Nash and Bayes–Nash equilibria in strategic-form games with intransitivities

Abstract

We study games with intransitive preferences that admit skew-symmetric representations. We introduce the notion of surrogate better-reply security for discontinuous skew-symmetric games and elucidate the relationship between surrogate better-reply security and other security concepts in the literature. We then prove existence of behavioral strategy equilibrium for discontinuous skew-symmetric games of incomplete information (and, in particular, existence of mixed-strategy equilibrium for discontinuous skew-symmetric games of complete information), generalizing extant results.

Appendix

1.1 Proof of Theorem 1

Theorem 1 is restated here for the convenience of the reader.

Theorem1. Suppose that$G=(X_{i},\varphi _{i})_{i=1}^{N}$is a compact SSYM game with$X_{i}$convex for eachi. IfGsatisfies surrogate correspondence security*, thenGpossesses a Nash equilibrium.

Proof

Suppose that G satisfies surrogate correspondence security* with surrogate function H. Suppose that G has no Nash equilibrium. We will adapt the proof of Theorem 4 in Nessah and Tian (2016) and construct a new game $G^{*}$ with two players $\alpha$ and $\beta$, each with the same strategy set X. We will then show that the game $G^{*}$ satisfies the hypotheses of Reny’s (2016a) Theorem 5.6, implying that $G^{*}$ has a Nash equilibrium, and that this implies that the game G has a Nash equilibrium. This contradiction establishes the result.

Suppose that G has no equilibrium.

Step 1 Define a new game $G^{*}=(X,X,\succsim _{\alpha },\succsim _{\beta })$ with two players, each of whom has strategy set X. The preferences of player $\alpha$ are defined as follows:

$$\begin{aligned} (\xi ,\eta )\succsim _{\alpha }(x,y)\text { if and only if }u_{\alpha }(\eta ,\xi )\ge u_{\alpha }(y,x), \end{aligned}$$

where $u_{\alpha }:X\times X\rightarrow \mathbb {R}$ is defined as

$$\begin{aligned} u_{\alpha }(\xi ,\eta ):= {\left\{ \begin{array}{ll} 1 &{}\quad \text { if }\xi =\eta , \\ 0 &{}\quad \text { if }\xi \ne \eta . \end{array}\right. } \end{aligned}$$

The preferences of player $\beta$ are defined as follows:

$$\begin{aligned} (\xi ,\eta )\succsim _{\beta }(x,y)\text { if and only if }f_{i}(\eta _{i},\xi )\ge f_{i}(y_{i},x)\text { for all }i, \end{aligned}$$

where $f_{i}:X_{i}\times X\rightarrow R$ is defined as follows:

$$\begin{aligned} f_{i}(\xi _{i},\eta )=\varphi _{i}((\xi _{i},\eta _{-i}),\eta )+H_{i}(\eta ) \text { for each }i\text { and each }(\xi _{i},\eta )\in X_{i}\times X. \end{aligned}$$

Applying Definition 5.4 in Reny (2016a), we claim that the game $G^{*}=(X,X,\succsim _{\alpha },\succsim _{\beta })$ is correspondence secure with respect to $I=\{\beta \}$. To see this, suppose that $G^{*}$ is not correspondence secure with respect to $I=\{\beta \}$. As per the definition of $B_{I}$ on p. 556 of Reny (2016a), note that

$$\begin{aligned} B_{\{\beta \}}=\{(x,y)\in X\times X:x=y\}. \end{aligned}$$

Then, there exists an $(x_{\alpha },x_{\beta })\in B_{\{\beta \}}$ such that $(x_{\alpha },x_{\beta })$ is not a Nash equilibrium in $G^{*}$ and the following holds: for every neighborhood U of $(x_{\alpha },x_{\beta })$ and every co-closed correspondence ($d_{\alpha },d_{\beta }):U\rightarrow X\times X$ with nonempty values, there exists a $(y_{\alpha },y_{\beta })\in U\cap$$B_{\{\beta \}}$ such that, for some $(x_{\alpha }^{\prime },x_{\beta }^{\prime })\in U\cap$$B_{\{\beta \}}$ and some $z\in d_{\beta }(x_{\alpha }^{\prime },x_{\beta }^{\prime }),$

$$\begin{aligned} y_{\beta }\in \text{ co }\{w\in X:(y_{\alpha },w)\succsim _{\beta }(x_{\alpha }^{\prime },z)\}. \end{aligned}$$

