Applied Mathematical Sciences, Vol. 10, 2016, no. 9, 403 - 416
HIKARI Ltd, www.m-hikari.com
http://dx.doi.org/10.12988/ams.2016.512734
GQVIs for Studying Competitive Equilibrium
Problem when Utilities are Locally Lipschitz
and Quasi-Concave
F. Rania
Department of Legal, Historical, Economic and Social Sciences
Magna Graecia University of Catanzaro
Campus loc. Germaneto, Viale Europa, 88100 Catanzaro, Italy
c 2015 F. Rania. This article is distributed under the Creative Commons
Copyright Attribution License, which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
Abstract
We establish an existence result on competitive equilibrium problem for an exchange economy when the consumers’ utilities are represented by a locally Lipschitz continuous and quasi-concave functions.
The consumer’s demand is found to be actually a multi-valued map.
Furthermore, any competitive equilibrium satisfies Walras’ law, too. To
achieve this goal, the theory of Nonsmooth Analysis combined with the
Generalized Quasi-Variational Inequalities (GQVIs) is used.
Mathematics Subject Classification: 91B50, 65K10, 58E17, 58E35
Keywords: Competitive equilibrium, Walras’ law, generalized quasi- variational inequalities, Clarke subdifferential
1
Introduction
Starting from the papers of Benedetti et al. [6] and of Rockafellar et al. [10],
our purpose here is to generalize the existence result on Arrow-Debreu equilibrium problem for the pure exchange economies (see for details [4, 9]). At this
aim, we weak the assumptions on how the consumer expresses the preferences
404
F. Rania
when has to chose a commodity bundle among the feasible ones. Specifically, the utility functions are assumed locally Lipschitz continuous and quasiconcave. These facts allow to treat with utility functions not differentiable
(but however, existing in real contexts) and to obtain a multi-valued demand
function, because quasi-concavity is weaker than concavity and than strictly
concavity, too. Non-differentiability implies the use of the Clarke subdifferential (see [7]).
Our main results are Theorems 7 and 10, which derive by variational approach, i.e. by using generalized quasi-variational inequalities (briefly GQVI).
In Theorem 7, we show that any solution to a suitable GQVI is a solution to
competitive economic equilibrium problem. In Theorem 10, recalling a result
of Cubiotti in [8], we show that GQVI problem admits at least one solution.
Furthermore, under the assumptions of non-local satiation and global desirability defined in a suitable compact set, in Proposition 5 we show that any
competitive equilibrium is a Walras competitive equilibrium.
As said above our existence result on competitive equilibrium is general
for the pure exchange economies because it includes the results in [1, 2] and,
probability, also the ones in [3, 12] related to the dynamic case introduced in
[11, 13] if we replace the concavity condition with quasi-concavity condition.
This paper is organized as follows: in Section 2, we shall describe the
model, formulate the competitive economic equilibrium problem and list the
assumptions; in Section 3, we shall establish that a solution of the competitive
economic equilibrium problem is a solution to a generalized quasi-variational
inequality defined on a suitable compact; in Section 4, we shall prove the
existence of a solution to GQVI and thus, as consequence, the existence of a
competitive equilibrium for the pure exchange economy.
2
Model and assumptions
We consider a pure exchange economy E with ` finite types of commodities
indexed in H = {1, ..., `} and n consumers indexed in I = {1, ..., n} (where
`, n ∈ N).1 We regard R`+ as the commodity space. Each consumer i ∈ I is
1
We denote by N and R, the set of natural number and the field of real numbers, respectively. We write Rn for the n-dimensional Euclidean space of the real
◦ n
n-vectors x = (x1 , ..., xn ); while, we write Rn+ , Rn0+ and R0+ for the cone of nonnegative, positive
positive vectors, respectively. Furthermore, the set ∆n−1 =
Pn and strongly
x ∈ Rn+ :
x
=
1
indicates
the unit simplex of Rn+ . We adopt the usual notation
i
i=1
for vector inequalities, that is: for any x, y ∈ Rn one has x ≥ y if x − y ∈ Rn+ ; x > y if
◦
n
x − y ∈ Rn0+ ; and x y if x − y ∈R0+ . We assume Rn equipped with the standard inner
Pn
1
product hx, yi := i=1 xi yi and with the norm |x|n := (hx, xi) 2 for any x, y ∈ Rn . Open
n
and closed balls of centre x ∈ R with radius ε > 0 are denoted by Bε (x) and B ε (x) . Given
X ⊆ Rn , its interior is int(X), its boundary is ∂X and its closure is X.
