Necessary and sufficient condition for equilibrium of the Hotelling model

TL;DR

This paper provides a precise mathematical condition that characterizes the equilibrium in Hotelling's linear city model, enhancing understanding of vendor competition and location strategies.

Contribution

It establishes a necessary and sufficient condition for equilibrium in the Hotelling model, offering a clear representation of equilibrium configurations.

Findings

Derived a mathematical condition for equilibrium

Proved the condition is both necessary and sufficient

Provided a new representation for equilibrium

Abstract

We study a model of vendors competing to sell a homogeneous product to customers spread evenly along a linear city. This model is based on Hotelling's celebrated paper in 1929. Our aim in this paper is to present a necessary and sufficient condition for the equilibrium. This yields a representation for the equilibrium. To achieve this, we first formulate the model mathematically. Next, we prove that the condition holds if and only if vendors are equilibrium.

Equations

$${\rho }_k(\mathbf{\xi },y)=\{\begin{array}{ll}0& \text{if}|{\xi }_k-y|>\underset{i}{\min }|{\xi }_i-y|\\ \frac{1}{|S(\mathbf{\xi },y)|}& \text{if}|{\xi }_k-y|=\underset{i}{\min }|{\xi }_i-y|\end{array},$$

$$f_k(\mathbf{\xi })={\int }_0^1{\rho }_k(\mathbf{\xi },y)dy,$$

$$f_k({\mathbf{x}}^{\ast })\ge f_k(x_1^{\ast },x_2^{\ast },x_3^{\ast },\dots ,x_{k-1}^{\ast },x_k,x_{k+1}^{\ast },\dots ,x_n^{\ast })$$

$$\mathbf{x}=\left(\frac{1}{10},\frac{1}{10},\frac{3}{10},\frac{3}{10},\frac{7}{10},\frac{7}{10},\frac{9}{10},\frac{9}{10}\right).$$

$$\mathbf{x}=\left(\frac{1}{2},\frac{1}{2}\right)\text{ is a unique equilibrium.}$$

$$\displaystyle\begin{array}[]{ll}&|I_{0}|=|I_{n}|>0\;(n\geq 2),\quad|I_{1}|=|I_{n-1}|=0\quad(n\geq 4),\\ &|I_{0}|:|I_{2}|=1:2,\quad|I_{n-2}|:|I_{n}|=2:1\quad(n\geq 4),\end{array}$$

$$|I_j|\le 2|I_0|(0\le j\le n),2|I_0|\le |I_k|+|I_{k+1}|(1\le k\le n-2).$$

$$x_1\ne 0\text{ and }{x}_n\ne 1.$$

$$x_1=x_2\text{ and }{x}_{n-1}=x_n.$$

$$f_1(x_1^{\prime },x_2,\cdots ,x_n)$$

$$f_1(x_1^{\prime },x_2,\cdots ,x_n)=\frac{1}{2}(|[0,x_1^{\prime }]|+|[x_1^{\prime },x_{i+1}]|)=\frac{1}{2}|[0,x_{i+1}]|=\frac{1}{2}|I_i|>\frac{1}{2i}|I_i|=f_1(\mathbf{x}).$$

$$f_1(x_1^{\prime },x_2,\cdots ,x_n)=\frac{1}{2}\left|\left[0,\frac{1}{2}\right]\right|+\left|\left[\frac{1}{2},1\right]\right|=\frac{1}{4}+\frac{1}{2}>\frac{1}{2}\ge \frac{1}{n}=f_1(\mathbf{x}).$$

$$f_k(x_1,\cdots ,x_{k-1},x_k^{\prime },x_{k+1},\cdots ,x_n)$$

$$=\frac{2}{3}|I_l|\ge f_k(\mathbf{x}).$$

$$f_k(x_1,\cdots ,x_{k-1},x_k^{\prime },x_{k+1},\cdots ,x_n)$$

$$=\frac{2}{5}\ge \frac{1}{n}=f_k(\mathbf{x}).$$

$$(x_1,x_2,,\dots ,x_{k-1},x_k^{\prime },x_{k+1},\dots ,x_n),$$

$$f_1(x_1^{\prime },x_2)=|[0,x_1^{\prime }]|+\frac{1}{2}|[x_1^{\prime },x_2]|<|[0,x_1^{\prime }]|+|[x_1^{\prime },x_2]|=|[0,x_2]|=\frac{1}{2}=f_1(\mathbf{x}).$$

$$f_1(x_1^{\prime },x_2)=|[0,x_1^{\prime }]|+\frac{1}{2}|[x_1^{\prime },x_2]|>|[0,x_1^{\prime }]|=\frac{1}{2}=f_1(\mathbf{x}).$$

$$f_1(x_1^{\prime },x_2,\cdots ,x_n)=\frac{1}{2}(2|I_0|+\delta )=|I_0|+\frac{1}{2}\delta >|I_0|+\frac{1}{4}\delta =f_1(\mathbf{x}).$$

