Ordinal potentials in smooth games

Abstract

In the class of smooth non-cooperative games, exact potential games and weighted potential games are known to admit a convenient characterization in terms of cross-derivatives (Monderer and Shapley in Games Econ Behav 14:124–143, 1996a). However, no analogous characterization is known for ordinal potential games. The present paper derives necessary conditions for a smooth game to admit an ordinal potential. First, any ordinal potential game must exhibit pairwise strategic complements or substitutes at any interior equilibrium. Second, in games with more than two players, a condition is obtained on the (modified) Jacobian at any interior equilibrium. Taken together, these conditions are shown to correspond to a local analogue of the Monderer–Shapley condition for weighted potential games. We identify two classes of economic games for which our necessary conditions are also sufficient.

Appendix: Proofs

Proof of Proposition 1

By contradiction. Suppose that, at some profile $x_{N}^{\ast}\in X_{N}$, and for some players i and j with $i\ne j$, we have

$$\begin{aligned}&x_{i}^{\ast}\in {\mathring{X}} _{i}, \quad x_{j}^{\ast}\in {\mathring{X}} _{j}, \quad \frac{\partial u_{i}(x_{N}^{\ast})}{\partial x_{i}} =\frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=0,\quad {\hbox {and}} \end{aligned}$$

(36)

$$\begin{aligned}&\frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}} \cdot \frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}} <0. \end{aligned}$$

(37)

By renaming players, if necessary, we may assume w.l.o.g. that

$$\begin{aligned} \frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}>0>\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j} }, \end{aligned}$$

(38)

which corresponds to the case shown in Fig. 1. It is claimed now that, for any sufficiently small $\varepsilon >0$, the payoff difference corresponding to the upper side of the square satisfies

$$\begin{aligned} \varDelta _{i}^{+}(\varepsilon )\equiv u_{i}\left( x_{i}^{\ast}+\varepsilon ,x_{j}^{\ast}+\varepsilon ,x_{-i,j}^{\ast}\right) -u_{i} \left( x_{i}^{\ast}-\varepsilon ,x_{j}^{\ast}+\varepsilon ,x_{-i,j}^{\ast}\right) >0. \end{aligned}$$

(39)

To prove this, we determine the second-order Taylor approximation of $\varDelta _{i}^{+}(\varepsilon )$ at $\varepsilon =0$. Writing $f(\varepsilon )$ for $\varDelta _{i}^{+}(\varepsilon )$, our differentiability assumptions combined with Taylor’s theorem imply that there is a remainder term r(.) with $\lim _{\varepsilon \rightarrow 0}r(\varepsilon )=0$ such that for any sufficiently small $\varepsilon >0$,

$$\begin{aligned} f(\varepsilon )=f(0)+f^{\prime }(0)\varepsilon +\frac{1}{2}f^{\prime \prime }(0)\varepsilon ^{2}+r(\varepsilon )\varepsilon ^{2}. \end{aligned}$$

(40)

Clearly, $f(0)=0$. As for the first derivative $f^{\prime }(0)$, one obtains

$$\begin{aligned} \frac{\partial \varDelta _{i}^{+}(\varepsilon )}{\partial \varepsilon }&=\left\{ \frac{\partial u_{i}(x_{i}^{\ast}+\varepsilon ,x_{j}^{\ast}+\varepsilon ,x_{-i,j}^{\ast})}{\partial x_{i}}+\frac{\partial u_{i}(x_{i}^{\ast}+\varepsilon ,x_{j}^{\ast}+\varepsilon ,x_{-i,j}^{\ast})}{\partial x_{j} }\right\} \nonumber \\&\quad -\left\{ -\frac{\partial u_{i}(x_{i}^{\ast}-\varepsilon ,x_{j}^{\ast}+\varepsilon ,x_{-i,j}^{\ast})}{\partial x_{i}}+\frac{\partial u_{i} (x_{i}^{\ast}-\varepsilon ,x_{j}^{\ast}+\varepsilon ,x_{-i,j}^{\ast})}{\partial x_{j}}\right\} . \end{aligned}$$

(41)

Evaluating at $\varepsilon =0$, and subsequently exploiting the necessary first-order condition for player i at the interior equilibrium $x_{N}^{\ast}$, we find

$$\begin{aligned} f^{\prime }(0)=\frac{\partial \varDelta _{i}^{+}(0)}{\partial \varepsilon } =2\cdot \frac{\partial u_{i}(x_{N}^{\ast})}{\partial x_{i}}=0. \end{aligned}$$

(42)

Next, consider the second derivative of $\varDelta _{i}^{+}(\varepsilon )$ at $\varepsilon =0$, i.e.,

$$\begin{aligned} \frac{\partial ^{2}\varDelta _{i}^{+}(0)}{\partial \varepsilon ^{2}}&=\left\{ \dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{i}^{2}}+\dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}+\dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}}+\dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}^{2}}\right\} \end{aligned}$$

(43)

$$\begin{aligned}&\qquad -\left\{ \dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{i}^{2}} -\dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}} -\dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}} +\dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}^{2}}\right\} \nonumber \\&\quad =2\cdot \dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}+2\cdot \dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}}. \end{aligned}$$

(44)

Invoking Schwarz’s theorem regarding the equality of cross-derivatives for twice continuously differentiable functions, and subsequently using (38), one obtains

$$\begin{aligned} f^{\prime \prime }(0)=\frac{\partial ^{2}\varDelta _{i}^{+}(0)}{\partial \varepsilon ^{2}}=4\cdot \dfrac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}>0. \end{aligned}$$

(45)

In sum, we have shown that $f^{\prime }(0)=f(0)=0$ and $f^{\prime \prime }(0)>0$. Thus, using (40), it follows that $\varDelta _{i}^{+}(\varepsilon )>0$ for any sufficiently small $\varepsilon >0$. Analogous arguments can be used to deal with the other three sides of the square. Specifically, one defines

$$\begin{aligned} \varDelta _{j}^{+}(\varepsilon )&=u_{j}(x_{j}^{\ast}-\varepsilon ,x_{i}^{\ast}+\varepsilon ,x_{-i,j}^{\ast})-u_{j}(x_{j}^{\ast}+\varepsilon ,x_{i}^{\ast}+\varepsilon ,x_{-i,j}^{\ast}), \end{aligned}$$

(46)

$$\begin{aligned} \varDelta _{i}^{-}(\varepsilon )&=u_{i}(x_{i}^{\ast}-\varepsilon ,x_{j}^{\ast}-\varepsilon ,x_{-i,j}^{\ast})-u_{i}(x_{i}^{\ast}+\varepsilon ,x_{j}^{\ast}-\varepsilon ,x_{-i,j}^{\ast}), \end{aligned}$$

(47)

$$\begin{aligned} \varDelta _{j}^{-}(\varepsilon )&=u_{j}(x_{j}^{\ast}+\varepsilon ,x_{i}^{\ast}-\varepsilon ,x_{-i,j}^{\ast})-u_{j}(x_{j}^{\ast}-\varepsilon ,x_{i}^{\ast}-\varepsilon ,x_{-i,j}^{\ast}), \end{aligned}$$