405
GQVIs for competitive equilibrium problem
characterized by his consumption set X (i) ⊆ R`+ , his preferences described by
an utility function ui : X (i) →P
R and of his initial endowment ω (i) ∈ X (i) . The
aggregate endowment is ω:= i∈I ω (i) . A pure exchange economy is the tuple
E = {X (i) , ui , ω (i) }i∈I , ω . An allocation is denoted by x = (x(1) , ..., x(n) ) ∈
(i)
(i)
(i)
Rn×`
= (x1 , ..., x` ) ∈ X (i) for any i ∈ I. An allocation x is feasible
+P where x
if i∈I x(i) ≤ ω. By assuming the vector p = (p1 , ..., p` ) ∈ P := ∆`−1 as a
price system, for any initial endowment ω (i) the consumer i’s budget set for p
`
corresponding to ω (i) is given by the multifunction B (i) : P → 2R+ defined by
B (i) (p) = x ∈ R`+ : hp, x(i) i ≤ hp, ω (i) i .
A state of E is a pair (p, x) ∈ R`×n`
where p is a price system and x is a
+
feasibile allocation.
In a pure exchange economy E where the consumers are considered as price
takers, for a suitable price p̄, the problem of any consumer i is to obtain the best
commodity bundle x̄(i) into his budget set B (i) (p̄) in according to own utility
function ui and, contextually, all together the consumers have to respect the
equality Q
between demand and supply for any commodity h ∈ H. Formally, set
B(p) := i∈I B (i) (p) one has
n×`
Problem 1. Find (p̄, x̄) ∈ P × R+
with x ∈ B(p̄) such that:
ui (x̄(i) ) = max ui (x(i) )
x(i) ∈B(i) (p̄)
X (i)
(i)
x̄h − ωh ≤ 0
∀i ∈ I
(1a)
∀h ∈ H
(1b)
i∈I
Definition 1. A pure exchange economy E ’s state (p̄, x̄) solving Problem 1
is said to be a competitive equilibrium for the E .
Definition 2. A competitive equilibrium (p̄, x̄) is said to be a Walras competitive equilibrium if in addition to (1a) and (1b) satisfies also the following
condition
hp̄, x̄(i) − ω (i) i = 0
∀i ∈ I
(1c)
We list below the assumptions that will use to solve Problem 1.
For any h ∈ H:
Assumption 1. There exists a price qh such that qh ≥
ph 6= 0, where ph is the hth entry of a price system p ∈ P .
1
`
and qh ≤ ph for
For any i ∈ I:
P
Assumption 2. ω (i) is such that i∈I ω (i) 0 and contains at least one
good h ∈ H with price qh > 0 or greater ph > qh ;
406
F. Rania
Assumption 3. X (i) is a closed, convex subset of R`+ ;
Assumption 4. Consider an open convex A ⊇ X (i)
a. ui (x(i) ) is a local Lipschitz continuous function on A;2
b. ui (x(i) ) is a quasi-concave function on A;3
Set K (i) =
Y
"
#!
o Y
[n
X
(i)
(i)
(a)
x(i) ∈ X (i) : xh ≤ ωh ∩
0,
ωb
i∈I
h∈H
b∈H
a∈I
c. 0Rl ∈
/ ∂ ◦ (−ui )(x(i) ) for all x(i) ∈ K (i) ;4
(i)
d. (−ui )◦ (x(i) , ih ) < 0 for all x(i) ∈ K (i) and h ∈ H such that xh = 0 where ih
is the unit vector of the hth axis.
Remark 3. The Assumption 1 guarantees the existence of a positive minimum price qk for any commodity h ∈ H. This fact, in addition with the
Assumption 2, where the initial endowment ω (i) ∈ R`0+ , permits to any consumer i to be always active in the trades. The Assumption 3 is standard and
from ω (i) ∈ X (i) and above considerations one has in particular X (i) ⊂ R`+ .
The Assumption 4 regards the alternative commodity bundles which each consumer i choices on the basis of own utility function. In details: 4.a extends
the condition of continuity of the preference relation considering, now, not differentiable utilities, too; 4.b assures that the indifference surfaces are convex;
4.c represents the non-satiation condition rewritten in terms of Clark subdifferential; 4.d indicates how the global desirability of a commodity bundle x(i)
changes through the increasing of the Clarke derivatives along each direction.