$$f_1(x_1^{\prime },x_2,\cdots ,x_n)$$

$$=|I_0|-\frac{1}{4}\delta =f_1(\mathbf{x}).$$

$$f_1(x_1^{\prime },x_2,\cdots ,x_n)=\frac{1}{2}|[x_n,x_1^{\prime }]|+|[x_1^{\prime },1]|>|[x_1^{\prime },1]|=|I_0|=f_1(\mathbf{x}).$$

$$\{\begin{array}{l}{f}_1(\mathbf{x})=f_2(\mathbf{x})=f_{n-1}(\mathbf{x})=f_n(\mathbf{x})=|I_0|.\\ f_k(\mathbf{x})=|I_0|\text{if}{x}_{k-1}=x_k\text{or}{x}_k=x_{k+1}(3\le k\le n-3).\\ f_k(\mathbf{x})\ge |I_0|\text{if}{x}_{k-1}\ne x_k\text{and}{x}_k\ne x_{k+1}(3\le k\le n-3).\end{array}$$

Full text

N. Tsuge’s research is partially supported by Grant-in-Aid for Scientific Research (C) 17K05315, Japan.

1. Introduction

We study a model in which a linear city of length $1$ on a line and customers are uniformly distributed with density $1$ along this interval. We consider $n$ vendors moving on this line. Let the location of the vendor $k\quad(k=1,2,3,\ldots,n)$ be $x_{k}\in[0,1]$. We assume that $x_{1}\leq x_{2}\leq\cdots\leq x_{n}$ and denote the location of $n$ vendors $(x_{1},x_{2},x_{3},\ldots,x_{n})$ by ${\bm{x}}$. Since we study the competition between vendors, we consider $n\geq 2$ in particular. The price of one unit of product for each vendor is identical. Moreover, we assume the following.

If there exist $l\quad(l=1,2,3,\ldots,n)$ vendors nearest to a customer, the customer purchases $1/l$ unit of product per unit of time from each of the $l$ vendors respectively.

Every vendor then seeks a location to maximize his profit.

We then represent the profit of vendor $k$ per unit of time by a mathematical notation. Given a vector ${\bm{\xi}}=(\xi_{1},\xi_{2},\ldots,\xi_{n})\in[0,1]^{n}$ and $0\leq y\leq 1$, we define a set $\displaystyle S({\bm{\xi}},y)=\{j\in\{1,2,3,\ldots,n\}:|\xi_{j}-y|=\min_{i}|\xi_{i}-y|\}$. By using a density function

[TABLE]

we define

[TABLE]

where $|A|$ represents a number of elements in a set $A$. We call $f_{k}({\bm{x}})$ the profit of vendor $k$ per unit of time for a location ${\bm{x}}$. We then define equilibrium as follows.

Definition 1.1**.**

A location ${\bm{x}}^{*}=(x^{*}_{1},x^{*}_{2},x^{*}_{3},\ldots,x^{*}_{n})\in[0,1]^{n}$ is called equilibrium, if

[TABLE]

holds for any $k\in\{1,2,3,\ldots,n\}$ and $x_{k}\in[0,1]$.

We review the known results. The present model is based on Hotelling’s model in [4]. Although we consider homogeneous vendors, Hotelling did heterogeneous vendors. In [1, Chaper 10], Alonso, W. treated with the same model as our problem for two vendors. He introduced this model as the competition between two vendors of ice cream along a beach. In [3], the model for $n$ vendors was studied. Furthermore, Eaton, B. C. and Lipsey, R. G. investigated a necessary and sufficient condition for equilibrium. More precisely, they claimed that (1.i) and (1.ii) in [3, p29] if and only if $n$ vendors are equilibrium (see also [p9][2]). Although this is an interesting approach from the point of mathematical view, unfortunately, it seems that (1.i) and (1.ii) are not sufficient conditions. In fact, we consider a location

[TABLE]

Then, we find that $\displaystyle f_{1}({\bm{x}})=f_{2}({\bm{x}})=f_{7}({\bm{x}})=f_{8}({\bm{x}})=1/{10},\;f_{3}({\bm{x}})=f_{4}({\bm{x}})=f_{5}({\bm{x}})=f_{6}({\bm{x}})=3/{20}$. Therefore, this example satisfies (1.i) and (1.ii).

Remark 1.1**.**

In [3], vendors in our problem are called firms and $f_{k}({\bm{x}})$ seems to be called a market of firm $k$. In addition, we regard two pairs of peripheral firms in (1.ii) of [3] as firms $1,2$ and $9,10$.

On the other hand, if $x_{3}$ moves from $3/10$ to $\displaystyle 1/2$, $x_{3}$ can obtain a profit $\displaystyle 2/{10}$ more than the original one $\displaystyle 3/{20}$. In addition, the definition of their terminologies seems not to be clear, such as market, peripheral, equilibrium, etc. Therefore, our goal in this paper is to formulate this model mathematically and present a revised necessary and sufficient for equilibrium.

For convenience, we set $x_{0}=0,\;x_{n+1}=1$ and denote a interval $[x_{k},x_{k+1}]$ by $I_{k}\;\;(k=0,1,2,\ldots,n)$. Then our main theorem is as follows.

