Applied Mathematical Sciences, Vol. 9, 2015, no. 103, 5117 - 5122
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2015.411954
An Equilibrium Theorem for Excess
Demand Correspondences
Christos E. Kountzakis
Department of Mathematics, University of the Aegean
Karlovassi, GR-83 200 Samos, Greece
c 2014 Christos E. Kountzakis. This article is distributed under the Creative
Copyright Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract
The aim of this brief paper is to give an equilibrium theorem for excess demand
correspondences in finite -commodity spaces, without necessarily lattice assumptions
for the cone of the partial ordering, and if the weak form of the Walras Law holds.
The proof relies both on the properties of the bases of cones in finite -dimensional
spaces and on the Browder Selection theorem.
Mathematics Subject Classification: 46B40; 91B50
Keywords: excess demand corespondence; fixed point; base of cone
1. Introduction
Equilibrium theorems for excess demand correspondences in finite -commodity
spaces is a common topic in mathematical economics, since Debreu’s first such
Theorem in [4]. In this paper we extend the equivalent result of M. Florenzano in [5], without necessarily lattice assumptions for the cone of the partial
ordering, without zero taking in account the zero price and by assuming the
same the weak form of the Walras Law. The proof relies both on the properties
of the bases of cones in finite -dimensional spaces and on both the Browder
Selection theorem [3, Th.1] and the Brouwer Fixed Point Theorem -see in [2].
We remark that the main point of the proof is the normalization of the prices
via a base of the cone C, which implies that C has to be generating (see also
Theorem 3.4), so that C 0 to be a cone. This implies that the ordering cone
C 0 of Rm as a commodity space has interior points.
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2. The main Theorem
Theorem 2.1. Consider a closed cone in Rm and a base of it Bf for some
strictly positive functional of it. Any upper hemicontinuous correspondence
m
z : Bf → 2R with non-empty, convex, compact values satifies the weak Walras
law:
If p ∈ Bf then there exists z ∈ z(p) such that p · z ≤ 0.
Then the set of equilibrium prices {p ∈ Bf |z(p) ∩ C 0 } is compact and nonempty.
Proof. The set of strictly positive functionals of C is non-empty in this case
(see Proposition 3.5 in the Appendix). Hence the existence of a strictly positive
functional f of C, that defies a closed, bounded base Bf of C is assured. Then,
we may follow the lines of the proof of the equilibrium theorem in [5], as it is
also presented in [1, Th.18.17], according to the Browder’s Selection Theorem
(see in [3, Th.1]). First of all the set of equilibrium prices is a closed subset of
the base Bf , from the upper -hemicontinuity of z. But since Bf is a compact
set of Rm , the set of equilibrium prices is also a compact set. In order to prove
that it is a non-empty set, we suppose that the set is empty, hence for any
p ∈ Bf , there exists a functional k 6= 0 such that
k · z > 0 ≥ k · g, z ∈ z(p), g ∈ C 0 ,
since C 0 is a wedge. Also, since k 6= 0, it may taken to lie on the base Bf .
m
Hence, a new correspondence is defined by K : Bf → 2R , such that
p 7→ K(p) = {h ∈ Bf |h · z > 0 > h · g, z ∈ z(p), g ∈ C 0 }.
We remind the Selection Theorem of Browder: Let E ⊂ Rn be a convex set
d
and r : E → 2R has convex values and r−1 (y) is open for any y. Then there
is a continuous selection f : E → Rd , such that f (x) ∈ r(x), x ∈ E. k has
non-empty, convex values for any p. Also, if p ∈ K −1 (h), this implies h · z >
0 > h · g, z ∈ z(p), g ∈ C 0 , where also we may define another correspondence
m
φ : Bf → 2R , with φ(p) = C 0 , p ∈ Bf . Since z is upper hemicontinuous
and the correspondence φ(p) = C 0 , p ∈ Bf is also upper hemicontinuous, the
set k u (z : h · z > 0) ∩ φu (g : h · g < 0) is a neighborhood of p, contained in
K −1 (p). Hence, by the Browder’s Selection Theorem, a continuous selection
k(p) ∈ K(p) exists, such that
k(p) · z > 0 > k(p) · g, z ∈ z(p), g ∈ C 0 , p ∈ Bf .
From Brouwer Fixed Point Theorem, there exists some fixed point of the continuous function k : Bf → Bf , since Bf is closed and bounded, namely compact. For this fixed point p0 such that p0 = k(p0 ), p0 · z > 0 for any z ∈ z(p0 ),
a contradiction, because weak Walras law is satisfied by z for any p ∈ Bf .