A function f : X → R is said to be locally Lipschitz continuous on X ⊆ Rn if, for each
x ∈ X, there exist constants L > 0 and ε > 0 such that y ∈ Bε (x) implies |f (y) − f (x)| ≤
L|y − x|n .
3
A function f : X → R is said to be quasi-concave on a convex X ⊆ Rn iff for every
r < supX f , the set {x ∈ X : f (x) ≥ r} is convex.
4
Given an open A ⊆ Rn , a function f : A → R, x0 ∈ A and z ∈ Rn , the Clarke derivative
(x)
of f at x0 along the direction z is defined by f ◦ (x0 , z) := lim supx→x0 ,σ→0+ f (x+σz)−f
and
σ
◦
n
◦
the Clarke subdifferential of f at x0 is defined by ∂ f (x0 ) := {T ∈ R : f (x0 , z) ≥ hT, zi,
for all z ∈ Rn }.
Note that, from the definition, the Clarke subdifferential is a convex set. Furthermore,
if f is a locally Lipschitz continuous function the Clarke derivative is finite and the Clarke
subdifferential is compact in Rn expressed by f ◦ (x0 , z) = max{hT, zi : T ∈ ∂ ◦ f (x0 )}, for all
z ∈ Rn .
2
GQVIs for competitive equilibrium problem
3
407
Variational approach
Now, the competitive economic equilibrium problem for a pure exchange economy E is revisited in key of generalized quasi-variational inequality problem.5
n×`
Problem 2. Find Q
(p̄, x̄) ∈ P × R+
, with x ∈ B(p̄), such that there exists
(1)
(n)
◦
T = (T , ..., T ) ∈ i∈I ∂ (−ui )(x̄(i) ) satisfying:
X
X
−
hT (i) , x(i) − x̄(i) i+h
x̄(i) − ω (i) , p− p̄i ≤ 0, ∀ (p, x) ∈ P ×B(p̄). (2)
i∈I
Let K =
i∈I
Q
i∈I
Ki , where K (i) is as in Assumptions 4.c and 4.d.
n×`
Proposition 4. Let (p̄, x̄) ∈ P ×R+
, with x̄ ∈ B(p̄), be satisfying condition
(1b) of Problem 1. Then x̄ ∈ K.
Proof. See proof of Proposition 1 of [1].
Proposition 5. Let Assumptions 4.c and 4.d be satisfied. Then, any competitive equilibrium is a Walras competitive equilibrium.
Proof. Let (p̄, x̄) ∈ P × B(p̄) be a competitive equilibrium and fix i ∈ I. From
Proposition 4, we have x̄(i) ∈ K (i) . So, by Assumption 4.c and condition (1a)
of Problem 1 it cannot be x̄(i) ∈ int(B (i) (p̄)).
Moreover, Assumption 4.d and again condition (1a) of Problem 1 imply
(i)
that xh > 0 for all h ∈ H. Therefore, the equation (1c) is verified, or equivalently, (p̄, x̄) is a Walras competitive equilibrium.
For the next result we need the following Propositions.
Proposition 6. Let Assumptions 4.a and 4.b be satisfied. Let i ∈ I and let
x(i) , z (i) ∈ Rl+ be such that ui (x(i) ) < ui (z (i) ). Then, (−ui )◦ (x(i) , z (i) −x(i) ) ≤ 0.
Proof. Let {(y (n) , t(n) )} be a sequence in A × (0, 1) (the set A is as in Assumptions 4.a and 4.b) such that (y (n) , t(n) ) → (x(i) , 0). From Assumption
4.a (and hence from continuity of ui ) and ui (x(i) ) < ui (z (i) ), we can suppose
ui (y (n) ) < ui (z (i) ) for all n ∈ N. From Assumption 4.b, one has ui (y (n) ) ≤
ui (y (n) + t(n) (z (i) − y (n) )) for all n ∈ N. Consequently,
lim sup
n→∞
5
−ui (y (n) + t(n) (z (i) − y (n) )) + ui (y (n) )
≤ 0.
t(n)
Given a set X, we write 2X for the family of all non-empty subsets of X. A correspondence or a multifunction between two sets X and Y is a function F : X → 2Y . The graph
of a multifunction F : X → 2Y is the subset of X × X defined by gr(F ) = {(x, y) ∈ X × Y :
n
x ∈ X ∧ y ∈ F (X)}. Given X ⊆ Rn and two multifunctions Γ : X → 2X , Φ : X → 2R , the
generalized quasi-variational inequality problem associated to X, Γ, Φ and denoted briefly
GQVI(X, Γ, Φ), is to find (x̄, z̄) ∈ X × Rn such that x̄ ∈ Γ(x̄), z̄ ∈ Φ(x̄) and hz̄, x̄ − yi ≤ 0
for all y ∈ Γ(x̄).