Given the definition of $B_{\{\beta \}},$ we conclude that there exists $x^{*}\in X$ such that $(x^{*},x^{*})$ is not a Nash equilibrium of $G^{*}$ and the following holds: for every neighborhood U of $(x^{*},x^{*})$ in $X\times X$ and every co-closed correspondence ($d_{\alpha },d_{\beta }):U\rightarrow X\times X$ with nonempty values, there exists a $y^{*}\in X$ with $(y^{*},y^{*})\in U$ such that for some $x^{\prime }\in X$ satisfying $(x^{\prime },x^{\prime })\in U$ and some $z\in d_{\beta }(x^{\prime },x^{\prime }),$

$$\begin{aligned} y^{*}\in \text{ co }\{w\in X:(y^{*},w)\succsim _{\beta }(x^{\prime },z)\}. \end{aligned}$$

Step 2 Note that $x^{*}$ is not a Nash equilibrium of G since $(x^{*},x^{*})$ is not a Nash equilibrium of $G^{*}$. Applying surrogate correspondence security* (Definition ), there exists an open set V containing $x^{*}$ and a co-closed correspondence $\delta :V\rightarrow X$ such that the following holds: for every $y\in V$ there exists a player j such that

$$\begin{aligned} y_{j}\notin \text{ co }\{w_{j}\in X_{j}:f_{j}(\zeta _{j},\xi )\le f_{j}(w_{j},y)\} \end{aligned}$$

whenever $\xi \in V$ and $\zeta _{j}\in \delta _{j}(\xi )$. Let $U:=V\times V$ and define $(d_{\alpha }(x,y),d_{\beta }(x,y)):=(\delta (x),\delta (y))$ for all $(x,y)\in V\times V$. Then, U is an open set in $X\times X$ containing $(x^{*},x^{*})$ and it is easily verified that $(d_{\alpha },d_{\beta }):U\rightarrow X\times X$ is co-closed with nonempty values. Applying Step 1, there exists $y^{*}\in V$ such that for some $x^{\prime }\in V$ and some $z\in d_{\beta }(x^{\prime },x^{\prime }),$

$$\begin{aligned} y^{*}\in \text{ co }\{w\in X:(y^{*},w)\succsim _{\beta }(x^{\prime },z)\}. \end{aligned}$$

(3)

Since $y^{*}\in V,$$x^{\prime }\in V,$ and $z_{i}\in \delta _{i}(x^{\prime })$ for each i, it follows that there exists a player j such that

$$\begin{aligned} y^{*}\notin \text{ co }\{w_{j}\in X_{j}:f_{j}(z_{j},x^{\prime })\le f_{j}(w_{j},y^{*})\}. \end{aligned}$$

(4)

Since

$$\begin{aligned} \text{ co }\{w\in X:(y^{*},w)\succsim _{\beta }(x^{\prime },z)\}&= \text{ co }\{w\in X:f_{i}(w_{i},y^{*})\ge f_{i}(z_{i},x^{\prime })\text { for each }i\} \\&=\text{ co }\left( \bigcap \limits _{i=1}^{N}\{w\in X:f_{i}(w_{i},y^{*})\ge f_{i}(z_{i},x^{\prime })\}\right) \\&\subseteq \bigcap \limits _{i=1}^{N}\text{ co }\{w\in X:f_{i}(w_{i},y^{*})\ge f_{i}(z_{i},x^{\prime })\}, \end{aligned}$$

(3) implies that

$$\begin{aligned} y^{*}\in \text{ co }\{w\in X:f_{j}(w_{j},y^{*})\ge f_{j}(z_{j},x^{\prime })\}, \end{aligned}$$

contradicting (4). This establishes that the game $G^{*}=(X,X,\succsim _{\alpha },\succsim _{\beta })$ is correspondence secure with respect to $I=\{\beta \}$ (according to Definition 5.4 in Reny (2016a)).