Theorem 1.1**.**

**

1. (i)

$n=2$**

[TABLE] 2. (ii)

$n=3$**

[TABLE] 3. (iii)

$n\geq 4$**

${\bm{x}}$* is equilibrium, if and only if the following conditions (1.7) and (1.8) hold.*

[TABLE]

2. Preliminary

In this section, we prepare some lemmas and a proposition to prove our main theorem in a next section. We first consider the profit of $i$ vendors which locate at one point. We have the following lemma.

Lemma 2.1**.**

We consider a location ${\bm{x}}=(x_{1},x_{2},x_{3},\ldots,x_{n})$ with $x_{1}\leq x_{2}\leq x_{3}\leq\dots\leq x_{n}$. We assume that $x_{l}<x_{l+1}=\cdots=x_{k}=\cdots=x_{l+i}<x_{l+i+1}\quad(n\geq 2,\;l\geq 0,\;l+i+1\leq n+1,\;1\leq i\leq n)$.

1. (i)

If $x_{l}\neq x_{0}$ and $x_{l+i+1}\neq x_{n+1}$, $\displaystyle f_{k}({\bm{x}})=\frac{1}{2i}(|I_{l}|+|I_{l+i}|)$. 2. (ii)

If $x_{l}=x_{0}$ and $x_{l+i+1}\neq x_{n+1}$, $\displaystyle f_{k}({\bm{x}})=\frac{1}{i}\left(|I_{l}|+\frac{1}{2}|I_{l+i}|\right)$. 3. (iii)

If $x_{l}\neq x_{0}$ and $x_{l+i+1}=x_{n+1}$, $\displaystyle f_{k}({\bm{x}})=\frac{1}{i}\left(\frac{1}{2}|I_{l}|+|I_{l+i}|\right)$. 4. (iv)

If $x_{l}=x_{0}$ and $x_{l+i+1}=x_{n+1}$, $\displaystyle f_{k}({\bm{x}})=\frac{1}{n}$.

Proof.

Proof of $(\mathrm{i})$

We have $f_{k}({\bm{x}})=\dfrac{1}{i}\left(\dfrac{1}{2}|I_{l}|+\dfrac{1}{2}|I_{l+i}|\right)=\dfrac{1}{2i}\left(|I_{l}|+|I_{l+i}|\right)$.

Proof of $(\mathrm{ii})$

We have $f_{k}({\bm{x}})=\dfrac{1}{i}\left(|I_{l}|+\dfrac{1}{2}|I_{l+i}|\right)=\dfrac{1}{i}\left(|I_{l}|+\dfrac{1}{2}|I_{l+i}|\right)$.

*Proof of $(\mathrm{iii})$

We have $f_{k}({\bm{x}})=\dfrac{1}{i}\left(\dfrac{1}{2}|I_{l}|+|I_{l+i}|\right)=\dfrac{1}{i}\left(\dfrac{1}{2}|I_{l}|+|I_{l+i}|\right)$.

Proof of $(\mathrm{iv})$

We have $f_{k}({\bm{x}})=\dfrac{1}{n}\left(|I_{0}|+|I_{n}|\right)=\dfrac{1}{n}$.

∎

Next, the following proposition play an important role.

Proposition 2.2**.**

If the location of $n$ vendors ${\bm{x}}=(x_{1},x_{2},x_{3},\ldots,x_{n})$ with $x_{1}\leq x_{2}\leq x_{3}\leq\dots\leq x_{n}\quad(n\geq 2)$ is equilibrium, the following holds.

[TABLE]

Proof.

Proof of (2.1)

If $x_{1}=0$, we show that ${\bm{x}}$ is not equilibrium.

1. (i)

$x_{1}=0$ and $x_{2}\neq 0$ We notice that $f_{1}({\bm{x}})=\dfrac{1}{2}|I_{1}|$. Setting $x_{1}^{\prime}=\dfrac{1}{2}|I_{1}|$, we then have

[TABLE]

where $|I|$ represents the length of a interval $I$. 2. (ii)

$x_{1}=\cdots=x_{i}=0$ and $x_{i+1}\neq 0$ $(2\leq i\leq n-1)$

We notice that $f_{1}({\bm{x}})=\dfrac{1}{2i}|I_{i}|$. Setting $x_{1}^{\prime}\in{(0,x_{i+1})}$, we have

[TABLE] 3. (iii)

$x_{1}=\cdots=x_{n}=0$ $(n\geq 2)$

We notice that $f_{1}({\bm{x}})=\dfrac{1}{n}$. Setting $x_{1}^{\prime}=\dfrac{1}{2}$, we have

[TABLE]

From (i)–(iii), if $x_{1}=0$, we have proved that ${\bm{x}}$ is not equilibrium. Furthermore, from the symmetry, if $x_{n}=1$, we can similarly prove that ${\bm{x}}$ is not equilibrium.