408
F. Rania
Taking into account the arbitrariness of the sequence {(y (n) , t(n) )}, conclusion
follows.
Theorem 7. Let Assumption 1 and 4 entirely be satisfied. Moreover, let
(p̄, x̄) ∈ P × Rn×`
+ , with x̄ ∈ B(p̄), be a solution of Problem 2. Then, (p̄, x̄) is
a solution of Problem 1.
Proof. Assume
(p̄, x̄) is a solution to Problem 2.
Q that
◦
Let T ∈ i∈I ∂P(−ui )(x̄(i) ) satisfying inequality (2). Testing (2) with (p, x̄),
p ∈ P , one has h i∈I (x̄(i) − ω (i)
P), p − p̄i ≤ 0 for all p ∈ P. Moreover, from
x̄ ∈ B(p̄), we promptly obtain h i∈I (x̄(i) − ω (i) ), p̄i ≤ 0 for all p ∈ P. Hence,
h
X
X
X
(x̄(i) − ω (i) ), pi = h (x̄(i) − ω (i) ), p − p̄i + h (x̄(i) − ω (i) ), p̄i ≤ 0
i∈I
i∈I
i∈I
for all p ∈ P . Now, from Assumption 1 choosing p = (0, .., 0, 1, 0, ..., 0) ∈ P ,
with 1 at the hth -position, we obtain condition (1b) of Problem 1.
At this point, we prove condition (1a) of Problem 1. Fix i ∈ I. Testing (2)
with (p̄, (x̄(1) , .., x̄(i−1) , x(i) , x̄(i+1) , .., x̄(n) )), xi ∈ B (i) (p̄), we obtain
hT (i) , x(i) − x̄(i) i ≥ 0.
(3)
From Assumption 4.c, there exists h ∈ Rn such that hT (i) , hi 6= 0. Without
loss of generality, we can suppose hT (i) , hi < 0. Now, let z (i) ∈ B (i) (p̄) and
(i)
(i)
put yθ,1 = (1 − θ)(x̄(i) + h) + θz (i) and yθ,2 = (1 − θ)(x̄(i) + h) + θx̄(i) for all
θ ∈ (0, 1). Then, we have
(i)
hT (i) , x̄(i) − yθ,2 i = −(1 − θ)hT (i) , hi > 0
and, taking (3) into account,
(i)
hT (i) , yθ,1 − x̄(i) i ≥ 0
for all θ ∈ (0, 1). Adding side to side the above inequality, we obtain:
(i)
(i)
0 < hT (i) , yθ,1 − yθ,2 i = hT (i) , z (i) − x̄(i) i ≤ (−ui )◦ (x̄(i) , z (i) − x̄(i) ).
From the above inequality and Proposition 6, it follows ui (x̄(i) ) ≥ ui (z (i) ).
From the arbitrariness of z (i) ∈ B (i) (p̄), condition (1a) of Problem 1 follows.
Remark 8. Under the assumptions of Theorem 7, condition (1b) actually
holds as equality. Indeed, fix i ∈ I and define
X
(i)
(i)
gi (x(i) ) := −
ph (xh − ωh ) for all x(i) ∈ B (i) (p̄).
(4)
h∈H
409
GQVIs for competitive equilibrium problem
We claim that gi (x̄(i) ) = 0. Indeed, if not, taking in mind that x̄i ∈ B (i) (p̄, ȳ),
it should be gi (x̄(i) ) > 0. Then, for each h = 1, ..., l, there exists ρh > 0 such
that
x̄(ρ) := (x̄(1) , ...., x̄(i−1) , x̄(i) +ρih0 , x̄(i+1) , .., x̄(n) ) ∈ B(p̄) for all ρ ∈ [0, ρh [. (5)
Testing (2) with (p̄, x̄(ρ)), we obtain
ρhT̄ (i) , ih i ≤ 0, for all ρ ∈ [0, ρh [.
