Location-Price Competition with Delivered Pricing and Elastic Demand

Abstract

In the past, implementing delivered pricing has been perceived as unrealistic because of practical difficulties in distinguishing between customers, determining an individual’s willingness to pay, and setting different prices to individuals. The rise of e-commerce has introduced the possibility of doing all three. Competitive location with delivered pricing was studied by Lederer and Hurter (1986) but only with inelastic customer demand. This paper extends the literature by allowing price elastic demand. A Nash equilibrium with inelastic demand always exists but examples show that it may not with price elasticity. General sufficient conditions guaranteeing existence of a Nash equilibrium are developed despite the fact that even with these conditions a firm’s profit function is generally not concave, quasi-concave, supermodular or even continuous in location choices. Examples demonstrate how violation of sufficient conditions result in lack of existence. Given price elasticity, equilibrium locations demonstrate properties unlike the inelastic case, for example, as transportation cost rises or firms’ production costs rise, each firm locates closer to its competitor. Given our sufficient conditions for equilibrium’s existence, the interval spanning firms’ ordered equilibrium locations always contains a social welfare optimum pair and a social welfare optimum pair is always contained by an ordered equilibrium.

Appendix

Proof of Lemma 2:

$$ {\displaystyle \begin{array}{l}\begin{array}{c}0={\Pi}^A\left({\overline{z}}_A,{\overline{z}}_B\right)-{\Pi}^A\left({\underline{z}}_A,{\overline{z}}_B\right)=\underset{-1}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz\\ {}\kern14.24em -\underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-\left|{\underline{z}}_A-z\right|\right)\rho (z) dz\end{array}\ \\ {}\begin{array}{c}\kern0.5em \left(\underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz+\underset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz\right)\\ {}\kern2.04em -\underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-\left|{\underline{z}}_A-z\right|\right)\rho (z) dz=0.\end{array}\end{array}} $$

(A1)

Combining the first and third terms in (A1) and cancelling:

$$ \underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left(|{\underline{z}}_A-z|-|{\overline{z}}_A-z|\right)\rho (z) dz+\underset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-|{\overline{z}}_A-z|\right)\rho (z) dz\Big)=0. $$

(A2)

The second term is positive as within the domain of integration any z is closer to ${\overline{z}}_A$ than to ${\overline{z}}_B$, thus the first integral is strictly negative:

$$ \underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\left(|{\underline{z}}_A-z|-|{\overline{z}}_A-z|\right)\rho (z) dz<0.\kern1em $$

(A3)

Next, I show (9) by differentiating (A2):

$$ {\displaystyle \begin{array}{c}{\varPi}_2^A\left({\overline{z}}_A,{\overline{z}}_B\right)-{\varPi}_2^A\left({\underline{z}}_A,{\overline{z}}_B\right)=\underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}{q}_1\left({\overline{z}}_B-z\right)\left(\left|{\underline{z}}_A-z\right|-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz+\\ {}+\underset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}{q}_1\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-|{\overline{z}}_A-z|\right)\rho (z) dz+\underset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\overline{z}}_B-z\right)\rho (z) dz.\end{array}} $$

(A4)

The sum of the second and third terms is the marginal profit earned in the interval [$\frac{{\underline{z}}_A+{\overline{z}}_B}{2},\frac{{\overline{z}}_A+{\overline{z}}_B}{2}\Big]$ as Z_B increases. As the local prices are below monopoly prices and the margins everywhere are positive, an increase in price raises profit in the interval. Now I claim that the first integral is non negative:

