t_0 \in {\mathbb R}\) such that

$$\begin{aligned} F_{X_1} (t)\le F_{X_2} (t) \quad if\; t<t_0 \;and \; F_{X_1} (t)\ge F_{X_2} (t) \quad if \;t>t_0, \end{aligned}$$

then

$$\begin{aligned} \int _{\Omega } G\big (X_2(\omega )\big ) dP \subset \int _{\Omega } G\big (X_1(\omega )\big ) dP \end{aligned}$$

for every convex integrable bounded set-valued map \(G:I\rightarrow cl(Y)\).

Proof

The proof is divided into two steps. First, we assume that \(Y={\mathbb R}\). Then, by Lemma 3 and the assumption that G is integrable bounded, we obtain that G is of the form \(G(x)= [g_1(x),g_2(x)]\), \(x\in I\), where \(g_1:I\rightarrow {\mathbb R}\) is convex and \(g_2:I\rightarrow {\mathbb R}\) is concave. By the Ohlin lemma (Lemma 1), we have

$$\begin{aligned} \mathbb {E}[g_1(X_1)]\le \mathbb {E}[g_1(X_2)] ~ \text {and}~ \mathbb {E}[g_2(X_1)] \ge \mathbb {E}[g_2(X_2)]. \end{aligned}$$
(3)

Hence, using Lemma 2(b), we get

$$\begin{aligned} \int _{\Omega } G\big (X_2(\omega )\big ) dP= & {} \big [\mathbb {E}[g_1(X_2)], \mathbb {E}[g_2(X_2)] \big ] \subset \big [\mathbb {E}[g_1(X_1)], \mathbb {E}[g_2(X_1)] \big ] \\= & {} \int _{\Omega } G\big (X_1(\omega )\big ) dP. \end{aligned}$$

Now, assume that Y is an arbitrary separable Banach space. Take a nonzero continuous linear functional \(y^*\in Y^*\). Since the set-valued map \(x\mapsto \overline{y^*(G(x))}\), \(x \in I\), is convex and has closed values in \({\mathbb R}\), by the previous step,

$$\begin{aligned} \int _{\Omega } \overline{y^*\Big (G\big (X_2(\omega )\big )\Big )} dP \subset \int _{\Omega } \overline{y^*\Big (G\big (X_1(\omega )\big )\Big )} dP. \end{aligned}$$
(4)

Take arbitrary \(z\in \int _{\Omega } G\big (X_2(\omega )\big ) dP\). By the definition of the Aumann integral, there exists an integrable selection \(g\circ X_2\) of the set-valued map \(G\circ X_2\) such that \(z= \int _{\Omega } g\big (X_2(\omega )\big ) dP\). Using (4), we obtain

$$\begin{aligned} y^*(z)= y^*\Big (\int _{\Omega } g\big (X_2(\omega )\big ) dP\Big )&= \int _{\Omega } y^*\Big (g\big (X_2(\omega )\big )\Big ) dP\in \int _{\Omega } \overline{y^*\Big (G\big (X_1(\omega )\big )\Big )} dP.\nonumber \\ \end{aligned}$$
(5)

Since G is integrable bounded and the values \(y^*\Big (G\big (X_1(\omega )\big )\Big )\) are convex, by Lemma 2(c), we get

$$\begin{aligned} \int _{\Omega } \overline{y^*\Big (G\big (X_1(\omega )\big )\Big )} dP= & {} \int _{\Omega } \overline{conv}\ y^*\Big (G\big (X_1(\omega )\big )\Big )dP= \overline{\int _{\Omega } y^* \Big ( G\big (X_1(\omega )\big ) dP\Big )}\nonumber \\= & {} \overline{y^*\Big (\int _{\Omega } G\big (X_1(\omega )\big ) dP\Big )} \subset \overline{y^*\Big (\overline{\int _{\Omega } G\big (X_1(\omega )\big ) dP}\Big )}. \end{aligned}$$
(6)

From (5) and (6),

$$\begin{aligned} y^*(z) \in \overline{y^*\Big (\overline{\int _{\Omega } G\big (X_1(\omega )\big ) dP}\Big )}. \end{aligned}$$

Since this condition holds for arbitrary \(y^*\in Y^*\) and, by Lemma 2(a) the set \(\overline{\int _{\Omega } G\big (X_1(\omega )\big ) dP}\) is convex and closed, by the separation theorem (see [16], Corollary 2.5.11), we obtain

$$\begin{aligned} z\in \overline{\int _{\Omega } G\big (X_1(\omega )\big ) dP} \end{aligned}$$

and hence, using once more Lemma 2(c),

$$\begin{aligned} z\in \int _{\Omega } G\big (X_1(\omega )\big ) dP. \end{aligned}$$