Hence the set of equilibrium prices is non-empty.
An equilibrium theorem for excess demand correspondences
5119
3. Appendix
In this Section we provide some essential notions about ordered spaces (see
also in [6]). Let E be some vector space, for example some Euclidean space
Rm . A set C ⊆ E satisfying C + C ⊆ C and λC ⊆ C for any λ ∈ R+ is called
wedge. A wedge for which C ∩ (−C) = {0} is called cone. If ≥ is a binary
relation on E satisfying the following properties:
(i) x ≥ x for any x ∈ E (reflexive)
(ii) If x ≥ y and y ≥ z then x ≥ z, where x, y, z ∈ E (transitive)
(iii) If x ≥ y then λx ≥ λy for any λ ∈ R+ and x + z ≥ y + z for any z ∈ E,
where x, y ∈ E (compatible with the linear structure of E),
then (E, ≥) is a partially ordered linear space. The binary relation ≥ in this
case is a partial ordering on Rm . The set P = {x ∈ E|x ≥ 0} is called
(positive) wedge of the partial ordering ≥ of E. Given a wedge C in E, the
binary relation ≥C defined as follows:
x ≥C y ⇐⇒ x − y ∈ C,
is a partial ordering on E, called partial ordering induced by C on E. If the
partial ordering ≥ of the space E is antisymmetric, namely if x ≥ y and y ≥ x
implies x = y, where x, y ∈ E, then P is a cone. If C is a cone, then a set
B ⊆ C is called base of C if for any x ∈ C \ {0} there exists a unique λx > 0
such that λx x ∈ B. The set Bf = {x ∈ C|f (x) = 1} where f is a strictly
positive functional of C is the base of C defined by f .
The partially ordered vector space E is a vector lattice if for any x, y ∈ E,
the supremum and the infimum of {x, y} with respect to the partial ordering
defined by P exist in E. In this case sup{x, y} and inf{x, y} are denoted by
x∨y, x∧y respectively. If so, |x| = sup{x, −x} is the absolute value of x and if
E is also a normed space such that k |x| k = kxk for any x ∈ E, then E is called
normed lattice. If a normed lattice is a Banach space, then it is called Banach
lattice. If F is a subspace of a vector lattice E and the partial ordering induced
on F by the cone F+ = F ∩E+ makes F a vector lattice, then F is called latticesubspace. Then for any x, y ∈ F , supF {x, y} = x ∨F y, inf F {x, y} = x ∧F y
exist in F . Their relation to equivalent x ∨ y, x ∧ y ∈ E is the following:
x ∧F y ≤ x ∧ y ≤ x ∨ y ≤ x ∨F y,
in terms of the partial ordering of E.
Relying on [7, Pr.1.1], [7, Th.3.6], it is known that the whole Rm has a
m
unique lattice-ordering, the one which implied by Rm
+ = {x ∈ R |x(i) ≥ 0, i =
1, 2, ...m}.
Lemma 3.1. If K is a wedge of Rm , then K − K is a subspace of Rm .
Proof. It suffices to prove that if x, y ∈ K − K, then x + y ∈ K − K and
λ·x ∈ K −K for any λ ∈ R. If x, y ∈ K −K, then there exist x1 , x2 , y1 , y2 ∈ K,
such that x = x1 − x2 , y = y1 − y2 . Since K is a wedge, x1 + y1 , x2 + y2 ∈ K,
hence x + y = (x1 + y1 ) − (x2 + y2 ) ∈ K − K. Moreover, if λ ≥ 0, since K is
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Christos E. Kountzakis
a wedge, λx1 , λx2 ∈ K, hence λx = λx1 − λx2 ∈ K − K. On the other side, if
λ < 0, −λ > 0 and this implies (−λ)x2 ∈ K, λx1 = −(−λx1 ) ∈ (−K), hence
λx = (−λ)x2 − (−λ)x1 ∈ K − K.
Definition 3.2. The wedge K generates Rm (or else K is generating) if
and only if K − K = Rm .
Definition 3.3. The dual wedge K 0 of K is the following subset of Rm :
K 0 = {y ∈ Rm |y · x ≥ 0, x ∈ K}.
We also remind of the following essential
Theorem 3.4. (Bipolar Theorem)
(i) If K is generates Rm , then K 0 is a cone.
(ii) K 00 = (K 0 )0 = K.