Proof of (2.2)

We prove that ${\bm{x}}$ is not equilibrium, provided that $i\;(3\leq i\leq n)$ vendors occupy at a point. We assume that $x_{l}<x_{l+1}=\cdots=x_{k}=\cdots=x_{l+i}<x_{l+i+1}\quad(l\geq 0,\;l+i+1\leq n+1,\;3\leq i\leq n)$. Here we recall that we set $x_{0}=0$ and $x_{n+1}=1$. Therefore there exist $x_{l}$ and $x_{l+i+1}$ at least one respectively. We notice that $x_{l}<x_{k}<x_{l+i+1}$ and there exists no vendor on $(x_{l},x_{k})$ and $(x_{k},x_{l+i+1})$. Dividing this proof into four cases, we prove (2.2).

1. (i)

$x_{l}\neq x_{0}$ and $x_{l+i+1}\neq x_{n+1}$ In this case, if $|I_{l}|\geq|I_{l+i}|$, we notice that $f_{k}({\bm{x}})=\dfrac{1}{2i}(|I_{l}|+|I_{l+i}|)\leq\dfrac{1}{6}(|I_{l}|+|I_{l+i}|)=\dfrac{1}{6}(|I_{l}|+|I_{l}|)=\dfrac{1}{3}|I_{l}|$. Setting $x_{k}^{\prime}\in{(x_{l},x_{l+1})}$, we have $f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})=\dfrac{1}{2}(|[x_{l},x_{k}^{\prime}]|+|[x_{k}^{\prime},x_{k}]|)=\dfrac{1}{2}|I_{l}|>f_{k}({\bm{x}})$.

For the other case $|I_{l}|<|I_{l+i}|$, from the symmetry, we can similarly show that there exists $x_{k}^{\prime}$ such that $f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})>f_{k}({\bm{x}})$. Thus, if $x_{l}\neq x_{0}$ and $x_{l+i+1}\neq x_{n+1}$, we can prove that ${\bm{x}}$ is not equilibrium. 2. (ii)

$x_{l}=x_{0}$ and $x_{l+i+1}\neq x_{n+1}$

In this case, if $|I_{l}|\geq|I_{l+i}|$, we find that $f_{k}({\bm{x}})=\dfrac{1}{i}(|I_{l}|+\dfrac{1}{2}|I_{l+i}|)\leq\dfrac{1}{3}(|I_{l}|+\dfrac{1}{2}|I_{l+i}|)\leq\dfrac{1}{3}(|I_{l}|+\dfrac{1}{2}|I_{l}|)=\dfrac{1}{2}|I_{l}|$. Setting $x_{k}^{\prime}=\dfrac{2}{3}|I_{l}|$, we have

[TABLE]

For the other case $|I_{l}|<|I_{l+i}|$, from the symmetry, we can similarly show that there exists $x_{k}^{\prime}$ such that $f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})>f_{k}({\bm{x}})$. Thus, if $x_{l}=x_{0}$ and $x_{l+i+1}\neq x_{n+1}$, we have showed that ${\bm{x}}$ is not equilibrium. 3. (iii)

$x_{l}\neq x_{0}$ and $x_{l+i+1}=x_{n+1}$

From the symmetry of $(\mathrm{ii})$, there exists $x_{k}^{\prime}$ satisfying

$f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})>f_{k}({\bm{x}})$. Thus if $x_{l}\neq x_{0}$ and $x_{l+i+1}=x_{n+1}$, we have showed that ${\bm{x}}$ is not equilibrium. 4. (iv)

$x_{l}=x_{0}$ and $x_{l+i+1}=x_{n+1}$

In this case, we find $f_{k}({\bm{x}})=\dfrac{1}{n}$. If $x_{k}\geq\dfrac{1}{2}$, for $x_{k}^{\prime}=\dfrac{2}{5}$, we have

[TABLE]

For the other case $x_{k}<\dfrac{1}{2}$, from the symmetry, we can similarly show that there exists $x_{k}^{\prime}$ such that $f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})>f_{k}({\bm{x}})$. Thus, if $x_{l}=x_{0}$ and $x_{l+i+1}=x_{n+1}$, we have showed that ${\bm{x}}$ is not equilibrium.

From (i)–(iv), we can complete the proof of (2.2).

Proof of (2.3)

If $x_{1}<x_{2}$, we show that ${\bm{x}}$ is not equilibrium. $(x_{1}\neq 0)$

We notice that $f_{1}({\bm{x}})=|I_{0}|+\dfrac{1}{2}|I_{1}|.$ Setting $x_{1}^{\prime}=|I_{0}|+\dfrac{1}{2}|I_{1}|$, we have

$f_{1}(x_{1}^{\prime},x_{2},\cdots,x_{n})=|[0,x_{1}^{\prime}]|+\dfrac{1}{2}|[x_{1}^{\prime},x_{2}]|>|[0,x_{1}^{\prime}]|=|I_{0}|+\dfrac{1}{2}|I_{1}|=f_{1}({\bm{x}})$. Thus ${\bm{x}}$ is not equilibrium. If $x_{n-1}<x_{n}$, we can similarly prove that ${\bm{x}}$ is not equilibrium $(x_{n}\neq 1)$. ∎

Finally, we compare a location after the movement of a vendor with the original one. To do this, we introduce the following notation.