Step 3 The game $G^{*}=(X,X,\succsim _{\alpha },\succsim _{\beta })$ satisfies the assumptions of Theorem 5.6 in Reny (2016a). Therefore, $G^{*}$ admits a Nash equilibrium $(\overline{x}, \overline{x})\in X\times X,$ i.e., for each i and for all $y_{i}\in X_{i},$

$$\begin{aligned} f_{i}(\overline{x}_{i},\overline{x})\ge f_{i}(y_{i},\overline{x})\text { for all }i\text {. } \end{aligned}$$

Therefore, $\varphi _{i}((y_{i},\overline{x}_{-i}),\overline{x})\le 0$ for each i and for all $y_{i}\in X_{i}$ implying that $\overline{x}$ is a Nash equilibrium in G. This last contradiction proves the theorem. $\square$

1.2 Proof of Lemma 3

1.2.1 Preliminary lemma

Lemma 6

Suppose that the SSYM Bayesian game $\left( (T_{i}, \mathcal {T}_{i}),X_{i},\psi _{i},p\right) _{i=1}^{N}$ satisfies uniform surrogate payoff security with surrogate function H. Suppose that $H_{i}:T\times X\rightarrow \mathbb {R}$ is bounded and $\left( [\otimes _{j=1}^{N}\mathcal {T}_{j}]\otimes [\otimes _{j=1}^{N}\mathcal {B} (X_{j})],\mathcal {B}(\mathbb {R})\right)$-measurable for each i. If p is absolutely continuous with respect to $p_{1}\otimes \cdots \otimes p_{N}$, then for each i, $\varepsilon >0$, and $s_{i}\in \mathcal {P}_{i}$, then there exists $s_{i}^{*}\in \mathcal {P}_{i}$ such that for every $\sigma \in \mathcal {Y}$, there exists a neighborhood $V_{\sigma }$ of $\sigma$ such that

$$\begin{aligned} \varvec{\psi }_{i}((s_{i}^{*},\nu _{-i}),\nu )+\varvec{H} _{i}(\nu )>\varvec{\psi }_{i}((s_{i},\sigma _{-i}),\sigma )+\varvec{H }_{i}(\sigma )-\varepsilon ,\quad \text {for all }\nu \in V_{\sigma }. \end{aligned}$$

(5)

Proof

Fix i, $\varepsilon >0$, and $s_{i}\in \mathcal {P}_{i}$. Let f be a density of p with respect to $p_{1}\otimes \cdots \otimes p_{N}$. To lighten the notation, let $P:=\otimes _{j=1}^{N}p_{j}$. Let $\mathcal {T}^{*}(P)$ denote the P-completion of $\mathcal {T}$ and let $P^{*}$ denote the unique extension of P to $\mathcal {T}^{*}(P)$. Let

$$\begin{aligned} \mathcal {T}^{*}:=\bigcap \limits _{P\in \Delta (T,\mathcal {T})}\mathcal {T} ^{*}(P) \end{aligned}$$

denote the universal completion of $\mathcal {T}$. Note that $\mathcal {T} \subseteq \mathcal {T}^{*}\subseteq \mathcal {T}^{*}(P)$ and, abusing notation slightly, we will use $P^{*}$ for the restriction of $P^{*}$ to $\mathcal {T}^{*}.$ Note that if $h:T\rightarrow \mathbb {R}$ is a bounded, $(\mathcal {T},\mathcal {B}(\mathbb {R}))$-measurable map, then h is a bounded $(\mathcal {T}^{*},\mathcal {B}(\mathbb {R}))$-measurable map and

$$\begin{aligned} \int _{T}h(t)P(\mathrm{d}t)=\int _{T}h(t)P^{*}(\mathrm{d}t). \end{aligned}$$

