Necessary and sufficient condition for equilibrium of the Hotelling model
aa r X i v : . [ ec on . T H ] J u l NECESSARY AND SUFFICIENT CONDITION FOREQUILIBRIUM OF THE HOTELLING MODEL
SATOSHI HAYASHI † AND NAOKI TSUGE ‡ Abstract.
We study a model of vendors competing to sell a homogeneousproduct to customers spread evenly along a linear city. This model is basedon Hotelling’s celebrated paper in 1929. Our aim in this paper is to presenta necessary and sufficient condition for the equilibrium. This yields a repre-sentation for the equilibrium. To achieve this, we first formulate the modelmathematically. Next, we prove that the condition holds if and only if vendorsare equilibrium. Introduction
We study a model in which a linear city of length 1 on a line and customers areuniformly distributed with density 1 along this interval. We consider n vendorsmoving on this line. Let the location of the vendor k ( k = 1 , , , . . . , n ) be x k ∈ [0 , x ≤ x ≤ · · · ≤ x n and denote the location of n vendors ( x , x , x , . . . , x n ) by x . Since we study the competition between vendors,we consider n ≥ l ( l = 1 , , , . . . , n ) vendors nearest to a customer, the customerpurchases 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 mathematicalnotation. Given a vector ξ = ( ξ , ξ , . . . , ξ n ) ∈ [0 , n and 0 ≤ y ≤
1, we define a set S ( ξ , y ) = { j ∈ { , , , . . . , n } : | ξ j − y | = min i | ξ i − y |} . By using a density function ρ k ( ξ , y ) = | ξ k − y | > min i | ξ i − y | | S ( ξ , y ) | if | ξ k − y | = min i | ξ i − y | , we define f k ( ξ ) = Z ρ k ( ξ , y ) dy, (1.1)where | A | represents a number of elements in a set A . We call f k ( x ) the profit ofvendor k per unit of time for a location x . We then define equilibrium as follows. Key words and phrases.
The Hotelling model, equilibrium, Mathematical formulation.N. Tsuge’s research is partially supported by Grant-in-Aid for Scientific Research (C)17K05315, Japan. † AND NAOKI TSUGE ‡ Definition 1.1.
A location x ∗ = ( x ∗ , x ∗ , x ∗ , . . . , x ∗ n ) ∈ [0 , n is called equilibrium,if f k ( x ∗ ) ≥ f k ( x ∗ , x ∗ , x ∗ , . . . , x ∗ k − , x k , x ∗ k +1 , . . . , x ∗ n ) (1.2) holds for any k ∈ { , , , . . . , n } and x k ∈ [0 , . We review the known results. The present model is based on Hotelling’s modelin [4]. Although we consider homogeneous vendors, Hotelling did heterogeneousvendors. In [1, Chaper 10], Alonso, W. treated with the same model as our prob-lem for two vendors. He introduced this model as the competition between twovendors 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 sufficientcondition 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 aninteresting approach from the point of mathematical view, unfortunately, it seemsthat (1.i) and (1.ii) are not sufficient conditions. In fact, we consider a location x = (cid:18) , , , , , , , (cid:19) . Then, we find that f ( x ) = f ( x ) = f ( x ) = f ( x ) = 1 / , f ( x ) = f ( x ) = f ( x ) = f ( 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 ( x ) seems tobe called a market of firm k . In addition, we regard two pairs of peripheral firmsin (1.ii) of [3] as firms 1 , , x moves from 3 /
10 to 1 / x can obtain a profit 2 / /
20. In addition, the definition of their terminologiesseems not to be clear, such as market, peripheral, equilibrium, etc. Therefore, ourgoal in this paper is to formulate this model mathematically and present a revisednecessary and sufficient for equilibrium.For convenience, we set x = 0 , x n +1 = 1 and denote a interval [ x k , x k +1 ] by I k ( k = 0 , , , . . . , n ). Then our main theorem is as follows. Theorem 1.1. (i) n = 2 x = (cid:18) , (cid:19) is a unique equilibrium. (1.3)(ii) n = 3 There exits no equilibrium. (1.4)(iii) n ≥ x is equilibrium, if and only if the following conditions (1.5) and (1.6) hold. | I | = | I n | > n ≥ , | I | = | I n − | = 0 ( n ≥ , | I | : | I | = 1 : 2 , | I n − | : | I n | = 2 : 1 ( n ≥ , (1.5) | I j | ≤ | I | (0 ≤ j ≤ n ) , | I | ≤ | I k | + | I k +1 | (1 ≤ k ≤ n − . (1.6) OTELLING’S MODEL 3 Preliminary
In this section, we prepare some lemmas and a proposition to prove our maintheorem in a next section. We first consider the profit of i vendors which locate atone point. We have the following lemma. Lemma 2.1.
