Dispersion equilibria in spatial Cournot competition

Abstract

We consider the standard model of spatial Cournot competition and show that for dispersion equilibria to exist, (a) a necessary condition is that the distribution be not unimodal, and (b) a sufficient condition is that the distribution be convex with a unique antimode and that asymmetry is not too strong.

Appendix

Proof of Proposition 1

The first-order conditions for profit maximization, given definition (2) and solving integrals, reduce to

$$\begin{aligned} A[1-2F(x_{1})]-\left[ 1+2F(x_{1})\right] \delta +x_{1}-\mu +2I(x_{1},x_{2})&= 0 \end{aligned}$$

(6a)

$$\begin{aligned} A[1-2F(x_{2})]+\left[ 1-2F(x_{2})\right] \delta +x_{2}-\mu +2I(x_{1},x_{2})&= 0 \end{aligned}$$

(6b)

where to ease notation, we let $\delta =x_{2}-x_{1}\ge 0$ and $I(x_{1},x_{2})=\int _{x_{1}}^{x_{2}}F(x)\mathrm{d}x$. Now, we may sum and subtract over (6) to get

$$\begin{aligned} A\Delta F-\frac{3}{2}\delta +\delta \Delta F=0 \end{aligned}$$

(7a)

$$\begin{aligned} A\left[ 1-\Sigma F\right] +\frac{x_{1}+x_{2}}{2}-\delta \Sigma F-\mu +2I(x_{1},x_{2})=0 \end{aligned}$$

(7b)

where $\Sigma F=F(x_{1})+F(x_{2})$ and $\Delta F=F(x_{2})-F(x_{1})\ge 0$. By summing over again, we obtain

$$\begin{aligned} \delta =\dfrac{1}{1+2F(x_{1})}\left( 2I(x_{1},x_{2})+x_{1}-\mu +\left[ 1-2F(x_{1})\right] A\right) \end{aligned}$$

(8)

which, letting equilibrium values $x_{1}^{*}=x^{*}$ and $x_{2}^{*}=x^{*}+\delta ^{*}$, amounts to Eq. (5a).

By subtracting (7b) from (7a), we now get

$$\begin{aligned} \left[ 1-2F(x_{2})\right] \delta =-\left\{ 2I(x_{1},x_{2})+x_{2}-\mu +\left[ 1-2F(x_{2})\right] A\right\} \end{aligned}$$

which, solving for $\delta$ from (8) and letting $x_{2}=x_{1}+\delta$, gives, after simplifications, the equilibrium value

$$\begin{aligned} \delta ^{*}=\frac{2A\Delta F}{3-2\Delta F} \end{aligned}$$

(9)

which can be solved for $\Delta F=F(x^{*}+\delta ^{*})-F(x^{*})$ to give Eq. (5b). $\square$

Proof of Proposition 2

If $(x^{*},x^{*}+\delta ^{*})$ is a pair of equilibrium dispersed locations, $\delta ^{*}\in (0,1-x^{*})$ must satisfy the twin conditions

$$\begin{aligned} \kappa (\delta ^{*})&= 0 \end{aligned}$$

(10a)

$$\begin{aligned} \lambda (\delta ^{*})&= 0 \end{aligned}$$

(10b)

where, using (5a, 5b), the following definitions apply:

$$\begin{aligned} \kappa (\delta )&= \delta [1+2F(x^{*})]-2\displaystyle \int \limits _{x^{*}}^{x^{*}+\delta }F(z)\mathrm{d}z-h(x^{*},A) \end{aligned}$$

(11a)

$$\begin{aligned} \lambda (\delta )&= \Delta _{F}(x^{*},\delta )-\frac{3}{2}\frac{\delta }{A+\delta } \end{aligned}$$

(11b)

here $h(x^{*},A)=x^{*}-\mu +\left[ 1-2F(x^{*})\right] A$ and, to ease notation, $\Delta _{F}(x^{*},\delta )=F(x^{*}+\delta )-F(x^{*})$, an increasing function of $\delta$. Given $x^{*}$, both $\kappa$ and $\lambda$ are continuous functions. We now proceed in three steps:

• (i) The function $\kappa (\delta )$ is strictly concave for any $\delta \in [0,1-x^{*}]$, increasing at least up to some $\widetilde{\delta }\in (\delta ^{*},1-x^{*}]$, and such that $\kappa (0)=-h(x^{*},A)<0$ and $\kappa ^{\prime }(\delta ^{*})<1$. Indeed, differentiation shows that $\kappa ^{\prime }(\delta )=1-2\Delta _{F}(x^{*},\delta )$, clearly decreasing and positive for $\delta \le \delta ^{*}$, as by Lemma 1 $\Delta _{F}(x^{*},\delta ^{*})<1/2$ : hence, $\kappa (0)=-h(x^{*},A)<0$ and that $\kappa ^{\prime }(\delta ^{*})>0$. Since (10a) is a necessary condition for equilibrium, this ensures that for the given $x^{*}$, there is only one $\delta ^{*}$ such that $(x^{*},x^{*}+\delta ^{*})$ is an equilibrium location pair.

• (ii) The function $\lambda (\delta )$ is such that:

  • (a) $\lambda (1-x^{*})>0$: from (i) it must be $h(x^{*},A)>0$: this directly implies that

    $$\begin{aligned} 1-F(x^{*})>\frac{1}{2}\left( 1-\frac{x^{*}}{A}\right) \end{aligned}$$

    (12)

    Using (11b) and noting that $\Delta _{F}(x^{*},1-x^{*})=1-F(x^{*})$, from (12) $\lambda (1-x^{*})>$ $\frac{1}{2}\left( 1- \frac{x^{*}}{A}\right) -\frac{3}{2}\frac{1-x^{*}}{A+1-x^{*}}$, so that $\lambda (1-x^{*})>0$ if

    $$\begin{aligned} \frac{1}{2}\left( 1-\frac{x^{*}}{A}\right) -\frac{3}{2}\frac{1-x^{*}}{A+1-x^{*}}>0 \end{aligned}$$

    (13)

    which amounts to $(A-x^{*})(A+1-x^{*})-3A(1-x^{*})>0$: this is surely true for any $x^{*}\in \left[ 0,1\right]$ and $A>2$.

  • (b) there are at least two values of $\delta$ in the interval $(0,1-x^{*})$ such that the derivative of $\lambda (\delta )$ vanishes. To see this, notice that $\lambda$ cannot be monotone as $\lambda (0)=\lambda (\delta ^{*})=0$. Also, $\lambda (1-x^{*})>0$ implies the existence of some $\widehat{\delta }\in (0,1-x^{*})$ such that $\widehat{\delta }\ge \delta ^{*}$, with $\lambda (\widehat{\delta } )=0<\lambda ^{\prime }(\widehat{\delta })$ and $\lambda (\delta )>0$ for all $\delta \in (\widehat{\delta },1-x^{*})$. Let now $\Gamma (\delta )=\kappa ^{\prime }(\delta )\lambda (\delta )+\kappa (\delta )$, such that $\Gamma (0)=\Gamma (\delta ^{*})=0$: there has to be some $\overline{ \delta }\in (0,\delta ^{*})$ such that $\Gamma ^{\prime }(\overline{ \delta })=\kappa ^{\prime \prime }(\overline{\delta })\lambda (\overline{ \delta })+\left[ \lambda ^{\prime }(\overline{\delta })+1\right] \kappa ^{\prime }(\overline{\delta })=0$ i.e., $\kappa ^{\prime \prime }(\overline{ \delta })\lambda (\overline{\delta })=-\left[ \lambda ^{\prime }(\overline{ \delta })+1\right] \kappa ^{\prime }(\overline{\delta })<0$: since $\lambda ^{\prime }(\delta )+1=f(x+\delta )+\frac{2(A+\delta )^{2}-3A}{2(A+\delta )^{2}}>0$ for all $\delta \in [0,1-x^{*}]$ and $\kappa$ is concave, it must be $\lambda (\overline{\delta })>0$. All of which implies that $\lambda ^{\prime }(\delta )$ changes sign at least twice over $[0, \widehat{\delta }]$: there is at least a pair $(\delta _{1},\delta _{2})$, $\delta _{1}<\delta _{2}$, say, $\delta _{i}\in (0,\widehat{\delta })$ for $i=1,2$, such that $\lambda ^{\prime }(\delta _{i})=0$.

