Social Welfare Functions for Sustainable Development
Thai Ha-Huy∗, Cuong Le Van†
September 9, 2015
Abstract
Keywords: anonymity; sustainable development, welfare function, Rawls
criterion.
JEL Classification: C62, D50, D81,D84,G1.
∗
University of Evry-Val-D’Essonne. E-mail: [email protected]
IPAG Business School, CNRS, Paris School of Economics, VCREME. E-mail: [email protected]
†
1
1
Introduction
For a sustainable development, one can distinguish two sets of criteria. The first
set imposes Anonymity (i.e. Equal Treatment) and Pareto (see e.g. Diamond
(1965), Basu and Mitra (2003)). The second set introduced by Chichilnisky
(1996), (1997) imposes the Social Welfare Function (SWF) to exhibit NoDictatorship of both the present and the future. Diamond (1965) shows that
there exists no SWF, continuous for the l∞ topology, which is anonymous
and Pareto. Basu and Mitra (2003) prove, for sequences of utilities uniformly
bounded in l∞ , that there exists no SWF which satisfies the Dominance Axiom (a weaker version of Pareto Axiom) and the Anonymity Axiom. The SWF
proposed by Rawls (1971) and by Chichilnisky (1996), (1997) are non-dictator.
While the Rawls criterion is anonymous, the Chichilnisky criterion is not. All
∞.
these authors consider sequences uniformly bounded in l+
This paper complements the papers by Diamond (1965) and by Basu and
Mitra (2003). We introduce a stronger notion of anonymity (Strong Anonymity).
When the SWF is linear, the two notions coincide. One of its purpose is to completely characterize the SWF which are strongly anonymous and continuous.
We distinguish three cases: (i) The set of utility sequences is endowed with the
product topology, (ii) this set is in l∞ , (iii) this set is in lp , p < +∞.
Cases (i) and (ii) are very usual in optimal growth models. Case (iii) is
considered, in some sense, in optimal growth models without discounting (see
e.g. Gale (1967), Brock (1970), Dana and Le Van (1990)). We prove that
(i) For the product topology, a SWF which is continuous and anonymous will
be constant.
(ii) In lp , with 1 < p ≤ +∞, it will be purely finitely additive if it satisfies
Strong Anonymity instead of Anonymity.
(iii) In l1 , for any Strongly Anonymous and continuous SWF, up to a continuous, real function, it is the sum over time of the instantaneous utilities.
Concerning the Pareto efficiency, we introduce a Weak Dominance Axiom which
is weaker than the Pareto Axiom and Dominance Axiom. We show there exists no continuous, anonymous SWF for the product topology, or for the l∞ topology, which is Paretian or Dominant or Weakly Dominant. In l1 , there
exist Social Welfare Functions which are continuous, strongly anonymous and
dominant. Moreover, they satisfy the No-Dictatorship properties.
The paper is organized as follows: In Section 2, we recall the different axioms.
In Section 3, we consider the set of sequences endowed with the product topology. In Section 4, we consider sequences in the l∞ space. In Section 5 we show
that one can define a SWF in lp , p ∈ N but larger than 1, which satisfies the
four axioms of sustainable development. In section 6, we consider the of l1 .
Since this space is not appropriate for growth models where the consumptions
2
path is uniformly bounded, we study in Section 7 the existence of SWF when
the state space is [0, 1]∞ . In Section 8, we study the existence of solutions to a
growth model with two kind of non-dictatorial criteria. And finally, in Section
9, we show that the Rawls model can be viewed as the limit of a sequence of
Ramsey models.
2
The Welfare Axioms
We will consider the set X of utility streams of real numbers. We suppose the
preferences on X are represented by a SWF.
Definition 1 (Anonymity) Given the space lp , p = 1, 2, . . . , +∞, a SWF
ϕ : lp → R satisfies the Anonymity Axiom if and only if:
for all x ∈ lp , all i, j positive integers, if we define y ∈ lp as follows: yi = xj ,
yj = xi , yk = xk for all k ̸= i, j , then we have: ϕ(x) = ϕ(y).
Remark 1: We use the definition given by of Basu and Mitra (2003).
Definition 2 (Strong Anonymity) Given the space X ⊆ lp , p = 1, 2, . . . , +∞,
a SWF ϕ : lp → R satisfies the Strong Anonymity Axiom if and only if:
for all x ∈ X, all a ∈ R, all i, j positive integers, if we define y, z ∈ lp as
follows: yi = xi + a, yk = xk for all k ̸= i, zj = xj + a, zk = xk for all k ̸= j ,
then we have: ϕ(y) = ϕ(z).
One can easily check that the two definitions coincide if the SWF is linear.
Proposition 1 Strong Anonymity implies Anonymity.
Proof : Let ϕ be strongly anonymous. We have, for any a ∈ R, any x ∈ X
ϕ(. . . , xi + a, . . . , xj , . . .) = ϕ(. . . , xi , . . . , xj + a, . . . , )
This implies
ϕ(. . . , xi + a − a, . . . , xj , . . . , ) = ϕ(. . . , xi + a, . . . , xj − a, . . .)
or equivalently
ϕ(. . . , xi , . . . , xj , . . . , ) = ϕ(. . . , xi + a, . . . , xj − a, . . .)