Consequently,

$$\begin{aligned} \int _{\Omega } G\big (X_2(\omega )\big ) dP \subset \int _{\Omega } G\big (X_1(\omega )\big ) dP, \end{aligned}$$

which finishes the proof. \(\square \)

Remark 5

In the above proof we use the Ohlin lemma (Lemma 1) to obtain the inequalities (3). Replacing in Theorem 4 the assumptions on \(X_1\) and \(X_2\) (the same as in Ohlin’s lemma) by any weaker conditions sufficient for (3) (for instance necessary and sufficient conditions such as in the Levin–Stečkin theorem [7]; cf. also [11]), we can obtain more general result. However, it should be emphasized that the assumptions in the Ohlin lemma are very convenient because they are simple and can be easy verified.

4 Applications

In this section, we present an application of the Ohlin-type lemma to obtain various inclusions related to convex set-valued maps in a simple and unified way. Some of these results (Corollaries 6–10) are known, but we present alternative proofs of them.

The first result is a counterpart of the classical integral Jensen inequality.

Corollary 6

(cf. [8]). Let \(G:I\rightarrow cl(Y)\) be integrable bounded set-valued map and \((\Omega ,\mathcal {A}, P)\) be a probability space with a non-atomic measure P. Then G is convex if and only if

$$\begin{aligned} \int _{\Omega }G\big (X(\omega )\big ) dP \subset G\Big (\int _{\Omega }X(\omega ) dP\Big ), \end{aligned}$$
(7)

for every integrable random variable \(X:\Omega \rightarrow I.\)

Proof

Assume first that \(G:I\rightarrow cl(Y)\) is a convex integrable bounded set-valued map and \(X:\Omega \rightarrow I\) is an integrable random variable. Take a random variable \(X_1:\Omega \rightarrow I\) with the distribution \(\mu _{X_1} = \delta _{\mathbb {E}[X]}.\) Then the distribution functions \(F_X, F_{X_1}\) satisfy condition (1) and \(\mathbb {E}[X]=\mathbb {E}[X_1]\). Therefore, by Theorem 4,

$$\begin{aligned} \int _{\Omega } G\big (X(\omega )\big ) dP \subset \int _{\Omega } G\big (X_1(\omega )\big ) dP =G(\mathbb {E}[X])= G\Big (\int _{\Omega }X(\omega ) dP\Big ). \end{aligned}$$

Now, assume that G satisfies condition (7) with every integrable random variable \(X:\Omega \rightarrow I.\) Fix \(x_1 , x_2 \in I\) and \(t\in (0,1)\), and take a random variable \(X:\Omega \rightarrow I\) with the distribution \(\mu _{X} = t \delta _{x_1} +(1-t)\delta {x_2}.\) Then \(\int _{\Omega } X(\omega )dP = tx_1 +(1-t)x_2\) and \(\int _{\Omega } G\big (X(\omega )\big ) dP = tG(x_1) +(1-t)G(x_2).\) Therefore by (7)

$$\begin{aligned} tG(x_1) +(1-t)G(x_2)\subset G(t x_1 +(1-t)x_2), \end{aligned}$$

which proves that G is convex. \(\square \)

If in the above corollary we take a random variable X with the distribution \(\mu _{X} = t_1\delta _{x_1} + \cdots + t_n \delta _{x_n},\) where \(x_1,\ldots ,x_n \in I \) and \(t_1,\ldots ,t_n >0\) are such that \(t_1+\cdots +t_n =1\), then we obtain a counterpart of the discrete Jensen inequality.

Corollary 7

(cf. [10]). If a set-valued map \(G:I\rightarrow cl(Y)\) is convex and integrable bounded, then

$$\begin{aligned} \sum _{i=1} ^{n} t_i G(x_i) \subset G\Big (\sum _{i=1} ^{n} t_i x_i\Big ), \end{aligned}$$

for all \(n\in {\mathbb N}\), \(x_1,\ldots ,x_n \in I \) and \(t_1,\ldots ,t_n >0\) with \(t_1+\cdots +t_n =1.\)

We have also the following converse Jensen inclusion for convex set-valued maps.