• (iii) By (11b), it is now easily seen that $f(x^{*}+\delta _{1})>f(x^{*}+\delta _{2})$, since $\lambda ^{\prime }(\delta _{i})=0$ is equivalent to $f(x^{*}+\delta _{i})=\frac{3}{2}\frac{A}{(A+\delta _{i})^{2}}$ and $\delta _{1}<\delta _{2}$. Hence, unimodality is ruled out if one can find some $\delta _{3}>\delta _{2}$ such that $f(x^{*}+\delta _{3})>f(x^{*}+\delta _{2})$. To do so, consider the function $\theta (\delta )=\frac{1-F(x^{*})}{1-x^{*}}\delta -\frac{3}{2}\frac{\delta }{A+\delta }$; this is positive for any $\delta \in (0,1-x^{*}]$: indeed, $\theta (1-x^{*})=$ $\lambda (1-x^{*})>0=\theta (0)$, while $\theta ^{\prime }(\delta )=\frac{1-F(x^{*})}{1-x^{*}}-\frac{3}{2} \frac{A}{(A+\delta )^{2}}>0$: as $\theta$ is strictly convex, $\theta ^{\prime }(\delta )>0$ if $\theta ^{\prime }(0)\ge 0$, i.e., $\frac{ 1-F(x^{*})}{1-x^{*}}\ge \frac{3}{2A}$, which is true for $A>2$. To see this, notice that $h(x^{*},A)>0$ implies $\frac{1-F(x^{*})}{ 1-x^{*}}>\frac{A-x^{*}+\mu }{2A(1-x^{*})}$, so that $\theta ^{\prime }(0)>0$ if $A+2x^{*}+\mu -3\ge 0$. That the latter is verified in equilibrium can be seen as follows:

  (a$^\prime$):

  $h(x+\delta ^{*},A+\delta ^{*})=x^{*}+\delta ^{*}-\mu +\left[ 1-2F(x^{*}+\delta ^{*})\right] (A+\delta ^{*})<0$: this follows from (10) and the definitions (11), by noting that $\delta ^{*}\left[ 1+2F(x^{*})\right] >h(x^{*},A)$ and substituting for $F(x^{*})$ from (10); there follows $\mu >x^{*}+\delta ^{*}+ \left[ 1-2F(x^{*}+\delta ^{*})\right] (A+\delta ^{*})$ and $A+2x^{*}+\mu -3>2A-3(1-x^{*})+2\delta ^{*}-2(A+\delta ^{*})F(x^{*}+\delta ^{*})$;

  (b$^\prime$):

  by substituting back $F(x^{*}+\delta ^{*})(A+\delta ^{*})$ from (10b) we get $A+2x^{*}+\mu -3>2A-3(1-x^{*})-\delta ^{*}+2(A+\delta ^{*})F(x^{*})>0$: which holds as $3(1-x^{*})+\delta ^{*}<4$, while $2A>4$.

  Now, notice that by construction $\widehat{\delta }>\delta _{2}>\delta _{1}$ and $\theta (\widehat{\delta })=\frac{1-F(x^{*})}{1-x^{*}}\widehat{ \delta }-\Delta _{F}(x^{*},\widehat{\delta })>0$. Since $\Delta _{F}(1-x^{*},x^{*})=1-F(x^{*})$, it is true at $\widehat{\delta }$ that $\frac{\Delta _{F}(x^{*},1-x^{*})}{1-x^{*}}>\frac{\Delta _{F}(x^{*},\widehat{\delta })}{\widehat{\delta }}$, so that there exists some $\delta _{3}\in [\widehat{\delta },1-x^{*}]$ such that $\frac{\Delta _{F}(x^{*},1-x^{*})}{1-x^{*}}<\frac{\partial }{ \partial \delta }\Delta _{F}(x^{*},\delta _{3})=f(x^{*}+\delta _{3})$ ; but $0<\theta ^{\prime }(\delta _{2})=\frac{\Delta _{F}(x^{*},1-x^{*})}{1-x^{*}}-f(x^{*}+\delta _{2})$: hence $f(x^{*}+\delta _{2})<\frac{\Delta _{F}(x^{*},1-x^{*})}{1-x^{*}} <f(x^{*}+\delta _{3})$. $\square$

Proof of Proposition 3

Observe that any solution to system (5) amounts to the first-order conditions for the firms’ maximization problem holding. In order to prove that one such solution exists, we need the following Lemmata. $\square$

Lemma 2

Given the assumptions of Proposition 3, (a) there exists a unique pair of solutions $\overline{x} _{i}\in (0,1),i=1,2$, to the equation $f(\cdot )=3/2A$ ; (b) there is a unique solution $x_{A}\in \left( 0,1\right)$ to the equation $h(\cdot ,A)=0$ ; (c) at any pair $(x,\delta )$ such that $\Delta _{F}\left( x,\delta \right) =\frac{3}{2}\frac{\delta }{ A+\delta }$, $\max \left\{ f(x),f(x+\delta )\right\} (A+\delta )\ge 1$.