Take a = xj − xi we then get
ϕ(. . . , xi , . . . , xj , . . . , ) = ϕ(. . . , xj , . . . , xi , . . .)
Now, let X ⊆ lp , and X ̸= ∅.
3
Definition 3 (Pareto Axiom) For all x,y ∈ X, if x > y (i.e. xi ≥ yi , ∀i,
and x ̸= y), then ϕ(x) > ϕ(y).
Definition 4 (Weak Pareto Axiom) For all x, y ∈ X, if there exists j ∈ N
such that xj > yj and, for all k ̸= j, xk = yk , then ϕ(x) > ϕ(y). For all
x,y ∈ X, if x >> y (i.e. xi > yi , ∀i), then ϕ(x) > ϕ(y).
Definition 5 (Dominance Axiom) (Basu and Mitra, 2003) Let ϕ : X → R.
ϕ satisfies the dominance property if and only if:
(1) for all x ∈ X, for all y ∈ X such that yi > xi for some i, and for all j ̸= i,
yj = xj , we have ϕ(y) > ϕ(x).
(2) For all x, y ∈ X, if x >> y (i.e. xi > yi , ∀i), then ϕ(x) ≥ ϕ(y).
Definition 6 (Weak Dominance Axiom) Let ϕ : X → R. ϕ satisfies the
weak dominance property if and only if:
ϕ(x, 0, 0, . . . , 0, . . .) > ϕ(y, 0, 0, . . . , 0, . . .) if (x, 0, 0, . . .) ∈ X, (y, 0, 0, . . . , ) ∈ X
and x > y.
The following Lemma is obvious
Lemma 1 Pareto Axiom ⇒ Weak Pareto Axiom ⇒ Dominance Axiom ⇒
Weak Dominance Axiom.
This following proposition shows that Weak Dominance Axiom is strictly weaker
than Dominance Axiom. Indeed, it exhibits a SWF which is anonymous and
weakly dominant in l∞ , while Basu and Mitra (2003) have proved there exists
no SWF in l∞ which is both anonymous and dominant.
Proposition 2 There exists a SWF in l∞ which is anonymous and weakly
dominant.
Proof : In l∞ define the equivalence relation R by: xRy iff xi ̸= yi for i in a
finite set. We denote by [|x|] the equivalence class of x. Define the function f :
f
: l∞ /R → l∞
[|x|] 7→ x ∈ [|x|]
Let
∀y ∈ l∞ , ϕ(y) =
∞
∑
(yi − f ([|y|])i )
i=0
This function is well-defined since y differs from f ([|y|]) for a finite number of
components. It is anonymous and satisfies the Weak Dominance Axiom.
4
Definition 7 (No Dictatorship of the Present) (Chichilnisky, 1996, 1997)
Let x, y ∈ X, with ϕ(x) > ϕ(y). The SWF ϕ exhibits no dictatorship of the
present if for any N , there exist k ≥ N , (z0 , z1 , . . .) ∈ X, (z0′ , z1′ , . . . , ) ∈ X
such that
′
′
ϕ(x0 , x1 , . . . , xk , zk+1 , zk+2 , . . .) ≤ ϕ(y0 , y1 , . . . , yk , zk+1
, zk+2
, . . .).
Definition 8 (No Dictatorship of the Future) (Chichilnisky, 1996, 1997)
Let x, y ∈ X, with ϕ(x) > ϕ(y). The SWF ϕ exhibits no dictatorship of the
future if for any N , there exist k ≥ N , (z0 , z1 , . . . , zk ), (z0′ , z1′ , . . . , zk′ ) such that
ϕ(z0 , z1 , . . . , zk , xk+1 , xk+2 , . . .) ≤ ϕ(z0′ , z1′ , . . . , zk′ , yk+1 , yk+2 , . . .).
3
Social Welfare Functions with the product topology
We now endow X with the product topology. The following results requires
neither Weak Dominance nor Pareto Axiom.
Theorem 1 A SWF ϕ is continuous for the product topology and anonymous
iff it is a constant function.
Proof :
Step 1 Let z ∈ R. For any n define zn ∈ X as follows:
ztn = 0, ∀t < n, ∀t > n + k,
n
n
znn = z0 , zn+1
= z1 , . . . , zn+k
= zk .
Then zn → 0, and hence ϕ(zn ) → ϕ(0). Since ϕ is anonymous, we have
∀n, ϕ(zn ) = ϕ(z0 , z1 , . . . , zk , 0, . . .)
Indeed,
ϕ(z0 , z1 , . . . , zk , 0, . . .) = ϕ(z0 , z1 , . . . , zk−1 , 0, . . . , 0, zk , 0, . . .)
where zk is at the (n + k + 1)th position. By induction, we have:
ϕ(z0 , z1 , . . . , zk , 0, . . .) = ϕ(0, 0, . . . , 0, z0 , . . . , zk , 0 . . .) = ϕ(zn )
where z0 is at the nth position. Hence
ϕ(z0 , . . . , zk , 0, . . . , 0, . . .) = ϕ(0).