Note that derivative of log[q(p)] is $\frac{q_1(p)}{q(p)}$. The second derivative $\frac{d^2\mathit{\log}\left[q(p)\right]}{dp^2}=\frac{d\left[\frac{q_1(p)}{q(p)}\right]}{dp}$ <0, which means that $\frac{q_1(p)}{q(p)}$ (a negative fraction which ×100% represents percent decline) decreases monotonically with p, and $\frac{q_1\left({\overline{z}}_B-z\right)}{q\left({\overline{z}}_B-z\right)}$ is negative and increases monotonically with $z\in \left[-1,\frac{{\overline{z}}_A-{\underline{z}}_A}{2}\right]$ as prices are falling with z. The term $\mid {\underline{z}}_A-z\mid -\mid {\overline{z}}_A-z\mid$ is monotone increasing in z, with range [−(${\overline{z}}_A-{\underline{z}}_A\left),{\overline{z}}_A-{\underline{z}}_A\right)\Big]$. Writing (A4) as

$$ {\displaystyle \begin{array}{l}\underset{-1}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_A}{2}}{\int }}q\left({\overline{z}}_B-z\right)\frac{q_1\left({\overline{z}}_B-z\right)}{q\left({\overline{z}}_B-z\right)}\left(\left|{z}_A-z\right|-\left(\left|{\overline{z}}_A-z\right|\right)\right)\rho (z) dz\kern0.5em \\ {}+\underset{\frac{{\underline{z}}_A+{\overline{z}}_A}{2}}{\overset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\int }}\frac{q_1\left({\overline{z}}_B-z\right)}{q\left({\overline{z}}_B-z\right)}\left(\left|{\underline{z}}_A-z\right|-\left(\left|{\overline{z}}_A-z\right|\right)\right)\rho (z) dz>0\end{array}} $$

(A5)

and comparing to (A2) shows that where the first integrand in (A2) was negative now (A5)‘s first integral is positive as it is multiplied by $\frac{q_1\left({\overline{z}}_B-z\right)}{q\left({\overline{z}}_B-z\right)}$which is negative but increasing in z and the second integral in (A2) was positive and now (A5)‘s second integral’s integrand is negative as it is multiplied by $\frac{q_1\left({\overline{z}}_B-z\right)}{q\left({\overline{z}}_B-z\right)}$ which continues to be negative and increasing in z. The fraction is everywhere less negative in the second integrand than in the first integrand. This causes the signs in (A5) to be the opposite to (A2) with more weight on what is now the positive term. So the sum of all three terms in (A3) is positive, and (9) is demonstrated. Q.E.D.

Proof of Lemma 3

The proof is by contradiction. Suppose to the contrary that at some ${\overline{z}}_B>{\hat{z}}_B$$F\left({\underline{z}}_A,{\overline{z}}_A,{\overline{z}}_B\right)\le 0$. Either $F\left({\underline{z}}_A,{\overline{z}}_A,{\hat{z}}_B\right)=0$ (in which case Lemma 2 applies) or $F\left({\underline{z}}_A,{\overline{z}}_A,{\hat{z}}_B\right)<0$ and in both cases a small neighborhood to the right of ${\hat{z}}_B$ exists such that $F\left({\underline{z}}_A,{\overline{z}}_A,{z}_B\right)>0$. By the continuity of the firms’ profit, there are one or more points z_B such that $F\left({\underline{z}}_A,{\overline{z}}_A,{z}_B\right)=0$ in the interval ${z}_B\in \left[{\hat{z}}_B,{\overline{z}}_B\right]$. Denote the largest of these by ${\overset{\check{} }{z}}_B\ \mathrm{such}\ \mathrm{that}$$F\left({\underline{z}}_A,{\overline{z}}_A,{\overset{\check{} }{z}}_B\right)=0$ . No point ${z}_B\in \left[{\overset{\check{} }{z}}_B,{\overline{z}}_B\right]$ can satisfy $F\left({\underline{z}}_A,{\overline{z}}_A,{z}_B\right)=0$, thus $F\left({\underline{z}}_A,{\overline{z}}_A,{z}_B\right)<0$ for all ${z}_B\in \left[{\overset{\check{} }{z}}_B,{\overline{z}}_B\right]$. Lemma 2 shows that ${F}_3\left({\underline{z}}_A,{\overline{z}}_A,{\overset{\check{} }{z}}_B\right)=0$and by continuity that implies that $F\left({\underline{z}}_A,{\overline{z}}_A,{z}_B\right)>0$everywhere in some interval N=$\left[{\overset{\check{} }{z}}_B,z^{\prime}\right]$ with ${z}^{\prime }>{\overset{\check{} }{z}}_B$which contradicts the definition of ${\overset{\check{} }{z}}_B$, so it must be that ${F}_3\left({\underline{z}}_A,{\overline{z}}_A,{\overset{\check{} }{z}}_B\right)>0$ for all ${z}_B\in \left({\hat{z}}_B,1\right)\Big]$. Q.E.D.