(6)
(i)
Thus, in view of Assumption 4.d, it must be x̄h > 0 for all h ∈ H. This
fact, together with gi (x̄(i) ) > 0, implies x̄(i) ∈ int(B (i) (p̄)). Testing (2) with
(p̄, x), where x = (x̄(1) , .., x̄(i−1) , x(i) , x̄(i+1) , .., x̄(n) ), being x(i) arbitrarily chosen
in B (i) (p̄), we obtain hT (i) , x(i) − x̄(i) i ≤ 0 for all x(i) ∈ B (i) (p̄). Since x̄(i) ∈
int(B (i) (p̄)), from this inequality it follows T (i) = θRl in contradiction with
Assumption 4.d.
4
Existence Results
Theorem 7 in Section 3 states that any solution to Problem 2 is a solution
to Problem 1. Thus, to find a solution of Problem 1, we will prove, in this
Section, that Problem 2 admits at least a solution.
First, we need the following Proposition:
`
Proposition 9. For any i ∈ I, the map ∂ ◦ (−ui ) : R`+ → 2R has closed
graph.
Proof. Fix i ∈ I. Let (x(i) , T (i) ) ∈ R`+ ×R` and let {(x(m) , T (m) )} be a sequence
in gr(∂ ◦ (−ui )) ⊂ R`+ × R` such that (x(m) , T (m) ) → (x(i) , T (i) ) ∈ R` × R` , as
m → +∞. Clearly, one has (x(i) , T (i) ) ∈ R`+ × R` . It remains to show that
(x(i) , T (i) ) ∈ gr(∂ ◦ (−ui )).
(7)
Fix z (i) ∈ R`+ . From (x(m) , T (m) ) ∈ gr(∂ ◦ (−ui )(x(m) , z (i) )), for all m ∈ N, one
has hT (m) , z (i) i ≤ (−ui )◦ (x(m) , z (i) ), for all m ∈ N. Then,
hT (i) , z (i) i = lim hT (m) , z (i) i = lim inf hT (m) , z (i) i ≤
m→+∞
m→+∞
◦
≤ lim inf (−ui ) (x
m→+∞
(m)
, z (i) ).
(8)
We claim that
lim inf (−ui )◦ (x(m) , z (i) ) ≤ (−ui )◦ (xi , z (i) ).
m→+∞
(9)
410
F. Rania
Indeed, assume that (9) does not hold. Then, it should exist m ∈ N and
M ∈ R such that
(−ui )◦ (x(m) , z (i) ) > M > (−ui )◦ (x(i) , z (i) ),
(10)
for all m ∈ N, with m ≥ m. From (10), for each fixed m ∈ N, with m ≥ m,
(m) (m)
we can find a sequence (yk , tk ) ∈ A × R+ , where A is as in Assumptions
(m) (m)
4.a and 4.b, such that (yk , tk ) → (x(m) , 0) as k → +∞, and
(m)
(−ui )(yk
(m)
(m)
+ tk z (i) ) − (−ui )(yk )
(m)
> M,
tk
for all k ∈ N. Moreover, again from (10), we can find δ > 0 such that
(−ui )(y (i) + tz (i) ) − (−ui )(y (i) )
< M,
t
for all y (i) ∈ A, with |y (i) −x(i) | < δ, and all t ∈]0, δ[. Thus, if we fix m ∈ N, with
(m)
m ≥ n, such that |x(m) −x(i) | < δ, we can find k 0 ∈ N such that |yk0 −x(i) | < δ
(m)
(m) (m)
and tk0 ∈]0, δ[. Then, the couple (yk0 , tk0 ) should satisfy
(m)
(m)
(m)
(−ui )(yk0 + tk0 z (i) ) − (−u(i) )(yk0 )
(m)
<M <
tk0
(m)
<
(m)
(m)
(−u(i) )(yk0 + tk0 z (i) ) − (−ui )(yk0 )
(m)
,
tk0
a contradiction. Therefore, inequality (9) holds. At this point, observe that
from inequalities (4) and (9) it easily follows that (x(i) , T (i) ) ∈ gr(∂ ◦ (−ui )).
Theorem 10. Let Assumptions 2, 4.a, 4.c, 4.d be satisfied. Then,
Problem 2 admits at least a solution in P × C × Y , where
)
(
X X (i) X X (i)
(i)
n×`
ωh .