We consider a location x = ( x , x , x , . . . , x n ) with x ≤ x ≤ x ≤· · · ≤ x n . We assume that x l < x l +1 = · · · = x k = · · · = x l + i < x l + i +1 ( n ≥ , l ≥ , l + i + 1 ≤ n + 1 , ≤ i ≤ n ) . (i) If x l = x and x l + i +1 = x n +1 , f k ( x ) = 12 i ( | I l | + | I l + i | ) . (ii) If x l = x and x l + i +1 = x n +1 , f k ( x ) = 1 i (cid:18) | I l | + 12 | I l + i | (cid:19) . (iii) If x l = x and x l + i +1 = x n +1 , f k ( x ) = 1 i (cid:18) | I l | + | I l + i | (cid:19) . (iv) If x l = x and x l + i +1 = x n +1 , f k ( x ) = 1 n .Proof.Proof of (i)We have f k ( x ) = 1 i (cid:18) | I l | + 12 | I l + i | (cid:19) = 12 i ( | I l | + | I l + i | ). Proof of (ii)We have f k ( x ) = 1 i (cid:18) | I l | + 12 | I l + i | (cid:19) = 1 i (cid:18) | I l | + 12 | I l + i | (cid:19) . Proof of (iii)We have f k ( x ) = 1 i (cid:18) | I l | + | I l + i | (cid:19) = 1 i (cid:18) | I l | + | I l + i | (cid:19) . Proof of (iv)We have f k ( x ) = 1 n ( | I | + | I n | ) = 1 n . (cid:3) Next, the following proposition play an important role.
Proposition 2.2.
If the location of n vendors x = ( x , x , x , . . . , x n ) with x ≤ x ≤ x ≤ · · · ≤ x n ( n ≥ is equilibrium, the following holds. x = 0 and x n = 1 . (2.1) No more than 2 vendors can occupy a location. (2.2) x = x and x n − = x n . (2.3) Proof.Proof of (2.1)If x = 0, we show that x is not equilibrium.(i) x = 0 and x = 0 We notice that f ( x ) = 12 | I | . Setting x ′ = 12 | I | , wethen have f ( x ′ , x , · · · , x n ) = | [0 , x ′ ] | + 12 | [ x ′ , x ] | > | [0 , x ′ ] | > | I | = f ( x ) , where | I | represents the length of a interval I . SATOSHI HAYASHI † AND NAOKI TSUGE ‡ (ii) x = · · · = x i = 0 and x i +1 = 0 (2 ≤ i ≤ n − f ( x ) = 12 i | I i | . Setting x ′ ∈ (0 , x i +1 ), we have f ( x ′ , x , · · · , x n ) = 12 ( | [0 , x ′ ] | + | [ x ′ , x i +1 ] | ) = 12 | [0 , x i +1 ] | = 12 | I i | > i | I i | = f ( x ) . (iii) x = · · · = x n = 0 ( n ≥ f ( x ) = 1 n . Setting x ′ = 12 , we have f ( x ′ , x , · · · , x n ) = 12 (cid:12)(cid:12)(cid:12)(cid:12)(cid:20) , (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:20) , (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) = 14 + 12 > ≥ n = f ( x ) . From (i)–(iii), if x = 0, we have proved that x is not equilibrium. Furthermore,from the symmetry, if x n = 1, we can similarly prove that x is not equilibrium. Proof of (2.2)We prove that x is not equilibrium, provided that i (3 ≤ i ≤ n ) vendors occupyat a point. We assume that x l < x l +1 = · · · = x k = · · · = x l + i < x l + i +1 ( l ≥ , l + i + 1 ≤ n + 1 , ≤ i ≤ n ). Here we recall that we set x = 0 and x n +1 = 1. Therefore there exist x l and x l + i +1 at least one respectively. We noticethat 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).(i) x l = x and x l + i +1 = x n +1 In this case, if | I l | ≥ | I l + i | , we notice that f k ( x ) = 12 i ( | I l | + | I l + i | ) ≤
16 ( | I l | + | I l + i | ) = 16 ( | I l | + | I l | ) = 13 | I l | . Setting x ′ k ∈ ( x l , x l +1 ), we have f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) = 12 ( | [ x l , x ′ k ] | + | [ x ′ k , x k ] | ) = 12 | I l | > f k ( x ).For the other case | I l | < | I l + i | , from the symmetry, we can similarlyshow that there exists x ′ k such that f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) >f k ( x ). Thus, if x l = x and x l + i +1 = x n +1 , we can prove that x is notequilibrium.(ii) x l = x and x l + i +1 = x n +1 In this case, if | I l | ≥ | I l + i | , we find that f k ( x ) = 1 i ( | I l | + 12 | I l + i | ) ≤
