Poor convexity and Nash equilibria in games

Abstract

This paper considers two-person non-zero-sum games on the unit square with payoff functions having a new property called poor convexity. This property describes “something between” the classical convexity and quasi-convexity. It is proved that various types of such games have Nash equilibria with a very simple structure, consisting of the players’ mixed strategies with at most two-element supports. Since poor convexity is a basic notion in the paper, also a theory of poorly convex functions is also developed.

1 Introduction

The problem of existence of the solution introduced by Nash (1951) for noncooperative games, is widely studied in the literature. In particular, many of such results have been obtained under assumptions related to different types of concavity or convexity of payoff functions, possibly discontinuous. We can mention here several recent papers like Bich (2009), Carmona (2009), Carmona (2010), McClendon (1986) and Połowczuk et al. (2007).

The aim of the paper is to study the problem of the existence of “simple” Nash equilibria formed by so-called two-point strategies, that is such mixed strategies whose supports consist of at most two of players’ pure strategies. We examine this problem for two-person non-zero-sum games on the unit square with payoff functions possessing some new properties called poor concavity and poor convexity. It is worth mentioning that several results of this type were found both for finite games (Połowczuk 2006, 2003; Połowczuk et al. 2007; Radzik 2000) and for games on the unit square (Parthasarathy and Raghavan 1975; Radzik 1991, 1993) with convex and concave payoff functions.

Five new “existence theorems” are proposed for games in which players’ payoff functions are poorly concave or poorly convex in various configurations. This new notion of poor convexity is a natural generalization and extension of the classical convexity. In the paper, we develop the theory of poorly convex functions, using these results further in the proofs of the theorems.

The organization of the paper is as follows. Section 2 is devoted to preliminary definitions and some background results which are a starting point for our study. In Sect. 3 we introduce the new basic notions of poorly convex functions and discuss them. In Sect. 4 we present our five main results about the existence of Nash equilibria in games. Poorly concave functions and their properties are studied in Sects. 5 and 6, while Sect. 7 contains the proofs of the theorems.

2 Preliminary results

In this section we introduce preliminary notions and quote three basic results from the literature (Theorems A–C) which are an inspiration for our considerations in the paper.

Throughout this paper we consider only two-person non-zero-sum games on the unit square, that is the games with normal form

$$\begin{aligned} \Gamma =\langle \{1,2\}, \{X_1, X_2\},\{F_1, F_2\}\rangle \end{aligned}$$

(1)

where

• 1. $\{1,2\}$ is the set of two players;

• 2. for $i = 1,2, X_i = [0, 1]$ is the interval space of pure strategies $x_i$ of Player $i$.

• 3. $F_1(x_1,x_2)$ and $F_2(x_1,x_2)$ are bounded functions on $X_1 \times X_2$, and for $i=1,2, F_i(x_1,x_2)$ describes the payoff function of Player $i$, in the situation when Players 1 and 2 use their pure strategies $x_1$ and $x_2$, respectively.

A mixed strategy for Player $i$ is any probability measure $\mu _i$ on $X_i, i=1,2$. We will also write $F_i(\mu _1,\mu _2)=\int \int F_i(x_1,x_2)d\mu _1(x_1)d\mu _2(x_2)$ for $i=1,2$. So $F_i(\mu _1,\mu _2)$ describes the expected payoff of Player $i$ when Players 1 and 2 use their mixed strategies $\mu _1$ and $\mu _2$, respectively, and the vector $(F_1(\mu _1,\mu _2),F_2(\mu _1,\mu _2))$ is the payoff vector corresponding to the pair $(\mu _1,\mu _2)$.

A (mixed) Nash equilibrium in game $\Gamma$ is any pair $(\mu _1^*,\mu _2^*)$ of players’ mixed strategies that satisfy the inequalities

$$\begin{aligned} F_1(\mu _1^*,\mu _2^*) \ge F_1(\mu _1,\mu _2^*)\quad \text{ and } \quad F_2(\mu _1^*,\mu _2^*) \ge F_2(\mu _1^*,\mu _2) \end{aligned}$$

for all mixed strategies $\mu _1$ and $\mu _2$ of Players 1 and 2, respectively. When these two inequalities hold up to an $\varepsilon >0$, the pair $(\mu _1^*,\mu _2^*)$ is called an $\varepsilon$-Nash equilibrium.

The main problem we consider in our paper is the existence of Nash equilibria consisting of two-point strategies in game $\Gamma$. By definition, a two-point strategy is any pure strategy or mixed one with support consisting of at most two pure strategies, and will be denoted by $\alpha \delta _x + (1\!-\!\alpha ) \delta _y$ with $0 \le \alpha ,x,y\le 1$. Here and throughout the paper, $\delta _t$ is a degenerate probability distribution concentrated at point $t$ and will be identified with a pure strategy $t$ , $0\le t\le 1$. So a two-point strategy $\alpha \delta _x + (1\!-\!\alpha ) \delta _y$ prescribes a player to choose pure strategy $x$ with probability $\alpha$ and pure strategy $y$ with probability $1\!-\!\alpha$. It appears that two-point strategies (in spite of their simple form) play an essential role in describing Nash equilibria in wide classes of games.

To begin with, we recall three results concerning games $\Gamma$ on the unit square in which the conditions for the existence of Nash equilibria in two-point strategies are described. These results are a starting point for our study.

First background theorem belongs to Glicksberg (1952) and relates to the existence of Nash equilibrium in pure strategies for general $n$-person games. We quote his result only in the version for a two-person game on the unit square of the form (1), better suited for our discussion. We recall here that a real-valued function $f$ on $[0, 1]$ is quasi-concave, when for each real $c$, the set $\{x:f(x)\ge c\}$ is convex. In the analogous way, a quasi-convex function is defined (after replacing “$\ge$” by “$\le$)”. Of course, every convex (concave) function is quasi-convex (quasi-concave).

Theorem A

Assume that the payoff functions $F_1(x_1,x_2)$ and $F_2(x_1,x_2)$ are continuous on the unit square $[0,1]^2$, and quasi-concave in $x_1$ and $x_2$, respectively. Then game $\Gamma$ possesses a Nash equilibrium in pure strategies.

The second theorem belongs to Parthasarathy and Raghavan (1975) and can be seen as complementary to Theorem A.

Theorem B

Assume that the payoff functions $F_1(x_1,x_2)$ and $F_2(x_1,x_2)$ are continuous on $[0, 1]^2$, and $F_1(x_1,x_2)$ is concave in $x_1$. Then game $\Gamma$ possesses a Nash equilibrium of the form $(\delta _a, \beta \delta _c+(1-\beta )\delta _d)$ with $0\le \beta ,a,c,d\le 1$.

The third result (Theorem 2.3 in Radzik 1993), also essential for our further considerations, is a “convex version” of Theorem B.

Theorem C

Assume that for $a=0, 1$ the payoff functions $F_1(a,x_2)$ and $F_2(a,x_2)$ are continuous in $x_2$, and $F_1(x_1,x_2)$ is convex in $x_1$. Then game $\Gamma$ possesses a Nash equilibrium of the form $(\alpha \delta _0 + (1-\alpha )\delta _1, \beta \delta _c+(1-\beta )\delta _d)$ with $0\le \alpha ,\beta ,c,d\le 1$.

When the quasi-concavity of the payoff functions is omitted in the assumption of Theorem A, then there still exists a Nash equilibrium, but generally in mixed strategies. Glicksberg (1952) showed it for general $n$-person games. However we quote it only in the form analogous to that of Theorem A.

Theorem D

Assume that the payoff functions $F_1(x_1,x_2)$ and $F_2(x_1,x_2)$ are continuous on the unit square $[0,1]^2$. Then game $\Gamma$ possesses a Nash equilibrium in mixed strategies.

Remark 1

It is known that the assumption about continuity of payoff functions in Theorem A cannot be weakened by upper semicontinuity (see Example 1 in Radzik and Ravindran 1989). However an open question is, if game $\Gamma$ with continuous payoff functions $F_1$ and $F_2$ and with function $F_1(x_1,x_2)$ quasi-concave in $x_1$ has a Nash equilibrium in two-point strategies. Note that Theorem B says that when quasi-concavity of both payoff functions in Theorem A is replaced by concavity of only one of them, game $\Gamma$ still has an “almost pure” Nash equilibrium. Further, Example 1 below shows that if we remove the continuity assumption for function $F_2$, both Theorems B and C are not longer true, and then game $\Gamma$ may have no Nash equilibrium at all. As far as the “concavity” assumption in Theorem B is concerned, Example 2 shows that it cannot be weakened by the assumption of quasi-concavity in $x_1$ for function $F_1(x_1,x_2)$. The answer is not known for the analogous question concerning Theorem C.

Remark 2

It is worth mentioning that Theorems B and C have their generalizations (see Theorems 2.1 and 2.3 in Radzik (1993)). Namely, it may be seen as rather surprising that if in Theorem B the assumption about continuity of payoff functions $F_1$ and $F_2$ is replaced by boundedness on $[0, 1]^2$, then game $\Gamma$ has an $\varepsilon$-Nash equilibria of the form $(\alpha \delta _a +(1-\alpha )\delta _b, \beta \delta _c+(1-\beta )\delta _d)$, for some ${0\le \alpha ,\,} \beta ,a,b,c,d\le 1$ with $|a-b|<\varepsilon$.

Similarly, after removing “continuity” in Theorem C, game $\Gamma$ also has an $\varepsilon$-Nash equilibrium of the form described there.

Example 1

Consider the game $\Gamma$ with the payoff functions $F_1$ and $F_2$ described by: $F_1(x_1,x_2) = 0$ for $0\le x_1,x_2\le 1$, and $F_2(x_1,x_2) = 0$ for $0\le x_1\le 1$ with $x_2=1$ and $F_2(x_1,x_2) = x_2$ otherwise. Since, as we easily see, each pure strategy $x_2\in [0,1]$ is strictly dominated by another one, there is no Nash equilibrium in this game. On the other hand, it is immediately seen that function $F_1(\,x_1,x_2)$ is both concave and convex in $x_1$, and function $F_2(\,x_1,x_2)$ is discontinuous. So, Theorems B and C are no longer true when the continuity assumptions are removed.