Uniform surrogate payoff security gives $s_{i}^{*}\in \mathcal {P}_{i}$ such that for every $(t,x,y)\in T\times X\times X$, there are neighborhoods $V_{x}$ and $V_{y}$ of x and y, respectively, such that

$$\begin{aligned} \begin{aligned} \psi _{i}(t,(s_{i}^{*}(t_{i}), x_{-i}^{\prime }),y^{\prime })&+H_{i}(t,y^{\prime }) \\&>\psi _{i}(t,(s_{i}(t_{i}),x_{-i}),y)+H_{i}(t,y)-\frac{\varepsilon }{2},\\&\quad \text {for all }(x^{\prime },y^{\prime })\in V_{x}\times V_{y}. \end{aligned} \end{aligned}$$

Therefore, for every $(t,x,y)\in T\times X\times X$, there are neighborhoods $V_{x}$ and $V_{y}$ of x and y, respectively, such that

$$\begin{aligned} \begin{aligned}{}[\psi _{i}(t,(s_{i}^{*}(t_{i}),x_{-i}^{\prime }),y^{\prime })&+H_{i}(t,y^{\prime })]f(t) \\&\ge \left[ \psi _{i}(t,(s_{i}(t_{i}),x_{-i}),y)+H_{i}(t,y)-\frac{ \varepsilon }{2}\right] f(t),\\&\text { for all }(x^{\prime },y^{\prime })\in V_{x}\times V_{y}. \end{aligned} \end{aligned}$$

(6)

Define $\xi :T\times X\times X\rightarrow \mathbb {R}$ by

$$\begin{aligned} \xi (t,x,y):=\sup _{n\in \mathbb {N}}\inf _{(x^{\prime },y^{\prime })\in N_{ \frac{1}{n}}(x)\times N_{\frac{1}{n}}(y)}[u_{i}(t,(s_{i}^{*}(t_{i}),x_{-i}^{\prime }))-u_{i}(t,y^{\prime })+H_{i}(t,y^{\prime })]f(t). \end{aligned}$$

Using an argument analogous to that in Step 3 of the proof of Lemma 5 in Carbonell-Nicolau and McLean (2018), one can show that there exists an open set $V_{\sigma }$ in $\mathcal {Y}$ (open with respect to the product topology generated by the $p_{i}$-narrow topology on each factor $\mathcal {Y} _{i}$) containing $\sigma$ such that

$$\begin{aligned}&\int _{T}\int _{X}\int _{X}\xi (t,x,y)\left[ \prod _{j=1}^{N}\nu _{j}(dx_{j}|t_{j})\right] \left[ \prod _{j=1}^{N}\nu _{j}(dy_{j}|t_{j})\right] P^{*}(\mathrm{d}t) \\&\quad >\int _{T}\int _{X}\int _{X}\xi (t,x,y)\left[ \prod _{j=1}^{N}\sigma _{j}(dx_{j}|t_{j})\right] \left[ \prod _{j=1}^{N}\sigma _{j}(dy_{j}|t_{j}) \right] P^{*}(\mathrm{d}t)-\frac{\varepsilon }{2} \end{aligned}$$

for all $(\nu _{1},\ldots ,\nu _{N})\in V_{\sigma }.$

Because for each $(t,x)\in T\times X$ there are neighborhoods $V_{x}$ and $V_y$ of x and y, respectively, such that (6) holds, $(t,x,y)\in T\times X\times X$ implies that

$$\begin{aligned} {[}\psi _{i}(t,(s_{i}^{*}(t_{i}),x_{-i}),y)&+H_i(t,y)]f(t) \\&\quad \ge \xi (t,x,y)\ge \left[ \psi _{i}(t,(s_{i}(t_{i}),x_{-i}),y)+H_i(t,y)- \frac{\varepsilon }{2}\right] f(t). \end{aligned}$$

This, together with the conclusion in the preceding paragraph, implies that for every $(\nu _{1},..,\nu _{N})\in V_{\sigma }$,