Proof

(a) The results follows trivially, given convexity and assumption ( i). (b) Convexity and assumptions (ii) and ( iii) imply that there is only one $x_{A}$ such that $h(\cdot ,A)=0$, since $h_{x}(x,A)=1-2Af(x)<0$ and $f(x)\ge f(x_{m})>1/2A$. By assumption ( ii), both $x_{A}$ and $x_{m}$ lie in $(\overline{x}_{1},\overline{x} _{2})$. (c) Convexity implies that at any pair $(x,\delta )$ such that $\Delta _{F}\left( x,\delta \right) =\frac{3}{2}\frac{\delta }{A+\delta }$,

$$\begin{aligned} \left[ f(x)+f(x+\delta )\right] \frac{\delta }{2}\ge \Delta _{F}\left( x,\delta \right) \end{aligned}$$

while $2\max \left\{ f(x),f(x+\delta )\right\} \ge$ $f(x)+f(x+\delta )\ge 2\min \left\{ f(x),f(x+\delta )\right\}$, so that $\max \left\{ f(x),f(x+\delta )\right\} \ge$ $\frac{3}{2}\frac{1}{A+\delta }>\frac{1}{ A+\delta }$. $\square$

Lemma 3

Let $(x^{*},x^{*}+\delta ^{*})$ be a solution to system (5) with $\delta ^{*}>0$ . Then, $h(x^{*},A)>0=h(x_{A},A)>h(x^{*}+\delta ^{*},A)$ , so that $x^{*}<x_{A}<x^{*}+\delta ^{*}$.

Proof

From the proof of Proposition 2, we already know that $\kappa (\delta )=\delta [1+2F(x^{*})]-2I(x^{*},\delta )-h(x^{*},A)$ is such that $\kappa (0)=-h(x^{*},A)<0$ at any solution of (5). There remains to show that $h(x^{*}+\delta ^{*},A)<0$. To do so, notice that by definition $h(x^{*}+\delta ^{*},A)=h(x^{*},A)+\delta ^{*}-2A\Delta _{F}(x^{*},\delta ^{*})$; at $(x^{*},x^{*}+\delta ^{*})$, we know that $\kappa (\delta ^{*})=0$, i.e., $h(x^{*},A)=\delta ^{*}[1+2F(x^{*})]-2I(x^{*},\delta ^{*})$, so that $h(x^{*}+\delta ^{*},A)=2\delta ^{*}[1+F(x^{*})]-2A\Delta _{F}(x^{*},\delta ^{*})-2I(x^{*},\delta ^{*})$, which—given (5a)—can be written as $h(x^{*}+\delta ^{*},A)=2\delta ^{*}[1+F(x^{*})]-\frac{3A\delta ^{*}}{A+\delta ^{*}}-2I(x^{*},\delta ^{*})=H(x^{*},\delta ^{*})$. Consider now the function $H(x^{*},\delta )$: surely $H(x^{*},0)=0$, and $H_{\delta }(x^{*},\delta )=2[1-\Delta _{F}(x^{*},\delta ^{*})]-\frac{3A^{2}}{\left( A+\delta \right) ^{2}}<0$ for all $\delta \in [0,\delta ^{*}]$. To see this, notice that $H_{\delta }(x^{*},0)<0$, so that $H(x^{*},\delta ^{*})\ge 0$ would require some $\delta ^{\prime }\in (0,\delta ^{*})$ such $H(x^{*},\delta ^{\prime })<0=H_{\delta }(x^{*},\delta ^{\prime })$, so that $\Delta _{F}(x^{*},\delta ^{\prime })=1-\frac{3A^{2}}{2\left( A+\delta ^{\prime }\right) ^{2}} <0$, which is impossible as $\Delta _{F}(x^{*},\delta )\ge 0$ for any $\delta \in [0,\delta ^{*}]$. $\square$

Lemma 4

      Let $G(\cdot )$ be the distribution such that $G(x)=1-F(1-x)$ for all $x\in \left[ 0,1 \right]$ , and let $(x^{*},x^{*}+\delta ^{*})$ be a solution to system (5) with $\delta ^{*}>0$ . Then $(z^{*}-\delta ^{*},z^{*})$ , with $z^{*}=1-x^{*}$ , is a solution of the same system where $F\left( \cdot \right)$ is replaced by $G\left( \cdot \right)$ . Moreover, if $(x^{*},x^{*}+\delta ^{*})$ is a dispersion equilibrium under $F$ , so is $(z^{*}-\delta ^{*},z^{*})$ under $G$.