13 ( | I l | + 12 | I l + i | ) ≤
13 ( | I l | + 12 | I l | ) = 12 | I l | . Setting x ′ k = 23 | I l | , we have f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) = | [0 , x ′ k ] | + 12 | [ x ′ k , x k ] | > | [0 , x ′ k ] | = 23 | I l | ≥ f k ( x ) . For the other case | I l | < | I l + i | , from the symmetry, we can similarly showthat there exists x ′ k such that f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) > f k ( x ).Thus, if x l = x and x l + i +1 = x n +1 , we have showed that x is not equi-librium.(iii) x l = x and x l + i +1 = x n +1 From the symmetry of (ii), there exists x ′ k satisfying f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) > f k ( x ). Thus if x l = x and x l + i +1 = x n +1 , we have showed that x is not equilibrium. OTELLING’S MODEL 5 (iv) x l = x and x l + i +1 = x n +1 In this case, we find f k ( x ) = 1 n . If x k ≥
12 , for x ′ k = 25 , we have f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) = | [0 , x ′ k ] | + 12 | [ x ′ k , x k ] | > | [0 , x ′ k ] | = 25 ≥ n = f k ( x ) . For the other case x k <
12 , from the symmetry, we can similarly showthat there exists x ′ k such that f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) > f k ( x ).Thus, if x l = x and x l + i +1 = x n +1 , we have showed that x is not equi-librium.From (i)–(iv), we can complete the proof of (2.2). Proof of (2.3)If x < x , we show that x is not equilibrium. ( x = 0)We notice that f ( x ) = | I | + 12 | I | . Setting x ′ = | I | + 12 | I | , we have f ( x ′ , x , · · · , x n ) = | [0 , x ′ ] | + 12 | [ x ′ , x ] | > | [0 , x ′ ] | = | I | + 12 | I | = f ( x ). Thus x is not equilibrium. If x n − < x n , we can similarly prove that x is not equilibrium( x n = 1). (cid:3) Finally, we compare a location after the movement of a vendor with the originalone. To do this, we introduce the following notation.For a given location of n vendors x = ( x , x , x , . . . , x n ) with x ≤ x ≤ · · · ≤ x n , we move vendor k from x k to a point in A ⊂ [0 , x k → A . We notice that x k → A represents the following vector( x , x , , . . . , x k − , x ′ k , x k +1 , . . . , x n ) , (2.4)where x ′ k is a location of vendor k after movement and x ′ k ∈ 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 x = ( x , x , x , . . . , x n ) satisfies (1.5) and (1.6) , | I | ≤ f k ( x ) (1 ≤ k ≤ n ) . Lemma 2.4.
If a location of n vendors x = ( x , x , x , . . . , x n ) satisfies (1.5) and (1.6) , f k ( x k → [0 , ≤ f k ( x ) (1 ≤ k ≤ n ) . Proof of Theorem 1.1
We are now position to prove our main theorem.
Proof of Theorem 1.1 (i)
We prove that if x = ( x , x ) = (cid:18) , (cid:19) , x is equilibrium. SATOSHI HAYASHI † AND NAOKI TSUGE ‡ Proof. (i) We consider vendor 1. We then notice that f ( x ) = 12 . For any x ′ ∈ (cid:20) , (cid:19) , we have f ( x ′ , x ) = | [0 , x ′ ] | + 12 | [ x ′ , x ] | < | [0 , x ′ ] | + | [ x ′ , x ] | = | [0 , x ] | = 12 = f ( x ) . For any x ′ ∈ (cid:18) , (cid:21) , from the symmetry of (a), we can similarly provethat f ( x ′ , x ) < f ( x ). Therefore, for any x ′ ∈ [0 , f ( x ) ≥ f ( x ′ , x ).(ii) Next, we consider vendor 2. For any x ′ ∈ [0 , f ( x ) ≥ f ( x , x ′ ) in a similar manner to (i).From (i) and (ii), we have showed that (cid:18) , (cid:19) is equilibrium. (cid:3) Next, we show that x = (cid:18) , (cid:19) is a necessary condition for equilibrium. There-fore, we prove that if x = (cid:18) , (cid:19) , then x is not equilibrium. Proof.