Step 2 Now, with any x, associate the sequences xn ∈ X defined as follows:
xnt = xt , t = 0, . . . , n; xt = 0, t ≥ n + 1
5
Then xn converges to x for the product topology and ϕ(xn ) → ϕ(x). But
∀n, ϕ(xn ) = ϕ(x0 , . . . , xn , 0, . . . , 0, . . .), by anonymity
= ϕ(0) from Step 1
Hence ϕ(x) = ϕ(0).
The converse is obvious.
Corollary 1 There exists no SWF ϕ, continuous for the product topology,
anonymous and weakly dominant
Proof : We have the contradiction
ϕ(1, 0, 0, 0, . . .) = ϕ(0), from theorem 1
> ϕ(0) by weak dominance
In view of Theorem 1, when X is endowed with the product topology, the
No-Dictatorship axioms do not make sense for the SWF which are anonymous
and continuous since they are constant functions.
4
Social Welfare Functions in l∞
We now endow X with the l∞ -topology. A function ϕ from X into R is called
purely finitely additive if for any x ∈ X, any y ∈ X, we have ϕ(x) = ϕ(y) if
xi = yi , ∀i ∈ N \ I, where I is a finite set of N.
Theorem 2 A SWF ϕ is strongly anonymous, continuous for the l∞ -topology
iff ϕ is purely finitely additive and continuous for the l∞ -topology. In particular,
(a) ϕ(x) = ϕ(0) if xi = 0, ∀i ∈ N \ I, where I is a finite set
(b) if xt → x when t → ∞, then ϕ(x) = ϕ(x, x, . . . , x, x, . . .).
Proof : Let x ∈ X and N be given. First, we will show that
ϕ(x) = ϕ(0, . . . , 0, xN +1 , . . . , xN +t , . . .).
Indeed, let δ =
∑N
t=0 xt .
Define the sequence x(T ) as follows
δ
,...,
N +T
δ
= xN +T −1 −
N +T
= xN +τ , ∀τ ≥ T.
x(T )0 = x0 −
x(T )N +T −1
x(T )N +τ
6
(1)
We have ∥x − x(T )∥∞ =
ϕ is continuous we have
δ
N +T .
This implies x(T ) → x when T → +∞. Since
lim ϕ(x(T )) = ϕ(x).
(2)
T →+∞
On the other hand, since ϕ is strongly anonymous, for any T we have successively
( N +T −1
)
∑
ϕ(x(T )) = ϕ (
xt ) − δ, 0, . . . , 0, xN +T , . . . , xN +τ , . . .
(
t=0
N +T
∑−1
= ϕ (
)
xt ), 0, 0, . . . , 0, xN +T , . . . , xN +τ , . . .
t=N +1
= ϕ(0, 0, . . . , 0, xN +1 , xN +2 , . . . , xN +τ , . . .).
Relation (5) implies
ϕ(x) = ϕ(0, 0, . . . , 0, xN +1 , xN +2 , . . . , xN +τ , . . .).
We now show that ϕ is purely finitely additive. Let x ∈ X, and y ∈ X satisfy
xi = yi , ∀i ∈ N \ I, where I is a finite set of N. This implies the existence of
N ∈ N such that xi = yi , ∀i ≥ N . We have
ϕ(x) = ϕ(0, . . . , 0, xN , xN +1 , . . . , xN +t , . . .)
ϕ(y) = ϕ(0, . . . , 0, xN , xN +1 , . . . , xN +t , . . .)
and hence
ϕ(x) = ϕ(y).
The converse is obvious.
The proof of (a) is obvious. We now prove (b). Let x satisfy xt → x when
t → +∞. Denote by x the sequence (x, . . . , x, . . .). Since ϕ is purely finitely
additive, we have
∀N, ϕ(x) = ϕ(x, . . . , x, xN +1 , . . . , xN +t , . . .).
Define the sequence x(N ) by
x(N )t = x, ∀t = 0, . . . , N, and x(N )t = xt , ∀t > N.
Since ϕ is purely finitely additive, we have
ϕ(x) = ϕ(x(N )), ∀N
(3)
Obviously, x(N ) → x for the l∞ -topology. Hence ϕ(x(N )) → ϕ(x). From (3),
ϕ(x) = ϕ(x).
7
Corollary 2 1. There is no strongly anonymous, continuous SWF in l∞ which
is Weakly Dominant (hence no SWF in l∞ is strongly anonymous, continuous
and either Paretian, or Weakly Paretian or Dominant).
2. If ϕ is a strongly anonymous, continuous SWF in l∞ then
(a) ϕ exhibits no dictatorship of the present.
(b) ϕ has the dictatorship of the future.
Proof : 1. We have, ∀a ∈ R++ , ϕ(a, 0, 0, . . . , 0, . . .) = ϕ(0): the Weak Dominance Axiom cannot hold.
2.a. Let ϕ(x) > ϕ(y). Then, for all N
ϕ(x0 . . . , xN , yN +1 , . . . , yN +t , . . .) = ϕ(y)
and ϕ has no dictatorship of the present.
2.b. Let ϕ(x) > ϕ(y). For any N , any (z0 , . . . , zN ), any (ζ0 , . . . , ζN ), we have
ϕ(x) = ϕ(z0 , . . . , zN , xN +1 , . . . , xN +t , . . .) > ϕ(y) = ϕ(ζ0 , . . . , ζN , yN +1 , . . . , yN +t , . . .).