Proof

Notice that $G$ is the “mirror image” of $F$, such that (with obvious notation and $z=1-x$), $\mu _{G}=1-\mu$, $-h_{G}(z,A)=h(x,A)$, and $G(z)-G(z-\delta )=F(x+\delta )-F(x)$. The result follows after some simplification by substituting accordingly within (5). Moreover, if the second-order conditions under $F$ hold, $\left( A+\delta ^{*}\right) \min \left\{ f(x^{*}),f(x^{*}\!+\!\delta ^{*})\right\} \ge 1$, it must be $\left( A+\delta ^{*}\right) \min \left\{ g(z^{*}-\delta ^{*}),g(z^{*})\right\} \ge 1$, as $f(x)=g(1-x)$ for all $x$. $\square$

The proof now proceed in four steps.

First step      Consider the equation

$$\begin{aligned} f(x+\delta )=f(x) \end{aligned}$$

(14)

Given convexity, this has only one solution $\widehat{x}$ for $\delta \in (0,\delta _{M})$, where $\delta _{M}\le 1$ solves either $f(\delta _{M})=f(0)$ when $f(0)<f(1)$, or $f(x+\delta _{M})=f(1)$ when $f(0)>f(1)$; $\delta _{M}=1$ and $f(0)=f(\delta _{M})=f(1)$ when $f(0)=f(1)$: so $f\left( \delta _{M}\right) =\min \left\{ f(0),f(1)\right\}$. (14) thus defines an implicit function $\widehat{x}\left( \delta \right)$ such that:

• (a) $\lim _{\delta \rightarrow 0}\widehat{x}\left( \delta \right) =x_{m}$;

• (b) $\lim _{\delta \rightarrow \delta _{M}}\widehat{x}\left( \delta \right) =x_{M}<x_{m}\,$, where $x_{M}=0$ when $f(0)\le f(1)$ or $\,x_{M}=1-\delta _{M}$ for $f(0)\ge f(1)$;

• (c) $\frac{\mathrm{d}}{\mathrm{d}\delta }\widehat{x}\left( \delta \right) =\frac{f^{\prime }(x+\delta )}{f^{\prime }(x)-f^{\prime }(x+\delta )}<0$, since at $f(x+\delta )-f(x)=0$ it must be $f^{\prime }(x+\delta )>0>f^{\prime }(x)$. Notice that $\frac{\mathrm{d}}{\mathrm{d}\delta }\widehat{x}\left( \delta \right) =-\frac{1}{2}$ if $f^{\prime }(x+\delta )=-f^{\prime }(x)$, such that under symmetry, $\widehat{x}\left( \delta \right) =\frac{1-\delta }{2}$; if there is symmetry around $x_{m}$, $\widehat{x}\left( \delta \right) =x_{m}-\frac{1}{2}\delta$;

• (d) $f(x+\delta )-f(x)=\Delta _{f}\left( x,\delta \right) >0\,$ for any $x> \widehat{x}\left( \delta \right)$, and $\Delta _{f}\left( x,\delta \right) <0$ for any $x<\widehat{x}\left( \delta \right)$.

Second step      Consider the equation

$$\begin{aligned} \dfrac{1}{1+2F(x)}\left( 2\displaystyle \int \limits _{x}^{x+\delta }F(z)\mathrm{d}z+x-\mu +\left[ 1-2F(x) \right] A\right) =\delta \end{aligned}$$

(5a)

which, solving the integral by parts, amounts to

$$\begin{aligned} 2\left( x+\delta \right) \Delta _{F}\left( x,\delta \right) -2\widetilde{\mu }\left( x,\delta \right) +h(x,A)=\delta \end{aligned}$$