Example 2

Consider the game $\Gamma$ with the payoff functions described by the following:

$$\begin{aligned} F_1(x_1,x_2) = \left\{ \begin{array}{ll} (1-2x_1)(1-2x_2) &{}\quad \text{ if } \quad 0\le x_1,x_2 < \frac{1}{2} \text{ or } \frac{1}{2} < x_1, x_2 \le 1 \\ 0 &{}\quad \mathrm{otherwise}\\ \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} F_2(x_1,x_2) = x_2^2-2x_1x_2\quad \text{ for } \quad 0\le x_1,x_2\le 1\,. \end{aligned}$$

One can easily verify that both payoff functions $F_1$ and $F_2$ are continuous on $[0, 1]^2$, and function $F_1(x_1,x_2)$ is quasi-concave in $x_1$ (but not concave), and function $F_2(x_1,x_2)$ is convex in $x_2$. On the other hand, it is easy to show that game $\Gamma$ does not have a Nash equilibrium of the form $\eta ^* = (\delta _a, \beta \delta _c+(1-\beta )\delta _d)$ with $c<d$ (the subcase $c=d$ is equivalent to $c<d$ with $\beta =0$ and thereby can be omitted). To prove it, suppose that game $\Gamma$ has a Nash equilibrium of the form $\eta ^*$, and assume first that $0<c<1$ or $0<d<1$. Then $F_2(\eta ^*)=\beta F_2(a,c)+(1-\beta ) F_2(a,d)< \max \{F_2(a,0),F_2(a,1)\}$, because function $F_2(x_1,x_2)$ is strictly convex in $x_2$. But this contradicts the fact that $\eta ^*$ is a Nash equilibrium. Therefore $c=0$ and $d=1$, and let $\mu _2^\beta :=\beta \delta _0+(1-\beta )\delta _1$.

Further, supposing that $a\not =\frac{1}{2}$ and $0<\beta <1$, we have: $F_2(a,\mu _2^\beta )$ $=\beta F_2(a,0)+(1-\beta ) F_2(a,1)< \max \{F_2(a,0),F_2(a,1)\}$, because $F_2(x_1,0)\not = F_2(x_1,1)$ for $x_1\not = \frac{1}{2}$. But this contradicts the optimality of strategy $\mu _2^\beta$ of Player 2.

Similarly, the case $[a=\frac{1}{2}$ and $0\le \beta \le 1$] is impossible, because for $\beta >0 F_1(\delta _{\frac{1}{2}},\mu _2^\beta )=\beta F_1\left( \frac{1}{2},0\right) +(1-\beta ) F_1\left( \frac{1}{2},1\right) =0 < \frac{\beta }{2}= F_1\left( \frac{1}{4},\mu _2^\beta \right)$, while for $\beta =0 F_1(\delta _{\frac{1}{2}},\mu _2^0)=F_1\left( \frac{1}{2},1\right) =0<\frac{1}{2}=F_1\left( \frac{3}{4},\mu _2^0\right)$, contradicting the optimality of strategy $\delta _{\frac{1}{2}}$ of Player 1.

The impossibility of the two remaining cases, (I) $[a\not =\frac{1}{2}$ and $\beta =0]$, and (II) $[a\not =\frac{1}{2}$ and $\beta =1]$, can be shown similarly. Namely, in case (I), if $a<1$ then we have: $F_1(\delta _a,\mu _2^0)= F_1(a,1) \le \max \{0,(1-2a)\cdot (-1)\} < 1= F_1(1,\mu _2^0)$, which contradicts the optimality of strategy $\delta _a$ of Player 1. On the other hand, if $a=1$ then $F_2(\delta _1,\mu _2^0)= F_2(1,1) = -1 <0= F_2(\delta _1,0)$, contradicting the optimality of strategy $\mu _2^0$ of Player 2. The impossibility of the case (II) can be shown in almost exactly the same way as for (I), and is left to the reader.

3 Definition of poorly convex functions

In this section we introduce two basic notions of poorly convex/concave functions and pairwise poorly convex/concave families, together with a wide discussion. They play a fundamental role in the paper. In particular, they arise in the assumptions of our five main theorems presented in Sect. 4. The theory of poorly convex functions, needed for the proofs of the theorems, is developed in Sect. 5.

We begin with the following definition.

Definition 1

A function $f$ on $[a,b]$ is poorly convex (poorly concave) if for every $(x_1,x_2,\lambda )$ with $a\le x_1<x_2\le b$ and $0<\lambda <1$ there is a real number $p=p(x_1,x_2,\lambda )$ with $0<p<1$ such that

$$\begin{aligned} f(\lambda x_1+(1-\lambda )x_2) \le (\ge ) p(x_1,x_2,\lambda )f(x_1) +[1-p(x_1,x_2,\lambda )]f(x_2). \end{aligned}$$

(2)

Remark 3

It is not difficult to see that the above definition can be straightforwardly generalized to poorly convex and poorly concave functions defined on convex subsets $U$ of an arbitrary vector space. It suffices only to replace the set of vectors $\{(x_1,x_2,\lambda ): a\le x_1<x_2\le b, 0<\lambda <1\}$ by the set $\{(x_1,x_2,\lambda ): x_1,x_2\in U, x_1\not = x_2, 0<\lambda <1\}$. However, in this paper we restrict our attention only to such functions on an interval of the real line which are needed to formulate our main theorems in Sect. 4.

Now we give a proposition which presents an equivalent but a little simpler characterization of poorly convex and poorly concave functions, more convenient for our further considerations. To formulate it, for brevity, we introduce the notation

$$\begin{aligned} \bar{u} : = (u_1,u_2,u_3), \end{aligned}$$

and, for a given interval $[a, b]$, we define the set

$$\begin{aligned} Q_{[a,b]} : =\{\bar{u}: a\le u_1<u_2<u_3\le b\}. \end{aligned}$$

(3)

Proposition 1

A function $f$ is poorly convex (poorly concave) on $[a, b]$ if and only if there is a positive function $T$ defined on the set $Q_{[a,b]}$, such that for all $\bar{u}\in Q_{[a,b]}$, the following inequality hold:

$$\begin{aligned} f(u_2)-f(u_1) \le (\ge ) T(\bar{u})[f(u_3)-f(u_2)]. \end{aligned}$$

(4)

Proof

($\Rightarrow$) Fix arbitrarily $(x_1,x_2,\lambda )$ with $a\le x_1<x_2\le b$ and $0<\lambda <1$. Let $u_1=x_1, u_3=x_2$ and $u_2=\lambda x_1+(1-\lambda )x_2$, whence $\lambda =\frac{u_3-u_2}{u_3-u_1}$. So we can write (2) as dependent only on $(u_1,u_2,u_3)$, in the form equivalent to (4) with

$$\begin{aligned} T(\bar{u}) = \frac{1-p\left( u_1,u_3,\frac{u_3-u_2}{u_3-u_1}\right) }{p\left( u_1,u_3,\frac{u_3-u_2}{u_3-u_1}\right) }, \end{aligned}$$

(5)

as it is easy to verify. This equality well defines a positive function $T$ on $Q_{[a,b]}$ because $\bar{u}\in Q_{[a,b]}$, whence $a\le u_1<u_3\le b$ and $0<\frac{u_3-u_2}{u_3-u_1}<1$.

($\Leftarrow$) Fix arbitrarily $\bar{u}\in Q_{[a,b]}$. Obviously, for some $0<\lambda <1, u_2= \lambda u_1+(1-\lambda )u_3$. Putting now $u_1=x_1, u_3=x_2$ and $u_2=\lambda x_1+(1-\lambda )x_2$ in (4), we easily check that this inequality is equivalent to (2) with

$$\begin{aligned} p(x_1,x_2,\lambda ) = \frac{1}{1+T(x_1,\lambda x_1+(1\!-\!\lambda )x_2,x_2)}. \end{aligned}$$

(6)

The value $p(x_1,x_2,\lambda )$ is well defined, because $a\le x_1<x_2\le b, 0<\lambda <1$ and $(x_1,\lambda x_1+(1\!-\!\lambda )x_2,x_2)\in Q_{[a,b]}$. This ends the proof. $\square$

Proposition 1 allows us to give an equivalent definition of poorly convex and poorly concave functions on an interval $[a,b]$, which is better suited for our further considerations.

Definition 2

A function $f$ on $[a, b]$ is poorly convex (poorly concave) if there is a positive function $T$ defined on the set $Q_{[a,b]}$ such that for all $\bar{u}\in Q_{[a,b]}$, inequality (4) holds. Then function $f$ is also called $T$-convex ($T$-concave) on $[a, b]$.

Now we are ready to give our second basic definition.

Definition 3

Let $\mathcal{X}$ be an index set. A family $\{f_\alpha : \alpha \in \mathcal{\ X}\}$ of functions on interval $[a, b]$ is pairwise poorly convex (pairwise poorly concave) if for every pair $\alpha , \beta \in \mathcal{\ X}$ there is a positive function $T$ on $Q_{[a,b]}$ such that the functions $f_\alpha$ and $f_\beta$ are $T$-convex ($T$-concave) on $[a, b]$.

Remark 4

One can easily see that function $T$ related to a poorly convex function (in Definition 1) neither has to be continuous nor unique. A wide analysis of poorly convex functions and pairwise poorly convex families of functions is presented in Sects. 5 and 6. In particular, in Proposition 3 we give necessary and sufficient conditions for a continuous function to be poorly convex, which shows that the family of poor convex (concave) functions on an interval is much richer than that of convex (concave) ones. We also mention here the relations between convex and poorly convex functions. Namely, every convex (concave) function $f$ on an interval $[a,b]$ is $T$-convex ($T$-concave) on $[a,b]$ with function $T$ of the form $T(\bar{u})=\frac{u_2-u_1}{u_3-u_2}$. Therefore every family of convex (concave) functions is a pairwise poorly convex (concave) one, but not conversely. The pairwise poorly convex (concave) families are substantially richer and play a fundamental role in our considerations. At the end of Sect. 6 we give two examples (Examples 3 and 4) of pairwise poorly convex families consisting of non-convex standard functions.

Remark 5

An easy analysis of inequality (4) shows that poor convexity of a function $f$ on an interval $[a, b]$ can also be defined by requiring for all $a \le u_1 < u_2 < u_3 \le b$ the following two implications: $[f(u_2) = f(u_1)] \Rightarrow [f(u_3) \ge f(u_2)]$, and $[f(u_2) > f(u_1)] \Rightarrow [f(u_3) > f(u_2)]$. This latter definition shows that poor convexity is indeed stronger than quasi-convexity. However, poor convexity can be also seen as a very minor strengthening of quasi-convexity, since every continuous quasi-convex function on an interval can be approximated by a sequence of continuous poorly convex functions in the topology of uniform convergence (see Proposition 4 in Sect. 5). Similar remarks can be made for poor concavity.¹

4 Main theorems

In this section we discuss possible generalizations of Theorems B and C from Sect. 2. As a result, five new Theorems 1–5 are proposed (their proofs will be given in Sect. 7). The basic question is how far we can weaken the assumptions of Theorems B and C, still having (in game $\Gamma$) the existence of a Nash equilibrium of the form described there. Example 2 given in Sect. 2 shows that Theorem B is false when we replace “concave” by “quasi-concave” in its assumptions. Hence a very intriguing question is whether Theorem B will remain true if we replace concavity assumption by “something between” concavity and quasi-concavity. The same question is essential for Theorem C. Just these two questions are basic for our study and we will show that the answer for both of them is positive.