Remark 2: In the converse part of Theorem 2, we have to assume the purely
finitely additive function is continuous for the l∞ -topology. Indeed, we now
exhibit a purely finitely additive function which is not continuous for the l∞ topology. Define ϕ as follows:
ϕ(0) = 0
ϕ(x) = ϕ(0), for any x different from 0 for a finite number of components
ϕ(x) = 1, otherwise
Then ϕ is purely finitely additive. But ϕ is not continuous. Indeed, let xn be
defined by xni = n1 , ∀n. Then xn → 0 in l∞ , but ϕ(xn ) = 1, ∀n.
Remark 3:
(i) The Rawls’ criterion ϕ(x) = inf t (xt ) is anonymous and continuous for the
l∞ -topology. It is not strongly anonymous. The proof of its anonymity is trivial.
To prove it is not strongly anonymous, consider the sequence {0, 1, . . . , 1, . . .}.
Then the infimum of this sequence is 0. Let {1, 1, . . . , 1, . . .} and {0, 2, 1, . . . , 1, . . .}.
Then inf{1, 1, . . . , 1, . . .} = 1 while inf{0, 2, 1, . . . , 1, . . .} = 0: strong anonymity
is not satisfied. The Rawls’ criterion is continuous for the l∞ -topology. For
that, let xn → x in this topology. Let ε be given. Then there exists N such
that |xnt − xt | ≤ ε, for all t, for all n > N . We then have
inf {xt } − ε ≤ inf {xnt } ≤ inf {xt } + ε, ∀n > N
t
t
t
8
In other words,
|ϕ(xn ) − ϕ(x)| ≤ ε, ∀n > N
(ii) The function ϕ(x) = lim inf t (xt ) is strongly anonymous, continuous for the
l∞ topology. It is purely finitely additive. Indeed, it is obvious that
lim inf(x0 , . . . , xi + a, xi+1 , . . .) = lim inf(x0 , . . . , xi , . . . , xj + a, xj+1 , . . .)
Hence strong anonymity is satisfied. We prove that this SWF is continuous for
the l∞ -topology. Let xn → x in this topology. Let ε be given. Then there
exists N such that |xnt − xt | ≤ ε, for all t, for all n > N . We have, given T
inf {xt } − ε ≤ inf {xnt } ≤ inf {xnt } + ε, ∀n > N
t≥T
t≥T
t≥T
Hence
lim{ inf {xt }} − ε ≤ lim{ inf {xnt }} ≤ lim{ inf {xt }} + ε, ∀n > N
T
t≥T
T
t≥T
T
t≥T
Equivalently,
|ϕ(xn ) − ϕ(x)| ≤ ε, ∀n > N
5
Social Welfare Functions in lp , with 1 < p < +∞
We now endow X with the lp -topology, with 1 < p < +∞.
Theorem 3 A SWF ϕ is strongly anonymous, continuous for the lp -topology,
with 1 < p < +∞ iff ϕ is constant function.
Proof :
We prove firstly that ϕ is purely finitely additive. Let x ∈ X and N be
given. Like in the previous section, we show that
ϕ(x) = ϕ(0, . . . , 0, xN +1 , . . . , xN +t , . . .).
Indeed, let δ =
∑N
t=0 xt .
Define the sequence x(T ) as follows
δ
,...,
N +T
δ
= xN +T −1 −
N +T
= xN +τ , ∀τ ≥ T.
x(T )0 = x0 −
x(T )N +T −1
x(T )N +τ
We have
∥x − x(T )∥pp =
N +T
∑−1
(
t=0
δ
δp
)p =
.
N +T
(N + T )p−1
9
(4)
This implies x(T ) → x when T → +∞. Since ϕ is continuous we have
lim ϕ(x(T )) = ϕ(x).
(5)
T →+∞
By using the same arguments like Section 4, we have ϕ is purely finitely additive.
Take any x ∈ lp . Define the sequence x(N ) by
x(N )t = 0, ∀t = 0, . . . , N, and x(N )t = xt , ∀t > N.
Since ϕ is purely additive, we have
ϕ(x(N )) = ϕ(x), ∀N.
Obviously, x(N ) → 0 for the lp -topology. Hence ϕ(x) = ϕ(0).
6
Social Welfare Functions in l1
We now take X = l1 .
Theorem 4 A SWF is strongly anonymous, continuous for the l1 -topology iff
it is expressed in the following form:
∞
∑
ϕ(x) = f (
xt )
t=0
with f : R → R continuous.
Proof : Suppose that ϕ : l1 → R is strongly anonymous, continuous. Define
f : R → R as:
f (x) = ϕ(x, 0, 0, ...)
By the continuity of ϕ, we have f is continuous.
For x ∈ l1 , define x(n) as follows:
x0 (n) =
n−1
∑
xt
t=0
xk (n) = 0 for all 1 ≤ k ≤ n − 1
xk (n) = xk for all k ≥ n
∑
and x = ( ∞
t=0 xt , 0, 0, ...). Notice that ϕ(x(n)) = ϕ(x) for all n by the strong
anonymity property.