Proof of Proposition 6

The following construction will find a pair of closed intervals lying below and above z_C. The best responses (over the unrestricted space) for each firm will be into the interval on the opposite side. Thus, if each firm locates in the respective interval, then a firm will never be “leapfrogged” by its competitor.

For −1≤ z_A ≤ z_B≤1, define

$$ {\hat{Z}}_A\left[{z}_B\right]=\underset{z_A:-1\le {z}_A\le 1}{\mathit{\operatorname{Min}}\Big\{ ArgMax}{\varPi}^A\left({z}_A,{z}_B\right)\left\};{\hat{Z}}_B\left[{z}_A\right]=\underset{z_B:-1\le {z}_B\le 1}{\mathit{\operatorname{Max}}\Big\{ ArgMax}{\varPi}^B\left({z}_A,{z}_B\right)\right\}. $$

(A6)

These are single valued increasing functions. Note that the previously defined reaction functions ${\hat{z}}_i\left[{z}_{-i}\right]$ (8) do not permit leapfrogging while ${\hat{Z}}_i\left[{z}_{-i}\right]$ do. Next, define the two pairs:

$$ {\displaystyle \begin{array}{l}{z}_A^0={\hat{Z}}_A\left[{z}_C\right],{z}_B^0={\hat{Z}}_B\left[{z}_C\right]\\ {}{z}_A^1=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_C\right\},{z}_B^1=\mathit{\operatorname{Max}}\left\{z|{\hat{Z}}_A\left[z\right]\le {z}_C\right\}\end{array}} $$

The first pair maps z_C into each firm’s optimal response. The second pair finds the value of z_A that maps B into z_C, similarly for the value of z_B that maps A into z_C. When leapfrogging is allowed and ${z}_A>{z}_A^1$, then firm B will optimally locate above z_C and if ${z}_B<{z}_B^1$, firm A will optimally locate below z_C.

To proceed, we note there are three cases that are analyzed separately.

Case i. ${z}_A^1\le {z}_A^0;{z}_B^1\ge {z}_B^0$. In this case,

$$ \psi :\left({z}_A,{z}_B\right)\in \left[{z}_A^1,{z}_C\right]\times \left[{z}_C,{z}_B^1\right]\to \left[{\hat{Z}}_A\left[{z}_B\right],{\hat{Z}}_B\left[{z}_A\right]\right]\in \left[{z}_A^0,{z}_C\right]\times \left[{z}_C,{z}_B^0,\right] $$

(A7)

is an into correspondence of the domain into a subset of the domain and the intervals are on opposite sides of z_C . The correspondence has been shown to be increasing. This is sufficient for our purposes as Zhou’s Theorem assures an equilibrium in locations. Because the mapping maps the right side of z_C to its left side, and vise-versa, Zhou’s Theorem shows that a Nash equilibrium exists.

Case ii. ${z}_A^1\ge {z}_A^0;{z}_B^1\ge {z}_B^0$ (or the parallel case ${z}_A^1\le {z}_A^0;{z}_B^1\ge {z}_B^0$). We consider the former, the proof of the latter follows the same process.

$\psi :\left({z}_A,{z}_B\right)\in \left[{z}_A^1,{z}_C\right]\times \left[{z}_C,{z}_B^1,\right]\to \left[{\hat{Z}}_A\left[{z}_B\right],{\hat{Z}}_B\left[{z}_A\right]\right]\in \left[{z}_A^0,{z}_C\right]\times \left[{z}_C,{z}_B^0,\right]$, so firm B maps A’s interval into a larger set. But firm A’s map is an into map. Now define:

$$ {z}_A^2={z}_C,{z}_B^2=\mathit{\operatorname{Min}}\left\{z\left|\hat{z}\left[z\right]\ge {z}_A^1\right.\right\} $$

Fig. 6

Locations of the firms in Case ii (panel a), and Case iii (panel b)

Figure 6, panel (a) shows the situation.