For a given location of $n$ vendors ${\bm{x}}=(x_{1},x_{2},x_{3},\ldots,x_{n})$ with $x_{1}\leq x_{2}\leq\cdots\leq x_{n}$, we move vendor $k$ from $x_{k}$ to a point in $A\subset[0,1]$. We denote the resultant location by $x_{k}\rightarrow A$. We notice that $x_{k}\rightarrow A$ represents the following vector

[TABLE]

where $x^{\prime}_{k}$ is a location of vendor $k$ after movement and $x^{\prime}_{k}\in A$.

Then we have the following lemmas. Since their proofs are a little complicated, they are postponed to Appendix.

Lemma 2.3**.**

If a location of $n$ vendors ${\bm{x}}=(x_{1},x_{2},x_{3},\ldots,x_{n})$ satisfies $\eqref{eqn:main1}$ and $\eqref{eqn:main2}$, $|I_{0}|\leq f_{k}({\bm{x}})\quad(1\leq k\leq n)$.

Lemma 2.4**.**

If a location of $n$ vendors ${\bm{x}}=(x_{1},x_{2},x_{3},\ldots,x_{n})$ satisfies $\eqref{eqn:main1}$ and $\eqref{eqn:main2}$, $f_{k}(x_{k}\rightarrow[0,1])\leq f_{k}({\bm{x}})\quad(1\leq k\leq n)$.

3. Proof of Theorem 1.1

We are now position to prove our main theorem.

Proof of Theorem 1.1 (i)

We prove that if ${\bm{x}}=(x_{1},x_{2})=\left(\dfrac{1}{2},\dfrac{1}{2}\right)$, ${\bm{x}}$ is equilibrium.

Proof.

1. (i)

We consider vendor $1$. We then notice that $f_{1}({\bm{x}})=\dfrac{1}{2}$. For any $x_{1}^{\prime}\in{\left[0,\dfrac{1}{2}\right)}$, we have

[TABLE]

For any $x_{1}^{\prime}\in{\left(\dfrac{1}{2},1\right]}$, from the symmetry of (a), we can similarly prove that $f_{1}(x_{1}^{\prime},x_{2})<f_{1}({\bm{x}})$. Therefore, for any $x_{1}^{\prime}\in{[0,1]}$, we have $f_{1}({\bm{x}})\geq f_{1}(x_{1}^{\prime},x_{2})$. 2. (ii)

Next, we consider vendor $2$. For any $x_{2}^{\prime}\in{[0,1]}$, we find that $f_{2}({\bm{x}})\geq f_{2}(x_{1},x_{2}^{\prime})$ in a similar manner to (i).

From $(\mathrm{i})$ and $(\mathrm{ii})$, we have showed that $\left(\dfrac{1}{2},\dfrac{1}{2}\right)$ is equilibrium. ∎

Next, we show that ${\bm{x}}=\left(\dfrac{1}{2},\dfrac{1}{2}\right)$ is a necessary condition for equilibrium. Therefore, we prove that if ${\bm{x}}\neq\left(\dfrac{1}{2},\dfrac{1}{2}\right)$, then ${\bm{x}}$ is not equilibrium.

Proof.

From (2.1), when $x_{1}=0$ or $x_{2}=1$, ${\bm{x}}$ is not equilibrium. Therefore, we treat with the case where $x_{1}\neq 0$ or $x_{2}\neq 1$.

1. (i)

$x_{1}\neq x_{2}$

From (2.3), ${\bm{x}}$ is not equilibrium in this case. 2. (ii)

$x_{1}=x_{2}$

We first notice that $f_{1}({\bm{x}})=\dfrac{1}{2}$ in this case. If $x_{1}>\dfrac{1}{2}$, setting $x_{1}^{\prime}=\dfrac{1}{2}$, we obtain

[TABLE]

If $x_{1}<\dfrac{1}{2}$, from symmetry, we can show that there exits $x_{1}^{\prime}\in{[0,1]}$ such that $f_{1}(x_{1}^{\prime},x_{2})>f_{1}({\bm{x}})$.

From the above, we have proved that $\left(\dfrac{1}{2},\dfrac{1}{2}\right)$ is a necessary condition for equilibrium. ∎

Proof of Theorem 1.1 (ii)

If $n=3$, (2.2) contradicts (2.3). Therefore, we conclude that there exists no equilibrium in this case.

Proof of Theorem 1.1 (iii)

Finally, we are concerned with the case where $n\geq 4$.

Proof.

First, it follows from Lemma 2.4 that (1.7) and (1.8) is a sufficient condition for equilibrium.

Next, we show that (1.7)–(1.8) is a necessary condition for equilibrium. Therefore, we prove that if (1.7)–(1.8) do not hold, then ${\bm{x}}$ is not equilibrium. Observing Proposition 2.2, we do not have to treat with the case where $x_{1}=0$ or $x_{n}=1$ or more than $2$ vendors occupy a location. From this reason, we assume that $|I_{0}|>0$. We prove the following cases.

1. (i)

Condition (1.7) does not hold.