We have:
∥x(n) − x∥ = |
∞
∑
xt | +
t=n
10
∞
∑
t=n
|xt |.
Since x ∈ l1 , the RHS of this equality converges to 0 when n conveges to ∞.
This implies
lim ∥x(n) − x∥ = 0.
n→∞
Since ϕ is continuous, we have
ϕ(x) =
∞
∑
xt ).
lim ϕ(x(n)) = ϕ(x) = f (
n→∞
t=0
∑
Conversely, it is obvious that if ϕ(x) = f ( ∞
t=0 xt ) with f continuous, then ϕ
1
is continuous for the l topology, strongly anonymous.
Let ϕ be a SWF which is strongly anonymous and continuous for l1 . From
Theorem 4, there exists a continuous real function defined on R, f , such that
∑
∀x ∈ lp , ϕ(x) = f ( ∞
t=0 xt ). In this case we say that f represents ϕ.
Lemma 2 ϕ is Weakly Dominant iff its representation f is strictly increasing.
Proof : : It is obvious.
The following corollary is immediate.
Corollary 3 If the function f which represents a SWF ϕ is increasing, then ϕ
is Paretian and hence Weakly Paretian, Dominant and Weakly Dominant.
Remark 4: (i) The lim inf SWF is continuous for the l1 -topology and strongly
anonymous. Its representation is the function 0. Indeed, for any x ∈ l1 , the
value of this sequence obtained with this SWF is 0. Actually, any purely finitely
additive SWF, which is strongly anonymous, continuous for the l1 -topology, has
the function 0 as representation.
(ii) The Rawls SWF is not strongly anonymous in l1 . Indeed, consider the sequence {−2, 0, 0, . . .} which is in l1 . Then min{−1, 0, 0, . . .} = −1 > −2 min{−2, 1, 0, . . .}.
Theorem 5 Suppose that ϕ is a strongly anonymous, continuous, weakly dominant SWF on l1 . Then ϕ satisfies the No-dictatorship properties.
Proof : (a) Let ϕ(x) > ϕ(y). From Theorem 4, it is equivalent to
∑∞
t=0 yt . For any N ∈ N, let K satisfy
∞
∑
t=0
xt <
N
∑
yt + K
t=0
Define z as follows:
zt = yt , ∀t = 0, . . . , N ; zN +1 = K; zt = 0, ∀t > N + 1
11
∑∞
t=0 xt
>
Then
∞
∑
xt <
t=0
N
∑
yt + K =
t=0
∞
∑
zt
t=0
Hence
ϕ(x) < ϕ(y0 , . . . , yN , K, 0, 0, . . .)
and ϕ satisfies the No-dictatorship of the present.
(b) Let ϕ(x) > ϕ(y). For any N ∈ N, let K satisfy
K>
∞
∑
xt −
t=0
∞
∑
yt
t=N
Define z as follows:
Then
z0 = K, zt = 0, ∀t = 1, . . . , N − 1, zt = yt , ∀t ≥ N
∑∞
t=0 xt <
t=0 zt and
∑∞
∞
∞
∑
∑
ϕ(x) = f (
xt ) < f (
zt ) = ϕ(z0 , . . . , zN −1 , yN , . . . , yN +t , . . .)
t=0
t=0
and ϕ satisfies the No-dictatorship of the future.
7
Social Welfare Functions in [0, 1]∞
The previous section deals with SWF in lp with p < +∞. But in growth models,
usually, the state space is [0, 1]∞ . This section focuses on the SWF defined on
∞ . The first result is quite ′′ pessimistic ′′ .
the unit ball of l+
Proposition 3 There exists no SWF W from X into R+ :
W (x) =
∞
∑
at xt + ϕ(x)
t=0
∑
with at ≥ 0, ∀t, ∞
t=0 at = 1, ϕ purely finitely additive, which satisfies the two
No-dictatorship properties.
Proof : Claim 1. If ϕ(1) > 1, then W does not satisfy the No-dictatorship of
the future property.
Indeed, let W (x) > W (y) where xt = 1, ∀t and yt = 0, ∀t. Let N be the
first index s.t. aN > 0. Then ∀k ≥ N, ∀(z0 , ..., zk ) ∈ Rk+ , ∀(z0′ , ..., zk′ ) ∈ Rk+ ,
W (z0 , ..., zk , xk+1 , ...xk+t ) =
k
∑
at z t +
t=0
>
∞
∑
t=0
=
∞
∑
at xt + ϕ(1) ≥ ϕ(1)
t=k+1
at ≥
k
∑
at zt′ +
∞
∑
t=0
t=k+1
′
W (z0 , ..., zk′ , yk+1 , ...).
12
at yt + ϕ(0)
Claim 2. If 0 ≤ ϕ(1) < 1, then W does not satisfy the No-dictatorship of
the present property.
To prove that, let ε > 0 satisfy 1 − ε > ϕ(1) and let xt = 1, ∀t and yt = 0,
∑
∑
∀t. Hence W (x) > W (y). There exists N s.t. ∀k ≥ N, kt=0 at xt = kt=0 at >
∑
∞
ϕ(1) + ∞
t=k+1 at . Then, for any k ≥ N, for any (zk+1 , ...) ∈ [0, 1] , we have
′
W (y0 , ..., yk , zk+1
, ...)