Below we formulate the main five results of the paper. Their proofs are given in Sect. 7 which is preceded by the two auxiliary Sects. 5 and 6, where a theory of poorly convex functions is developed. The first two theorems generalize Theorems B and C.

Theorem 1

Assume that the payoff functions $F_1(x_1,x_2)$ and $F_2(x_1,x_2)$ are continuous on the unit square $[0, 1]^2$, and $\{F_1(\,\cdot \,,x_2): x_2\in [0,1]\}$ is a pairwise poorly concave family of functions on $[0,1]$. Then game $\Gamma$ possesses a Nash equilibrium of the form $(\delta _a, \beta \delta _c+(1-\beta )\delta _d)$, for some $0\le \beta ,a,c,d\le 1$.

Theorem 2

Assume for $a=0,1$ that the payoff functions $F_1(a,\,\cdot \,)$ and $F_2(a,\,\cdot \,)$ are continuous, and $\{F_1(\,\cdot \,,x_2): x_2\in [0,1]\}$ is a pairwise poorly convex family of functions on $[0,1]$. Then game $\Gamma$ possesses a Nash equilibrium of the form $(\alpha \delta _0 + (1-\alpha )\delta _1, \beta \delta _c+(1-\beta )\delta _d)$, for some $0\le \alpha , \beta ,c,d\le 1$.

The next two results can be seen as a completion of Theorem 1. They show that players’ “simple strategies” (with two-point supports) arising in Theorems 1 and 2 are quite satisfactory for them in many game cases.

Theorem 3

Assume that $F_2(x_1,x_2)$ is continuous on the unit square $[0, 1]^2$, and $\{F_1(\,\cdot \,,x_2): x_2\in [0,1]\}$ is a pairwise poorly concave family of continuous functions on $[0,1]$. If game $\Gamma$ possesses a Nash equilibrium $(\mu ,\nu )$ with a payoff vector $(f_1,f_2)$ then it also has a Nash equilibrium of the form $(\mu \,, \beta \delta _c + (1-\beta )\delta _d)$ for some $0\le \beta ,c,d\le 1$, with a payoff vector $(f_1^{^{\prime }},f_2^{^{\prime }})$ having the same second component $f_2^{^{\prime }}=f_2$.

Remark 6

We do not know if the assumptions of Theorem 3 guarantee the existence of a Nash equilibrium in game $\Gamma$. However, it can facilitate solving this problem in concrete cases of $\Gamma$.

Theorem 3 leads immediately to the next one.

Theorem 4

Assume that the payoff functions $F_1(x_1,x_2)$ and $F_2(x_1,x_2)$ are continuous on the unit square $[0, 1]^2$, and $\{F_1(\,\cdot \,,x_2): x_2\in [0,1]\}$ and $\{F_2(x_1\,,\cdot ): x_1\in [0,1]\}$ are pairwise poorly concave families of functions on $[0,1]$. Then for any Nash equilibrium in game $\Gamma$ there is also a Nash equilibrium of the form $(\alpha \delta _a + (1-\alpha )\delta _b\,, \beta \delta _c + (1-\beta )\delta _d)$ for some $0\le \alpha ,\beta ,a,b,c,d\le 1$, with the same payoff vector.

Remark 7

Theorem D from Sect. 2 implies that under the assumption of Theorem 4, game $\Gamma$ always has a Nash equilibrium in mixed strategies. Hence, the result of Theorem 4 can be interpreted that any Nash equilibrium in mixed strategies has an “equivalent” Nash equilibrium in two-point strategies in this game.

The last theorem is a “discontinuous modification” of Theorem 2.

Theorem 5

Assume that $\{F_1(\,\cdot \,,x_2): x_2\in [0,1]\}$ and $\{F_2(x_1\,,\cdot ): x_1\in [0,1]\}$ are pairwise poorly convex families of functions on $[0,1]$. Then game $\Gamma$ has a Nash equilibrium of the form $({ \alpha }\delta _0 + (1-\alpha )\delta _1\,, \beta \delta _0 + (1-\beta )\delta _1)$ with $0\le \alpha ,\beta \le 1$.

Remark 8

In general, Theorems 1 and 2 are no longer true when we admit for payoff functions $F_1$ and $F_2$ to be discontinuous (the same was shown for Theorems B and C in Example 1 in Sect. 2). Moreover, it may happen then that there are no Nash equilibria at all. However the author does not know whether after removing the continuity assumption in Theorems 1 and 2, the games still have $\varepsilon$-Nash equilibria for all $\varepsilon >0$ [we recall that this is true for Theorems B and C (see Remark 2)]. The next question is whether Theorems 3 and 4 still remain true after changing “poorly concave” with “poorly convex”.

Remark 9

Some “discrete” versions of poorly concave games considered in this paper have been also studied in the literature, where the spaces of players’ pure strategies are finite, and payoff functions are assumed to satisfy a discrete version of the pairwise poor concavity property. In paper Radzik (2000) some counterparts of Theorems 1–4 are presented for matrix games, while in Połowczuk (2006) for bimatrix games. A discrete versions of these theorems for poorly concave $n$-person games can be found in Połowczuk et al. (2007).

5 Theory of poorly convex functions

Pairwise poor convexity and poor concavity are basic properties of families of functions considered in the assumptions of our main Theorems 1–5. Having this in mind, we devote this section to develop a basic theory of such functions. Among other things, we find necessary and sufficient conditions for families of continuous functions to be pairwise poorly convex. This gives us tools for constructing pairwise poorly convex families and thereby to show that the class of such families is substantially richer than the family of convex functions (Examples 3 and 4 in Sect. 6). We mainly study poorly convex functions, because the results obtained can be trivially modified to analogous ones for poorly concave functions. We do it in Propositions 2–7.

In the first proposition, functions $f$ and $g$ are considered to be defined on an interval $[a,b]$. We list here five basic properties of poorly convex functions. Obviously, the proposition remains true after replacing “convex” by “concave” in it.

Proposition 2

The following statements hold:

• (a) a function $f$ is poorly convex if and only if $-f$ is poorly concave;

• (b) every convex function $f$ is poorly convex;

• (c) every poorly convex function $f$ is quasi-convex;

• (d) if $f$ is $T_1$- and $T_2$-convex then $f$ is ($\alpha T_1\!+\! \beta T_2$)-convex for all $\alpha , \beta \ge 0$ with $\alpha + \beta =1$;

• (e) if $f$ and $g$ are $T$-convex functions then for every $\alpha , \beta \ge 0$ the function $\alpha f+ \beta g$ is also $T$-convex.

Proof

Statements (a), (d) and (e) are simple consequences of Definition 1. Statement (b) was justified in Remark 4. Therefore, it suffices to show statement (c).

By assumption, function $f$ satisfies inequality (4). Let $c\in R, a\le u_1<u_2<u_3\le b$, and assume that $f(u_1)\le c$ and $f(u_3)\le c$. Hence, we can conclude with the help of inequality (4) as follows:

$$\begin{aligned} f(u_2)(1+T(\bar{u}))\le f(u_1)+T(\bar{u})f(u_3) \le c+cT(\bar{u}) = c(1+T(\bar{u})), \end{aligned}$$

and thereby $f(u_2)\le c$, because $T(\bar{u})>0$. Therefore the set $\{u: f(u)\le c\}$ is convex which, in view of arbitrarity of $c$, implies that function $f$ is quasi-convex.$\square$

The next proposition characterizes an arbitrary poorly convex continuous function. To express it, we need to define the following two quantities for a function $f$ on an interval $[a, b]$:

$$\begin{aligned} \tau _{f}: = { \sup } \{x\in [a,b]: f \text{ is } \text{ strictly } \text{ decreasing } \text{ on } [a,x]\} \end{aligned}$$

(7)

and

$$\begin{aligned} \tau ^{f}:= { \inf } \{x\in [a,b]: f \text{ is } \text{ strictly } \text{ increasing } \text{ on } [x,b]\}. \end{aligned}$$

(8)

Obviously, $\tau _f \le \tau ^f$. Both these quantities are basic for our subsequent considerations.

Proposition 3

A continuous function $f$ on $[a,b]$ is poorly convex if and only if there are constants $c\le d$ in $[a,b]$ such that $f$ is strictly decreasing on $[a, c]$, strictly increasing on $[d,b]$, and constant on $[c, d]$. Then $c=\tau _f$ and $d=\tau ^f$.

Proof

($\Rightarrow$) Since this part of the proposition is obvious in the case $\tau _f = \tau ^f$, we can assume $\tau _f < \tau ^f$. Suppose first that function $f$ is not constant on $[\tau _f, \tau ^f]$. Hence there are $u_1<u_2$ in the interval $(\tau _f, \tau ^f)$ such that $f(u_1)>f(u_2)$ or $f(u_1)<f(u_2)$. We consider these two cases.

Case 1: $f(u_1)>f(u_2)$.

By Definition 2, function $f$ is $T$-convex for some positive function $T$ defined on the set $Q_{[a,b]}$. Let $a\le u^{\prime }<u^{\prime \prime }< u_1<u_2\le b$ and denote $T_1:=T(u^{\prime },u^{\prime \prime },u_1)$ and $T_2:=T(u^{\prime \prime },u_1,u_2)$. Then (4) implies that

$$\begin{aligned} f(u^{\prime \prime })-f(u^{\prime }) \le T_1[f(u_1)-f(u^{\prime \prime })] \le T_1T_2[f(u_2)-f(u_1)], \end{aligned}$$

whence we easily deduce that $f(u^{\prime }) > f(u^{\prime \prime })$ and $f(u^{\prime \prime }) > f(u_1)$, because of $f(u_1) > f(u_2)$ and $T_1, T_2>0$. But this, in view of the arbitrarity of $u^{\prime }, u^{\prime \prime }$, proves that function $f$ is strictly decreasing in the interval $[a,u_1]$. However this contradicts (7) because of $u_1\in (\tau _f, \tau ^f)$. Therefore Case 1 cannot hold.

Case 2: $f(u_1)<f(u_2)$.

This case is also impossible. To show it, it suffices to repeat the reasoning of Case 1, replacing parameters $u^{\prime }, u^{\prime \prime }, u_1$ and $u_2$ by $u_1, u_2, u^{\prime }$ and $u^{\prime \prime }$, respectively. The clear details are omitted.

Therefore $f(u_1)=f(u_2)$. But this, in view of the arbitrarity of $u_1, u_2$, proves that function $f$ is constant in the interval $[\tau _f, \tau ^f]$. This completes the proof of part ($\Rightarrow$).

($\Leftarrow$) We can directly verify that function $f$ satisfies (4) for positive function $T$ defined on $Q_{[a,b]}$ in the following way:

$$\begin{aligned} T(\bar{u}): = \left\{ \begin{array}{crl} \frac{f(u_2)-f(u_1)}{f(u_3)-f(u_2)} &{} \text{ if } &{} [f(u_2)-f(u_1)][f(u_3)-f(u_2)]>0\\ 1 &{} &{} \text{ otherwise }. \end{array}\right. \end{aligned}$$

This completes the proof of Proposition 3.$\square$

Now we give a proposition showing that poor convexity is a very minor strengthening of quasi-convexity.