(48)

and now readily verifies that

$$\begin{aligned} \frac{\partial \varDelta _{j}^{+}(0)}{\partial \varepsilon }&=(-\,2)\cdot \frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=0, \end{aligned}$$

(49)

$$\begin{aligned} \frac{\partial \varDelta _{i}^{-}(0)}{\partial \varepsilon }&=(-\,2)\cdot \frac{\partial u_{i}(x_{N}^{\ast})}{\partial x_{i}}=0, \end{aligned}$$

(50)

$$\begin{aligned} \frac{\partial \varDelta _{j}^{-}(0)}{\partial \varepsilon }&=2\cdot \frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=0, \end{aligned}$$

(51)

and that

$$\begin{aligned} \frac{\partial ^{2}\varDelta _{j}^{+}\left( 0\right) }{\partial \varepsilon ^{2}}&=\left( -\,4\right) \cdot \dfrac{\partial ^{2}u_{j}\left( x_{N}^{\ast}\right) }{\partial x_{i}\partial x_{j}}>0, \end{aligned}$$

(52)

$$\begin{aligned} \frac{\partial ^{2}\varDelta _{i}^{-}\left( 0\right) }{\partial \varepsilon ^{2}}&=4\cdot \dfrac{\partial ^{2}u_{i}\left( x_{N}^{\ast}\right) }{\partial x_{j}\partial x_{i} }>0, \end{aligned}$$

(53)

$$\begin{aligned} \frac{\partial ^{2}\varDelta _{j}^{-}\left( 0\right) }{\partial \varepsilon ^{2}}&=\left( -\,4\right) \cdot \dfrac{\partial ^{2}u_{j}\left( x_{N}^{\ast}\right) }{\partial x_{i}\partial x_{j}}>0. \end{aligned}$$

(54)

It follows that $\varDelta _{i}^{+}(\varepsilon )>0$, $\varDelta _{j}^{+} (\varepsilon )>0$, $\varDelta _{i}^{-}(\varepsilon )>0$, and $\varDelta _{j} ^{-}(\varepsilon )>0$ all hold for $\varepsilon >0$ small enough. But then, the finite sequence (6) is a strict improvement cycle, which is incompatible with the existence of a generalized ordinal potential by Lemma 1. □

Proof of Proposition 2

Suppose that the modified Jacobian $J=J(x_{N})$ is semipositive. Then, by definition, there exists a vector $\lambda _{N}=(\lambda _{1},\ldots ,\lambda _{n})^{T}\in {\mathbb{R}} ^{n}$ with $\lambda _{N}>0$ such that $J\lambda _{N}>0$. Consider the finite sequence

$$\begin{aligned}&\cdots \rightarrow x_{N}^{(1,+)}(\varepsilon )=(x_{1}^{\ast}+\lambda _{1}\varepsilon ,x_{2}^{\ast}-\lambda _{2}\varepsilon ,x_{3}^{\ast}-\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast}-\lambda _{n-1}\varepsilon ,x_{n}^{\ast}-\lambda _{n}\varepsilon )\nonumber \\&\quad \rightarrow x_{N}^{(2,+)}(\varepsilon )=(x_{1}^{\ast}+\lambda _{1} \varepsilon ,x_{2}^{\ast}+\lambda _{2}\varepsilon ,x_{3}^{\ast}-\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast}-\lambda _{n-1}\varepsilon ,x_{n}^{\ast}-\lambda _{n}\varepsilon )\rightarrow \nonumber \\&\quad \vdots \nonumber \\&\quad \rightarrow x_{N}^{(n,+)}(\varepsilon )=(x_{1}^{\ast}+\lambda _{1} \varepsilon ,x_{2}^{\ast}+\lambda _{2}\varepsilon ,x_{3}^{\ast}+\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast}+\lambda _{n-1}\varepsilon ,x_{n}^{\ast}+\lambda _{n}\varepsilon )\nonumber \\&\quad \rightarrow x_{N}^{(1,-)}(\varepsilon )=(x_{1}^{\ast}-\lambda _{1} \varepsilon ,x_{2}^{\ast}+\lambda _{2}\varepsilon ,x_{3}^{\ast}+\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast}+\lambda _{n-1}\varepsilon ,x_{n}^{\ast}+\lambda _{n}\varepsilon )\nonumber \\&\quad \rightarrow x_{N}^{(2,-)}(\varepsilon )=(x_{1}^{\ast}-\lambda _{1} \varepsilon ,x_{2}^{\ast}-\lambda _{2}\varepsilon ,x_{3}^{\ast}+\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast}+\lambda _{n-1}\varepsilon ,x_{n}^{\ast}+\lambda _{n}\varepsilon )\rightarrow \nonumber \\&\quad \vdots \nonumber \\&\quad \rightarrow x_{N}^{(n,-)}(\varepsilon )=(x_{1}^{\ast}-\lambda _{1} \varepsilon ,x_{2}^{\ast}-\lambda _{2}\varepsilon ,x_{3}^{\ast}-\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast}-\lambda _{n-1}\varepsilon ,x_{n}^{\ast}-\lambda _{n}\varepsilon )\rightarrow \cdots , \end{aligned}$$

(55)

where $\varepsilon >0$ is a small constant as before. Figure 2 illustrates this path for $n=3$, where the rectangular-shaped box has sides of respective length $\varepsilon _{i}=\lambda _{i}\varepsilon$ for $i=1,2,3$. It is claimed that, for any $\varepsilon >0$ sufficiently small, the following four conditions hold:

• 1. player 1’s payoff at $x_{N}^{(1,+)}(\varepsilon )$ is strictly higher than at $x_{N}^{(n,-)}(\varepsilon )$;

• 2. for $i=2,\ldots ,n$, player i’s payoff at $x_{N}^{(i,+)} (\varepsilon )$ is strictly higher than at $x_{N}^{(i-1,+)}(\varepsilon )$;

• 3. player 1’s payoff at $x_{N}^{(1,-)}(\varepsilon )$ is strictly higher than at $x_{N}^{(n,+)}(\varepsilon )$;

• 4. for $i=2,\ldots ,n$, player i’s payoff at $x_{N}^{(i,-)} (\varepsilon )$ is strictly higher than at $x_{N}^{(i-1,-)}(\varepsilon )$.