C := x = xh
∈ R+
:
xh ≤
i∈I, h∈H
i∈I h∈H
i∈I h∈H
Proof. Let us divide the proof in several steps.
and define
Step 1. Put X = P × Rn×`
+
- u(x) = u1 (x(1) ), ..., un (x(n) ) , for all x = (x(1) , ..., x(n) ) ∈ Rn×`
+ ,
- Γ(p, x) = P × B(p), for all (p, x) ∈ X,
P
- Φ(p, x) = − i∈I (x(i) − ω (i) ), ∂ ◦ (−u)(x) , for all (p, x) ∈ X, where
n×`
∂ ◦ (−u)(x) = (∂ ◦ (−u1 )(x(1) ), ..., ∂ ◦ (−un )(x(n) )), for all x ∈ R+
.
GQVIs for competitive equilibrium problem
411
By means of these notations, we can rewrite Problem 2 and the variational
inequality (2) as follows:
n×`
Problem 2 bis. find (p̄, x̄) ∈ P × R+
, with (p̄, x̄) ∈ Γ(p̄, x̄), and (ẑ, T ) ∈
Φ(p̄, x̄) such that
h(ẑ, T ), (p̄, x̄) − (p, x)i ≤ 0, for all (p, x) ∈ Γ(p̄, x̄).
(11)
Step 2. Note that:
- the set X is nonempty closed and convex in R` × Rn×` ;
- the set K := P × C ⊂ X is nonempty and compact in R` × Rn×` ;
- Γ(p, x) is a nonempty convex subset of X, for all (p, x) ∈ X.
Moreover, recalling that ∂ ◦ (−ui )(x(i) ) is (nonempty) convex and compact in
R` , for all i ∈ I and for all x(i) ∈ R`+ , we also have that
- Φ(p, x) is a nonempty convex and compact subset of R` × Rn×` , for all
(p, x) ∈ X.
In the next Steps we check the conditions of Theorem 3.2 of [8]6 .
Step 3. Prove that the below condition holds true:
(a1 ) the set Λ(ρ, τ ) := (p, x) ∈ X :
inf
h(z, T ), (ρ, τ )i ≤ 0 is closed,
(z,T )∈Φ(p,x)
for each (ρ, τ ) ∈ X − X.
6
For convenience of reader we report below the following
Theorem 3.2 (of [8]). Let X be a closed convex subset of Rn , K ⊆ X a nonempty compact set,
n
Φ : X → 2R and Γ : X → 2X two multifunctions. Assume that:
(i) the set Φ(x) is convex for each x ∈ K, with x ∈ Γ(x);
(ii) the set Φ(x) is nonempty and compact for each x ∈ X;
(iii) for each y ∈ X − X, the set x ∈ X : inf z∈Φ(x) hz, yi ≤ 0 is closed;
(iv) Γ is a lower semicontinuous multifunction (i.e. {x ∈ X : Γ(x) ∩ A 6= ∅} is open in X, for each open
set A in X) with closed graph and convex values.
Moreover, assume that there exists an increasing sequence k of positive real numbers, with X ∩ B̄(0, 1 ) 6= ∅
and limk→∞ k = +∞ such that, if one puts Dk = B̄(0, k ), for each k ∈ N one has:
(v) Γ(x) ∩ Dk 6= ∅, for all x ∈ X ∩ Dk ;
(vi) supy∈Γ(x)∩Dk inf z∈Φ(x) hz, x − yi > 0, for each x ∈ (X ∩ Dk )\K, with x ∈ Γ(x),
Then, there exists at least one solution to GQVI(X, Γ, Φ) belonging to K × Rn .
412
F. Rania
Fix (ρ, τ ) ∈ X − X and let {(pk , xk )} be a sequence in Λ(ρ, τ ) such that
(pk , xk ) → (p∗ , x∗ ) as k → ∞. Let us to show that (p∗ , x∗ ) ∈ Λ(ρ, τ ). At first
observe that, since X is closed, one has (p∗ , x∗ ) ∈ X. Moreover, since Φ(pk , xk )
is compact for each k ∈ N, and the function (z, T ) ∈ R` ×Rn×` → h(z, T ), (ρ, τ )i
is continuous in R` ×Rn×` , then, for each k ∈ N, we can find (z k , Tk ) ∈ Φ(pk , xk )
such that
h(z k , Tk ), (ρ, τ )i ≤ 0.
Note that, from the definition of Φ, for each k ∈ N, one has:
X (i)
zk = −
(xk − ω (i) );
(12)
(13)
i∈I
Tk ∈ ∂(−u)(xk ).