(15)

where $\widetilde{\mu }\left( x,\delta \right) =\int _{x}^{x+\delta }zf(z)\mathrm{d}z>0$. For any given $\delta >0$, (15) solves for some $x=x^{a}\left( \delta \right)$. This will be well defined so long as the derivative wrt $x$ of

$$\begin{aligned} \kappa (x,\delta )=2\left( x+\delta \right) \Delta _{F}\left( x,\delta \right) -2\widetilde{\mu }\left( x,\delta \right) +h(x,A)-\delta \end{aligned}$$

does not vanishes at $\kappa =0$. At any such point, we have

$$\begin{aligned} \kappa _{x}(x,\delta )=1-2\left( A+\delta \right) f(x)<0 \end{aligned}$$

for all $x\in \left[ 0,1\right]$ as $f(x)\ge f(x_{m})>1/2A$. Then, $x^{a}\left( \delta \right)$ is such that:

• (a) $\lim _{\delta \rightarrow 0}x^{a}\left( \delta \right) =x_{A}$;

• (b) $\frac{\mathrm{d}}{\mathrm{d}\delta }x^{a}\left( \delta \right) =\frac{1-2\Delta _{F}\left( x,\delta \right) }{1-2\left( A+\delta \right) f(x)}<0$ for $\Delta _{F}\left( x,\delta \right) <1/2\,$;

• (c) $\lim _{\delta \rightarrow 0}\frac{\mathrm{d}}{\mathrm{d}\delta }x^{a}\left( \delta \right) =\frac{1}{1-2Af(x(A))}<0$.

Third step      Consider the equation

$$\begin{aligned} \Delta _{F}\left( x,\delta \right) -\frac{3}{2}\frac{\delta }{A+\delta }=0 \end{aligned}$$

(5b)

Under the stated assumptions, this has two solutions, $x_{1}^{b}\left( \delta \right)$ and $x_{2}^{b}\left( \delta \right)$. Indeed:

• (a) $\Delta _{F}\left( x,\delta \right) -\frac{3}{2}\frac{\delta }{A+\delta } =0$ can be written as

  $$\begin{aligned} \frac{\Delta _{F}\left( x,\delta \right) }{\delta }-\frac{3}{2}\frac{1}{ A+\delta }=0 \end{aligned}$$

  so that $\lim _{\delta \rightarrow 0}\left( \frac{\Delta _{F}\left( x,\delta \right) }{\delta }-\frac{3}{2}\frac{1}{A+\delta }\right) =f(x)-\frac{3}{2A}$ : since our condition must hold also in the limit, it must be $f(x)=\frac{3}{ 2A}$ which by Lemma 2 gives two solutions in $\left( 0,1\right)$, $\overline{x}_{1}<\overline{x}_{2}$, say. Hence: $\lim _{\delta \rightarrow 0}x_{1}^{b}\left( \delta \right) =\overline{x}_{1}<\lim _{\delta \rightarrow 0}x_{2}^{b}\left( \delta \right) =\overline{x}_{2}$.

• (b) More generally, for $\delta >0$, there exists some $z\in \left( x,x+\delta \right)$ such that $\frac{\Delta _{F}\left( x,\delta \right) }{ \delta }=f(z)$, and $f(z)-\frac{3}{2}\frac{1}{A+\delta }=0$ gives a pair $\left( x_{1}^{b}\left( \delta \right) ,x_{2}^{b}\left( \delta \right) \right)$ such that $\Delta _{F}\left( x,\delta \right) -\frac{3}{2}\frac{ \delta }{A+\delta }=0$.

• (c) $x_{i}^{b}\left( \delta \right)$, $i=1,2$ are well-defined functions for all $\delta$, such that

  $$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\delta }x_{i}^{b}\left( \delta \right) =-\frac{f(x+\delta )-\frac{3 }{2}\frac{A}{\left( A+\delta \right) ^{2}}}{\Delta _{f}\left( x,\delta \right) } \end{aligned}$$

  Indeed, since $\lim _{\delta \rightarrow 0}\widehat{x}\left( \delta \right) =x_{m}\in \left( x_{1},x_{2}\right)$, and using (d) of the first step, we have that so long as $x_{1}^{b}\left( \delta \right) <\widehat{x}\left( \delta \right)$, $\Delta _{f}\left( x,\delta \right) <0$, while so long as $x_{2}^{b}\left( \delta \right) >\widehat{x}\left( \delta \right)$, $\Delta _{f}\left( x,\delta \right) >0$.