1. (a)

$|I_{1}|\neq 0\quad($resp. $|I_{n-1}|\neq 0)$. 2. (b)

$|I_{1}|=|I_{n-1}|=0$ and $|I_{0}|:|I_{2}|\neq 1:2\quad($resp. $|I_{1}|=|I_{n-1}|=0$ and $|I_{n-2}|:|I_{n}|\neq 2:1)$. 3. (c)

$|I_{1}|=|I_{n-1}|=0$ and $|I_{0}|:|I_{2}|=1:2$ and $|I_{n-2}|:|I_{n}|=2:1$ and $|I_{0}|\neq|I_{n}|$. 2. (ii)

Condition (1.7) holds and condition (1.8) does not hold.

1. (a)

Condition (1.7) holds and $|I_{j}|>2|I_{0}|$. 2. (b)

Condition (1.7) holds and $|I_{j}|\leq 2|I_{0}|$ and $2|I_{0}|>|I_{k}|+|I_{k+1}|$.

Dividing this proof into the above cases, we prove our main theorem.

1. (i)

2. (a)

$|I_{1}|\neq 0$ $($resp. $|I_{n-1}|\neq 0)$

From (2.3) and $x_{1}<x_{2}$ $($resp. $x_{n-1}<x_{n})$, ${\bm{x}}$ is not equilibrium. 2. (b)

$|I_{1}|=|I_{n-1}|=0$ and $|I_{0}|:|I_{2}|\neq 1:2\;($resp. $|I_{1}|=|I_{n-1}|=0$ and $|I_{n-2}|:|I_{n}|\neq 2:1$)

1. (1)

$|I_{0}|:|I_{2}|=|I_{0}|:(2|I_{0}|+\delta)$ $(\delta>0)$

We notice that $f_{1}({\bm{x}})=\dfrac{1}{2}\left\{|I_{0}|+\dfrac{1}{2}(2|I_{0}|+\delta)\right\}=|I_{0}|+\dfrac{1}{4}\delta$. Setting $x_{1}^{\prime}\in{(x_{2},x_{3})}$, we have

[TABLE] 2. (2)

$|I_{0}|:|I_{2}|=|I_{0}|:(2|I_{0}|-\delta)$ $(\delta>0)$

We notice that $f_{1}({\bm{x}})=\dfrac{1}{2}\left\{|I_{0}|+\dfrac{1}{2}\left(2|I_{0}|-\delta\right)\right\}=|I_{0}|-\dfrac{1}{4}\delta$. Setting $x_{1}^{\prime}=|I_{0}|-\dfrac{1}{4}\delta$, we have

[TABLE]

From the symmetry of (1)–(2), we can similarly show in the case where $|I_{n-2}|:|I_{n}|\neq 2:1$. Thus, ${\bm{x}}$ is not equilibrium in the case of (b). 3. (c)

$|I_{1}|=|I_{n-1}|=0$ and $|I_{0}|:|I_{2}|=1:2$ and $|I_{n-2}|:|I_{n}|=2:1$ and $|I_{0}|\neq|I_{n}|$

1. (1)

$|I_{0}|<|I_{n}|$

We notice that $f_{1}({\bm{x}})=\dfrac{1}{2}\left(|I_{0}|+\dfrac{1}{2}\cdot 2|I_{0}|\right)=|I_{0}|$. Setting $x_{1}^{\prime}=1-|I_{0}|$, we have

[TABLE] 2. (2)

$|I_{0}|>|I_{n}|$

From the symmetry, we can similarly show that there exists $x_{n}^{\prime}$ such that $f_{n}(x_{1},\cdots,x_{n-1},x_{n}^{\prime})>f_{n}({\bm{x}})$.

From (1)–(2), ${\bm{x}}$ is not equilibrium in the case of (c). 2. (ii)

Condition (1.7) holds and condition (1.8) does not hold.

1. (a)

(1.7) holds and $|I_{j}|>2|I_{0}|$

We notice that $f_{1}({\bm{x}})=\dfrac{1}{2}\left(|I_{0}|+\dfrac{1}{2}\cdot 2|I_{0}|\right)=|I_{0}|$. Setting $x_{1}^{\prime}\in{(x_{j},x_{j+1})}$, we have

$f_{1}(x_{1}^{\prime},x_{2},\cdots,x_{n})=\dfrac{1}{2}|I_{j}|>\dfrac{1}{2}\cdot 2|I_{0}|=|I_{0}|=f_{1}({\bm{x}})$.

Thus, ${\bm{x}}$ is not equilibrium in this case. 2. (b)

(1.7) holds and $|I_{j}|\leq 2|I_{0}|$ and $2|I_{0}|>|I_{k}|+|I_{k+1}|$

When $k=1,2,n-1,n$, we notice that $|I_{1}|+|I_{2}|=2|I_{0}|$ and $|I_{2}|+|I_{3}|\geq 2|I_{0}|$. Thus, we devote to considering $3\leq k\leq n-2$. In view of (2.2), we divide (b) into the following three parts.