(14)
Moreover, in force of Assumption 4.a, then, for each k ∈ N, there exist an open
neighborhood Ak of xk in Rn×` and a constant Lk ≥ 0 such that:
sup
T ∈∂ ◦ (−u)(x)
|T | ≤ Lk , for each x ∈ Ak ∩ Rn×`
+ .
(15)
Furthermore, there exist an open neighborhood A0 of x∗ in Rn×` , and a constant L0 ≥ 0 such that
sup
T ∈∂ ◦ (−u)(x)
|T | ≤ L0 , for each x ∈ A0 ∩ Rn×`
+ .
(16)
At this point, observe that the family of open sets {Ak }k∈N∪{0} is a covering of
the compact set {xk }k∈N ∪ {x∗ }. Therefore, from (15) and (16), we infer that
there exists a constant L ≥ 0 such that
sup
T ∈∂ ◦ (−u)(xk )
|T | ≤ L, for each k ∈ N.
Consequently, from (14), up to a subsequence, we can suppose that the sequence {Tk }k∈N converges to some T∗ ∈ Rn×` . Now, from (13), we infer that
X
(i)
z k → z ∗ := −
(x(i)
(17)
∗ − ω ), as k → ∞;
i∈I
and, from Proposition 9 and (14), we also infer that
T∗ ∈ ∂(−u)(x∗ ).
(18)
Furthermore, from (12), passing to the limit as k → ∞, one has
h(z ∗ , T∗ ), (ρ, τ )i ≤ 0.
(19)
At this point, observe that conditions (17) and (18) mean that (z ∗ , T∗ ) ∈
Φ(p∗ , x∗ ) and this latter condition, together with (19), gives (p∗ , x∗ ) ∈ Λ(ρ, τ ).
Therefore, condition (a1 ) is proved.
Step 4. Now, let us to show that the below condition holds true
413
GQVIs for competitive equilibrium problem
(a2 ) the map Γ : X → 2X is lower semicontinuous with closed graph;
To this end, it is sufficient to prove that for every (p0 , x0 ) ∈ X, every
(p, x) ∈ Γ(p0 , x0 ), and every sequence {(p̂k , x̂k )} in X such that (p̂k , x̂k ) →
(p0 , x0 ) as k → +∞, there exists a sequence {(pk , xk )} in X such that (pk , xk ) ∈
Γ(p̂k , x̂k ) for all k ∈ N, and (pk , xk ) → (p, x) as k → +∞ (see [5] at page 39,
for instance).
So, let (p0 , x0 ), (p, x) and {(p̂k , x̂k )} be as above. For each i ∈ I, using the
fact that (p, x) ∈ Γ(p0 , x0 ), we have the following two situations:
either
hp0 , x(i) i < hp0 , ω (i) i,
(20)
hp0 , x(i) i = hp0 , ω (i) i.
(21)
or
Suppose that (20) holds. Then, since p̂k → p0 , there exists k0 ∈ N such that
hp̂k , x(i) i < hp̂k , ω (i) i, for all k ∈ N, with k ≥ k0 . So, in this case, if we put
(i)
(i)
xk = x(i) for k ≥ k0 , and xk = 0 for k = 1, ..., k0 − 1, it is easy to check that
(i)
(i)
xk ∈ B (i) (p̂k ), for all k ∈ N. Moreover, it is clear that xk → x(i) as k → +∞.
Suppose that (21) holds. Then, from the Assumption 2, we have hp0 , x(i) i =
hp0 , ω (i) i > 0. Consequently, since p̂k → p0 , we have
hp0 , ω (i) i
hp̂k , ω (i) i
=
= 1.
k→∞ hp̂k , x(i) i
hp0 , x(i) i
n
o
k ,ω (i) i
(i)
Therefore, if we put ak = max 0, 1 − hp̂
, for all k ∈ N, and xk =
hp̂k ,x(i) i
lim
(i)
(1 − ak )x(i) for all k ∈ N, it is easy to check that xk ∈ B (i) (p̂k ), for all k ∈ N,
(i)
and that xk → x(i) as k → +∞.