Proposition 4

Any continuous quasi-convex function $f$ on $[a,b]$ can be approximated by a sequence $(f_n)$ of poorly convex continuous functions in the topology of uniform convergence.

Proof

For $n=1,2,\ldots$, let us choose real numbers $\Delta _n$ to satisfy: $|f(x^{\prime })-f(x^{\prime \prime })|\le \Delta _n$ if $|x^{\prime }-x^{\prime \prime }|\le \frac{1}{n}$ in $[a,b]$. Since every continuous function on an interval is also uniformly continuous, we may choose $\Delta _n\rightarrow 0$ as $n\rightarrow \infty$.

Let us fix $c$ satisfying $f(c)=\min _{u\in [a,b]}f(u)$, and firstly assume that $a<c<b$. For a natural $n$, we define two sequences $(x_0,x_1,\ldots , x_n)$ and $(y_0,y_1,\ldots , y_n)$ by the following:

$$\begin{aligned} x_k=a+k\frac{c-a}{n}\quad \text{ and } \quad y_k=c-k\frac{b-c}{n}\,,\quad k=0,1,\ldots , n. \end{aligned}$$

Note that $a=x_0<x_1<\ldots < x_n=c=y_n<y_{n-1}<\ldots <y_0=b$. Now, let $f_n$ be a function on $[a,b]$ determined by the following two conditions:

• (i) $f_n(x_k)=f_n(y_k)=f(x_k)-\sum _{i=0}^k \frac{1}{n^{i+1}}$ for $k=0,1,\ldots , n\,$, and

• (ii) function $f_n$ is linear on all the intervals of the form $[x_k,x_{k+1}]$ and $[y_{k+1},y_k], k=0,1,\ldots , n-1\,$.

By Proposition 3, function $f$ is nonincreasing on $[a,c]$ and nondecreasing on $[c,b]$. Hence, one can easily conclude with the help of conditions (i) and (ii) that function $f_n$ is continuous on $[a,b]$, strictly decreasing on $[a,c]$ and strictly increasing on $[c,b]$. Therefore, Proposition 3 can be used again to conclude that function $f_n$ is poorly convex on $[a,b]$. Besides we easily deduce that

$$\begin{aligned} \max _{u\in [a,b]}|f(u)-f_n(u)| \le \Delta _n+ \sum \limits _{i=0}^n \frac{1}{n^{i+1}} < \Delta _n+\sum \limits _{i=0}^ \infty \frac{1}{n^{i+1}} = \Delta _n+\frac{1}{n-1}\,, \end{aligned}$$

which immediately completes the proof in the case $a<c<b$. When $a=c$ or $b=c$, the reasoning can be repeated with only the sequence $\{y_k\}$ or $\{x_k\}$, respectively.$\square$

In the next three propositions we discuss the question of the conditions guaranteeing for a pair of continuous poorly convex functions to be pairwise poorly convex. However, for their proofs, we need the following lemma.

Lemma 1

Let $f$ and $g$ be poorly convex functions on $[a,b]$. Then the pair $\{f, g\}$ is pairwise poorly convex if and only if for every $\bar{u}\in Q_{[a,b]}$ satisfying

$$\begin{aligned} f(u_1)<f(u_2)<f(u_3) \quad \text{ and } \quad g(u_1)>g(u_2)>g(u_3)\,, \end{aligned}$$

(9)

or

$$\begin{aligned} f(u_1)>f(u_2)>f(u_3) \quad \text{ and } \quad g(u_1)<g(u_2)<g(u_3)\,, \end{aligned}$$

(10)

there is an $h>0$ such that

$$\begin{aligned} f(u_2)-f(u_1) \le h[f(u_3)-f(u_2{ )]} \end{aligned}$$

(11)

and

$$\begin{aligned} g(u_2)-g(u_1) \le h[g(u_3)-g(u_2)]. \end{aligned}$$

(12)

Proof

By Definition 2, there are some positive functions $T_1(\bar{u})$ and $T_2(\bar{u})$ on $Q_{[a,b]}$ such that function $f$ is $T_1$-convex and function $g$ is $T_2$-convex on $[a, b]$.

($\Rightarrow$) In view of Definitions 2 and 3, this part of the lemma is obvious.

($\Leftarrow$) Let us arbitrarily fix $\bar{u}\in Q_{[a,b]}$. If (9) or (10) are satisfied, pairwise poor convexity of the pair $\{f, g\}$ follows by the assumption. Therefore, for the rest of our considerations, we can assume that (9) and (10) do not hold. To complete the proof of the lemma, it suffices to show that there is a number $h>0$ for which (11) and (12) hold. We consider several cases.

Case 1: $f(u_1)\le f(u_2)$ and $g(u_1)\le g(u_2)$.

Hence, with the help of (4), we can conclude as follows:

$$\begin{aligned} f(u_3)-f(u_2) \ge \frac{1}{T_1(\bar{u})}[f(u_2)-f(u_1)]\ge \frac{1}{\max \{T_1(\bar{u}),T_2(\bar{u})\}}[f(u_2)-f(u_1)], \end{aligned}$$

and thereby (11) follows for $h=\frac{1}{\max \{T_1(\bar{u}),T_2(\bar{u})\}}$. Exactly in the same way, replacing $f$ and $T_1$ by $g$ and $T_2$, respectively, we show (12) for the same $h$.

Case 2: $f(u_1)\ge f(u_2)$ and $g(u_1)\ge g(u_2)$.

Now we can repeat the reasoning of Case 1 to get the validity of (11) and (12) for $h=\frac{1}{\min \{T_1(\bar{u}),T_2(\bar{u})\}}$.

Case 3: $f(u_2)\le f(u_3)$ and $g(u_2)\le g(u_3)$.

Then we have:

$$\begin{aligned} f(u_2)-f(u_1) \le T_1(\bar{u})[f(u_3)-f(u_2)]\le \max \{T_1(\bar{u}),T_2(\bar{u})\}[f(u_3)-f(u_2)] \end{aligned}$$

and

$$\begin{aligned} g(u_2)-g(u_1) \le T_2(\bar{u})[g(u_3)-g(u_2)]\le \max \{T_1(\bar{u}),T_2(\bar{u})\}[g(u_3)-g(u_2)]. \end{aligned}$$

Consequently, (11) and (12) hold for $h=\max \{T_1(\bar{u}),T_2(\bar{u})\}$.

Case 4: $f(u_2)\ge f(u_3)$ and $g(u_2)\ge g(u_3)$.

Now we can repeat the reasoning of Case 3 to get the validity of (11) and (12) for $h=\min \{T_1(\bar{u}),T_2(\bar{u})\}$.

Case 5: $f(u_1)\ge f(u_2)\le f(u_3{ )}$.

Then (11) and (12) hold for $h=T_2(\bar{u}{ )}$.

Case 6: $g(u_1)\ge g(u_2)\le g(u_3$).

Then (11) and (12) hold for $h=T_1(\bar{u}{ )}$.

One can easily deduce that if (9) and (10) do not hold, then one of cases 1–6 must hold. This ends the proof of the lemma. $\square$

Proposition 5

Let $f$ and $g$ be continuous and poorly convex functions on $[a,b]$, satisfying $[\tau _f,\tau ^f]\cap [\tau _g,\tau ^g]\not =\emptyset$. Then the pair $\{f,\, g\}$ is pairwise poorly convex on $[a,b]$.

Proof

Let $\bar{u}\in Q_{[a,b]}$ and suppose that (9) holds. Then Proposition 3 applied to functions $f$ and $g$ implies that $u_2>\tau ^f$ and $u_2<\tau _g$, whence $\tau ^f<\tau _g$. But this contradicts the assumption $[\tau _f,\tau ^f]\cap [\tau _g,\tau ^g]\not =\emptyset$.

On the other hand, when we suppose that (10) holds, in a similar way we get $\tau ^g<\tau _f$, contradicting the assumption again. Therefore, (9) and (10) do not hold for $\bar{u}\in Q_{[a,b]}$, which by Lemma 1 ends the proof. $\square$

The next proposition considers the second case $[\tau _f,\tau ^f]\cap [\tau _g,\tau ^g]= \emptyset$. To formulate it, we need to introduce the following notation:

$$\begin{aligned} Q^f_g: = \{\bar{u}: \tau ^f \le u_1< u_2<u_3 \le \tau _g\}. \end{aligned}$$

Proposition 6

Let $f$ and $g$ be continuous and poorly convex functions on $[a,b]$ with $\tau ^f < \tau _g$. Then the pair $\{f, g\}$ is pairwise poorly convex on $[a,b]$ if and only if

$$\begin{aligned} \frac{f(u_2)-f(u_1)}{f(u_3)-f(u_2)}\le \frac{g(u_2)-g(u_1)}{g(u_3)-g(u_2)}\quad \text{ for } \, \bar{u} \in Q^f_g. \end{aligned}$$

(13)

Proof

In view of $\tau ^f < \tau _g$, Proposition 3 implies that function $f$ is strictly increasing in $[\tau ^f, \tau _g]$, and function $g$ is strictly decreasing in $[\tau ^f, \tau _g]$. Hence, it follows that

$$\begin{aligned} f(u_1)<f(u_2)<f(u_3) \quad \text{ and } \quad g(u_1)>g(u_2)>g(u_3) \quad \text{ for } \, \bar{u} \in Q^f_g. \end{aligned}$$

(14)

($\Rightarrow$) By Definitions 2 and 3, it follows that there is a positive function $T$ on $Q_{[a,b]}$ such that for $\bar{u} \in Q_{[a,b]}$

$$\begin{aligned} f(u_2)-f(u_1) \le T(\bar{u})[f(u_3)-f(u_2)] \end{aligned}$$

(15)

and

$$\begin{aligned} g(u_2)-g(u_1) \le T(\bar{u})[g(u_3)-g(u_2)], \end{aligned}$$

(16)

which, by (14), immediately implies (13).

($\Leftarrow$) Assume now that (13) holds. To end the proof it suffices to show that there is a positive function $T$ defined on the set $Q_{[a,b]}$ such that inequalities (15) and 16) hold for all $\bar{u} \in Q_{[a,b]}$.

Case 1: $\bar{u}\in Q^f_g$.

Let us define $T(\bar{u})$ as any positive value satisfying the inequalities,

$$\begin{aligned} \frac{f(u_2)-f(u_1)}{f(u_3)-f(u_2)}\le T(\bar{u}) \le \frac{g(u_2)-f(u_1)}{g(u_3)-g(u_2)}, \end{aligned}$$

(17)

which is possible because of (13). But this together with (14) imply (15) and (16).