• (d) there is a unique $\overline{\delta }\in \left( 0,\delta _{M}\right)$ such that $\lim _{\delta \rightarrow \overline{\delta }}x_{1}^{b}(\delta )=\lim _{\delta \rightarrow \overline{\delta }}x_{2}^{b}(\delta )=\widehat{x}( \overline{\delta })$ and $\lim _{\delta \rightarrow \overline{\delta }}\frac{\mathrm{d} }{\mathrm{d}\delta }x_{1}^{b}(\delta )=\lim _{\delta \rightarrow \overline{\delta }} \frac{\mathrm{d}}{\mathrm{d}\delta }x_{2}^{b}(\delta )=\infty$. To see this, we use the first step to define

  $$\begin{aligned} \Phi \left( \delta \right) =\Delta _{F}\left( \widehat{x}\left( \delta \right) ,\delta \right) -\frac{3}{2}\frac{\delta }{A+\delta } \end{aligned}$$

  (16)

  such that $\Phi \left( 0\right) =0$ and $\Phi \left( \delta _{M}\right) >0$. The former is obvious, while the latter can be shown as follows. Either $\Phi \left( \delta _{M}\right) =F(\delta _{M})-\frac{3}{2}\frac{\delta _{M}}{ A+\delta _{M}}$, or $\Phi \left( \delta _{M}\right) =1-F(1-\delta _{M})- \frac{3}{2}\frac{\delta _{M}}{A+\delta _{M}}$; since by Lemma 4, one can always define $F(\delta _{M})=1-G(1-\delta _{M})$, and one can concentrate simply on $1-F(1-\delta _{M})-\frac{3}{2}\frac{\delta _{M}}{A+\delta _{M}}$, which is positive for all $\delta _{M}>0$. Indeed, letting $\rho (\delta )=1-F(1-\delta )-\frac{3}{2}\frac{\delta }{A+\delta }$, we have $\rho (0)=0$, $\rho (1)=1-\frac{3}{2}\frac{1}{A+1}>0$, and $\rho ^{\prime }\left( \delta \right) =f(1-\delta )-\frac{3}{2}\frac{A}{\left( A+\delta \right) ^{2}}$, so that $\rho ^{\prime }\left( 0\right) =f(1)-\frac{3}{2A}>0$ and $\rho ^{\prime }\left( 1\right) =f(0)-\frac{3}{2}\frac{A}{\left( A+1\right) ^{2}} >0$; hence, a negative $\rho$ would require at least two values of $\delta$, $\delta _{1}<\delta _{2}$, such that $\rho ^{\prime }\left( \delta _{1}\right) =0<\rho ^{\prime \prime }\left( \delta _{1}\right)$ and $\rho ^{\prime }\left( \delta _{2}\right) =0>\rho ^{\prime \prime }\left( \delta _{2}\right)$, which is ruled out by convexity, as $-f^{\prime }(1-\delta )~$ is increasing in $\delta$ while $-\frac{\mathrm{d}}{\mathrm{d}\delta }\frac{3}{2}\frac{A}{ \left( A+\delta \right) ^{2}}=\frac{3A}{\left( A+\delta \right) ^{3}}$ is decreasing. Observe now that $\Phi ^{\prime }\left( 0\right) =f(x_{m})-\frac{3}{2A}<0$: $\Phi \left( \delta _{M}\right) >0$ then implies the existence of some $\overline{\delta }\in \left( 0,\delta _{M}\right)$ such that $\Phi \left( \overline{\delta }\right) =0<\Phi ^{\prime }\left( \overline{\delta }\right) =f(x+\overline{\delta })-$ $\frac{3}{2}\frac{1}{A+ \overline{\delta }}$, and the latter is unique, as convexity implies that $\max \left\{ f(x),f(x+\overline{\delta })\right\} =f(x+\overline{\delta } )\ge$ $\frac{3}{2}\frac{1}{A+\delta }$ for all pairs $\left( x,\delta \right)$ such that $\Phi =0$ (Lemma 2(c)). By construction, $\lim _{\delta \rightarrow \overline{\delta }}x_{1}(\delta )=\lim _{\delta \rightarrow \overline{\delta }}x_{2}(\delta )=\widehat{x}(\overline{\delta } )$ since $\Delta _{F}\left( \widehat{x}\left( \overline{\delta }\right) , \overline{\delta }\right) =\frac{3}{2}\frac{\overline{\delta }}{A+\overline{ \delta }}$, while $\lim _{\delta \rightarrow \overline{\delta }}\frac{\mathrm{d}}{ \mathrm{d}\delta }x_{1}(\delta )=\frac{\mathrm{d}}{\mathrm{d}\delta }\lim _{\delta \rightarrow \overline{ \delta }}x_{2}(\delta )=\infty$ as $f\left( \widehat{x}(\overline{\delta })+ \overline{\delta }\right) =f\left( \widehat{x}\left( \overline{\delta } \right) \right)$.