Corollary 8

(cf. [6]). Let \(m, M \in I\), \(m<M\). If \(G:I\rightarrow cl(Y)\) is convex and integrable bounded, then

$$\begin{aligned} \frac{M-\bar{x}}{M-m}G(m) + \frac{{\bar{x}}-m}{M-m}G(M)\subset \sum _{i=1} ^{n} t_i G(x_i), \end{aligned}$$
(8)

for all \(x_1,\ldots ,x_n \in [m, M] \), \(t_1,\ldots ,t_n >0\) with \(t_1+\cdots +t_n =1\) and \(\bar{x}= t_1x_1+\cdots +t_n x_n .\)

Proof

Take random variables \(X_1, X_2:\Omega \rightarrow I\) with the distributions

$$\begin{aligned} \mu _{X_1} = \sum _{i=1}^n t_i \delta _{x_i} ~\text {and}~ \mu _{X_2} = \frac{M-{\bar{x}}}{M-m}\delta _{m} + \frac{{\bar{x}}-m}{M-m}\delta _{M}. \end{aligned}$$

Then the distribution functions \(F_X, F_Y\) satisfy condition (1) and

$$\begin{aligned} \mathbb {E}[X_1]= \sum _{i=1}^n t_i x_i = \bar{x} = \frac{M-\bar{x}}{M-m}m +\frac{\bar{x}-m}{M-m}M= \mathbb {E}[X_2]. \end{aligned}$$

Moreover

$$\begin{aligned} \int _{\Omega }G\big (X_1(\omega )\big ) dP= \sum _{i=1}^n t_i G(x_i), \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega }G\big (X_2(\omega )\big )dP=\frac{M-\bar{x}}{M-m}G(m) +\frac{\bar{x}-m}{M-m}G(M). \end{aligned}$$

Therefore, by Theorem 4, we obtain (8). \(\square \)

The next two corollaries are versions of the Hermite–Hadamard inequalities for convex set-valued maps.

Corollary 9

(cf. [9, 17]). If \(G:I\rightarrow cl(Y)\) is convex and integrable bounded, then

$$\begin{aligned} \frac{G(a) + G(b)}{2} \subset \frac{1}{b-a} \int _a^b G(x) \, dx \subset G\Big ( \frac{a+b}{2}\Big ), \end{aligned}$$
(9)

for all \(a, b \in I,\ a < b.\)

Proof

Let \(X_1, X_2:\Omega \rightarrow I\) be random variables with the distributions \(\mu _{X_1} = \delta _{(a+b)/2}\), \(\mu _{X_2} = \frac{1}{2}(\delta _a + \delta _b)\) and let \(X_3:\Omega \rightarrow I\) has the uniform distribution on [a, b]. Then the pairs \(X_1, X_3\) and \(X_3, X_2\) satisfy the assumptions of Theorem 4. Moreover,

$$\begin{aligned} \int _{\Omega }G\big (X_1(\omega )\big ) dP = G\Big (\frac{a+b}{2}\Big ), \int _{\Omega }G\big (X_2(\omega )\big ) dP= \frac{G(a)+G(b)}{2} \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega }G\big (X_3(\omega )\big ) dP=\frac{1}{b-a} \int _a^{b}G(x)dx. \end{aligned}$$

Therefore, by Theorem 4, we obtain (9). \(\square \)

Corollary 10

(cf. [9]) If \(G:I\rightarrow cl(Y)\) is convex and integrable bounded, \([a,b]\subset I\) and \(\mu \) is a Borel measure on [a, b] with \(\mu ([a,b])>0\), then

$$\begin{aligned} \frac{b-x_{\mu }}{b-a}G(a) + \frac{x_{\mu }-a}{b-a}G(b)\subset \frac{1}{\mu ([a,b])} \int _a^b G(x) \, d\mu (x) \subset G(x_{\mu }), \end{aligned}$$
(10)

where \(x_{\mu }=\frac{1}{\mu ([a,b])} \int _a^b x \, d\mu (x)\) is the barycenter of \(\mu \) on [a, b].

Proof

By the mean value theorem \(x_{\mu } \in [a,b]\). Let \(X_1, X_2, X_3:\Omega \rightarrow [a,b]\) be random variables with the distributions

$$\begin{aligned} \mu _{X_1} = \delta _{x_{\mu }}, \quad \mu _{X_2} = \frac{b-x_{\mu }}{b-a}\delta _a + \frac{x_{\mu }-a}{b-a}\delta _b, \quad \mu _{X_3}=\frac{1}{\mu ([a,b])}\,\mu . \end{aligned}$$