Case 2: $\bar{u}\notin Q^f_g, f(u_1)< f(u_2)< f(u_3)$ and $g(u_1)> g(u_2)> g(u_3)$.

One can easily see with the help of Proposition 3 that

$$\begin{aligned} \tau ^f < u_2 < \tau _g. \end{aligned}$$

(18)

However, this implies that

$$\begin{aligned} u_1<\tau ^f \text{ or } u_3>\tau _g \,, \end{aligned}$$

since otherwise, it would contradict $\bar{u}\notin Q^f_g$.

Consider now the first subcase

$$\begin{aligned} ( \text{ A } ):\, u_1\le \tau ^f \quad \text{ and } \quad u_3\ge \tau _g \,, \end{aligned}$$

(19)

where one of the inequalities is strict. By Proposition 3 and the assumption $\tau ^f<\tau _g$, function $f$ is nonincreasing on $[u_1, \tau ^f]$ and strictly increasing on $[\tau _g,u_3]$, and function $g$ is strictly decreasing on $[u_1, \tau ^f]$ and nondecreasing on $[\tau _g,u_3]$. It implies that for $u_1^{^{\prime }}=\tau ^f$ and $u_3^{^{\prime }}=\tau _g$ we have

$$\begin{aligned} f(u_2)-f(u_1)&\le f(u_2)-f(u_1^{^{\prime }})\quad \text{ and } \quad g(u_2)-g(u_1) \le g(u_2)-g(u_1^{^{\prime }}),\qquad \end{aligned}$$

(20)

$$\begin{aligned} f(u_3)-f(u_2)&\ge f(u_3^{^{\prime }})-f(u_2) \quad \text{ and } \quad g(u_3)-g(u_2) \ge g(u_3^{^{\prime }})-g(u_2) \,, \qquad \end{aligned}$$

(21)

and $u_1^{^{\prime }}<u_2<u_3^{^{\prime }}$, because of (18) and (19).

Therefore the vector $\bar{u}^*=(u_1^{^{\prime }},u_2,u_3^{^{\prime }})$ satisfies $\bar{u}^*\in Q^f_g$ and consequently, by Case 1, for some $T(\bar{u}^*)>0$,

$$\begin{aligned} f(u_2)-f(u_1^{^{\prime }}) \le T(\bar{u}^*)[f(u_3^{^{\prime }})-f(u_2)] \end{aligned}$$

and

$$\begin{aligned} g(u_2)-g(u_1^{^{\prime }}) \le T(\bar{u^*})[g(u_3^{^{\prime }})-g(u_2)]. \end{aligned}$$

But this together with (20) and (21) imply that inequalities (15) and (16) hold for $T(\bar{u})= T(\bar{u}^*)$ in the first subcase (19) considered.

The two remaining subcases can be described by the conditions, (B): $u_1\le \tau ^f<u_2<u_3<\tau _g$, and (C): $\tau ^f<u_1<u_2<\tau _g<u_3$. We can analyze them, by the exact repetition of reasoning in subcase (A), changing only $u_3^{^{\prime }}$ with $u_3$ for subcase (B), and $u_1^{^{\prime }}$ with $u_1$ for subcase (C). In both subcases, the results obtained will be the same as in subcase (A). The clear details are omitted.

Case 3: $\bar{u}\notin Q^f_g, f(u_1)> f(u_2)> f(u_3)$ and $g(u_1)< g(u_2)< g(u_3)$.

This case cannot occur. Namely, then Proposition 3 would imply $u_1<u_2\le \tau ^f$ and $\tau _g \le u_2<u_3$ which is impossible, because of the assumption $\tau ^f<\tau _g$.

Now, taking into account Lemma 1 and conditions describing Cases 1–3, we easily deduce that the pair of functions $\{f, g\}$ is pairwise poorly convex on $[a,b]$, completing the proof of the proposition. $\square$

One can see the verification of condition (13) as somewhat complex. The next proposition is a special version of the previous one where verification of (13) is replaced by a much easier condition.

Proposition 7

Let $f$ and $g$ be continuous and poorly convex functions on $[a,b]$ with $\tau ^f < \tau _g$. Assume that there are continuous derivatives $f^{^{\prime }}$ and $g^{^{\prime }}$ on $(\tau ^f, \tau _g)$, with $g^{^{\prime }}\not =0$. Then the pair of functions $\{f, g\}$ is pairwise poorly convex on $[a,b]$ if and only if the function $G(u)=f^{^{\prime }}(u)/g^{^{\prime }}(u)$ is nonincreasing on $(\tau ^f,\tau _g)$.

Proof

($\Rightarrow$) Let $\tau ^f < u_1< u_2<u_3< u_4 <\tau _g$. By Proposition 3, functions $f$ and $g$ are strictly increasing and strictly decreasing in interval $[\tau ^f,\tau _g]$, respectively. Hence,

$$\begin{aligned} f(u_2)>f(u_1),\quad f(u_4)>f(u_3),\quad g(u_2)<g(u_1) \quad \text{ and } \quad g(u_4)<g(u_3). \end{aligned}$$

(22)

The assumption that the pair $\{f, g\}$ is pairwise poorly convex on $[a,b]$, implies that for some positive numbers $h$ and $h^{\prime }$ we have

$$\begin{aligned} f(u_2)-f(u_1)\le h[f(u_3)-f(u_2)]\le hh^{\prime }[f(u_4)-f(u_3)] \end{aligned}$$

and, similarly, the inequality $g(u_2)-g(u_1)\le hh^{\prime }[g(u_4)-g(u_3)]$. Hence, in view of (22), we get $\frac{f(u_2)-f({ u_1})}{f(u_4)-f(u_3)}\le hh^{\prime } \le \frac{g(u_2)-g(u_1)}{g(u_4)-g(u_3)},$ and thereby, $\frac{f(u_2)-f(u_1)}{g(u_2)-g(u_1)}\ge \frac{f(u_4)-f(u_3)}{g(u_4)-g(u_3)}.$ But this, by the classical Cauchy’s theorem, leads to the inequality, $\frac{f^{^{\prime }}(\theta _{12})}{g^{^{\prime }}(\theta _{12})} \ge \frac{f^{^{\prime }}(\theta _{34})}{g^{^{\prime }}(\theta _{34})}$ for some $\theta _{12}\in (u_1,u_2)$ and $\theta _{34}\in (u_3,u_4)$. Hence, taking into account that $\theta _{12}\rightarrow u_1$ as $u_2\rightarrow u_1$, and $\theta _{34}\rightarrow u_4$ as $u_3\rightarrow u_4$, we get $\frac{f^{^{\prime }}(u_1)}{g^{^{\prime }}(u_1)} \ge \frac{f^{^{\prime }}(u_4)}{g^{^{\prime }}(u_4)}$. In view of the arbitrarity of $u_1$ and $u_4$, it follows that the function $G(u)=f^{^{\prime }}(u)/g^{^{\prime }}(u)$ is nonincreasing on interval $(\tau ^f,\tau _g)$.

($\Leftarrow$) Assume now that the function $G(u)=f^{^{\prime }}(u)/g^{^{\prime }}(u)$ is nonincreasing in $(\tau ^f,\tau _g)$, and let us arbitrarily choose $\bar{u} \in Q^f_g$. Then Proposition 3 implies that

$$\begin{aligned} f(u_1)<f(u_2)<f(u_3) \quad \text{ and } \quad g(u_1)>g(u_2)>g(u_3). \end{aligned}$$

Suppose now that inequality (13) does not hold. Then, in view of the above inequalities, we would have $\frac{f(u_2)-f({ u_1})}{g(u_2)-g(u_1)}< \frac{f(u_3)-f(u_2)}{g(u_3)-g(u_2)}.$ But this, by Cauchy’s theorem, leads to the inequality $\frac{f^{^{\prime }}(\theta _{12})}{g^{^{\prime }}(\theta _{12})} < \frac{f^{^{\prime }}(\theta _{23})}{g^{^{\prime }}(\theta _{23})}$ for some $\theta _{12}\in (u_1,u_2)$ and $\theta _{23}\in (u_2,u_3)$, contradicting the assumption that function $G$ is nonincreasing in $(\tau ^f,\tau _g)$. Therefore (13) holds, which by Proposition 6 completes the proof.$\square$

6 On two properties of poorly convex functions

In this section we use the theory from the previous section to present two main results describing some properties of poorly convex functions which are basic for the proofs of Theorems 1–5. We also give two examples illustrating the fact that the family of such functions are much richer than the class of convex functions.

In the first theorem we show that pairwise poor convexity of a finite family of functions can be remarkably strengthened.

Theorem 6

Let a finite family $\mathcal{F}=\{f_i : i=1,2,\ldots ,n\}$ of functions on $[a,b]$ be pairwise poorly convex. Then there is a positive function $T$ on $Q_{[a,b]}$ such that all the functions $f_i, { i=1,\ldots , n}$, are $T$-convex on $[a, b]$.

Proof

Let us arbitrarily fix $\bar{u}=(u_1,u_2,u_3)$ in $Q_{[a,b]}$. It suffices to show that there is an $h>0$ such that

$$\begin{aligned} f_i(u_2)-f_i(u_1)\le h[f_i(u_3)-f_i(u_2)],\quad i=1,2,\ldots ,n. \end{aligned}$$

(23)

Now, let $1\le k\le n$. Since function $f_k$ is poorly convex, there is an $h_{k}>0$ such that

$$\begin{aligned} f_k(u_2)-f_k(u_1)\le h_k[f_k(u_3)-f_k(u_2)]. \end{aligned}$$

(24)

On the other hand, one can easily see that if the inequality (24) holds for two positive numbers $h_k=h^{^{\prime }}_k$ and $h_k=h^{^{\prime \prime }}_k$, then it also holds for every convex combination $h_k=\alpha h^{^{\prime }}_k\! + \!(1\!-\!\alpha )h^{^{\prime \prime }}_k$, $0\le \alpha \le 1$. Hence, it follows that all the sets

$$\begin{aligned} A_k = \{h_k: h_k>0 \text{ and } \text{ inequality } (24) \text{ holds } \}\,, k=1,\ldots , n\,, \end{aligned}$$

are nonempty and convex subsets of $R^1$. Now, taking into account the assumption about the pairwise poor convexity of family $\mathcal{F}$, we easily conclude that the class $\{A_1, A_2,\ldots , A_n\}$ of nonempty and convex subsets of $R^1$ has as property that $A_j\cap A_k\not = \emptyset$ for any $1\le j,k \le n, j\not = k$. Hence, by Helly’s theorem (Eckhoff 1993), $\cap _{i=1}^nA_i\not = \emptyset$. But this implies that there is an $h>0$ such that (23) holds, completing the proof. $\square$

The last theorem we give in this section completes our considerations about poorly convex functions. It describes a nice property of families of such functions.