Fourth step       We are now ready to prove existence. To do so, notice that by the second step, $x^{a}\left( \delta \right)$ is defined for all $\delta \in \left[ 0,1\right]$, and obeys $\lim _{\delta \rightarrow 0}x^{a}\left( \delta \right) =x_{A}\in (\overline{x}_{1}, \overline{x}_{2})$; since $\lim _{\delta \rightarrow \overline{\delta } }x_{1}^{b}\left( \delta \right) =\lim _{\delta \rightarrow \overline{\delta } }x_{2}^{b}\left( \delta \right)$, there exists some $\delta ^{*}\in \left( 0,\overline{\delta }\right]$, at which either $x^{a}\left( \delta ^{*}\right) =x_{1}^{b}\left( \delta ^{*}\right)$, or $x^{a}\left( \delta ^{*}\right) =x_{2}^{b}\left( \delta ^{*}\right)$, or both (in which case $\delta ^{*}=\overline{\delta }$). The pair $\left( x_{1}^{*},x_{2}^{*}\right) = \left( x^{*},x^{*}+\delta ^{*}\right)$ such that $x^{*}=x^{a}\left( \delta ^{*}\right)$ is a dispersion equilibrium. Indeed, by construction (Proposition 1), it gives both firms’ first-order conditions for maximum profits. Also, the second-order conditions $\left( A+\delta ^{*}\right) \min \left\{ f(x^{*}),f(x^{*}+\delta ^{*})\right\} \ge 1$ are satisfied. First, by convexity, $\max \left\{ f(x),f(x+\delta )\right\} \ge$ $\frac{3}{ 2}\frac{1}{A+\delta }>\frac{1}{A+\delta }$ for all $\left( x,x+\delta \right) \,$(Lemma 2(c)), which gives those conditions when $\delta ^{*}=\overline{\delta }$, i.e., whenever $\Delta _{f}\left( x^{*},\delta ^{*}\right) =0$ (which includes the case of symmetry) and ensures that if $\Delta _{f}\left( x^{*},\delta ^{*}\right) \ne 0$, $\max \left\{ f(x^{*}),f(x^{*}+\delta ^{*})\right\} >\frac{1}{ A+\delta ^{*}}$. Secondly, in the latter case, we have $\delta ^{*}< \overline{\delta }$ and equilibrium is given by $x^{a}\left( \delta ^{*}\right) =x_{i}^{b}\left( \delta ^{*}\right)$, $i=1$ for $x_{A}<x_{m}$, with $\Delta _{f}\left( x^{*},\delta ^{*}\right) <0$; and $i=2$ for $x_{A}>x_{m}$ with $\Delta _{f}\left( x^{*},\delta ^{*}\right) >0$. Take the case $x_{A}<x_{m}$: then $\Delta _{f}\left( x^{*},\delta ^{*}\right) <0$ and $\min \left\{ f(x^{*}),f\left( x^{*}+\delta ^{*}\right) \right\} =f\left( x^{*}+\delta ^{*}\right)$; by Lemma 3, $x^{*}+\delta ^{*}>x_{A}$, and by convexity $f\left( x^{*}+\delta ^{*}\right) +f\left( x_{A}\right) \ge 2\frac{\Delta _{F}\left( x^{*}+\delta ^{*},x_{A}\right) }{x^{*}+\delta ^{*}-x_{A}}=2f(z_{1})$, say; surely $z_{1}\in \left( x^{*}+\delta ^{*},x_{A}\right)$ and $f(z_{1})\ge f(x_{m})$: hence, $f\left( x^{*}+\delta ^{*}\right) \ge 2f(x_{m})-f(x_{A})\ge \frac{1}{A}>\frac{1}{A+\delta ^{*}}$ by assumption (iii). The same goes for $x_{A}>x_{m}$, in which case by the same Lemma 3 $x^{*}<x_{A}$ and $f\left( x^{*}\right) +f\left( x_{A}\right) \ge 2\frac{\Delta _{F}\left( x_{A},x^{*}\right) }{ x_{A}-x^{*}}=2f(z_{2})$, say, so that $f\left( x^{*}\right) \ge 2f(x_{m})-f(x_{A})>\frac{1}{A+\delta ^{*}}$.