1. (1)

$x_{k-1}\neq x_{k}$ and $x_{k}\neq x_{k+1}$

We notice that $f_{k}({\bm{x}})=\dfrac{1}{2}(|I_{k}|+|I_{k+1}|)<\dfrac{1}{2}\cdot 2|I_{0}|=|I_{0}|$. Setting $x_{k}^{\prime}\in{(x_{2},x_{3})}$, we have $f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})=\dfrac{1}{2}\cdot 2|I_{0}|=|I_{0}|>f_{k}({\bm{x}})$. Thus ${\bm{x}}$ is not equilibrium in this case. 2. (2)

$x_{k-1}=x_{k}$ $(x_{k-2}\neq x_{k-1}$ and $x_{k}\neq x_{k+1})$

From $|I_{k}|+|I_{k+1}|<2|I_{0}|$, we find $|I_{k}|<2|I_{0}|$. From $|I_{j}|\leq 2|I_{0}|$, we notice that $|I_{k-2}|\leq 2|I_{0}|$. It follows that $f_{k}({\bm{x}})=\dfrac{1}{4}(|I_{k-2}|+|I_{k}|)<\dfrac{1}{4}(2|I_{0}|+2|I_{0}|)=|I_{0}|$. Therefore, for $x_{k}^{\prime}\in{(x_{2},x_{3})}$, we have $f_{k}(x_{1},\cdots,x_{k-1},x_{k}^{\prime},x_{k+1},\cdots,x_{n})=\dfrac{1}{2}\cdot 2|I_{0}|=|I_{0}|>f_{k}({\bm{x}})$. This means that ${\bm{x}}$ is not equilibrium in this case. 3. (3)

$x_{k}=x_{k+1}$ $(x_{k-1}\neq x_{k}$ and $x_{k+1}\neq x_{k+2})$

We can prove this case in a similar manner to (2).

From (i)–(ii), we have showed that if condition (1.7) or condition (1.8) do not hold, then ${\bm{x}}$ is not equilibrium.

We can complete the proof of Theorem 1.1. ∎

Appendix A Proof of Lemma 2.3

Proof.

We estimate the profit of each vendor, $f_{k}({\bm{x}})\quad(k=1,2,\ldots,n)$.

1. (i)

Vendor $1$

We have $f_{1}({\bm{x}})=\dfrac{1}{2}(|I_{0}|+\dfrac{1}{2}\cdot 2|I_{0}|)=|I_{0}|$. 2. (ii)

Vendor $2$

We can similarly deduce $f_{2}({\bm{x}})=|I_{0}|$. 3. (iii)

Vendor $n-1$ and $n$

From the symmetry with vendors $1$ and $2$, we have $f_{n-1}({\bm{x}})=f_{n}({\bm{x}})=|I_{0}|$. 4. (iv)

Vendor $k$ $(3\leq k\leq n-2)$

1. (a)

$x_{k-1}\neq x_{k}$ and $x_{k}\neq x_{k+1}$

We have $f_{k}({\bm{x}})=\dfrac{1}{2}(|I_{k-1}|+|I_{k}|)\geq\dfrac{1}{2}\cdot 2|I_{0}|=|I_{0}|$. 2. (b)

$x_{k-1}=x_{k},x_{k}\neq x_{k+1}$ $(k\neq 3)$

From (1.8), we notice that $x_{k-2}\neq x_{k-1}$. On the other hand, from $|I_{k-1}|=0$ and (1.8), we have $|I_{k-2}|\leq 2|I_{0}|$ and $2|I_{0}|\leq|I_{k-2}|$. Thus we have $|I_{k-2}|=2|I_{0}|$. Similarly, we have $|I_{k}|=2|I_{0}|$. As a consequence, we have $f_{k}({\bm{x}})=\dfrac{1}{4}(|I_{k-2}|+|I_{k}|)=\dfrac{1}{4}\cdot 4|I_{0}|=|I_{0}|$. 3. (c)

$x_{k-1}\neq x_{k},x_{k}=x_{k+1}$ $(k\neq n-2)$

We can show that $f_{k}({\bm{x}})=|I_{0}|$ in a similar manner to (b).

Combining (i)–(iv), we obtain

[TABLE]

Thus, for ${\bm{x}}$ satisfying $\eqref{eqn:main1}$ and $\eqref{eqn:main2}$ and any $k$ $(1\leq k\leq n)$, we have showed that $f_{k}({\bm{x}})\geq|I_{0}|$.

∎

Appendix B Proof of Lemma 2.4

Proof.

Dividing the proof into three parts, we prove this lemma.

1. (i)

$f_{k}(x_{k}\rightarrow(x_{l},x_{l+1}))\leq f_{k}({\bm{x}})\quad(x_{l}<x_{l+1}$ and $l\neq k-1,k$ and $0\leq l\leq n)$. 2. (ii)

$f_{k}(x_{k}\rightarrow(x_{k-1},x_{k+1}))\leq f_{k}({\bm{x}})$ $(1\leq k\leq n)$. 3. (iii)

$f_{k}(x_{k}\rightarrow\{x_{l}\})\leq f_{k}({\bm{x}})$ $(1\leq k\leq n$ and $0\leq l\leq n+1)$.