From (2.1), when x = 0 or x = 1, x is not equilibrium. Therefore, wetreat with the case where x = 0 or x = 1.(i) x = x From (2.3), x is not equilibrium in this case.(ii) x = x We first notice that f ( x ) = 12 in this case. If x >
12 , setting x ′ = 12 ,we obtain f ( x ′ , x ) = | [0 , x ′ ] | + 12 | [ x ′ , x ] | > | [0 , x ′ ] | = 12 = f ( x ) . If x <
12 , from symmetry, we can show that there exits x ′ ∈ [0 ,
1] suchthat f ( x ′ , x ) > f ( x ).From the above, we have proved that (cid:18) , (cid:19) is a necessary condition for equilib-rium. (cid:3) Proof of Theorem 1.1 (ii) If n = 3, (2.2) contradicts (2.3). Therefore, we conclude that there exists noequilibrium in this case. Proof of Theorem 1.1 (iii)
Finally, we are concerned with the case where n ≥ Proof.
First, it follows from Lemma 2.4 that (1.5) and (1.6) is a sufficient conditionfor equilibrium.Next, we show that (1.5)–(1.6) is a necessary condition for equilibrium. There-fore, we prove that if (1.5)–(1.6) do not hold, then x is not equilibrium. ObservingProposition 2.2, we do not have to treat with the case where x = 0 or x n = 1 or OTELLING’S MODEL 7 more than 2 vendors occupy a location. From this reason, we assume that | I | > | I | 6 = 0 (resp. | I n − | 6 = 0).(b) | I | = | I n − | = 0 and | I | : | I | 6 = 1 : 2 (resp. | I | = | I n − | = 0 and | I n − | : | I n | 6 = 2 : 1).(c) | I | = | I n − | = 0 and | I | : | I | = 1 : 2 and | I n − | : | I n | = 2 : 1 and | I | 6 = | I n | .(ii) Condition (1.5) holds and condition (1.6) does not hold.(a) Condition (1.5) holds and | I j | > | I | .(b) Condition (1.5) holds and | I j | ≤ | I | and 2 | I | > | I k | + | I k +1 | .Dividing this proof into the above cases, we prove our main theorem.(i) (a) | I | 6 = 0 (resp. | I n − | 6 = 0)From (2.3) and x < x (resp. x n − < x n ), x is not equilibrium.(b) | I | = | I n − | = 0 and | I | : | I | 6 = 1 : 2 (resp. | I | = | I n − | = 0 and | I n − | : | I n | 6 = 2 : 1)(1) | I | : | I | = | I | : (2 | I | + δ ) ( δ > f ( x ) = 12 (cid:26) | I | + 12 (2 | I | + δ ) (cid:27) = | I | + 14 δ .Setting x ′ ∈ ( x , x ), we have f ( x ′ , x , · · · , x n ) = 12 (2 | I | + δ ) = | I | + 12 δ > | I | + 14 δ = f ( x ) . (2) | I | : | I | = | I | : (2 | I | − δ ) ( δ > f ( x ) = 12 (cid:26) | I | + 12 (2 | I | − δ ) (cid:27) = | I | − δ .Setting x ′ = | I | − δ , we have f ( x ′ , x , · · · , x n ) = | [0 , x ′ ] | + 12 | [ x ′ , x ] | > | [0 , x ′ ] | = | I | − δ = f ( x ) . From the symmetry of (1)–(2), we can similarly show in the casewhere | I n − | : | I n | 6 = 2 : 1. Thus, x is not equilibrium in the case of(b).(c) | I | = | I n − | = 0 and | I | : | I | = 1 : 2 and | I n − | : | I n | = 2 : 1 and | I | 6 = | I n | (1) | I | < | I n | We notice that f ( x ) = 12 (cid:18) | I | + 12 · | I | (cid:19) = | I | . Setting x ′ = 1 − | I | , we have f ( x ′ , x , · · · , x n ) = 12 | [ x n , x ′ ] | + | [ x ′ , | > | [ x ′ , | = | I | = f ( x ) . (2) | I | > | I n | From the symmetry, we can similarly show that there exists x ′ n such that f n ( x , · · · , x n − , x ′ n ) > f n ( x ).From (1)–(2), x is not equilibrium in the case of (c).