To establish claim (1), we proceed as in the proof of Proposition 1, and consider the first two derivatives of the payoff difference

$$\begin{aligned} \varDelta ^{(1,+)}(\varepsilon )&=u_{1}(x_{N}^{(1,+)}(\varepsilon ))-u_{1} (x_{N}^{(n,-)}(\varepsilon )) \end{aligned}$$

(56)

$$\begin{aligned}&=u_{1}(x_{1}^{\ast}+\lambda _{1}\varepsilon ,x_{2}^{\ast}-\lambda _{2}\varepsilon ,x_{3}^{\ast}-\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast} -\lambda _{n-1}\varepsilon ,x_{n}^{\ast}-\lambda _{n}\varepsilon )\nonumber \\&\quad -\,u_{1}(x_{1}^{\ast}-\lambda _{1}\varepsilon ,x_{2}^{\ast}-\lambda _{2}\varepsilon ,x_{3}^{\ast}-\lambda _{3}\varepsilon ,\ldots ,x_{n-1}^{\ast} -\lambda _{n-1}\varepsilon ,x_{n}^{\ast}-\lambda _{n}\varepsilon ) \end{aligned}$$

(57)

at $\varepsilon =0$. The first derivative of $\varDelta ^{(1,+)}(\varepsilon )$ at $\varepsilon =0$ is given by

$$\begin{aligned} \frac{\partial \varDelta ^{(1,+)}(0)}{\partial \varepsilon }&=\left( \lambda _{1}\frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{1}}-\lambda _{2} \frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{2}}-\cdots -\lambda _{n} \frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{n}}\right) \nonumber \\&\quad -\left( -\lambda _{1}\frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{1} }-\lambda _{2}\frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{2}} -\cdots -\lambda _{n}\frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{n}}\right) \end{aligned}$$

(58)

$$\begin{aligned}&=2\lambda _{1}\frac{\partial u_{1}(x_{N}^{\ast})}{\partial x_{1}}. \end{aligned}$$

(59)

Hence, from player 1’s first-order condition,

$$\begin{aligned} \frac{\partial \varDelta ^{(1,+)}(0)}{\partial \varepsilon }=0. \end{aligned}$$

(60)

Next, one considers the second derivative of $\varDelta ^{(1,+)}(\varepsilon )$ at $\varepsilon =0$, i.e.,

$$\begin{aligned} \frac{\partial ^{2}\varDelta ^{(1,+)}(0)}{\partial \varepsilon ^{2}}&=\left\{ (\lambda _{1})^{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{1}^{2} }-\lambda _{2}\lambda _{1}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{2}\partial x_{1}}-\cdots -\lambda _{n}\lambda _{1}\dfrac{\partial ^{2}u_{1} (x_{N}^{\ast})}{\partial x_{n}\partial x_{1}}\right. \nonumber \\&\qquad -\lambda _{1}\lambda _{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{1}\partial x_{2}}+(\lambda _{2})^{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast} )}{\partial x_{2}^{2}}+\cdots +\lambda _{n}\lambda _{2}\dfrac{\partial ^{2} u_{1}(x_{N}^{\ast})}{\partial x_{n}\partial x_{2}}\nonumber \\&\qquad \vdots \nonumber \\&\qquad \left. -\lambda _{1}\lambda _{n}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast} )}{\partial x_{1}\partial x_{n}}+\lambda _{2}\lambda _{n}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{2}\partial x_{n}}+\cdots +(\lambda _{n} )^{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{n}^{2}}\right\} \nonumber \\&\qquad -\left\{ (\lambda _{1})^{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{1}^{2}}+\lambda _{2}\lambda _{1}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast} )}{\partial x_{2}\partial x_{1}}+\cdots +\lambda _{n}\lambda _{1}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{n}\partial x_{1}}\right. \nonumber \\&\qquad +\lambda _{1}\lambda _{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{1}\partial x_{2}}+(\lambda _{2}^{2})\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast} )}{\partial x_{2}^{2}}+\cdots +\lambda _{n}\lambda _{2}\dfrac{\partial ^{2} u_{1}(x_{N}^{\ast})}{\partial x_{n}\partial x_{2}}\nonumber \\&\qquad \vdots \nonumber \\&\qquad +\left. \lambda _{1}\lambda _{n}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast} )}{\partial x_{1}\partial x_{n}}+\lambda _{2}\lambda _{n}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{2}\partial x_{n}}+\cdots +(\lambda _{n} )^{2}\dfrac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{n}^{2}}\right\} . \end{aligned}$$

(61)

Collecting terms, one obtains

$$\begin{aligned} \frac{\partial ^{2}\varDelta ^{(1,+)}(0)}{\partial \varepsilon ^{2}}=2\lambda _{1}\left\{ -\lambda _{2}\frac{\partial ^{2}u_{1}(x_{N}^{\ast})}{\partial x_{2}\partial x_{1}}-\lambda _{3}\frac{\partial ^{2}u_{1}(x_{N}^{\ast} )}{\partial x_{3}\partial x_{1}}-\cdots -\lambda _{n}\frac{\partial ^{2}u_{1} (x_{N}^{\ast})}{\partial x_{n}\partial x_{1}}\right\} . \end{aligned}$$

(62)

Thus, using $\lambda _{1}>0$, and noting that the expression in the curly brackets corresponds to the first component of the vector $J\lambda _{N}>0$, one arrives at

$$\begin{aligned} \frac{\partial ^{2}\varDelta ^{(1,+)}(0)}{\partial \varepsilon ^{2}}>0. \end{aligned}$$

(63)

It follows that $\varDelta ^{(1,+)}(\varepsilon )>0$ for any $\varepsilon >0$ sufficiently small, which proves claim (1). To verify claims (2) through (4), define payoff differences

$$\begin{aligned} \varDelta ^{(i,+)}(\varepsilon )&=u_{i}(x_{N}^{(i,+)}(\varepsilon ))-u_{i} (x_{N}^{(i-1,+)}(\varepsilon ))\quad (i=2,\ldots ,n), \end{aligned}$$

(64)

$$\begin{aligned} \varDelta ^{(1,-)}(\varepsilon )&=u_{1}(x_{N}^{(1,-)}(\varepsilon ))-u_{1} (x_{N}^{(n,+)}(\varepsilon )), \end{aligned}$$

(65)

$$\begin{aligned} \varDelta ^{(i,-)}(\varepsilon )&=u_{i}(x_{N}^{(i,-)}(\varepsilon ))-u_{i} (x_{N}^{(i-1,+)}(\varepsilon ))\quad (i=2,\ldots ,n). \end{aligned}$$

(66)

Using players’ necessary first-order conditions, it is straightforward to validate that

$$\begin{aligned} \frac{\partial \varDelta ^{(i,+)}(0)}{\partial \varepsilon }&=0\quad (i=2,\ldots ,n), \end{aligned}$$

(67)

$$\begin{aligned} \frac{\partial \varDelta ^{(1,-)}(0)}{\partial \varepsilon }&=0, \end{aligned}$$

(68)

$$\begin{aligned} \frac{\partial \varDelta ^{(i,-)}(0)}{\partial \varepsilon }&=0\quad (i=2,\ldots ,n). \end{aligned}$$

(69)

Moreover, for $i=2,\ldots ,n$, calculations analogous to (61, 62) yield

$$\begin{aligned}&\frac{\partial ^{2}\varDelta ^{(i,+)}(0)}{\partial \varepsilon ^{2}} =2\lambda _{i}\left\{ \lambda _{1}\frac{\partial ^{2}u_{i}(x_{N}^{\ast} )}{\partial x_{1}\partial x_{i}}+\cdots +\lambda _{i-1}\frac{\partial ^{2} u_{i}(x_{N}^{\ast})}{\partial x_{i-1}\partial x_{i}}\right. \end{aligned}$$