$$\begin{aligned}&\varvec{\psi }_i((s^*_i,\nu _{-i}),\nu )+\varvec{H}_i(\nu ) \\&\quad =\int _{T}\int _X\int _X [\psi _{i}(t,(s_{i}^{*}(t_{i}),x_{-i}),y)\\&\qquad +H_i(t,y)]f(t)\left[ \prod _{j=1}^N\nu _{j}(dx_{j}|t_{j}) \right] \left[ \prod _{j=1}^N\nu _{j}(dy_{j}|t_{j})\right] P(\mathrm{d}t) \\&\quad =\int _{T}\int _X\int _X [\psi _{i}(t,(s_{i}^{*}(t_{i}),x_{-i}),y)\\&\qquad +H_i(t,y)]f(t)\left[ \prod _{j=1}^N\nu _{j}(dx_{j}|t_{j}) \right] \left[ \prod _{j=1}^N\nu _{j}(dy_{j}|t_{j})\right] P^*(\mathrm{d}t) \\&\quad \ge \int _{T}\int _{X}\int _X\xi (t,x,y)\left[ \prod _{j=1}^N \nu _{j}(dx_{j}|t_{j})\right] \left[ \prod _{j=1}^N\nu _{j}(dy_{j}|t_{j})\right] P^{*}(\mathrm{d}t) \\&\quad >\int _{T}\int _{X}\int _X\xi (t,x,y) \left[ \prod _{j=1}^N \sigma _{j}(dx_{j}|t_{j})\right] \left[ \prod _{j=1}^N\sigma _{j}(dy_{j}|t_{j}) \right] P^{*}(\mathrm{d}t)-\frac{\varepsilon }{2} \\&\quad \ge \int _{T}\int _{X}\int _X\left[ \psi _{i}(t,(s_{i}(t_{i}),x_{-i}),y)\right. \\&\qquad \left. +\,H_i(t,y)\right] f(t) \left[ \prod _{j=1}^N\sigma _{j}(dx_{j}|t_{j})\right] \left[ \prod _{j=1}^N \sigma _{j}(dy_{j}|t_{j})\right] P^{*}(\mathrm{d}t)-\varepsilon \\&\quad =\int _{T}\int _{X}\int _X[\psi _{i}(t,(s_{i}(t_{i}),x_{-i}),y)\\&\qquad +H_i(t,y)]f(t) \left[ \prod _{j=1}^N\sigma _{j}(dx_{j}|t_{j})\right] \left[ \prod _{j=1}^N\sigma _{j}(dy_{j}|t_{j})\right] P(\mathrm{d}t)-\varepsilon \\&\quad =\varvec{\psi }_i((s_i,\sigma _{-i}),\sigma )+\varvec{H} _i(\sigma )-\varepsilon . \end{aligned}$$

This establishes (5). $\square$

1.2.2 Proof of Lemma 3

We restate Lemma 3 here for the convenience of the reader.

Lemma3. Suppose that the SSYM Bayesian game$\left( (T_{i},\mathcal {T}_{i}),X_{i},\psi _{i},p\right) _{i=1}^{N}$satisfies uniform surrogate payoff security with surrogate functionH. If$H_{i}:T\times X\rightarrow \mathbb {R}$is bounded and$\left( [\otimes _{j=1}^{N}\mathcal {T}_{j}]\otimes [\otimes _{j=1}^{N}\mathcal {B} (X_{j})],\mathcal {B}(\mathbb {R})\right)$-measurable for eachi, and ifpis absolutely continuous with respect to$p_{1}\otimes \cdots \otimes p_{N}$ , then the game$G_{\Gamma }$defined in (2) is surrogate payoff secure with surrogate function$\varvec{H}$.

Proof

Fix $(\sigma ,\mu )\in \mathcal {Y}\times \mathcal {Y}$ , i, and $\varepsilon >0$. Let f be a density of p with respect to $P:=p_{1}\otimes \cdots \otimes p_{N}$. We must show that there exist $\sigma _{i}^{*}\in \mathcal {Y}_{i}$ and a neighborhood $V_{\sigma }$ of $\sigma$ such that

$$\begin{aligned} \begin{aligned}&\varvec{\psi }_{i}((\sigma _{i}^{*},\nu _{-i}),\nu )+\varvec{H} _i(\nu ) \\&\quad >\varvec{\psi }_{i}((\mu _i,\sigma _{-i}),\sigma )+\varvec{H} _i(\sigma )-\varepsilon ,\quad \text {for every }\nu \in V_{\sigma }. \end{aligned} \end{aligned}$$

(7)