As the mapping is increasing ${\hat{Z}}_A\left[{z}_C\right]$=${z}_A^0$≤${z}_A^1=$${\hat{Z}}_A\left[{z}_B^2\right]$ so ${z}_C<{z}_B^2$. As ${\hat{Z}}_A\left[{z}_B^2\right]=$${z}_A^1<{z}_C={\hat{Z}}_A\left({z}_B^1\right)$, we conclude that ${z}_B^2<{z}_B^1$ . Thus, ${\hat{Z}}_A:\left[{z}_B^2,{z}_B^1\right]\subset \left[{z}_A^1,{z}_C\right]$ and ${\hat{Z}}_B:\left[{z}_A^1,{z}_C\right]\subset \left[{z}_C,{z}_B^0\right]$ . So the mapping $\left({\hat{Z}}_A,{\hat{Z}}_B\right):\left[{z}_B^2,{z}_B^1\right]\times \left[{z}_A^1,{z}_C\right]$ is not into because ${z}_B^1<{z}_B^0$ but it maps the closed interval on each side to the other side. The next task is to construct intervals so that the mapping is into.

Define ${z}_A^3=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_B\left[z\right]>{z}_B^2\right\}\ge {z}_A^1$ and ${z}_B^3={z}_B^1$. Then, ${\hat{Z}}_A:\left[{z}_B^2,{z}_B^3\right]\subset \left[{z}_A^1,{z}_C\right]$ and ${\hat{Z}}_B:\left[{z}_A^3,{z}_A^2\right]\subset \left[{z}_B^2,{z}_B^0\right]$. The ordering is ${z}_A^1<{z}_A^2$ =${z}_C<{z}_B^2<{z}_B^1={z}_B^3$ and $, {z}_A^2>{z}_A^1.$ Now define =${z}_A^4={z}_A^2$and ${z}_B^4=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_A\left[z\right]>{z}_A^3\right\}\ge \mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_A\left[z\right]>{z}_A^1\right\}={z}_B^2.$

Proceeding: for n ≥ 2 ${z}_A^{2n}={z}_c$ and ${z}_B^{2n+1}={z}_B^1$ and

$$ {\displaystyle \begin{array}{l}{z}_A^{2n+1}=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_B^{2n}\right\}>{z}_A^{2n-1};\\ {}{z}_B^{2n}=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_A\left[z\right]\ge {z}_A^{2n-1}\right\}\ge {z}_B^{2\left(n-1\right)}\end{array}} $$

As ${z}_A^{2n+1}$and ${z}_B^{2n}$ are both increasing sequences that are bounded, the Monotone Convergence theorem states there are limit points $\underset{n\to \infty }{\lim }{z}_A^{2n+1}={z}_A^{\prime }$, $\underset{n\to \infty }{\lim }{z}_B^{2n}={z}_B^{\prime }$. Note that ${z}_A^{\prime}\ge {z}_A^{2n+1}$ and ${\hat{Z}}_B\left[{z}_A^{\prime}\right]\ge {\hat{Z}}_B\left[{z}_A^{2n}\right]>{z}_B^{2n}.\kern0.5em$ Taking the limit of this expression: ${\hat{Z}}_B\left[{z}_A^{\prime}\right]\ge {z}_B^{\prime }$. That argument can be repeated resulting in: ${\hat{Z}}_A\left[{z}_B^{\prime}\right]\ge {z}_A^{\prime }$ . Thus: ${\hat{Z}}_B$ maps $\left[{z}_A^{\prime },{z}_C\right]$into$\left[{z}_B^{\prime },{z}_C\right]$ while ${\hat{Z}}_A$ maps $\left[{z}_B^{\prime },{z}_B^1\right]$into$\left[{z}_A^{\prime },{z}_C\right]$. Thus we have an into mapping, which maps two intervals on opposite sides of z_C into each other and the mapping is monotone. Thus by Zhao’s Theorem, there is a Nash equilibrium.