(70)

$$\begin{aligned}&\quad \quad \left. -\lambda _{i+1}\frac{\partial ^{2}u_{i} (x_{N}^{\ast})}{\partial x_{i+1}\partial x_{i}}-\cdots -\lambda _{n}\frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{n}\partial x_{i}}\right\} \nonumber \\&\quad >0, \end{aligned}$$

(71)

where the expression in the curly brackets corresponds to the i’s component of the vector $J\lambda _{N}>0$. Finally, one notes that, from $d(-\,\varepsilon )^{2}=d\varepsilon ^{2}$, it follows that

$$\begin{aligned} \frac{\partial ^{2}\varDelta ^{(i,-)}(0)}{\partial \varepsilon ^{2}}=\frac{\partial ^{2}\varDelta ^{(i,+)}(0)}{\partial \varepsilon ^{2}} \quad (i=1,\ldots ,n). \end{aligned}$$

(72)

In sum, this proves claims (2) through (4). Thus, there exists a strict improvement cycle in the generalized ordinal potential game $\varGamma$. Since this is impossible, the proposition follows. □

The three lemmas below will be used in the proof of Theorem 1. Following the literature, we will call a square matrix $A\in {\mathbb{R}} ^{n\times n}$ inverse nonnegative if the matrix inverse $A^{-1}$ exists and if, in addition, all entries of $A^{-1}$ are nonnegative. The following lemma provides a useful recursive characterization of semipositivity.

Lemma 2

(Johnson et al. 1994). A square matrix $A\in {\mathbb{R}} ^{n\times n}$ is semipositive if and only if at least one of the following two conditions holds:

• 1. A is inverse nonnegative;

• 2. there exists $m\in \{1,\ldots ,n-1\}$ and a submatrix ${\widehat{A}}\in {\mathbb{R}} ^{n\times m}$ obtained from A via deletion of $n-m$ columns, such that all $m\times m$ submatrices of ${\widehat{A}}$ are semipositive.

Using the lemma above, we may derive the following implication of Proposition 2.

Lemma 3

Suppose that the smooth n-player game $\varGamma$ admits a generalized ordinal potential. Then, at any profile $x_{N}^{\ast}$, and for any set $\{i,j,k\}\subseteq N$ of pairwise different players,

$$\begin{aligned}&\left\{ x_{i}^{\ast}\in {\mathring{X}} _{i}{,}\, x_{j}^{\ast}\in {\mathring{X}} _{j}{,}\, x_{k}^{\ast}\in {\mathring{X}} _{k}{,} \quad{{and}}\quad \frac{\partial u_{i}(x_{N}^{\ast} )}{\partial x_{i}}=\frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}} =\frac{\partial u_{k}(x_{N}^{\ast})}{\partial x_{k}}=0\right\} \nonumber \\&\quad \Rightarrow \frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}\cdot \frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{k}\partial x_{j}}\cdot \frac{\partial ^{2}u_{k}(x_{N}^{\ast})}{\partial x_{i}\partial x_{k}}=\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j} }\cdot \frac{\partial ^{2}u_{k}(x_{N}^{\ast})}{\partial x_{j}\partial x_{k} }\cdot \frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{k}\partial x_{i} }. \end{aligned}$$

(73)

Proof

Fix some profile $x_{N}^{\ast}$ and a set $\{i,j,k\}\subseteq N$ of pairwise different players such that

$$\begin{aligned}&x_{i}^{\ast} \in {\mathring{X}} _{i},\quad x_{j}^{\ast}\in {\mathring{X}} _{j},\quad x_{k}^{\ast}\in {\mathring{X}} _{k}, \quad {\text {and}} \end{aligned}$$

(74)

$$\begin{aligned}&\frac{\partial u_{i}(x_{N}^{\ast})}{\partial x_{i}} =\frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=\frac{\partial u_{k}(x_{N}^{\ast} )}{\partial x_{k}}=0. \end{aligned}$$

(75)

Using the notation

$$\begin{aligned} \chi _{ij}=\frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i} },\quad \chi _{jk}=\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{k}\partial x_{j}},\ldots \end{aligned}$$

(76)

for cross-derivatives, it needs to be shown that

$$\begin{aligned} \chi _{ij}\cdot \chi _{jk}\cdot \chi _{ki}=\chi _{ji}\cdot \chi _{kj}\cdot \chi _{ik}. \end{aligned}$$

(77)

Suppose first that $\varGamma$ exhibits weak strategic complements at $x_{N}^{\ast}$, i.e.,

$$\begin{aligned} \chi _{ij}\ge 0, \quad \chi _{ji}\ge 0, \quad \chi _{ik}\ge 0,\quad \chi _{ki}\ge 0,\quad \chi _{jk}\ge 0,\quad {\text {and}}\quad \chi _{kj}\ge 0. \end{aligned}$$

(78)

Consider now a small circular path along the edges of a small three-dimensional rectangular-shaped box around $x_{N}^{\ast}$. Along the path, players i, j, and k move in this order, with i and k initially increasing their strategies, while j initially decreases her strategy. Since this corresponds to flipping around player j’s strategy space, all cross-derivatives involving player j change sign, so that the corresponding modified Jacobian reads

$$\begin{aligned} J_{3}=\left( \begin{array}{ccc} 0 &{}\quad \chi _{ij} &{}\quad -\chi _{ik}\\ -\chi _{ji} &{}\quad 0 &{}\, \chi _{jk}\\ \chi _{ki} &{}\quad -\chi _{kj} &{}\, 0 \end{array} \right) . \end{aligned}$$

(79)

By Proposition 2, $J_{3}$ cannot be inverse nonnegative. To prove (77), it suffices to show that the determinant of $J_{3}$,

$$\begin{aligned} \left| J_{3}\right| =\chi _{ij}\chi _{jk}\chi _{ki}-\chi _{ji}\chi _{kj}\chi _{ik}, \end{aligned}$$

(80)

vanishes. To provoke a contradiction, suppose first that $\left| J_{3}\right| >0$. Then, from weak strategic complements at $x_{N}^{\ast}$, all the entries of the matrix inverse of $J_{3}$,

$$\begin{aligned} (J_{3})^{-1}=\frac{1}{\left| J_{3}\right| } \left( \begin{array}{ccc} \chi _{jk}\chi _{kj} &{}\quad \chi _{ik}\chi _{kj} &{}\quad \chi _{ij}\chi _{jk}\\ \chi _{ki}\chi _{jk} &{}\quad \chi _{ik}\chi _{ki} &{}\quad \chi _{ji}\chi _{ik}\\ \chi _{ji}\chi _{kj} &{}\quad \chi _{ij}\chi _{ki} &{}\quad \chi _{ij}\chi _{ji} \end{array} \right) , \end{aligned}$$