By an argument analogous to that in the proof of Lemma 2 of Carbonell-Nicolau and McLean (2018), there exists $s_i\in \mathcal {P}_i$ such that $\varvec{\psi }_{i}((s_i,\sigma _{-i}),\sigma )\ge \varvec{\psi }_{i}((\mu _i,\sigma _{-i}),\sigma )-\frac{\varepsilon }{2}$. Consequently, there exists $s_{i}\in \mathcal {P}_{i}$ such that

$$\begin{aligned} \varvec{\psi }_{i}((s_i,\sigma _{-i}),\sigma )+\varvec{H}_i(\sigma )\ge \varvec{\psi }_{i}((\mu _i,\sigma _{-i}),\sigma )+\varvec{H}_i(\sigma )- \frac{\varepsilon }{2}. \end{aligned}$$

(8)

By Lemma 6, there exist $s_{i}^{*}\in \mathcal {P}_{i}$ and a neighborhood $V_{\sigma }$ of $\sigma$ such that

$$\begin{aligned} \varvec{\psi }_i((s^*_i,\nu _{-i}),\nu )+\varvec{H}_i(\nu )>\varvec{\psi }_i((s_i,\sigma _{-i}),\sigma )+\varvec{H}_i(\sigma )- \frac{\varepsilon }{2},\quad \text {for all }\nu \in V_{\sigma }. \end{aligned}$$

This, together with (8), gives (7) for some $\sigma _{i}^{*}\in \mathcal {Y}_{i}$. $\square$

1.3 Proof that the game in Example 2 satisfies surrogate better-reply security

The game was defined as $G=(X_{i},u_{i})_{i=1}^{2}$, where for each i, $X_{i}:=[0,2]$, and for $x=(x_{1},x_{2})\in [0,1]^{2}$, the payoff $u_{i}$ is defined as follows:

$$\begin{aligned} u_{i}(x_{1},x_{2}):= {\left\{ \begin{array}{ll} 1-x_{i} &{}\quad \text {if }x_{i}>x_{-i}, \\ 0 &{}\quad \text {if }x_{i}\le x_{-i}. \end{array}\right. } \end{aligned}$$

For $x\in [1,2]^{2}\setminus \{(1,1)\}$, define $u_{2}\equiv 0$ and

$$\begin{aligned} u_{1}(x_{1},x_{2}):= {\left\{ \begin{array}{ll} 1 &{}\quad \text {if }1\le x_{1}<2\text { and }x_{2}=2, \\ 0 &{}\quad \text {if }1\le x_{1}<2\text { and }1\le x_{2}<2, \\ 2 &{}\quad \text {if }x_{1}=2\text { and }x_{2}=2, \\ x_{2}-1 &{}\quad \text {if }x_{1}=2\text { and }1\le x_{2}<2. \end{array}\right. } \end{aligned}$$

Everywhere else in [0, 2], the payoffs are identically zero.

Suppose that $H(x):=u(x)$ for all $x\in [0,1]^2$ and $H(x):=0$ elsewhere. Suppose that $x=(x_1,x_2)$ is not a Nash equilibrium. We consider six cases.

Case 1$x_{2}=2$, $0\le x_{1}<2$. Let $\overline{x}_{1}:=2.$ Choose $\varepsilon$ so that $z_{1}<2$ and $z_{2}> \frac{3}{2}$ whenever $z\in B_{\varepsilon }(x)$ (here $B_\varepsilon (x)$ represents the open neighborhood of x with radius $\varepsilon$). Note that $H(z)=0$ for all $z\in B_{\varepsilon }(x)$ and that $(x,\alpha )\in \overline{\Gamma }_{H}$ implies that $\alpha =0.$ Suppose that $z\in B_{\varepsilon }(x).$ Then

$$\begin{aligned} z_{2}<2\Rightarrow u_{1}(2,z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})=(z_{2}-1)-0+0>\frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} z_{2}=2\Rightarrow u_{1}(2,z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})=2-\max \{0,1\}+0\ge 1, \end{aligned}$$

implying that

$$\begin{aligned} \inf _{z\in B_{\varepsilon }(x)}[u_{1}(\overline{x} _{1},z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})]>0. \end{aligned}$$