Proof of $(\mathrm{i})$

1. (a)

$l\neq 0$ and $l\neq n$

We have $f_{k}(x_{k}\rightarrow(x_{l},x_{l+1}))=\dfrac{1}{2}|I_{l}|\leq\dfrac{2|I_{0}|}{2}=|I_{0}|\leq f_{k}({\bm{x}})$. 2. (b)

$l=0$

We have $f_{k}(x_{k}\rightarrow(x_{0},x_{1}))<|I_{0}|\leq f_{k}({\bm{x}})$. 3. (c)

$l=n$

We have $f_{k}(x_{k}\rightarrow(x_{n},x_{n+1}))<|I_{0}|\leq f_{k}({\bm{x}})$.

Proof of $(\mathrm{ii})$

1. (a)

$k\neq 1$ and $k\neq n$

1. (1)

If $x_{k-1}\neq x_{k}$ and $x_{k}\neq x_{k+1}$, we have $f_{k}(x_{k}\rightarrow(x_{k-1},x_{k+1}))=f_{k}({\bm{x}})$; 2. (2)

If $x_{k-1}=x_{k}\;(x_{k}\neq x_{k+1})$, since $|I_{k}|=2|I_{0}|$ from (1.8) and $|I_{k-1}|=0$, we have $f_{k}(x_{k}\rightarrow(x_{k-1},x_{k+1}))=\dfrac{1}{2}\cdot 2|I_{0}|=|I_{0}|\leq f_{k}({\bm{x}})$; 3. (3)

If $x_{k}=x_{k+1}\quad(x_{k-1}\neq x_{k})$, we can deduce $f_{k}(x_{k}\rightarrow(x_{k-1},x_{k+1}))=|I_{0}|\leq f_{k}({\bm{x}})$ in a similar manner to (2). 2. (b)

$k=1$

We have $f_{1}(x_{1}\rightarrow(x_{0},x_{2}))<|I_{0}|\leq f_{1}({\bm{x}})$. 3. (c)

$k=n$

We have $f_{n}(x_{n}\rightarrow(x_{n-1},x_{n+1}))<|I_{0}|\leq f_{n}({\bm{x}})$.

Proof of $(\mathrm{iii})$

1. (a)

$l\neq 0,1,2,n-1,n,n+1$

1. (1)

$l=k$

It clearly holds that $f_{k}(x_{k}\rightarrow\{x_{l}\})=f_{k}({\bm{x}})$. 2. (2)

$l=k-1,k+1$ and there exists another vendor at $x_{k}$ except for vendor $k$, or $l\neq k-1,k,k+1$

We deduce from $\eqref{eqn:main2}_{1}$ that $f_{k}(x_{k}\rightarrow\{x_{l}\})\leq\dfrac{1}{4}\cdot 4|I_{0}|=|I_{0}|\leq f_{k}({\bm{x}})$. 3. (3)

$l=k-1,k+1$ and there exists no vendor at $x_{k}$ except for vendor $k$.

We deduce from $\eqref{eqn:main2}_{1}$ that $f_{k}(x_{k}\rightarrow\{x_{l}\})\leq\dfrac{1}{4}(2|I_{0}|+2f_{k}({\bm{x}}))\leq\dfrac{1}{4}(2f_{k}({\bm{x}})+2f_{k}({\bm{x}}))=f_{k}({\bm{x}})$. 2. (b)

$l=0\;($resp.$\;l=n+1)$

We have $f_{k}(x_{k}\rightarrow\{x_{0}\})=\dfrac{1}{2}|I_{0}|<|I_{0}|\leq f_{k}({\bm{x}}).$

(resp.$\;f_{k}(x_{k}\rightarrow\{x_{n+1}\})=\dfrac{1}{2}|I_{0}|<|I_{0}|\leq f_{k}({\bm{x}}))$ 3. (c)

$l=1,2\;($resp.$\;l=n-1,n)$

1. (1)

$k=3\;($resp.$\;k=n-2)$

We have $f_{3}(x_{3}\rightarrow\{x_{l}\})=\dfrac{1}{3}(|I_{0}|+\dfrac{1}{2}\left(|I_{2}|+|I_{3}|)\right)\leq\dfrac{1}{3}(|I_{0}|+\dfrac{1}{2}(2|I_{0}|+2|I_{0}|))=|I_{0}|\leq f_{3}({\bm{x}}).$

(resp.$\;f_{n-2}(x_{n-2}\rightarrow\{x_{l}\})=\dfrac{1}{3}(|I_{n}|+\dfrac{1}{2}(|I_{n-3}|+|I_{n-2}|)\leq f_{n-2}({\bm{x}}))$ 2. (2)

$k\neq 3\;($resp.$\;k\neq n-2)$

We have $f_{k}(x_{k}\rightarrow\{x_{l}\})\leq\dfrac{1}{2}(|I_{0}|+\dfrac{1}{2}|I_{2}|)=\dfrac{1}{2}(|I_{0}|+\dfrac{1}{2}\cdot 2|I_{0}|)=|I_{0}|\leq f_{k}({\bm{x}}).$

(resp.$\;f_{k}(x_{k}\rightarrow\{x_{l}\})\leq\dfrac{1}{2}(|I_{0}|+\dfrac{1}{2}|I_{n-2}|)\leq f_{k}({\bm{x}}))$

We can complete the proof of Lemma 2.4. ∎

Acknowledgements

The authors would like to thank Prof. Suzuki for his kind help and comments.