(81)

are nonnegative, in contradiction to the fact that $J_{3}$ is not inverse nonnegative. Hence, $\left| J_{3}\right| \le 0$. Suppose next that $\left| J_{3}\right| <0$. Then, by running through the above-considered path in the opposite direction (i.e., by exchanging the roles of players i and k), Proposition 2 implies that

$$\begin{aligned} {\widehat{J}}_{3}=\left( \begin{array}{ccc} 0 &{}\quad \chi _{kj} &{}\, -\chi _{ki}\\ -\,\chi _{jk} &{}\, 0 &{}\, \chi _{ji}\\ \chi _{ik} &{}\, -\chi _{ij} &{}\, 0 \end{array} \right) \end{aligned}$$

(82)

is not inverse nonnegative. Clearly, the determinant of ${\widehat{J}}_{3}$ is

$$\begin{aligned} \left| {\widehat{J}}_{3}\right| =\chi _{kj}\chi _{ji}\chi _{ik}-\chi _{jk}\chi _{ij}\chi _{ki}=-\left| J_{3}\right| >0. \end{aligned}$$

(83)

Hence, using an expression for the matrix inverse analogous to (81), all entries of $({\widehat{J}}_{3})^{-1}$ are seen to be nonnegative, in contradiction to the fact that ${\widehat{J}}_{3}$ is not inverse nonnegative. It follows that $\left| J_{3}\right| =0$, which proves the claim in the case where $\varGamma$ exhibits weak strategic complements at $x_{N}^{\ast}$. The case of weak strategic substitutes, where

$$\begin{aligned} \chi _{ij}\le 0,\quad \chi _{ji}\le 0, \quad \chi _{ik}\le 0,\quad \chi _{ki}\le 0,\quad \chi _{jk}\le 0,\quad {\text {and}}\quad \chi _{kj}\le 0, \end{aligned}$$

(84)

is entirely analogous, and therefore omitted. We are now in the position to address the general case. From Proposition 1, we know that $\varGamma$ exhibits pairwise weak strategic complements or substitutes at $x_{N}^{\ast}$. Hence, up to a renaming of the players, there are only two remaining cases:

• 1. Weak strategic complements at $x_{N}^{\ast}$ between player i and each of players j and k, as well as strategic substitutes at $x_{N}^{\ast}$ between players j and k;

• 2. Weak strategic substitutes at $x_{N}^{\ast}$ between player i and each of players j and k, as well as weak strategic complements at $x_{N}^{\ast}$ between players j and k.

In either case, by flipping around the strategy space of player i, the game may be transformed into a game that exhibits either weak strategic substitutes at $x_{N}^{\ast}$ or weak strategic complements at $x_{N}^{\ast}$. Since the operation of flipping around individual strategy spaces does not affect the validity of Eq. (77), we find that the conclusion indeed holds in the general case. □

Lemma 4

Suppose that the smooth n-player game $\varGamma$ admits a generalized ordinal potential. Then, at any interior strategy profile $x_{N}^{\ast}$ at which all first-order conditions hold, and for any set of pairwise distinct players $\{i_{1},\ldots ,i_{m}\}\subseteq N$ with $m\ge 3$, using the notation introduced in (76), it holds that

$$\begin{aligned} \chi _{i_{1}i_{2}}\cdot \chi _{i_{2}i_{3}}\cdot \ldots \cdot \chi _{i_{m-1}i_{m}} \cdot \chi _{i_{m}i_{1}}=\chi _{i_{2}i_{1}}\cdot \chi _{i_{3}i_{2}}\cdot \ldots \cdot \chi _{i_{m}i_{m-1}}\cdot \chi _{i_{1}i_{m}}{,} \end{aligned}$$

(85)

provided that $\chi _{i_{1}i_{3}}\cdot \chi _{i_{3}i_{1}}\ne 0,\ldots ,\chi _{i_{1}i_{m-1}}\cdot \chi _{i_{m-i}i_{1}}\ne 0 .$

Proof

The proof proceeds by induction. The case $m=3$ follows directly from Lemma 3. Suppose that $m\ge 4$, and let $\{i_{1} ,i_{2},\ldots ,i_{m}\}$ be an arbitrary set of pairwise different players. Suppose that the claim has been shown for any $m^{\prime }\in \{3,4,\ldots ,m-1\}$. Then, a consideration of the two subsets $\{i_{1},i_{2},\ldots ,i_{m-1}\}$ and $\{i_{1},i_{m-1},i_{m}\}$ = $\{i_{m-1},i_{m},i_{1}\}$ shows that

$$\begin{aligned}&\chi _{i_{1}i_{2}}\cdot \ldots \cdot \chi _{i_{m-2}i_{m-1}}\cdot \chi _{i_{m-1}i_{1}} =\chi _{i_{2}i_{1}}\cdot \ldots \cdot \chi _{i_{m-1}i_{m-2}}\cdot \chi _{i_{1} i_{m-1}}, \quad {\text {and}} \end{aligned}$$

(86)

$$\begin{aligned}&\chi _{i_{m-1}i_{m}}\cdot \chi _{i_{m}i_{1}}\cdot \chi _{i_{1}i_{m-1}} =\chi _{i_{m}i_{m-1}}\cdot \chi _{i_{1}i_{m}}\cdot \chi _{i_{m-1}i_{1}}. \end{aligned}$$

(87)

Combining the two equations via multiplication yields

$$\begin{aligned}&\left( \chi _{i_{1}i_{2}}\cdot \ldots \cdot \chi _{i_{m-2}i_{m-1}}\cdot \chi _{i_{m-1}i_{1}}\right) \cdot \left( \chi _{i_{m-1}i_{m}}\cdot \chi _{i_{m}i_{1}}\cdot \chi _{i_{1}i_{m-1}}\right) \nonumber \\&\quad =\left( \chi _{i_{2}i_{1}}\cdot \ldots \cdot \chi _{i_{m-1}i_{m-2}}\cdot \chi _{i_{1}i_{m-1}}\right) \cdot \left( \chi _{i_{m}i_{m-1}}\cdot \chi _{i_{1}i_{m}}\cdot \chi _{i_{m-1}i_{1}}\right) . \end{aligned}$$

(88)

By assumption, $\chi _{i_{1}i_{m-1}}\cdot \chi _{i_{m-1}i_{1}}\ne 0$. Hence, eliminating these common nonzero factors, (88) implies

$$\begin{aligned} \chi _{i_{1}i_{2}}\cdot \ldots \cdot \chi _{i_{m-1}i_{m}}\cdot \chi _{i_{m}i_{1}} =\chi _{i_{2}i_{1}}\cdot \ldots \cdot \chi _{i_{m}i_{m-1}}\cdot \chi _{i_{1}i_{m} }, \end{aligned}$$

(89)

as claimed. This concludes the induction step, and therefore proves the lemma. □

Proof of Theorem 1

Let $x_{N}^{\ast}$ be an interior strategy profile such that all first-order conditions hold at $x_{N}^{\ast}$ and such that $\chi _{ij}\ne 0$ for all $i\ne j$. We need to find positive constants $w_{1}>0,\ldots ,w_{n}>0$ such that

$$\begin{aligned} \chi _{ij}w_{i}=\chi _{ji}w_{j}\quad (i,j\in N,j\ne i). \end{aligned}$$