Then the pairs \(X_1, X_3\) and \(X_3, X_2\) satisfy the assumptions of Theorem 4. Moreover,

$$\begin{aligned} \int _{\Omega }G\big (X_1(\omega )\big ) dP = G(x_{\mu }), \int _{\Omega }G\big (X_2(\omega )\big ) dP= \frac{b-x_{\mu }}{b-a}G(a) + \frac{x_{\mu }-a}{b-a}G(b) \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega }G\big (X_3(\omega )\big ) dP=\frac{1}{\mu ([a,b])} \int _a^b G(x) \, d\mu (x). \end{aligned}$$

Therefore, by Theorem 4, we obtain (10). \(\square \)

The next two corollaries are counterparts for convex set-valued maps of the following inequalities concerning convex functions \(f: I \rightarrow \mathbb {R}\) (cf. [5, 15]):

$$\begin{aligned} \frac{f(c)+f(d)}{2}-f\left( \frac{c+d}{2}\right) \le \frac{f(a)+f(b)}{2} - f\left( \frac{a+b}{2}\right) \end{aligned}$$

for all \(a, b, c, d \in I \) such that \(a<c<d<b\),

and the Popoviciu inequality

$$\begin{aligned} \frac{2}{3}\left[ f\left( \frac{x+y}{2}\right) +f\left( \frac{y+z}{2}\right) +f\left( \frac{z+x}{2}\right) \right]\le & {} \frac{f(x)+f(y)+f(z)}{3} \\&+ f\left( \frac{x+y+z}{3}\right) , \end{aligned}$$

for all \(x,y,z \in I\).

Corollary 11

If \(G:I\rightarrow cl(Y)\) is convex and integrable bounded, then

$$\begin{aligned} \frac{G(a)+G(b)}{2} +G\left( \frac{c+d}{2}\right) \subset \frac{G(c)+G(d)}{2}+G\left( \frac{a+b}{2}\right) , \end{aligned}$$
(11)

for all \(a, b, c, d \in I \) such that \(a<c<d<b\).

Proof

Let \(X_1, X_2:\Omega \rightarrow I\) be random variables with the distributions

$$\begin{aligned} \mu _{X_1} = \frac{1}{4}\left( \delta _c+\delta _d\right) +\frac{1}{2} \delta _{(a+b)/2}, \quad \mu _{X_2} = \frac{1}{4}\left( \delta _a+\delta _b\right) +\frac{1}{2}\delta _{\delta _{(c+d)/2}}. \end{aligned}$$

Then the pair \(X_1, X_2\) satisfies the assumptions of Theorem 4. Moreover,

$$\begin{aligned} \int _{\Omega }G\big (X_1(\omega )\big ) dP= & {} \frac{G(c)+G(d)}{4}+ \frac{1}{2}G\left( \frac{a+b}{2}\right) ,\\ \int _{\Omega }G\big (X_2(\omega )\big ) dP= & {} \frac{G(a)+G(b)}{4} + \frac{1}{2} G\left( \frac{c+d}{2}\right) . \end{aligned}$$

Therefore, by Theorem 4, we obtain (11). \(\square \)

Corollary 12

If \(G:I\rightarrow cl(Y)\) is convex and integrable bounded, then

$$\begin{aligned} \frac{G(x)+G(y)+G(z)}{3} + G\left( \frac{x+y+z}{3}\right)\subset & {} \frac{2}{3}\left[ G\left( \frac{x+y}{2}\right) +G\left( \frac{y+z}{2}\right) \right. \\&\left. +G\left( \frac{z+x}{2}\right) \right] \end{aligned}$$

for all \(x,y,z \in I\).

Proof

Let \(X_1, X_2:\Omega \rightarrow I\) be random variables with the distributions

$$\begin{aligned}&\mu _{X_1} =\frac{1}{3}\left( \delta _{(x+y)/2}+\delta _{(y+z)/2}+\delta _{(z+x)/2}\right) ,\\&\quad \mu _{X_2} =\frac{1}{6}\left( \delta _x+\delta _y+\delta _z\right) +\frac{1}{2} \delta _{(x+y+z)/3}. \end{aligned}$$

Then the pair \(X_1, X_2\) satisfies the assumptions of Theorem 4. Moreover,

$$\begin{aligned} \int _{\Omega }G\big (X_1(\omega )\big ) dP = \frac{1}{3}\left[ G\left( \frac{x+y}{2}\right) +G\left( \frac{y+z}{2}\right) +G\left( \frac{z+x}{2}\right) \right] ,\\ \int _{\Omega }G\big (X_2(\omega )\big ) dP= \frac{G(x)+G(y)+G(z)}{6} + \frac{1}{2}G\left( \frac{x+y+z}{3}\right) . \end{aligned}$$

Therefore, by Theorem 4, the corollary is proved. \(\square \)