Theorem 7

Let $\mathcal{T}=\{f_\alpha : \alpha \in \mathcal{X}\}$ be a family of continuous functions on an interval $[a, b]$ such that every pair of functions in $\mathcal{T}$ is pairwise poorly convex. Let $\sup _{\alpha \in \mathcal{X}}f_{\alpha }(u) > 0$ for each $u\in [a,b]$. Then there exist $\alpha , \beta \in \mathcal{X}$ and a vector ($\lambda _1, \lambda _2)$ with $\lambda _1, \lambda _2\ge 0$ and $\lambda _1+ \lambda _2=1$, such that $\lambda _1f_\alpha (u)+ \lambda _2f_\beta (u) > 0$ for all $u\in [a,b]$.

Proof

Let $A_\alpha = \{u\!\in \! [a,b]\!: f_\alpha (u)>0\}$ and $B_\alpha = \{u\!\in \! [a,b]\!: f_\alpha (u)\le 0\}$. If $B_{\alpha ^{\prime }}=\emptyset$ for some $\alpha ^{\prime }\in \mathcal{X}$, then the theorem is satisfied by $\alpha =\alpha ^{\prime }, \lambda _1=1, \lambda _2=0$ and arbitrary $\beta$. Therefore, for the rest of the proof we can assume that $B_{\alpha }\not =\emptyset$ for every $\alpha \in \mathcal{X}$

By assumption, every function $f_\alpha$ in $\mathcal{T}$ is continuous and poorly convex on $[a,b]$. Therefore, we can easily deduce with the help of Proposition 3 that for every $\alpha \in \mathcal{X}, B_\alpha$ has the form of a nonempty closed interval, say $B_\alpha =[a_{\alpha }, b_{\alpha }]$, and thereby, $A_\alpha =[a,a_{\alpha })\cup (b_{\alpha },b]$.

The assumption of the theorem implies that $\cup _{\alpha \in \mathcal{X}} A_\alpha = [a,b]$. Since each $A_\alpha$ is an open set in $[a,b]$ (as a space), therefore there exists a finite cover $A_{\alpha _1},A_{\alpha _2},\ldots , A_{\alpha _n}$ of interval $[a,b]$ with minimal $n$. Consequently, for $i=1,2,\ldots , n$ we have

$$\begin{aligned} f_{\alpha _i} (u) \le 0 \quad \text{ for } u\in [a_{\alpha _i}, b_{\alpha _i}] \end{aligned}$$

(25)

and

$$\begin{aligned} f_{\alpha _i}(u) > 0\quad \text{ for } u\in [a,a_{\alpha _i})\cup (b_{\alpha _i},b]. \end{aligned}$$

(26)

Without loss of generality we can assume that $a_{\alpha _1}\le a_{\alpha _2}\le \ldots \le a_{\alpha _n}$, and let $b_{\alpha _q} = \min _{1\le i\le n}b_{\alpha _i}$. The case $a_{\alpha _n} \le b_{\alpha _q}$ cannot occur because it would imply that $f_{\alpha _i}(u)\le 0$ for $u\in [a_{\alpha _n}, b_{\alpha _q}]$ and $i=1,2,\ldots , n$, contradicting the equality $\cup _{i=1}^n A_{\alpha _i} = [a,b]$. Therefore, $b_{\alpha _q}<a_{\alpha _n}$. We will show that the two functions $f_\alpha =f_{\alpha _q}$ and $f_\beta =f_{\alpha _n}$ satisfy the theorem with some positive $\lambda _1=\lambda ^*$ and $\lambda _2=1-\lambda ^*, 0<\lambda ^*<1$.

One can easily deduce with the help of (25), (26) and Proposition 3 that continuous functions $f_{\alpha _q}$ and $f_{\alpha _n}$ are strictly increasing and strictly decreasing on interval $[b_{\alpha _q},a_{\alpha _n}]$, respectively, and besides,

$$\begin{aligned} f_{\alpha _q}(b_{\alpha _q})= 0 \quad \text{ and } \quad f_{\alpha _n}(a_{\alpha _n})= 0. \end{aligned}$$

(27)

Hence, one can easily deduce that there exists a $c$ such that

$$\begin{aligned}&b_{\alpha _q}<c<a_{\alpha _n}\,, f_{\alpha _q}(c)=f_{\alpha _n}(c)>0,\end{aligned}$$

(28)

$$\begin{aligned}&f_{\alpha _q}(u)>0\quad \mathrm{for}\quad u\in (b_{\alpha _q},c] \quad \text{ and } \quad f_{\alpha _n}(u)>0 \quad \text{ for } u\in [b_{\alpha _q},c], \end{aligned}$$

(29)

and

$$\begin{aligned} f_{\alpha _n}(b_{\alpha _q})>f_{\alpha _n}(c)\quad \text{ and } \quad f_{\alpha _q}(b_{\alpha _q})<f_{\alpha _q}(c). \end{aligned}$$

(30)

Consider now the continuous function $G(u,\lambda )=\lambda f_{\alpha _q}(u)+(1-\lambda )f_{\alpha _n}(u)$ defined on rectangle: $a\le u\le b, 0\le \lambda \le 1$. By assumption, the pair of functions $\{f_{\alpha _q}, f_{\alpha _n}\}$ is pairwise poorly convex. Therefore there is a positive function $T$ on $Q_{[a,b]}$ such that functions $f_{\alpha _q}$ and $f_{\alpha _n}$ are $T$-convex. Hence, by statement (e) of Proposition 2, for any fixed $0\le \lambda \le 1, G(u,\lambda )$ is a $T$-convex function of variable $u$, and thereby also poorly convex on $[a,b]$.

In view of inequalities (30), we have $G(c,0)-G(b_{\alpha _q},0) = f_{\alpha _n}(c)-f_{\alpha _n}(b_{\alpha _q})<0$ and $G(c,1)-G(b_{\alpha _q},1) =f_{\alpha _q}(c)-f_{\alpha _q}(b_{\alpha _q})>0$. Hence, by continuity of the function $F(\lambda )= G(c,\lambda )-G(b_{\alpha _q},\lambda )$, there exists a $\lambda ^*, 0<\lambda ^*<1$, such that $F(\lambda ^*)=0$. But this, with the help of (27), (28) and (29) implies that

$$\begin{aligned} G(b_{\alpha _q},\lambda ^*)=G(c,\lambda ^*)>0 \end{aligned}$$

and

$$\begin{aligned} G(u,\lambda ^*)>0\quad \text{ for } \quad u\in [b_{\alpha _q},c]. \end{aligned}$$

Hence, in view of the inequality $b_{\alpha _q}<c$, Proposition 3 applied to function $G(u,\lambda ^*)$ of variable $u$ easily implies that for all $u\in [a,b]$ we have $G(u,\lambda ^*)>0$. Thus we have proved that $\lambda ^* f_{\alpha _q}(u) + (1-\lambda ^*)f_{\alpha _n}(u)>0$ for all $u\in [a,b]$, completing the proof. $\square$

Remark 10

Theorem 7 obviously holds for any family $\mathcal{T}$ of convex continuous functions on interval $[a,b]$ because every such family is pairwise poorly convex (with respect to function $T(\bar{u})=\frac{u_2-u_1}{u_3-u_2}$ (see Remark 4 in Section 3)). Moreover, in the literature one can find the following more general result of Bohnenblust et al. (1950) (applied by them in game theory): If $\mathcal{T}$ is a family of convex continuous functions defined on a compact convex set $K$ in $R^n$ with the property that $\sup _{f \in \mathcal{T}}f(u) > 0$ for each $u\in K$, then there are $n+1$ functions $f_1,\ldots ,f_{n+1}$ in $\mathcal{T}$ and a vector $(\lambda _1,\ldots ,\lambda _{n+1})$ with nonnegative components such that $\sum _{i=1}^{n+1} \lambda _if_i(u)>0$ for all $u\in K$. With this correlation, the natural question arises, if Theorem 7 can be generalized to families of poorly convex functions on $R^n$? Of course, this problem is closely related with the theory of poorly convex functions on $R^n$ which is not studied in this paper.

Pairwise poorly convex and concave families of functions are essential in the assumptions of Theorems 1–5. Therefore, we end this section with two examples presenting some of such families consisting of non-convex functions described with the help of very standard formulas. Proposition 7 appears to be a very useful tool in the analysis of such families. It is worth mentioning here that these examples show that the class of pairwise poorly convex functions is substantially richer than that of convex ones.

Example 3

Consider the family $\mathcal{F}= \{f_\alpha (x): \alpha \in [0,\sqrt{2}]\}$ of non-convex continuous functions with common domain $x\in [0,\sqrt{2}]$, where $f_\alpha (x)=-\exp \{ {-(x\!-\!\alpha )^2}\}$. It can be easily verified that each function $f_\alpha (x)$ is strictly convex on interval $[c_\alpha ,d_\alpha ] = [0,\sqrt{2}]\cap [\alpha \!-\!\frac{\sqrt{2}}{2},\alpha \!+\!\frac{\sqrt{2}}{2}]$ and strictly concave on interval $[0, \alpha \!-\!\frac{\sqrt{2}}{2}]$ or on $[\alpha \!+\!\frac{\sqrt{2}}{2}, \sqrt{2}]$ if $\alpha \ge \frac{\sqrt{2}}{2}$ or $\alpha < \frac{\sqrt{2}}{2}$, respectively. Besides, each function $f_\alpha (x)$ is strictly decreasing in $[0,\alpha ]$ and strictly increasing in $[\alpha ,\sqrt{2}]$. Therefore, by Proposition 3, $\mathcal{F}$ is a family of poorly convex functions on $[0,\sqrt{2}]$.

Now, let us arbitrarily choose two different functions $f(x)=f_{\alpha _1} (x)$ and $g(x)=f_{\alpha _2} (x)$ from $\mathcal{F}$. So We assume that $0\le {\alpha _1}<{\alpha _2}\le \sqrt{2}$. Obviously, $\tau ^{\!f}={\alpha _1} < {\alpha _2}=\tau _{\!g}$. By direct computation we get that

$$\begin{aligned} \left[ f^{^{\prime }}(x)/g^{^{\prime }}(x)\right] ^{^{\prime }} = \frac{{\alpha _2}-{\alpha _1}}{(x-{\alpha _2})^2}\,\Big [2(x-{\alpha _1})({\alpha _2}-x) -1\Big ] e^{[({\alpha _2}-{\alpha _1})({\alpha _2}+{\alpha _1})-2x]}. \end{aligned}$$

But $2(x-{\alpha _1})({\alpha _2}-x) -1\le 0$ for $x\in ({\alpha _1},{\alpha _2})$ (because $({\alpha _1},{\alpha _2})\subset [0,\sqrt{2}]$). Therefore this implies that $[f^{^{\prime }}(x)/{ g^{^{\prime }}(x)}]^{^{\prime }} \le 0$ for $x\in ({\alpha _1},{\alpha _2})$, and thereby function $G(x) = f^{^{\prime }}(x)/g^{^{\prime }}(x)$ is nonincreasing in interval $(\tau ^{\!f}, \tau _{\!g})$. Hence, by Proposition 7, every pair of functions on family $\mathcal{F}$ is pairwise poorly convex.