(90)

It is claimed that

$$\begin{aligned} w_{i}=\left( \left| \chi _{12}\right| \cdot \ldots \cdot \left| \chi _{i-1i}\right| \right) \cdot \left( \left| \chi _{i+1i}\right| \cdot \ldots \cdot \left| \chi _{nn-1}\right| \right) \quad (i\in N) \end{aligned}$$

(91)

does the job, with the usual convention that any empty product equals one. Clearly, it suffices to check (90) for $i<j$, because if $i>j$, one may just exchange the two sides of Eq. (90). Suppose first that $n=2$. Then, (91) implies $w_{1}=\left| \chi _{21}\right|$ and $w_{2}=\left| \chi _{12}\right|$, and the claim follows directly from Proposition 1. Suppose next that $n\ge 3$. Splitting the product in the second bracket of (91), one obtains

$$\begin{aligned} \chi _{ij}w_{i}&=\left( \left| \chi _{12}\right| \cdot \ldots \cdot \left| \chi _{i-1i}\right| \right) \nonumber \\&\quad \cdot {\text {sgn}}(\chi _{ij})\cdot \left( \left| \chi _{i+1i}\right| \cdot \ldots \cdot \left| \chi _{jj-1}\right| \cdot \left| \chi _{ij}\right| \right) \nonumber \\&\quad \cdot \left( \left| \chi _{j+1j}\right| \cdot \ldots \cdot \left| \chi _{nn-1}\right| \right) . \end{aligned}$$

(92)

From Proposition 1,

$$\begin{aligned} {\text {sgn}}(\chi _{ij})={\text {sgn}}(\chi _{ji}). \end{aligned}$$

(93)

Moreover, from Lemma 4,

$$\begin{aligned} \chi _{i+1i}\cdot \ldots \cdot \chi _{jj-1}\cdot \chi _{ij}=\chi _{ii+1}\cdot \ldots \cdot \chi _{j-1j}\cdot \chi _{ji}. \end{aligned}$$

(94)

Hence, using (93) and (94) in relationship (92) delivers

$$\begin{aligned} \chi _{ij}w_{i}&=\left( \left| \chi _{12}\right| \cdot \ldots \cdot \left| \chi _{i-1i}\right| \right) \end{aligned}$$

(95)

$$\begin{aligned}&\quad \cdot {\text {sgn}}(\chi _{ji})\cdot \left( \left| \chi _{ii+1}\right| \cdot \ldots \cdot \left| \chi _{j-1j}\right| \cdot \left| \chi _{ji}\right| \right) \end{aligned}$$

$$\begin{aligned}&\quad \cdot \left( \left| \chi _{j+1j}\right| \cdot \ldots \cdot \left| \chi _{nn-1}\right| \right) \end{aligned}$$

$$\begin{aligned}&=\chi _{ji}\cdot \left( \left| \chi _{12}\right| \cdot \ldots \cdot \left| \chi _{j-1j}\right| \right) \cdot \left( \left| \chi _{j+1j}\right| \cdot \ldots \cdot \left| \chi _{nn-1}\right| \right) \end{aligned}$$

(96)

$$\begin{aligned}&=\chi _{ji}w_{j}. \end{aligned}$$

(97)

This proves the claim and, hence, the theorem. □

Proof of Proposition 3

To construct an ordinal potential locally at $x_{N}^{\ast}$, start by letting

$$\begin{aligned} S_{j}\equiv {\text {sgn}}\left( \frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}\right) \in \{1,-1\}\quad (j\in N\backslash \{i\}), \end{aligned}$$

(98)

where the sign function sgn(.) is defined in Sect. 7. Next, choose a constant $b_{j}$, for each $j\in N\backslash \{i\}$, such that $\left| \partial u_{i}(x_{N}^{\ast})/\partial x_{j}\right| <b_{j}$. Then, since $\varGamma$ is smooth, there is a small rectangular neighborhood $U_{N}(x_{N}^{\ast})$ of $x_{N}^{\ast}$ such that

$$\begin{aligned} U_{N}(x_{N}^{\ast})\subseteq \left\{ x_{N}\in X_{N}\,{\text {s.t. sgn}}\left( \frac{\partial u_{j}(x_{N})}{\partial x_{j}}\right) =S_{j}\quad {\text {and}}\quad \left| \frac{\partial u_{i}(x_{N})}{\partial x_{j}}\right| <b _{j}\quad {\text {for all}} \quad j\ne i\right\} . \end{aligned}$$

(99)

We claim that

$$\begin{aligned} P(x_{N})=u_{i}(x_{N})+\sum _{j\in N\backslash \{i\}}S_{j}b _{j} x_{j} \end{aligned}$$

(100)

is an ordinal potential for $\varGamma$ when joint strategy choices are restricted to $U_{N}(x_{N}^{\ast})$. To see this, note first that

$$\begin{aligned} P(x_{i},x_{-i})-P({\widehat{x}}_{i},x_{-i})=u_{i}(x_{i},x_{-i})-u_{i} ({\widehat{x}}_{i},x_{-i}), \end{aligned}$$

(101)

for any $x_{i}\in X_{i}$, ${\widehat{x}}_{i}\in X_{i}$, and $x_{-i}\in X_{-i}$. Next, let $j\in N\backslash \{i\}$. Take any $x_{j}\in X_{j}$, ${\widehat{x}} _{j}\in X_{j}$, and $x_{-j}\in X_{-j}$ such that $(x_{j},x_{-j})\in U_{N}(x_{N}^{\ast})$ and $({\widehat{x}}_{j},x_{-j})\in U_{N}(x_{N}^{\ast})$. Then, from (100),

$$\begin{aligned} P(x_{j},x_{-j})-P({\widehat{x}}_{j},x_{-j})=u_{i}(x_{j},x_{-j})-u_{i} ({\widehat{x}}_{j},x_{-j})+b_{j}S_{j}(x_{j}-{\widehat{x}}_{j}). \end{aligned}$$

(102)

Invoking the mean value theorem, there exists a strict convex combination $\widetilde{x}_{j}\$ of ${\widehat{x}}_{j}$ and $x_{j}$ such that

$$\begin{aligned} u_{i}(x_{j},x_{-j})-u_{i}({\widehat{x}}_{j},x_{-j})=\frac{\partial u_{i}(\widetilde{x}_{j},x_{-j})}{\partial x_{j}}(x_{j}-{\widehat{x}} _{j}). \end{aligned}$$

(103)

Hence,

$$\begin{aligned} P(x_{j},x_{-j})-P({\widehat{x}}_{j},x_{-j})=\left\{ \frac{\partial u_{i}(\widetilde{x}_{j},x_{-j})}{\partial x_{j}}+b_{j}S_{j}\right\} (x_{j}-{\widehat{x}}_{j}). \end{aligned}$$

(104)