Case iii. ${z}_A^1\ge {z}_A^0;{z}_B^1\le {z}_B^0$. Then set ${z}_B^2=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_A\left[z\right]\le {z}_B^1\right\}$; ${z}_A^2=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_A^1\right\}$ . Figure A, panel (b) shows the situation. We know that ${z}_A^2\ge {z}_A^1\ge {z}_A^0;\kern0.5em {z}_B^2\le {z}_B^1\le {z}_B^0$. We consider the intervals: $\left[{z}_A^1,{z}_A^2\right]\subseteq \Big[{z}_A^0,{z}_C$ ]; $\left[{z}_C,{z}_B^0\right]$$\subseteq \left[{z}_B^2,{z}_B^1\right]$.

Construct the sequence of mappings for n ≥ 1

$$ {\displaystyle \begin{array}{l}{z}_A^{2n}=\mathit{\operatorname{Max}}\left\{z|{\hat{Z}}_B\left[z\right]\le {z}_B^{2n-1}\right\};{z}_B^{2n}=\operatorname{Min}\left\{z|{Z}_A\left[z\right]\ge {z}_A^{2n-1}\right\}\\ {}{z}_A^{2n+1}=\operatorname{Min}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_B^{2n}\right\}{z}_B^{2n+1}=\operatorname{Max}\left\{z|{\hat{Z}}_A\left[z\right]\le {z}_A^{2n}\right\}\end{array}} $$

In particular for ${z}_A^3,{z}_B^3$ in the above definition $\left[{z}_A^3,{z}_A^2\right]\subseteq \left[{z}_A^{1,}{z}_A^2\right]$ ; $\left[{z}_B^2,{z}_B^3\right]$$\subseteq \left[{z}_B^2,{z}_B^1\right]$. As the reaction function is increasing, all mappings are closed and this process generates an infinite series of ever shrinking closed segments with the set inclusions as follows:

$$ {\displaystyle \begin{array}{l}\left[{z}_A^{2n+1},{z}_A^{2n}\right]\subseteq \left[{z}_A^{2n-1},{z}_A^{2n}\right]\kern1em \mathrm{and}\kern1em \left[{z}_B^{2n},{z}_B^{2n+1}\right]\subseteq \left[{z}_B^{2n},{z}_B^{2n-1}\right]\\ {}\left[{z}_A^{2n+1},{z}_A^{2\left(n+1\right)}\right]\subseteq \left[{z}_A^{2n+1},{z}_A^{2n}\right]\kern1em \mathrm{and}\kern1em \left[{z}_B^{2\left(n+1\right)},{z}_B^{2n}\right]\subseteq \left[{z}_B^{2n},{z}_B^{2n+1}\right]\end{array}}. $$

Because they are shrinking and bounded this sequence of intervals will converge to a closed interval according to the Monotone Convergence Theorem as applied to the endpoints:

$$ \underset{n\to \infty }{\lim}\left[{z}_A^{2n+1},{z}_A^{2n}\right]=\left[{z}_A^{\prime },{z}_A^{\prime \prime}\right];\underset{n\to \infty }{\lim}\left[{z}_B^{2n},{z}_B^{2n+1}\right]=\left[{z}_B^{\prime },{z}_B^{\prime \prime}\right]. $$

Because the best reply correspondence maps the interval to the right of z_C into the interval to its left side, and vise-versa, this equilibrium is also a Nash equilibrium. Q.E.D.