Example 4

Consider the family $\mathcal{U}= \{f_{\alpha \beta \gamma } (x): \alpha \in { {R}}, \beta <0, \gamma \in [-\frac{\pi }{2},\frac{\pi }{2}]\}$ of non-convex continuous functions with common domain $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$, where $f_{\alpha \beta \gamma } (x)= \alpha + \beta \sin (x-\gamma )$. The fact that each function in family $\mathcal{U}$ is poorly convex easily follows from Proposition 3.

Let us arbitrarily choose two different functions $f(x)=f_{\alpha _1\beta _1\gamma _1}(x)$ and $g(x)=f_{\alpha _2\beta _2\gamma _2}(x)$ from $\mathcal{U}$. We easily deduce that $\tau _{\!f}=\tau ^{\!f}=\gamma _1-\frac{\pi }{2}$ and $\tau _{\!g}=\tau ^{\!g}=\gamma _2-\frac{\pi }{2}$. When $\gamma _1=\gamma _2$, the pair of functions $f(x)$ and $g(x)$ are pairwise poorly convex because of Proposition 5. For the second case we can assume that $-\frac{\pi }{2}\le \gamma _1<\gamma _2\le \frac{\pi }{2}$, whence $\tau ^{\!f} < \tau _{\!g}$. Then we easily get

$$\begin{aligned} \left[ f^{^{\prime }}(x)/g^{^{\prime }}(x)\right] ^{^{\prime }} = - \frac{\beta _1\sin (\gamma _2-\gamma _1)}{\beta _2\cos ^2 (x-\gamma _2)}. \end{aligned}$$

Hence, $[f^{^{\prime }}(x)/g^{^{\prime }}(x)]^{^{\prime }}\le 0$ for $x\in (\tau ^{\!f}, \tau _{\!g})$, because of $0< \gamma _2-\gamma _1\le \pi$. Therefore, by Proposition 7, every pair of functions in family $\mathcal{U}$ is pairwise poorly convex.

7 Proofs of the main theorems

In this section we prove our main Theorems 1–5 from Sect. 4.

Proof of Theorem 1

We construct two matrices $A^n$ and $B^n$ of dimension $n\times n$ with their elements denoted by $a^n_{ij}$ and $b^n_{ij}$, and defined by the following: $a^n_{ij}=F_1\left( \frac{i}{n},\frac{j}{n}\right)$ and $b^n_{ij}=F_2\left( \frac{i}{n},\frac{j}{n}\right) , i,j=1,2,\ldots ,n$. Further, let $\Gamma (A^n,B^n)$ be the bimatrix game with payoff matrices $A^n$ and $B^n$ for Players 1 and 2, respectively.

Let $n$ be a natural number and consider the finite family $F=\{f_i(u)\!: i=1,\ldots , n\}$ of functions on $[0, 1]$, where $f_i(u)=F_1(u,\frac{i}{n})$ for $1\le i\le n$ and $0\le u\le 1$. By assumption, family $F$ is pairwise poorly concave. Therefore Theorem 6 and statement (a) of Proposition 2 imply that there is a positive function $H(u_1,u_2,u_3)$ on $Q_{[0, 1]}$ such that for all $0\le u_1<u_2<u_3\le 1$,

$$\begin{aligned} F_1\left( u_2,\frac{i}{n}\right) -F_1\left( u_1,\frac{i}{n}\right) \ge H(u_1,u_2,u_3)[F_1\left( u_3,\frac{i}{n}\right) -F_1\left( u_2,\frac{i}{n}\right) ] \quad \text{ for } i=1,\ldots ,n. \end{aligned}$$

But this implies that for the positive constants $h_i=H\left( \frac{i}{n},\frac{i+1}{n},\frac{i+2}{n}\right) , 1\le i\le n-2$, and for all $j, 1\le j \le n$, the inequalities hold: $a^n_{i+1,j}-a^n_{ij} \ge h_{i}(a^n_{i+2,j}-a^n_{i+1,j})$ for $i=1,2,\ldots , n-2$. Hence, we can easily see that there is a sequence $\theta _1, \theta _2,\ldots ,\theta _{n-1}$ of positive numbers such that for all $j, 1\le j\le n$,

$$\begin{aligned} \theta _{1}(a^n_{2j}-a^n_{1j}) \ge \theta _{2}(a^n_{3j}-a^n_{2j})\ge \ldots \ge \theta _{n-1}(a^n_{nj}-a^n_{n-1,j}). \end{aligned}$$

Therefore the bimatrix game $\Gamma (A^n,B^n)$ is column-concave [see Definition 1 and Theorem 4 in Połowczuk (2006)], and consequently, it follows [by Theorem 7 in Połowczuk (2006)] that game $\Gamma (A^n,B^n)$ has a Nash equilibrium $(\mu _n,\nu _n)$ of the form $\mu _n= \lambda _n\delta _{s_n} + (1-\lambda _n)\delta _{s_n+1}$ and $\nu _n= \gamma _n\delta _{r_n} + (1-\gamma _n)\delta _{u_n}$ for some reals $0\le \lambda _n, \gamma _n\le 1$ and naturals $1\le s_n<n, 1\le r_n,u_n\le n$. Therefore, by the Nash equilibrium inequality,

$$\begin{aligned}&\lambda _n \gamma _n a_{s_n,r_n} + (1\!-\!\lambda _n)\gamma _na_{s_n+1,r_n} \!\!+\!\! \lambda _n (1\!-\!\gamma _n) a_{s_n,u_n} + (1\!-\!\lambda _n) (1-\gamma _n) a_{s_n+1,u_n} \\&\quad \ge \gamma _n a_{i,r_n} + (1\!-\!\gamma _n) a_{i,u_n}\quad \text{ for } i=1,\ldots ,n, \end{aligned}$$

and

$$\begin{aligned}&\lambda _n \gamma _n b_{s_n,r_n} + (1\!-\!\lambda _n)\gamma _nb_{s_n+1,r_n} \!\!+\!\! \lambda _n (1\!-\!\gamma _n) { b}_{s_n,u_n} + (1\!-\!\lambda _n) (1-\gamma _n) b_{s_n+1,u_n}\\&\quad \ge \lambda _n b_{s_n,j} + (1\!-\!\lambda _n) b_{s_n+1,j} \quad \text{ for } j=1,\ldots ,n. \end{aligned}$$

But these inequalities are equivalent to

$$\begin{aligned}&\lambda _n \gamma _n F_1\left( \frac{s_n}{n},\frac{r_n}{n}\right) + (1-\lambda _n)\gamma _nF_1\left( \frac{s_n+1}{n},{ \frac{r_n}{n}}\right) + \lambda _n (1-\gamma _n) F_1\left( \frac{s_n}{n},\frac{u_n}{n}\right) \\&\quad + (1-\lambda _n) (1-\gamma _n) F_1\left( \frac{s_n+1}{n},\frac{u_n}{n}\right) \ge \gamma _n F_1\left( \frac{i}{n},\frac{r_n}{n}\right) \\&\quad + (1-\gamma _n) F_1\left( \frac{i}{n},\frac{u_n}{n}\right) \quad \text{ for } i=1,\ldots ,n, \end{aligned}$$

and

$$\begin{aligned}&\lambda _n \gamma _n F_2\left( \frac{s_n}{n},\frac{r_n}{n}\right) + (1-\lambda _n)\gamma _nF_2\left( \frac{s_n+1}{n},{ \frac{r_n}{n}}\right) + \lambda _n (1-\gamma _n) F_2\left( \frac{s_n}{n},\frac{u_n}{n}\right) \\&\quad + (1-\lambda _n) (1-\gamma _n) F_2\left( \frac{s_n+1}{n},\frac{u_n}{n}\right) \ge { \lambda _n} F_2\left( { \frac{s_n}{n},\frac{j}{n}}\right) \\&\quad + (1-{ \lambda _n}) F_2\left( { \frac{s_n+1}{n}},\frac{j}{n}\right) \quad \text{ for } i=1,\ldots ,n. \end{aligned}$$

Now, letting $n\rightarrow \infty$, we can obviously choose a subsequence $n^{^{\prime }}\rightarrow \infty$ to have the convergence $\lambda _{n^{^{\prime }}}\rightarrow \alpha , \gamma _{n^{^{\prime }}}\rightarrow \beta , {s_{n^{^{\prime }}}}/{n^{^{\prime }}}\rightarrow a, {r_{n^{^{\prime }}}}/{n^{^{\prime }}}\rightarrow c$ and ${u_{n^{^{\prime }}}}/{n^{^{\prime }}}\rightarrow d$ for some $0\le \alpha , \beta , a, c, d\le 1$. Therefore also ${(s_{n^{^{\prime }}}+1)}/{n^{^{\prime }}}\rightarrow a$.

Further, let us arbitrarily choose $0\le x,y\le 1$. Of course, there are sequences $i^{^{\prime }},j^{^{\prime }}\rightarrow \infty$ such that ${i^{^{\prime }}}/{n^{^{\prime }}}\rightarrow x$ and ${j^{^{\prime }}}/{n^{^{\prime }}}\rightarrow y$.

Hence, taking into account the continuity of functions $F_1$ and $F_2$, the last two inequalities imply (after taking $n=n^{^{\prime }}\rightarrow \infty$) the following: for any $0\le x,y\le 1$

$$\begin{aligned} \beta F_1(a,c) + (1-\beta ) F_1(a,d) \ge \beta F_1(x,c) + (1-\beta ) F_1(x,d) \end{aligned}$$

and

$$\begin{aligned} \beta F_2(a,c) + (1-\beta ) F_2(a,d) \ge F_2(a,y). \end{aligned}$$

After defining two strategies $\mu ^*=\delta _a$ and $\nu ^*=\beta \delta _c + (1-\beta )\delta _d$ for Players 1 and 2 in game $\Gamma$, respectively, the last inequalities can be equivalently rewritten as

$$\begin{aligned} F_1(\mu ^*, \nu ^*) \ge F_1(x, \nu ^*) \quad \text{ and } \quad F_2(\mu ^*, \nu ^*) \ge F_2(\mu ^*, y),\quad 0\le x,y \le 1. \end{aligned}$$

Thus $(\mu ^*, \nu ^*)$ is a Nash equilibrium in game $\Gamma$, which ends the proof of Theorem 1. $\square$

Proof of Theorem 2

Let us define an auxiliary non-zero-sum two-person game $\tilde{\Gamma }$ on the unit square with payoff functions $\tilde{F}_1$ and $\tilde{F}_2$ of the form

$$\begin{aligned} \tilde{F}_1(p,x_2) = p F_1(0,x_2) + (1-p) F_1(1,x_2), \quad 0\le p, x_2 \le 1, \end{aligned}$$

and

$$\begin{aligned} \tilde{F}_2(p,x_2) = p F_2(0,x_2) + (1-p) F_2(1,x_2). \quad 0\le p, x_2 \le 1. \end{aligned}$$