=
∞
∑
at zt′
′
ϕ(y0 , ..., yk , zk+1
, ...)
+
t=k+1
<
k
∑
≤ ϕ(1) +
at
t=k+1
∞
∑
at x t +
t=0
∞
∑
at zt + ϕ(x0 , ..., xk , zk+1 , ...)
t=k+1
= W (x0 , ..., xk , zk+1 , ...).
Claim 3. Assume ϕ(1) = 1. Then W does not satisfy the No-dictatorship of
the present property.
Again let xt = 1, ∀t and yt = 0, ∀t. Hence W (x) > W (y). For any k, any
(z0 , ..., zk ) ∈ Rk+ , any (z0′ , ..., zk′ ) ∈ Rk+ ,we always have
W (z0 , ..., zk , xk+1 , ...) ≥
>
k
∑
t=0
k
∑
t=0
at zt + ϕ(1) ≥ 1
at zt′ =
k
∑
at zt′ + ϕ(0) = W (z0′ , ..., zk′ , yk+1 , ...).
t=0
The proof of the proposition is over.
However, observe that the previous results are based on the assumption
∑∞
t=0 at = 1.
∑
We now assume ∞
t=0 at = +∞ and, for simplicity, that at > 0, ∀t.
∑
Let Z = {x ∈ X : ∞
t=0 at xt < +∞} . If (at ) satisfies ∀t, 0 < a ≤ at ≤ a,
then (x) ∈ Z ⇒ xt → 0. In this case, ϕ(x) = 0. We consider thus, for any
x ∈ Z, the SWF W with the following form:
W (x) =
∞
∑
at x t .
t=0
Observe that W is upper semi-continuous for the product topology: W : Z →
∑∞
t=0 at xt .
Proposition 4 Let (at ) satisfy
(i) ∀t, 0 < a ≤ at ≤ a
∑
(ii) ∞
t=0 at = +∞.
∑∞
∑
Let Z = {x ∈ X : ∞
t=0 at xt for x ∈ Z.
t=0 at xt < +∞} , and W (x) =
Then W satisfies the two No-dictatorship axioms.
13
Proof : (a) No-Dictatorship of the present
Let W (x) > W (y). We know that yt → 0. Let T be given. Define the
following sequences: zt′n = ε ∈ (0, 1), for t = T + 1, ..., T + n, zt′n = 0,
∀t > T + n. The sequence yn = (y0 , ..., yT , zT′n+1 , ...) belongs to Z for any n.
T∑
+n
W (yn ) = a0 y0 + ... + at yT + ε
at < +∞
t=T +1
and yn ∈ Z. But W (yn ) → +∞.
(b) No-Dictatorship of the future
∑
∑T
Let W (x) > W (y). There exists N s.t. ∀T ≥ N, ∞
T +1 at xt <
0 at since
∑∞
′
0 at = +∞. Let zt = 0, ∀t = 0, ..., T and zt = 1, ∀t = 0, ..., T. In this case,
W (z0 , ..., zT , xT +1 , ...) =
∞
∑
at xt <
T
∑
t=0
at
t=0
t=T +1
≤
T
∑
∞
∑
at +
at yt = W (z0′ , ..., zT′ , yT +1 , ...).
t=T +1
More generally, let γ be a correspondence from R into R : x ∈ [0, 1] ⊂ R →
γ(x) ⊂ [0, 1], the graph of which has a non-empty interior.
Let us assume there exists x > 0 such that (x, x) ∈ intgraphγ.
∑
We consider as before a sequence a which satisfies at > 0, ∀t ≥ 0 ∞
t=0 at =
+∞, 0 < a ≤ at ≤ a, ∀t.
Let
{
}
∞
∑
Z= x∈X:
at (xt − x) exists in R, xt+1 ∈ γ(xt ), ∀t ≥ 0
t=0
Assume also that x ∈ Z ⇒ xt → x when t tends to ∞.
∑
Let W (x) = ∞
t=0 at (xt − x), for any x ∈ Z.
Proposition 5 W satisfies the two No-dictatorship axioms.
Proof : (a) No-Dictatorship of the present
Let W (x) > W (y) with x ∈ Z, y ∈ Z. Since yt → x and (x, x) ∈ intgraphγ,
there exists x
b close enough to x, and T large enough, s.t. for any t ≥ T,
x
b ∈ γ(xt+1 ), x
b ∈ γ(b
x), x ∈ γ(b
x) and x
b < x.
Take N ≥ T.
14
Define zn as follows: ztn = xt , for t = 0, ..., N + 1, ztn = x
b, for t = N +
n
n
n
2, ..., N + n + 1, zt = x, for t > N + n + 1. Then zt+1 ∈ γ(zt ), ∀t > 0, and
n
W (z ) =
N
+1
∑
at (xt − x) +
t=0
=
N
+1
∑
N∑
+n+1
at (b
x − x)
t=N +2
at (xt − x) + (b
x − x)
t=0
N∑
+n+1
at ∈ R.
t=N +2
We have zn ∈ Z for any n, and W (zn ) → −∞ when n → +∞ since (b
x−
∑N +n+1
n
n
x) t=N +2 at ≤ na(b
x−x) → −∞. Therefore, for any (zN +2 , ...) s.t. W (y0 , ..., yN +1 , zN
+2 , ...) ∈
R, we have
n
n
W (x0 , ..., xN +1 , zN
+2 , ...) < W (y0 , ..., yN +1 , zN +2 , ...)
for n large enough.