(ii) Condition (1.5) holds and condition (1.6) does not hold. SATOSHI HAYASHI † AND NAOKI TSUGE ‡ (a) (1.5) holds and | I j | > | I | We notice that f ( x ) = 12 (cid:18) | I | + 12 · | I | (cid:19) = | I | . Setting x ′ ∈ ( x j , x j +1 ), we have f ( x ′ , x , · · · , x n ) = 12 | I j | > · | I | = | I | = f ( x ).Thus, x is not equilibrium in this case.(b) (1.5) holds and | I j | ≤ | I | and 2 | I | > | I k | + | I k +1 | When k = 1 , , n − , n , we notice that | I | + | I | = 2 | I | and | I | + | I | ≥ | I | . Thus, we devote to considering 3 ≤ k ≤ n −
2. In view of (2.2),we divide (b) into the following three parts.(1) x k − = x k and x k = x k +1 We notice that f k ( x ) = 12 ( | I k | + | I k +1 | ) < · | I | = | I | . Setting x ′ k ∈ ( x , x ), we have f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) = 12 · | I | = | I | > f k ( x ). Thus x is not equilibrium in this case.(2) x k − = x k ( x k − = x k − and x k = x k +1 )From | I k | + | I k +1 | < | I | , we find | I k | < | I | . From | I j | ≤ | I | ,we notice that | I k − | ≤ | I | . It follows that f k ( x ) = 14 ( | I k − | + | I k | ) <
14 (2 | I | + 2 | I | ) = | I | . Therefore, for x ′ k ∈ ( x , x ), wehave f k ( x , · · · , x k − , x ′ k , x k +1 , · · · , x n ) = 12 · | I | = | I | > f k ( x ).This means that x is not equilibrium in this case.(3) x k = x k +1 ( x k − = x k and x k +1 = x k +2 )We can prove this case in a similar manner to (2).From (i)–(ii), we have showed that if condition (1.5) or condition (1.6) do nothold, then x is not equilibrium.We can complete the proof of Theorem 1.1. (cid:3) Appendix A. Proof of Lemma 2.3
Proof.
We estimate the profit of each vendor, f k ( x ) ( k = 1 , , . . . , n ).(i) Vendor 1We have f ( x ) = 12 ( | I | + 12 · | I | ) = | I | .(ii) Vendor 2We can similarly deduce f ( x ) = | I | .(iii) Vendor n − n From the symmetry with vendors 1 and 2, we have f n − ( x ) = f n ( x ) = | I | .(iv) Vendor k (3 ≤ k ≤ n − x k − = x k and x k = x k +1 We have f k ( x ) = 12 ( | I k − | + | I k | ) ≥ · | I | = | I | .(b) x k − = x k , x k = x k +1 ( k = 3)From (1.6), we notice that x k − = x k − . On the other hand, from | I k − | = 0 and (1.6), we have | I k − | ≤ | I | and 2 | I | ≤ | I k − | . Thus OTELLING’S MODEL 9 we have | I k − | = 2 | I | . Similarly, we have | I k | = 2 | I | . As a conse-quence, we have f k ( x ) = 14 ( | I k − | + | I k | ) = 14 · | I | = | I | .(c) x k − = x k , x k = x k +1 ( k = n − f k ( x ) = | I | in a similar manner to (b).Combining (i)–(iv), we obtain f ( x ) = f ( x ) = f n − ( x ) = f n ( x ) = | I | .f k ( x ) = | I | if x k − = x k or x k = x k +1 (3 ≤ k ≤ n − .f k ( x ) ≥ | I | if x k − = x k and x k = x k +1 (3 ≤ k ≤ n − . Thus, for x satisfying (1.5) and (1.6) and any k (1 ≤ k ≤ n ), we have showed that f k ( x ) ≥ | I | . (cid:3) Appendix B. Proof of Lemma 2.4
Proof.