Case 2$1<x_{2}<2$, $0\le x_{1}<2.$ Let $\overline{x}_{1}:=2.$ Choose $\varepsilon$ so that $z_{1}<2$ and $z_{2}> \frac{x_{2}+1}{2}$ whenever $z\in B_{\varepsilon }(x).$ Note that $H(z)=0$ for all $z\in B_{\varepsilon }(x)$ and that $(x,\alpha )\in \overline{\Gamma }_{H}$ implies that $\alpha =0.$ Suppose that $z\in B_{\varepsilon }(x).$ Then

$$\begin{aligned} u_{1}(2,z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})=(z_{2}-1)-0+0>\frac{ x_{2}+1}{2}-1=\frac{x_{2}-1}{2}, \end{aligned}$$

implying that

$$\begin{aligned} \inf _{z\in B_{\varepsilon }(x)}[u_{1}(\overline{x} _{1},z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})]>0. \end{aligned}$$

Case 3$x_{2}=1$, $0\le x_{1}<1.$ Choose $\varepsilon$ so that $0<\varepsilon <\frac{1-x_{1}}{2}$ and $z_{2}>z_{1}$ whenever $z\in B_{\varepsilon }(x).$ Next, we claim that $\alpha _{2}=0$ if $(x,\alpha )\in \overline{\Gamma }_{H}.$ To see this, suppose that $(x^{k},u(x^{k}))\rightarrow (x,\alpha ).$ Then $x_{2}^{k}>x_{1}^{k}$ for all sufficiently large k. If $x_{2}^{k}\ge 1,$ then $u_{2}(x^{k})=0.$ If $x_{2}^{k}<1,$ then $u_{2}(x^{k})=1-x_{2}^{k}.$ Therefore, $u_{2}(x^{k})\rightarrow 0.$ Now let $\overline{x}_{2}=x_{1}+\varepsilon .$ Suppose that $z\in B_{\varepsilon }(x).$ Then $1>x_{1}+\varepsilon >z_{1}.$ Therefore,

$$\begin{aligned} z_{2}&\ge 1\ and x_{1}+\varepsilon>z_{1} \\&\Rightarrow u_{2}(z_{1},x_{1}+\varepsilon )-u_{2}(z_{1},z_{2})+H_{2}(z_{1},z_{2})=[1-(x_{1}+\varepsilon )]-0+0>\frac{ 1-x_{1}}{2} \end{aligned}$$

and

$$\begin{aligned} z_{2}&<1\ and x_{1}+\varepsilon>z_{1} \\&\Rightarrow u_{2}(z_{1},x_{1}+\varepsilon )-u_{2}(z_{1},z_{2})+H_{2}(z_{1},z_{2}) \\&\quad \quad \quad \quad \quad =[1-(x_{1}+\varepsilon )]-(1-z_{2})+(1-z_{2})=1-(x_{1}+\varepsilon )>\frac{1-x_{1}}{2}, \end{aligned}$$

implying that

$$\begin{aligned} \inf _{z\in B_{\varepsilon }(x)}[u_{2}(2,\overline{x} _{2})-u_{1}(z_{1},z_{2})+H(z_{1},z_{2})]>0. \end{aligned}$$

Case 4$0\le x_{2}<1$, $1<x_{1}\le 2$. Choose $\varepsilon$ so that $0<\varepsilon <\frac{1-x_{2}}{2}$ and $0\le z_{2}<1$ and $1<z_{1}\le 2$ whenever $z\in B_{\varepsilon }(x).$ Note that $H(z)=0$ for all $z\in B_{\varepsilon }(x)$ and that $(x,\alpha )\in \overline{\Gamma }_{H}$ implies that $\alpha =0.$ Suppose that $z\in B_{\varepsilon }(x).$ Then $z_{2}<x_{2}+\varepsilon$ and $z_{2}<1.$ Consequently,

$$\begin{aligned} u_{1}(x_{2}+\varepsilon ,z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})=[(1-(x_{2}+\varepsilon )]-0+0> \frac{1-x_{2}}{2}, \end{aligned}$$

implying that

$$\begin{aligned} \inf _{z\in B_{\varepsilon }(x)}[u_{1}(\overline{x} _{1},z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})]>0. \end{aligned}$$