(i)
So, for each i ∈ I, in both cases (20) and (21), we can find a sequence xk
(i)
which converges to x(i) and such that xk ∈ B (i) (p̂k ), for all k ∈ N. Conse(1)
(n)
quently, if we put xk = (xk , ..., xk ), for all k ∈ N, the sequence {(pk , xk )},
where pk = p, for all k ∈ N, satisfies (pk , xk ) ∈ Γ(p̂k , x̂k ) for all k ∈ N, and
(pk , xk ) → (p, x) as k → +∞, as desired. Therefore, Γ is lower semicontinuous
in X.
To show condition (a2 ) holds true, it remains to prove that Γ has closed
graph. To this end, let {(p̂k , x̂k )} and {(pk , xk )} be two sequences in X such
that (pk , xk ) ∈ Γ(p̂k , x̂k ), for all k ∈ N, and suppose that (p̂k , x̂k ) → (p0 , x0 ),
(pk , xk ) → (p, x), as k → ∞. Let us to show that (p, x) ∈ Γ(p0 , x0 ).
414
F. Rania
Since P is closed, one has p ∈ P . Moreover, being (pk , xk ) ∈ Γ(p̂k , x̂k ) for all
(i)
k ∈ N, one has hp̂k , xk i ≤ hp̂k , ω (i) i, for all k ∈ N, and i ∈ I. Passing to the
limit as k → ∞, we obtain hp0 , x(i) i ≤ hp0 , ω (i) i, for all i ∈ I. Thus, x ∈ B(p0 )
which, together p ∈ P , gives (p, x) ∈ Γ(p0 , x0 ), as desired.
Step 5. It remains to show that condition
(a3 ) there exists R0 such that, if for each R ∈ [R0 , ∞[ one has B R (0) ∩ X 6= ∅,
and:
(i) Γ(p, x) ∩ B R (0) 6= ∅, for all (p, x) ∈ X ∩ B R (0);
(ii)
inf
sup
h(z, T ), (p, x) − (p0 , x0 )i > 0,
(p0 ,x0 )∈Γ(p,x)∩B R (0) (z,T )∈Φ(p,x)
for all (p, x) ∈ X ∩ B R (0) \ K, with (p, x) ∈ Γ(p, x).
holds true as well.
Let R0 > 0 be such that B R0 (0) ⊂ Rn × Rn×` contains the compact set K.
Then, for each R ∈ [R0 , ∞[, one K ⊂ X ∩ B R (0) and P × {0} ⊂ Γ(p, x) ∩
B R (0), for all (p, x) ∈ X. Therefore, condition (i) of (a3 ) holds. Suppose that
condition (ii) of (a3 ) does not hold. Then, it should exist (p̄, x̄) ∈ X∩B R (0)\K,
with (p̄, x̄) ∈ Γ(p̄, x̄), such that:
inf
h(z, T ), (p̄, x̄) − (p0 , x0 )i ≤ 0, for all (p0 , x0 ) ∈ Γ(p̄, x̄) ∩ B R (0). (22)
(z,T )∈Φ(p̄,x̄)
Now, let us put p∗ := (1/l, ..., 1/l) ∈ P . Then, (p∗ , x̄) ∈ B R (0) ∩ X. Moreover,
| {z }
n−times
from (p̄, x̄) ∈ Γ(p̄, x̄), it trivially follows (p∗ , x̄) ∈ Γ(p̄, x̄). Thus, we can test
(22) with (p0 , x0 ) = (p∗ , x̄). Doing so, we get inf (z,T )∈Φ(p̄,x̄) h(z, T ), (p̄ − p∗ , 0)i ≤
0. Therefore, being Φ(p̄, x̄) a compact set, it should exist (z̄, T̄ ) ∈ Φ(p̄, x̄) such
that h(z̄, T̄ ), (p̄ − p∗ , 0)i ≤ 0. From the definition of Φ, the previous inequality
is equivalent to
*
+
+
*
X
X
x̄(i) − ω (i) , p∗ − p̄ ≤ 0 =⇒
x̄(i) − ω (i) , p∗ ≤ 0
i∈I
i∈I
taking in mind that x̄ ∈ B(p̄). Consequently,
*
+
X X (i)
X
(i)
x̄h − ωh = `
x̄(i) − ω (i) , p∗ ≤ 0.
i∈I h∈H
i∈I
Therefore, x̄ ∈ C. But this contradicts the fact that (p̄, x̄) ∈ X ∩ B R (0) \ K =
(P × Rn×`
+ ) ∩ B R (0) \ (P × C).
GQVIs for competitive equilibrium problem
415
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Received: December 30, 2015; Published: February 5, 2016