(i) Consider some y 6= 0 such that y ∈ K 0 ∩ (−K 0 ). Namely, there
exists some y 6= 0, such that y · x ≥ 0 for any x ∈ K and (−y) · x ≥ 0,
for any x ∈ K. Hence, y · x = 0, for any x ∈ K and y · h = 0, for any
h ∈ K − K. But since (K − K) ⊕ (K − K)⊥ = Rm and y 6= 0, this
implies dim(K − K)⊥ ≥ 1, a contradiction since in this case K − K
does not generate Rm .
(ii) We notice that K ⊆ K 00 . Since K ◦◦ is closed, K ⊆ K 00 Consider
some x ∈ K ◦◦ \ K. Then from Convex Sets’ Separation Theorem, there
exists some f 6= 0 strictly separating {x}, K, such that f (x) < 0 and
f (k) ≥ 0 for any k ∈ K, since K is a cone. But this is a contradiction,
because we suppoed that x ∈ K 00 , hence f (x) ≥ 0, since f ∈ K 0 .
Proof.
The elements of K 0 are called positive functionals of K. If for some
y ∈ K 0 , y · x > 0 for any x ∈ K \ {0}, then y is called strictly positive
functional of K. The set of strictly positive functionals of K is denoted by
0
K.
Proposition 3.5. (This proof was given by I.A. Polyrakis) The set of strictly
0
positive functionals K of K, where K is a closed cone of Rm is non-empty.
Proof. Since K ∩ (−K) = {0}, hence x ∈ K, kxk = 1, we have x ∈
/ −K. Since
K is closed, the set S+ = {x ∈ K|kxk = 1} is compact. If xi ∈ (−K), xi 6= 0,
then there exist ri > 0 such that B(xi , ri ) ∩ K = ∅, where B(xi , ri ) is the
open ball (under any norm in Rm ), being centered at xi whose radius equal
to ri , where i ∈ I. Consider the set S− = {x ∈ −K|kxk = 1}. The set
F = {B(xi , r2i ), i ∈ I} is an open cover of S− which is compact. Hence there is
a finite subset of F , being also a cover of S− . Namely, there exist xi ∈ S− , i =
1, 2, ..., k such that S− ⊆ ∪ki=1 B(xi , r2i . If Di is the closed ball centered at xi ,
whose radius is equal to r2i , then Di ∩ K = ∅, i = 1, 2, ..., k. Then from the
Convex Sets’ Separation Theorem, a yi ∈ Rm , yi 6= 0 exists such that strictly
An equilibrium theorem for excess demand correspondences
5121
separates K and Di , i = 1, 2, ..., k. Then yi · z < ai < yi · x, for any z ∈ Di and
any x ∈ K. We notice that yi · z < 0 for any z ∈ Di , since 0 ∈ K and moreover
yi · x ≥ 0, for any x ∈ K. This holds because if we suppose that yi · x < 0
for some x ∈ K, then yi · λx → −∞, if λ → −∞, which is a contradiction
since the values of yi on K dominate its values on Di . Consider the functional
y = y1 + y2 + ... + yk . Then yi is positive on K and since yi · x < 0, x ∈ Di ,
consequently y · x < 0, x ∈ ∪ki=1 B(xi , r2i ). Finally y · x < 0, x ∈ S− , hence
y · x > 0, x ∈ S+ , which implies y · x > 0, x ∈ K \ {0}.
0
Proposition 3.6. For any f ∈ K , the set Bf = {x ∈ K|f (x) = 1} is a base
of the cone K.
Proof. Let x ∈ K \ {0}. Since f is a strictly positive functional of K, f (x) > 0.
1
Then f (x)
x ∈ Bf , hence the set Bf is actually a base of the cone K.
Theorem 3.7. ([8, Pr.3])
Every strictly positive functional of a closed cone K of Rm , defines a bounded
base on K.
Proof. Consider a sequence {xn }n∈N of a base B = {c ∈ Rm |y · c = 1} which
is unbounded. Then, vn = kxxnn k , where kvn k = 1 for any n, hence vn has a
convergent subsequence, denoted again by vn . vn → v, kvk = 1, while since
y·xn
K is closed, v ∈ K. On the other hand, y(vn ) = kx
= kx1n k → 0, while
nk
from the continuity of the evaluation map (the usual inner product) we obtain
y ·vn → y ·v, which implies y ·v = 0, which is a contradiction because v 6= 0 and
y is a strictly positive functional of K. Hence such an unbounded sequence
does not exist for any strictly positive functional of K.
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Received: November 25, 2014; Published: August 1, 2015