(b) No-Dictatorship of the future
Again, let W (x) > W (y) with x ∈ Z, y ∈ Z. As in (a), there exists x
b < x,
close enough to x, s.t. for any t ≥ k, xt ∈ γ(b
x).
∑
x − x) = −∞. Let zt = x
We have limT →∞ Tt=0 at (b
b for t = 0, ..., T ≥ k.
Then
T
∞
∞
∑
∑
∑
at (zt − x) +
at (xt − x) ≤
at (yt − x),
t=0
for any T large enough, since
when T → +∞.
8
8.1
t=0
t=T +1
∑T
t=0 at (zt
− x) → −∞,
∑∞
t=T +1 at (xt
− x) → 0
Solving the one sector Ramsey model with two
non-dictatorial criteria
Criterion 1
Consider two problems:
R : max
∞
∑
β t u(ct )
t=0
C : max
∞
∑
β t u(ct ) + ϕ(u(c1 ), u(c2 ), ...)
t=0
under the constraint ct + kt+1 ≤ f (kt ) for all t, with k0 > 0 given.
Let Π(k0 ) denote the set of feasible capital stocks. We make the following
assumptions.
15
H0 0 < β < 1.
H1 The function u : R → R is twice continuously differentiable and satisfies
u(0) = 0. Moreover, its derivatives satisfy u′ > 0 (strictly increasing) and
u′′ < 0 (strictly concave).
H2 Inada condition u′ (0) = +∞.
H3 The function f : R → R is twice continuously differentiable and satisfies
f (0) = 0. Its derivatives satisfy f ′ > 0 (strictly increasing), f ′′ < 0, strictly
concave, limx→∞ f ′ (x) < 1.
H4 ϕ is purely finitely additive, continuous for the l∞ -topology, homogeneous
of degree 1, non decreasing, ϕ(0) = 0, ϕ(1) > 0.
Lemma 3 Let u ∈ l∞ be such that limt→∞ ut = u, then
ϕ(u) = ϕ(u, u, ...)
Proof : Denote u = (u, u, ...). Construct un as:
unt = u for 0 ≤ t ≤ n
unt = ut for t > n
Since ϕ is purely finitely additive, so for all n, ϕ(un ) = ϕ(u). On the other
hand, ∥un − u∥ = supt>n |ut − u| ⇒ limn→∞ ∥un − u∥ = 0, then
ϕ(u) =
lim ϕ(un )
n→∞
= ϕ(u)
For each k ∈ Π(k0 ), define:
Φ(k) =
∞
∑
β t u(f (kt ) − kt+1 ) + ϕ(u(f (k0 ) − k1 ), u(f (k1 ) − k2 ), ...)
t=0
and
W (k) =
∞
∑
β t u(f (kt ) − kt+1 )
t=0
Theorem 6 Given k0 > 0, suppose that there exists a solution to problem C,
then this solution k∗ satisfies:
∗
i: For all t > 0, 0 < kt+1
< f (kt∗ ).
ii: (Euler equation), for all t, we have:
∗
∗
∗
∗
u′ (f (kt∗ ) − kt+1
) = βu′ (f (kt+1
) − kt+2
)f ′ (kt+1
)
16
Proof : The proof is easy.
Theorem 7 If the sequence k∗ is solution to problem C, then it is also solution
to problem R.
Proof : For each k, denote f t (k) = f (f (...(k)...), t times.
First case f ′ (0) ≤ 1. In this case, for all feasible sequence k, we have
limt→∞ kt = 0, then ϕ(u(c0 ), u(c1 ), ...) = 0 for all k ∈ Π(k0 ), and the problem is over.
Second case f ′ (0) > β1 . Denote k the unique positive solution to the equation
f (k) = k. For all x > 0, we have limt→∞ f t (x) = k.
∗
< f (kt∗ ), ∀t.
(a) Assume k0 ≤ k. Then for all t > 0, kt∗ < k since kt+1
For each ϵ > 0, choose T sufficiently large such that for all k ∈ Π(k0 ), we have
∞
∑
β t u(f (kt ) − kt+1 ) <
t=T
ϵ
2
Choose arbitrarily k ∈ Π(k0 ). We know that limt→∞ f t (kT ) = k. We will prove
that there exists τ > 0 such that kT∗ +τ < f τ (kT ). Suppose not, then for all
∗ ) = 0.
t > 0, f t (kT ) ≤ kT∗ +t < k ⇒ limt→∞ kt∗ = k ⇒ limt→∞ u(f (kt∗ ) − kt+1
From Lemma 3 we have:
Φ(k∗ ) =
∞
∑
β t u(c∗t )
t=0
= W (k∗ )
⇒ for all k′ ∈ Π(k0 ), we have:
W (k∗ ) = Φ(k∗ )
≥ Φ(k′ )
∞
∑
=
β t u(c′t ) + ϕ(u(c′0 ), u(c′1 ), ...)
t=0
≥
∞
∑
β t u(c′t ) = W (k′ )
t=0
Then k∗ is the solution to problem R. We know that in this case limt→∞ kt∗ =
k s < k: a contradiction.