Proof of Corollary 3

By Proposition 4, each firm’s best response is an increasing correspondence. The argument of the proof follows that of the previous proofs but now two distinct values corresponding to z_C must be defined. Let z_CA, z_CB have the property:

$$ {\displaystyle \begin{array}{l}\underset{-1\le {z}_A\le {z}_{CA}}{\max }{\Pi}^A\left({z}_A,{z}_{CA}\right)=\underset{z_{CA}\le {z}_A\le 1}{\max }{\Pi}^A\left({z}_{CA},{z}_A\right)\\ {}\underset{-1\le {z}_B\le {z}_{CB}}{\max }{\Pi}^B\left({z}_A,{z}_{CB}\right)=\underset{z_{CB}\le {z}_B\le 1}{\max }{\Pi}^B\left({z}_{CB},{z}_B\right).\end{array}} $$

We can assume that z_CA < z_CB. Define

$$ {\displaystyle \begin{array}{l}{\mathrm{z}}_A^0=\underset{z_A:-1\le {z}_A\le {z}_{CA}}{\mathit{\operatorname{Min}}\Big\{ ArgMax}{\Pi}^A\left({z}_A,{z}_B\right)\left\};{\mathrm{z}}_B^0=\underset{z_B:{z}_{CB}\le {z}_B\le 1}{\mathit{\operatorname{Max}}\Big\{ ArgMax}{\Pi}^B\left({z}_A,{z}_B\right)\right\}\\ {}{z}_A^1=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_{CB}\right\},{z}_B^1=\mathit{\operatorname{Max}}\left\{z|{\hat{Z}}_A\left[z\right]\le {z}_{CA}\right\}\end{array}} $$

There are three cases again.

Case i. ${z}_A^1\le {z}_A^0;{z}_B^1\ge {z}_B^0$ . In this case,

$$ \psi :\left({z}_A,{z}_B\right)\in \left[{z}_A^1,{z}_{CB}\right]\times \left[{z}_{CA},{z}_B^1,\right]\to \left[{z}_A^0,{z}_{CA}\right]\times \left[{z}_{CB},{z}_B^0\right] $$

which again is an into mapping. A Nash equilibrium exists by Zhou’s Theorem.

Case ii ${z}_A^1\ge {z}_A^0;{z}_B^1\ge {z}_B^0$ (or the parallel case ${z}_A^1\le {z}_A^0;{z}_B^1\ge {z}_B^0$)

$$ \psi :\left({z}_A,{z}_B\right)\in \left[{z}_A^1,{z}_{CB}\right]\times \left[{z}_{CA},{z}_B^1,\right]\to \left[{\hat{Z}}_A\left[{z}_B\right],{\hat{Z}}_B\left[{z}_A\right]\right]\in \left[{z}_A^0,{z}_{CA}\right]\times \left[{z}_{CB},{z}_B^0\right] $$

where ${\hat{Z}}_i\left[{z}_{-i}\right]$ are as in (A6) and $\left[{z}_A^1,{z}_{CB}\right]\subseteq \left[{z}_A^0,{z}_{CB}\right]$, so ${\hat{z}}_B\left[\cdotp \right]$ maps its domain into a set containing ${\hat{Z}}_A\left[\cdotp \right]$‘s domain, and thus is not an into map. But ${\hat{Z}}_A\left[\cdotp \right]$ maps its domain into${\hat{Z}}_B\left[\cdotp \right]$domain so that component is into. Now define for n ≥ 1

$$ {\displaystyle \begin{array}{l}{z}_A^{2n}={z}_{CB}\\ {}{z}_A^{2n+1}=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_B^{2n}\right\}\\ {}{z}_B^{2n+1}={z}_B^1\\ {}{z}_B^{2n}=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_A\left[z\right]\ge {z}_A^{2n-1}\right\}.\end{array}} $$