Dividing the proof into three parts, we prove this lemma.(i) f k ( x k → ( x l , x l +1 )) ≤ f k ( x ) ( x l < x l +1 and l = k − , k and 0 ≤ l ≤ n ).(ii) f k ( x k → ( x k − , x k +1 )) ≤ f k ( x ) (1 ≤ k ≤ n ).(iii) f k ( x k → { x l } ) ≤ f k ( x ) (1 ≤ k ≤ n and 0 ≤ l ≤ n + 1). Proof of (i)(a) l = 0 and l = n We have f k ( x k → ( x l , x l +1 )) = 12 | I l | ≤ | I | | I | ≤ f k ( x ).(b) l = 0We have f k ( x k → ( x , x )) < | I | ≤ f k ( x ).(c) l = n We have f k ( x k → ( x n , x n +1 )) < | I | ≤ f k ( x ). Proof of (ii)(a) k = 1 and k = n (1) If x k − = x k and x k = x k +1 , we have f k ( x k → ( x k − , x k +1 )) = f k ( x );(2) If x k − = x k ( x k = x k +1 ), since | I k | = 2 | I | from (1.6) and | I k − | = 0,we have f k ( x k → ( x k − , x k +1 )) = 12 · | I | = | I | ≤ f k ( x );(3) If x k = x k +1 ( x k − = x k ), we can deduce f k ( x k → ( x k − , x k +1 )) = | I | ≤ f k ( x ) in a similar manner to (2).(b) k = 1We have f ( x → ( x , x )) < | I | ≤ f ( x ).(c) k = n We have f n ( x n → ( x n − , x n +1 )) < | I | ≤ f n ( x ). Proof of (iii)(a) l = 0 , , , n − , n, n + 1(1) l = k It clearly holds that f k ( x k → { x l } ) = f k ( x ).(2) l = k − , k +1 and there exists another vendor at x k except for vendor k , or l = k − , k, k + 1 † AND NAOKI TSUGE ‡ We deduce from (1.6) that f k ( x k → { x l } ) ≤ · | I | = | I | ≤ f k ( x ).(3) l = k − , k + 1 and there exists no vendor at x k except for vendor k .We deduce from (1.6) that f k ( x k → { x l } ) ≤
14 (2 | I | + 2 f k ( x )) ≤
14 (2 f k ( x ) + 2 f k ( x )) = f k ( x ).(b) l = 0 (resp. l = n + 1)We have f k ( x k → { x } ) = 12 | I | < | I | ≤ f k ( x ) . (resp. f k ( x k → { x n +1 } ) = 12 | I | < | I | ≤ f k ( x ))(c) l = 1 , l = n − , n )(1) k = 3 (resp. k = n − f ( x → { x l } ) = 13 ( | I | + 12 ( | I | + | I | )) ≤
13 ( | I | + 12 (2 | I | +2 | I | )) = | I | ≤ f ( x ) . (resp. f n − ( x n − → { x l } ) = 13 ( | I n | + 12 ( | I n − | + | I n − | ) ≤ f n − ( x ))(2) k = 3 (resp. k = n − f k ( x k → { x l } ) ≤
12 ( | I | + 12 | I | ) = 12 ( | I | + 12 · | I | ) = | I | ≤ f k ( x ) . (resp. f k ( x k → { x l } ) ≤
12 ( | I | + 12 | I n − | ) ≤ f k ( x ))We can complete the proof of Lemma 2.4. (cid:3) Acknowledgements
The authors would like to thank Prof. Suzuki for his kind help and comments.
References [1] Alonso, W., 1964, Location theory, in J. Friedmann and A. W. Alonso (eds.), RegionalDevelopment and Planning: A Reader, Cambridge University Press (Cambridge)[2] Eaton, B. C. and Lipsey, R. G., 1972, The Principle of Minimum Differentiation Recon-sidered: Some New Developments in the Theory of Spatial Competition, Queen’s Institutefor Economic Research, Discussion Paper No. 87[3] Eaton, B. C. and Lipsey, R. G., 1975, The principle of minimum differentiation reconsidered: some new developments in the theory of spatial cometition, Review of Economic Studies,42, 27–49.[4] Hotelling, H., 1929, Stability in competition, Economic Journal, 39, 41-57. †‡ Department of Mathematics Education, Faculty of Education, Gifu University, 1-1Yanagido, Gifu Gifu 501-1193 Japan.
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