Case 5$0\le x_{2}<1$, $x_{1}=1.$ Choose $\varepsilon$ so that $0<\varepsilon <\frac{1-x_{2}}{2}$ and $z_{1}>z_{2}$ whenever $z\in B_{\varepsilon }(x).$ Next, we claim that $\alpha _{1}=0$ if $(x,\alpha )\in \overline{\Gamma }_{H}.$ To see this, suppose that $(x^{k},u(x^{k}))\rightarrow (x,\alpha ).$ Then $x_{1}^{k}>x_{2}^{k}$ for all sufficiently large k. If $x_{1}^{k}\ge 1,$ then $u_{1}(x^{k})=0.$ If $x_{1}^{k}<1,$ then $u_{1}(x^{k})=1-x_{1}^{k}.$ Therefore, $u_{1}(x^{k})\rightarrow 0.$ Now let $\overline{x}_{1}=x_{2}+\varepsilon$ and note that $x_{2}+\varepsilon <1.$ Suppose that $z\in B_{\varepsilon }(x).$ Then $x_{2}+\varepsilon >z_{2}$. Consequently,

$$\begin{aligned} z_{1}&\ge 1\ and x_{2}+\varepsilon>z_{2} \\&\Rightarrow u_{1}(x_{2}+\varepsilon ,z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})=[1-(x_{2}+\varepsilon )]-0+0> \frac{1-x_{2}}{2} \end{aligned}$$

and

$$\begin{aligned} z_{1}<1\ and&x_{2}+\varepsilon>z_{2} \\&\Rightarrow u_{1}(x_{2}+\varepsilon ,z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2}) \\&\quad \quad \quad \quad \quad \quad \quad \quad =[1-(x_{2}+\varepsilon )]-(1-z_{1})+(1-z_{1})>\frac{1-x_{2}}{2}, \end{aligned}$$

implying that

$$\begin{aligned} \inf _{z\in B_{\varepsilon }(x)}[u_{1}(\overline{x} _{1},z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})]>0. \end{aligned}$$

Case 6$0\le x_{1}<1,0\le x_{2}<1$. To begin, we claim that for each $(x,\alpha )\in \overline{\Gamma }_{H},$ there exist an i such that $\alpha _{i}=0.$ To see this, suppose that $(x^{k},u(x^{k}))\rightarrow (x,\alpha ).$ If $x_{1}^{k}>x_{2}^{k}$ for all sufficiently large k, then $u_{2}(x^{k})=0$ for all sufficiently large k , implying that $\alpha _{2}=0.$ Otherwise, there exists a subsequence $(x^{k_{m}},u(x^{k_{m}}))$ with $x_{1}^{k_{m}}\le x_{2}^{k_{m}}$ for all m . Consequently, $u_{1}(x^{k_{m}})=0$ for all m implying that $\alpha _{1}=0.$ So suppose that $\alpha _{1}=0.$

Choose $\varepsilon$ so that $0<\varepsilon <\frac{1-x_{2}}{2}$. Note that $H(z)=u(z)$ for all $z\in B_{\varepsilon }(x).$ Now let $\overline{x} _{1}=x_{2}+\varepsilon$ and note that $x_{2}+\varepsilon <1.$ Suppose that $z\in B_{\varepsilon }(x)$. Then $x_{2}+\varepsilon >z_{2}.$ Consequently,

$$\begin{aligned}&u_{1}(x_{2}+\varepsilon ,z_{2})-u_{2}(z_{1},z_{2})+H_{2}(z_{1},z_{2})\\&\qquad =[1-(x_{2}+\varepsilon )]-u_{2}(z_{1},z_{2})+u_{2}(z_{1},z_{2})>\frac{1-x_{2}}{2}, \end{aligned}$$

implying that

$$\begin{aligned} \inf _{z\in B_{\varepsilon }(x)}[u_{1}(\overline{x} _{1},z_{2})-u_{1}(z_{1},z_{2})+H_{1}(z_{1},z_{2})]>0. \end{aligned}$$

A completely symmetric argument applies if $\alpha _{2}=0$, and the proof is complete.