To check the definition of an ordinal potential, suppose that $u_{j} (x_{j},x_{-j})-u_{j}({\widehat{x}}_{j},x_{-j})>0$. Then, using (99), $S_{j}(x_{j}-{\widehat{x}}_{j})>0$. Moreover, since $U_{N}(X_{N}^{\ast})$ is convex, $\left| \partial u_{i}(\widetilde{x}_{j},x_{-j})/\partial x_{j}\right| <b_{j}$. It follows that $P(x_{j},x_{-j})-P({\widehat{x}} _{j},x_{-j})>0$. Since these steps may be reversed, P is indeed an ordinal potential for $\varGamma$ in the neighborhood $U_{N}(x_{N}^{\ast})$. This proves the proposition. □

Proof of Proposition 4

We adapt the proof of Proposition 1. To provoke a contradiction, suppose that

$$\begin{aligned}&x^{\ast}_{i}=\underline{x}_{i},x^{\ast}_{j} \in \mathring{X}_{j},\quad \frac{\partial u_{i} (x_{N}^{\ast})}{\partial x_{i}}=\frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=0, \end{aligned}$$

(105)

$$\begin{aligned}&\frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}} \cdot \frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j} }<0,\quad {\text {and}} \end{aligned}$$

(106)

$$\begin{aligned}&\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}} \cdot \frac{\partial ^{3}u_{j}(x_{N}^{\ast})}{\partial x_{j}^{3}}<0. \end{aligned}$$

(107)

There are two cases. Suppose first that, as illustrated in Fig. 3,

$$\begin{aligned} \frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}>0>\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j} }. \end{aligned}$$

(108)

Then, condition (107) implies $\partial ^{3}u_{j}(x_{N}^{\ast} )/\partial x_{j}^{3}>0$. Given constants $\lambda _{i}>0$ and $\lambda _{j} >0$, let

$$\begin{aligned} {\widehat{\Delta }}_{i}^{+}\left( \varepsilon \right)&=u_{i}\left( x_{i}^{\ast}+\lambda _{i}\varepsilon ,x_{j}^{\ast}+\lambda _{j}\varepsilon ,x_{-i,j}^{\ast} \right) -u_{i}\left( x_{i}^{\ast},x_{j}^{\ast}+\lambda _{j}\varepsilon ,x_{-i,j}^{\ast}\right) , \end{aligned}$$

(109)

$$\begin{aligned} {\widehat{\Delta }}_{j}^{+}\left( \varepsilon \right)&=u_{j}\left( x_{j}^{\ast}-\lambda _{j}\varepsilon ,x_{i}^{\ast}+\lambda _{i}\varepsilon ,x_{-i,j}^{\ast} \right) -u_{j}\left( x_{j}^{\ast}+\lambda _{j}\varepsilon ,x_{i}^{\ast}+\lambda _{i}\varepsilon ,x_{-i,j}^{\ast}\right) , \end{aligned}$$

(110)

$$\begin{aligned} {\widehat{\Delta }}_{i}^{-}\left( \varepsilon \right)&=u_{i}\left( x_{i}^{\ast},x_{j}^{\ast}-\lambda _{j}\varepsilon ,x_{-i,j}^{\ast}\right) -u_{i}\left( x_{i}^{\ast}+\lambda _{i}\varepsilon ,x_{j}^{\ast}-\lambda _{j}\varepsilon ,x_{-i,j}^{\ast}\right) , \end{aligned}$$

(111)

$$\begin{aligned} {\widehat{\Delta }}_{j}^{-}\left( \varepsilon \right)&=u_{j}\left( x_{j}^{\ast}+\lambda _{j}\varepsilon ,x_{i}^{\ast},x_{-i,j}^{\ast}\right) -u_{j}\left( x_{j}^{\ast}-\lambda _{j}\varepsilon ,x_{i}^{\ast},x_{-i,j}^{\ast}\right) . \end{aligned}$$

(112)

Straightforward calculations deliver

$$\begin{aligned} \frac{\partial {\widehat{\Delta }}_{i}^{+}(0)}{\partial \varepsilon }&=\lambda _{i}\frac{\partial u_{i}(x_{N}^{\ast})}{\partial x_{i}}=0, \end{aligned}$$

(113)

$$\begin{aligned} \frac{\partial ^{2}{\widehat{\Delta }}_{i}^{+}(0)}{\partial \varepsilon ^{2}}&=\lambda _{i}\left( \lambda _{i}\frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{i}^{2}}+2\lambda _{j}\frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}\right) , \end{aligned}$$

(114)

and

$$\begin{aligned} \frac{\partial {\widehat{\Delta }}_{j}^{+}(0)}{\partial \varepsilon }&=-\,2\lambda _{j}\frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=0, \end{aligned}$$

(115)

$$\begin{aligned} \frac{\partial ^{2}\varDelta _{j}^{+}(0)}{\partial \varepsilon ^{2}}&=-\,4\lambda _{j}\lambda _{i}\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}}>0. \end{aligned}$$

(116)

Similarly,

$$\begin{aligned} \frac{\partial {\widehat{\Delta }}_{i}^{-}(0)}{\partial \varepsilon }&=-\,\lambda _{i}\frac{\partial u_{i}(x_{N}^{\ast})}{\partial x_{i}}=0, \end{aligned}$$

(117)

$$\begin{aligned} \frac{\partial ^{2}{\widehat{\Delta }}_{i}^{-}(0)}{\partial \varepsilon ^{2}}&=\lambda _{i}\left( -\lambda _{i}\frac{\partial ^{2}u_{i}(x_{N}^{\ast} )}{\partial x_{i}^{2}}+2\lambda _{j}\frac{\partial ^{2}u_{i}(x_{N}^{\ast} )}{\partial x_{j}\partial x_{i}}\right) , \end{aligned}$$

(118)

and

$$\begin{aligned} \frac{\partial {\widehat{\Delta }}_{j}^{-}(0)}{\partial \varepsilon }&=2\lambda _{j}\frac{\partial u_{j}(x_{N}^{\ast})}{\partial x_{j}}=0,\end{aligned}$$

(119)

$$\begin{aligned} \frac{\partial ^{2}{\widehat{\Delta }}_{j}^{-}(0)}{\partial \varepsilon ^{2}}&=0, \end{aligned}$$

(120)

$$\begin{aligned} \frac{\partial ^{3}{\widehat{\Delta }}_{j}^{-}(0)}{\partial \varepsilon ^{3}}&=2\lambda _{j}^{3}\frac{\partial ^{3}u_{j}(x_{N}^{\ast})}{\partial x_{j}^{3} }>0. \end{aligned}$$

(121)