As in the proof of Proposition 6 ${z}_A^{2n+1}$and ${z}_B^{2n}$ are both increasing sequences that are bounded, Then Monotone Convergence theorem states there are limit points $\underset{n\to \infty }{\lim }{z}_A^{2n+1}={z}_A^{\prime }$, $\underset{n\to \infty }{\lim }{z}_B^{2n}={z}_B^{\prime }$. Note that ${z}_A^{\prime}\ge {z}_A^{2n+1}$ and ${\hat{Z}}_B\left[{z}_A^{\prime}\right]\ge {\hat{Z}}_B\left[{z}_A^{2n}\right]>{z}_B^{2n}.\kern0.5em$ Taking the limit of this expression: ${\hat{Z}}_B\left[{z}_A^{\prime}\right]\ge {z}_B^{\prime }$. That argument can be repeated resulting in: ${\hat{Z}}_A\left[{z}_B^{\prime}\right]\ge {z}_A^{\prime }$ . Thus: ${\hat{Z}}_B$ maps $\left[{z}_A^{\prime },{z}_{CB}\right]$into$\left[{z}_B^{\prime },{z}_B^0\right]$ while ${\hat{Z}}_A$ maps $\left[{z}_B^{\prime },{z}_B^1\right]$into$\left[{z}_A^{\prime },{z}_{CB}\right]$. Thus we have an into mapping, which maps two intervals on opposite sides of z_CA and z_CB into each other and the mapping is monotone. Thus by Zhao’s Theorem, there is a Nash equilibrium.

Case iii. ${z}_A^1\ge {z}_A^0;{z}_B^1\le {z}_B^0$. Set ${z}_A^2=\mathit{\operatorname{Max}}\left\{z|{\hat{Z}}_B\left[z\right]\le {z}_B^1\right\}$; ${z}_B^2=\mathit{\operatorname{Min}}\left\{z|{\hat{Z}}_A\left[z\right]\ge {z}_A^1\right\}$ . Figure A, panel (b) shows the situation. We know that ${z}_A^2\ge {z}_A^1\ge {z}_A^0;\kern0.5em {z}_B^2\le {z}_B^1\le {z}_B^0$. We consider the intervals: $\left[{z}_A^1,{z}_A^2\right]\subseteq \left[{z}_A^0,{z}_{CB}\right]$ ]; $\left[{z}_{CA},{z}_B^0\right]$$\subseteq \left[{z}_B^2,{z}_B^1\right]$.

Construct the sequence of mappings for n ≥ 1

$$ {\displaystyle \begin{array}{l}{z}_A^{2n}=\mathit{\operatorname{Max}}\left\{z|{\hat{Z}}_B\left[z\right]\le {z}_B^{2n-1}\right\};{z}_B^{2n}=\operatorname{Min}\left\{z|{\hat{Z}}_A\left[z\right]\ge {z}_A^{2n-1}\right\}\\ {}{z}_A^{2n+1}=\operatorname{Min}\left\{z|{\hat{Z}}_B\left[z\right]\ge {z}_B^{2n}\right\}{z}_B^{2n+1}=\operatorname{Max}\left\{z|{\hat{Z}}_A\left[z\right]\le {z}_A^{2n}\right\}\end{array}}; $$

In particular for ${z}_A^3,{z}_B^3$ in the above definition $\left[{z}_A^3,{z}_A^2\right]\subseteq \left[{z}_A^{1,}{z}_A^2\right]$ ]; $\left[{z}_B^2,{z}_B^3\right]$$\subseteq \left[{z}_B^2,{z}_B^1\right]$. As the reaction function is increasing, all mappings are closed and this process generates an infinite series of ever shrinking closed segments with the set inclusions as follows:

$$ {\displaystyle \begin{array}{l}\left[{z}_A^{2n+1},{z}_A^{2n}\right]\subseteq \left[{z}_A^{2n-1},{z}_A^{2n}\right]\kern1em \mathrm{and}\kern1em \left[{z}_B^{2n},{z}_B^{2n+1}\right]\subseteq \left[{z}_B^{2n},{z}_B^{2n-1}\right]\\ {}\left[{z}_A^{2n+1},{z}_A^{2\left(n+1\right)}\right]\subseteq \left[{z}_A^{2n+1},{z}_A^{2n}\right]\kern1em \mathrm{and}\kern1em \left[{z}_B^{2\left(n+1\right)},{z}_B^{2n}\right]\subseteq \left[{z}_B^{2n},{z}_B^{2n+1}\right]\end{array}}. $$