By assumption, $\tilde{F}_1(p,x_2)$ and $\tilde{F}_2(p,x_2)$ are continuous functions on $[0,1]^2$. Additionally, function $\tilde{F}_1(p,x_2)$ is linear in variable $p$, and thereby concave in $p$. Hence, the family of functions $\{\tilde{F}_1(\cdot ,x_2): 0\le x_2\le 1\}$ is pairwise poorly concave (as stated in Remark 4 in Sect. 3). Therefore game $\tilde{\Gamma }$ satisfies the assumptions of Theorem 1 and consequently, there are $0\le \alpha , \beta , c,d\le 1$ such that the pair of strategies $\mu _1^0=\delta _{\alpha }$ and $\mu _2^0=\beta {\delta _c} + (1-\beta ){\delta _d}$ creates a Nash equilibrium in game $\tilde{\Gamma }$. This implies the following inequalities:

$$\begin{aligned} \tilde{F}_1(x_1,\mu ^0_2) \le \tilde{F}_1(\mu ^0_1,\mu ^0_2) \quad \text{ and } \quad \tilde{F}_2(\mu ^0_1,x_2) \le \tilde{F}_2(\mu ^0_1,\mu ^0_2) \quad \text{ for } \quad 0\le x_1, x_2\le 1. \nonumber \\ \end{aligned}$$

(31)

We will show that the pair $(\mu _1^*,\mu _2^*)$ with $\mu _1^*=\alpha {\delta _0} + (1-\alpha ){\delta _1}$ and $\mu _2^*=\mu _2^0$ is a Nash equilibrium in game $\Gamma$.

First notice that

$$\begin{aligned} \tilde{F}_i(\mu _1^0,\mu _2^0) = F_i(\mu _1^*,\mu _2^*),\quad i=1,2. \end{aligned}$$

(32)

By assumption, the family of functions $\{F_1(\,\cdot \,,x_2): 0\le x_2\le 1\}$ is pairwise poorly convex. Hence, for every $0<x_1<1$ there is a positive number $h(x_1)$ such that

$$\begin{aligned} F_1(x_1,c) - F_1(0,c) \le h(x_1)[F_1(1,c) - F_1(x_1,c)] \end{aligned}$$

(33)

and

$$\begin{aligned} F_1(x_1,d) - F_1(0,d) \le h(x_1)[F_1(1,d) - F_1(x_1,d)]. \end{aligned}$$

(34)

Using (33) and the definition of $\tilde{F}_1$, for $0 < x_1 < 1$, we have

$$\begin{aligned} F_1(x_1,c) \le \frac{1}{1+h(x_1)}F_1(0,c) + \frac{{ h(x_1)}}{1+h(x_1)}F_1(1,c) = \tilde{F}_1\left( \frac{1}{1+h(x_1)},c\right) . \end{aligned}$$

Besides, $F_1(0,c)=\tilde{F}_1(1,c)$ and $F_1(1,c)=\tilde{F}_1(0,c)$. Therefore, for any $0\le x_1\le 1$ there is a number $p_{x_1}\in [0, 1]$ such that

$$\begin{aligned} F_1(x_1,c) \le \tilde{F}_1(p_{x_1},c) \quad \text{ for } \quad 0\le x_1\le 1. \end{aligned}$$

Exactly in the same way, using (34), we also get

$$\begin{aligned} F_1(x_1,d) \le \tilde{F}_1(p_{x_1},d)\quad \text{ for } \quad 0\le x_1\le 1. \end{aligned}$$

But the last two inequalities imply that $F_1(x_1,\mu ^0_2) \le \tilde{F}_1(p_{x_1},\mu _2^0)$ for $0\le x_1\le 1$.

Now, using this inequality, (31), (32) and the equality $\mu _2^*=\mu _2^0$, we can conclude as follows:

$$\begin{aligned} F_1(x_1,\mu ^*_2) = F_1(x_1,\mu ^0_2) \le \tilde{F}_1(p_{x_1},\mu _2^0) \le \tilde{F}_1(\mu _1^0,\mu _2^0) = F_1(\mu _1^*,\mu _2^*). \end{aligned}$$

Consequently,

$$\begin{aligned} F_1(x_1,\mu ^*_2) \le F_1(\mu _1^*,\mu _2^*) \quad \text{ for } \quad 0\le x_1 \le 1. \end{aligned}$$

(35)

Similarly, using (31), (32) and the definition of strategy $\mu ^*_1$, for all $x_2\in [0,1]$ we have

$$\begin{aligned} F_2(\mu ^*_1,x_2)&= \alpha F_2(0,x_2) + (1-\alpha ) F_2(1,x_2) =\tilde{F}_2(\mu _1^0,x_2) \le \tilde{F}_2(\mu _1^0,\mu _2^0) \\&= F_2(\mu _1^*,\mu _2^*). \end{aligned}$$

Therefore, $F_2(\mu ^*_1,x_2) \le F_2(\mu _1^*,\mu _2^*)$ for $0\le x_2\le 1$, which together with (35) completes the proof of Theorem 2. $\square$

Proof of Theorem 3

Assume that $(\mu ,\nu )$ is a Nash equilibrium in game $\Gamma$. Let $x_1\in [0,1]$ and suppose that $F_1(\mu ,x^{^{\prime }}_2)<F_1(x_1,x^{^{\prime }}_2)$ for all $x^{^{\prime }}_2\in \text{ supp } (\nu )$. But then this inequality would imply that $F_1(\mu ,\nu )<F_1(x_1,\nu )$ which contradicts the fact that $(\mu ,\nu )$ is a Nash equilibrium.

Let us fix $\varepsilon >0$. Therefore for all $x_1\in [0,1]$

$$\begin{aligned} \sup _{x_2\in \mathrm{supp}(\nu )}[F_1(\mu ,x_2) - F_1(x_1,x_2)+\varepsilon ] > 0. \end{aligned}$$

Let us denote $G_{x_2}(x_1) = F_1(\mu ,x_2) - F_1(x_1,x_2)+\varepsilon$. By assumption, the family of functions $\{F_1(\,\cdot \,,x_2): 0\le x_2\le 1\}$ is pairwise poorly concave. Hence, we can easily state with the help of statement (a) of Proposition 2 that the family of functions $\{G_{x_2}(x_1): x_2 \in \text{ supp } (\nu )\}$ of variable $x_1$ is pairwise poorly convex on $[0,1]$ and satisfies the assumptions of Theorem 7. Therefore, there are $c,d\in$ supp($\nu$) and a probability vector $(\beta ,1-\beta )$ such that for all $x_1\in [0,1]$ we have $\beta G_{c}(x_1) + (1-\beta )G_{d}(x_1)>0$. But this, in view of the arbitrarity of $\varepsilon$, implies that for each $x_1\in [0,1]$

$$\begin{aligned} \beta F_1(\mu ,c) + (1-\beta ) F_1(\mu ,d) \ge \beta F_1(x_1,c) + (1-\beta ) F_1(x_1,d), \end{aligned}$$

or equivalently,

$$\begin{aligned} F_1(\mu \,, \beta \delta _c + (1-\beta )\delta _d) \ge F_1(x_1, \beta \delta _c + (1-\beta )\delta _d). \end{aligned}$$

(36)

On the other hand, since $(\mu ,\nu )$ is a Nash equilibrium,

$$\begin{aligned} F_2({ \mu ,\nu }) \ge F_2(\mu ,x_2) \quad \text{ for } \quad 0\le x_2\le 1. \end{aligned}$$

(37)

Besides, $c,d\in$ supp($\nu$) and function $F_2(x_1,x_2)$ is continuous on $[0,1]^2$. This and inequality (37) imply that $F_2(\mu ,c)=F_2(\mu ,d)=F_2(\mu ,\nu )$, and consequently

$$\begin{aligned} F_2(\mu ,\nu ) = F_2(\mu \,, \, \beta \delta _c + (1-\beta )\delta _d). \end{aligned}$$

Now, taking into account this equality, (37) and (36), we easily deduce that the pair $(\mu \,, \, \beta \delta _c + (1-\beta )\delta _d)$ is also a Nash equilibrium in game $\Gamma$, and it satisfies Theorem 3. Thus the proof has been completed. $\square$

Proof of Theorem 4

It is an immediate consequence of Theorem 3. $\square$

Proof of Theorem 5

Let us define an auxiliary non-zero-sum two-person game $\hat{\Gamma }$ on the unit square with payoff functions ${ \hat{F}_1}(p,q)$ and ${ \hat{F}_2}{(p,q)}, 0\le p, q \le 1$, of the form

$$\begin{aligned} \hat{F}_i(p,q) =&pq F_i(0,0) + (1-p)q F_i(1,0) + p(1-q) F_i(0,1) \\&+ (1-p)(1-q) F_i(1,1), \end{aligned}$$

for $i=1,2$.

We easily see that payoff functions $\hat{F}_1(p,q)$ and $\hat{F}_2(p,q)$ are continuous on $[0,1]^2$ and linear in each variable. Therefore they satisfy the assumptions of Theorem A in Sect. 2. Consequently, there are $0\le \alpha , \beta \le 1$ such that the pair of strategies $\mu _1^0=\delta _{\alpha }$ and $\mu _2^0=\delta _{\beta }$ is a Nash equilibrium in game $\hat{\Gamma }$. We will show that the pair $(\mu _1^*,\mu _2^*)$ with $\mu _1^*=\alpha {\delta _0} + (1-\alpha ){\delta _1}$ and $\mu _1^*=\beta {\delta _0} + (1-\beta ){\delta _1}$ is a Nash equilibrium in game $\Gamma$.

First notice that

$$\begin{aligned} \hat{F}_i(\mu _1^0,\mu _2^0) = F_i(\mu _1^*,\mu _2^*),\quad i=1,2. \end{aligned}$$

(38)

By assumption, the family of functions $\{F_1(\,\cdot \,,x_2): 0\le x_2\le 1\}$ is pairwise poorly convex. Now, using this and (38), one can easily see that after changing $\tilde{F}_1$ with $\hat{F}_1$, the reasoning given between formulae (33) and (35) can be repeated with $c=0, d=1$ to get that $F_1(x_1,\mu ^*_2) \le F_1(\mu _1^*,\mu _2^*)$ for $0\le x_1 \le 1$. On the other hand, since also the family of functions $\{F_2(x_1,\,\cdot ): 0\le x_1\le 1\}$ is pairwise poorly convex, by the same way $F_2(\mu ^*_1,x_2) \le F_2(\mu _1^*,\mu _2^*)$ for $0\le x_2\le 1$. Therefore the pair $(\mu _1^*,\mu _2^*)$ is a Nash equilibrium in game $\Gamma$, which completes the proof. $\square$