Choose now $\lambda _{i}=1$, and $\lambda _{j}>0$ sufficiently large so that the bracket terms in (114) and (118) are positive. Then, second-order and third-order Taylor approximations at $\varepsilon =0$ show that ${\widehat{\Delta }}_{i}^{+}(\varepsilon )>0$, ${\widehat{\Delta }}_{j} ^{+}(\varepsilon )>0$, ${\widehat{\Delta }}_{i}^{-}(\varepsilon )>0$, and ${\widehat{\Delta }}_{j}^{-}(\varepsilon )>0$ all hold for $\varepsilon >0$ small enough, which yields the desired contradiction. In the second case,

$$\begin{aligned} \frac{\partial ^{2}u_{i}(x_{N}^{\ast})}{\partial x_{j}\partial x_{i}}<0<\frac{\partial ^{2}u_{j}(x_{N}^{\ast})}{\partial x_{i}\partial x_{j}}, \end{aligned}$$

(122)

and consequently $\partial ^{3}u_{j}(x_{N}^{\ast})/\partial x_{j}^{3}<0$. It is now straightforward to check that, again for $\lambda _{i}=1$ and $\lambda _{j}>0$ sufficiently large, inequalities ${\widehat{\Delta }}_{i}^{+} (\varepsilon )<0$, ${\widehat{\Delta }}_{j}^{+}(\varepsilon )<0$, $\widehat{\Delta }_{i}^{-}(\varepsilon )<0$, and ${\widehat{\Delta }}_{j}^{-}(\varepsilon )<0$ all hold for $\varepsilon >0$ small enough. The remainder of the argument stays as before. This completes the proof. □

Proof of Proposition 5

It will be shown that P is a weighted potential. Since any weighted potential is, in particular, a generalized ordinal potential, this is sufficient to prove the proposition. So consider some player $i\in N$, prices $p_{i}^{\prime }\ge 0$ and $p_{i} ^{\prime \prime }\ge 0$, as well as a price vector $p_{-i}\in {\mathbb{R}} _{+}^{n-1}$. It is claimed that

$$\begin{aligned} P(p_{i}^{\prime },p_{-i})-P(p_{i}^{\prime \prime },p_{-i})=\frac{{\widehat{w}}_{i} }{1+2s_{i}c_{i}}\cdot \left( u_{i}(p_{i}^{\prime },p_{-i})-u_{i}(p_{i} ^{\prime \prime },p_{-i})\right) . \end{aligned}$$

(123)

To see this, note first that

$$\begin{aligned} u_{i}(p_{i},p_{-i})&=\left( p_{i}-c_{i}\left( Q_{i}-s_{i}p_{i} +\sum _{j\ne i}\theta _{ij}p_{j}\right) \right) \left( Q_{i}-s_{i}p_{i} +\sum _{j\ne i}\theta _{ij}p_{j}\right) \end{aligned}$$

(124)

$$\begin{aligned}&=(1+2s_{i}c_{i})Q_{i}p_{i}-(1+s_{i}c_{i})s_{i}p_{i}^{2}+(1+2s_{i}c_{i} )\sum _{j\ne i}\theta _{ij}p_{i}p_{j}\nonumber \\&\quad +\left\{ {\text {terms constant in}} \,p_{i}\right\} \end{aligned}$$

(125)

$$\begin{aligned}&=(1+2s_{i}c_{i})p_{i}\left\{ Q_{i}\left( 1-\frac{p_{i}}{2p_{i}^{0} }\right) +\sum _{j\ne i}\theta _{ij}p_{j}\right\} \nonumber \\&\quad +\left\{ {\text {terms constant in}} \,p_{i}\right\} . \end{aligned}$$

(126)

Therefore,

$$\begin{aligned} u_{i}(p_{i}^{\prime },p_{-i})-u_{i}(p_{i}^{\prime \prime },p_{-i})&=(1+2s_{i}c_{i})Q_{i}\left\{ p_{i}^{\prime }\left( 1-\frac{p_{i}^{\prime } }{2p_{i}^{0}}\right) -p_{i}^{\prime \prime }\left( 1-\frac{p_{i}^{\prime \prime }}{2p_{i}^{0}}\right) \right\} \nonumber \\&\quad +(1+2s_{i}c_{i})\sum _{j\ne i}\theta _{ij}\left( p_{i}^{\prime }-p_{i}^{\prime \prime }\right) p_{j}. \end{aligned}$$

(127)

On the other hand,

$$\begin{aligned} P(p_{i}^{\prime },p_{-i})-P(p_{i}^{\prime \prime },p_{-i})&={\widehat{w}} _{i}Q_{i}\left\{ p_{i}^{\prime }\left( 1-\frac{p_{i}^{\prime }}{2p_{i}^{0}} \right) -p_{i}^{\prime \prime }\left( 1-\frac{p_{i}^{\prime \prime }}{2p_{i}^{0}}\right) \right\} \nonumber \\&\ \ \ +{\widehat{w}}_{i}\sum _{j=1}^{i-1}\theta _{ij}\left( p_{i}^{\prime }-p_{i}^{\prime \prime }\right) p_{j}+\sum _{j=i+1}^{n}{\widehat{w}}_{j} \theta _{ji}p_{j}(p_{i}^{\prime }-p_{i}^{\prime \prime }). \end{aligned}$$

(128)

We will show below that

$$\begin{aligned} {\widehat{w}}_{j}\theta _{ji}={\widehat{w}}_{i}\theta _{ij}\quad (j=i+1,\ldots ,n). \end{aligned}$$

(129)

Using this in (128), and recalling (127), we get

$$\begin{aligned} P(p_{i}^{\prime },p_{-i})-P(p_{i}^{\prime \prime },p_{-i})&={\widehat{w}} _{i}Q_{i}\left\{ p_{i}^{\prime }\left( 1-\frac{p_{i}^{\prime }}{2p_{i}^{0}} \right) -p_{i}^{\prime \prime }\left( 1-\frac{p_{i}^{\prime \prime }}{2p_{i}^{0}}\right) \right\} \nonumber \\&\quad +{\widehat{w}}_{i}\sum _{j\ne i}\theta _{ij}\left( p_{i} ^{\prime }-p_{i}^{\prime \prime }\right) p_{j}\end{aligned}$$

(130)

$$\begin{aligned}&=\frac{{\widehat{w}}_{i}}{1+2s_{i}c_{i}}\left( u_{i}(p_{i}^{\prime } ,p_{-i})-u_{i}(p_{i}^{\prime \prime },p_{-i})\right) , \end{aligned}$$

(131)

as claimed. So it remains to be checked that (129) holds true, or equivalently, that

$$\begin{aligned}&\left| \theta _{12}\right| \cdot \ldots \cdot \left| \theta _{j-1j}\right| \cdot \left| \theta _{j+1j}\right| \cdot \ldots \cdot \left| \theta _{nn-1}\right| \cdot \theta _{ji}\nonumber \\&\quad =\left| \theta _{12}\right| \cdot \ldots \cdot \left| \theta _{i-1i}\right| \cdot \left| \theta _{i+1i}\right| \cdot \ldots \cdot \left| \theta _{nn-1}\right| \cdot \theta _{ij}\quad (j=i+1,\ldots ,n). \end{aligned}$$

(132)

But this was shown in the proof of Theorem 1. Since ${\widehat{w}}_{i}>0$ for $i\in N$, we conclude that P is indeed a weighted potential for the differentiated Bertrand game. The proposition follows. □