Because they are shrinking and bounded this sequence of intervals will converge to a closed interval according to the Monotone Convergence Theorem as applied to the endpoints:

$$ \underset{n\to \infty }{\lim}\left[{z}_A^{2n+1},{z}_A^{2n}\right]=\left[{z}_A^{\prime },{z}_A^{\prime \prime}\right];\underset{n\to \infty }{\lim}\left[{z}_B^{2n},{z}_B^{2n+1}\right]=\left[{z}_B^{\prime },{z}_B^{\prime \prime}\right]. $$

Because the best reply correspondence maps the interval to the right of z_CA into the interval to its left side, and the limiting interval on the left side of z_CB into its right, this equilibrium is also a Nash equilibrium. Q.E.D.

Proof of Proposition 8

Following the argument found in the proof of Lemma 2, suppose that if firm B is at ${\overline{z}}_B$ firm A generates equal social welfare at either location, $SW\left({\overline{z}}_A,{\overline{z}}_B\right)= SW\left({\underline{z}}_A,{\overline{z}}_B\right)$with ${\underline{z}}_A\le {\overline{z}}_A\le {\overline{z}}_B$. As in the proof of Lemma 2, differentiate $SW\left({\overline{z}}_A,{\overline{z}}_B\right)- SW\left({\underline{z}}_A,{\overline{z}}_B\right)$ with respect to${\overline{z}}_B$ . If the following expression is positive then the social welfare optimizing best response functions are increasing:

$$ {\displaystyle \begin{array}{l}\frac{d\left( SW\right(\left({\overline{z}}_A,{\overline{z}}_B\right)- SW\left(\left({\underline{z}}_A,{\overline{z}}_B\right)\right)}{dz_B}=\underset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left(z-{\underline{z}}_A\right)\rho (z) dz\\ {}\underset{\frac{{\underline{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}{q}_1\left({\overline{z}}_B-z\right)\left({\overline{z}}_B-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz+\underset{-1}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}{q}_1\left({\overline{z}}_B-z\right)\left({\underline{z}}_A-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz.\end{array}} $$

It has previously been shown in the proof of Lemma 2 that the last term is positive. Now consider the derivative in profit for A located at ${\overline{z}}_A$ with respect to B located at ${\overline{z}}_B$ . The derivative

$\frac{d{\Pi}_A\left({\overline{z}}_A,{\overline{z}}_B\right)}{d{z}_B}=\underset{-1}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left(z-{\overline{z}}_B\right)\rho (z) dz+\underset{-1}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}{q}_1\left(z-{z}_B\right)\left({\overline{z}}_B-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz+\frac{1}{2}q\Big(\frac{{\overline{z}}_B-{\overline{z}}_A}{2}$) >0 as profit rises at every point when firm B moves away from A. Note that it is also true that

$\underset{\frac{{\underset{\_}{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left(z-{\overline{z}}_B\right)\rho (z) dz+\underset{\frac{{\underset{\_}{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}{q}_1\left(z-{z}_B\right)\left({\overline{z}}_B-z-\left|{\overline{z}}_A-z\right|\right)\rho (z) dz$) >0 because for every point within the interval $z\in \left[\frac{{\underset{\_}{z}}_A+{\overline{z}}_B}{2},\frac{{\overline{z}}_A+{\overline{z}}_B}{2}\right]$rising price increases local profit as prices are assumed below monopoly levels. But $\underset{\frac{{\underset{\_}{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left(z-{\underset{\_}{z}}_A\right)\rho (z) dz>\underset{\frac{{\underset{\_}{z}}_A+{\overline{z}}_B}{2}}{\overset{\frac{{\overline{z}}_A+{\overline{z}}_B}{2}}{\int }}q\left({\underset{\_}{z}}_B-z\right)\rho (z) dz\kern0.5em$because in the domain of integration $z-{\overline{z}}_B>z-{\underset{\_}{z}}_A$. Thus, $\frac{d\left( SW\right(\left({\overline{z}}_A,{\overline{z}}_B\right)- SW\left(\left({\underset{\_}{z}}_A,{\overline{z}}_B\right)\right)}{d{z}_B}$. Q.E.D.