So there exists τ > 0 such that kT∗ +τ < f τ (kT ). We construct the sequence k′
as follows:
kt′ = kt for all 0 ≤ t ≤ T
kT′ +t = f t (kT ) for all 1 ≤ t ≤ τ − 1
kT′ +τ +t = kT∗ +τ +t for all t ≥ 0
17
Since kT∗ +τ < f τ (kT ), we can easily check that k′ is feasible. Observe that for all
∗ , c′ = f (k ′ ) − k ′ .
t ≥ T + τ , kt′ = kt∗ ⇒ ϕ(c∗ ) = ϕ(c′ ), where c∗t = f (kt∗ ) − kt+1
t
t
t+1
0 ≤ Φ(k∗ ) − Φ(k′ )
∞
∞
∑
∑
=
β t u(c∗t ) −
β t u(c′t )
t=0
∞
∑
≤
≤
β t u(c∗t ) −
t=0
t=0
∞
∑
∞
∑
β t u(c∗t ) −
t=0
because
∑∞
T
t=0
T
−1
∑
β t u(ct ) +
ϵ
2
β t u(ct ) + ϵ
t=0
β t u(ct ) < 2ϵ . Then we have:
∞
∑
β t u(c∗t ) ≥
t=0
∞
∑
β t u(ct ) − ϵ
t=0
That is true for all ϵ > 0, hence k∗ is solution to problem R.
(b) Now assume k0 > k. Suppose such that for all t ≥ 0, kt∗ ≥ k. Then for all t
∗
we have kt+1
< f (kt∗ ) ≤ kt∗ ⇒ kt∗ ↓ k ∗ ≥ k. We have k ∗ ≤ f (k ∗ ) ≤ k ∗ ⇒ k ∗ = k.
∗ ) → 0 ⇒ ϕ(u(c∗ ), u(c∗ ), ...) = 0 ⇒ W (k∗ ) = Φ(k∗ ).
Then u(f (kt∗ ) − kt+1
0
1
Consider k ∈ Π(k0 ), then
W (k∗ ) = Φ(k∗ )
≥ Φ(k)
∞
∑
=
β t u(ct ) + ϕ(u(c0 ), u(c1 ), ...)
t=0
≥
∞
∑
β t u(ct ) = W (k)
t=0
So k ∗ is the solution to problem R ⇒ limt→∞ kt∗ = k s < k: a contradiction.
Hence, there exist T0 such that kT∗ 0 < k. We then go back to case (a). So, if
f ′ (0) > β1 , the solutions to the two problems coincide.
Third case 1 < f ′ (0) ≤ β1 . Then f ′ (k) < β1 for all k > 0. From the Euler’s
equation we have for all t:
∗
u′ (c∗t ) = βu′ (c∗t+1 )f ′ (kt+1
)
≤ u′ (c∗t+1 )
⇒ c∗t ≥ c∗t+1 and c∗t ↓ c∗ . If c∗ > 0, then we have βf ′ (kt∗ ) → 1. If f ′ (0) < β1 , we
have obviously a contradiction, since f ′ is decreasing. If f ′ (0) = β1 then kt∗ → 0:
a contradiction. Thus c∗ = 0 and ϕ(u(c∗0 ), u(c∗1 ), ...) = 0 ⇒ W (k∗ ) = Φ(k∗ ).
18
Consider k ∈ Π(k0 ), then
W (k∗ ) = Φ(k∗ )
≥ Φ(k)
∞
∑
=
β t u(ct ) + ϕ(u(c0 ), u(c1 ), ...)
t=0
≥
∞
∑
β t u(ct ) = W (k)
t=0
k∗ is solution to problem R.
Theorem 8 Suppose that f ′ (0) > 1, then the solution to problem R is NOT a
solution to problem C.
Proof : Suppose that k∗ is solution to problem R. We know that kt∗ → k s , the
steady state. Let x be the solution to equation f ′ (x) = 1. Observe that for all
x ̸= k > 0, f (k) − k < f (x) − x. Denote c = f (x) − x. Then u(cs ) < u(c) ⇒
u(cs )ϕ(1) = ϕ(u(cs ), u(cs ), ...) < u(c)ϕ(1) = ϕ(u(c), u(c), ...)
Choose ϵ > 0 such that:
0 < ϵ < ϕ(u(c), u(c), ...) − ϕ(u(cs ), u(cs ), ...)
Choose T > 0 such that for any feasible path k ∈ Π(k0 ) we have:
∞
∑
β t u(ct ) < ϵ, where ct = f (kt ) − kt+1
t=T +1
Denote, as above, by k the unique solution to equation f (x) = x. Notice that
x < k. We know also limt→∞ f t (kT∗ ) = k. Then there exists τ > 0 such that
f τ (kT∗ ) > x. We construct the sequence k as follows:
kt = kt∗ for all 0 ≤ t ≤ T
kT +t = f t (kT∗ ) for all 1 ≤ t ≤ τ − 1
kT +τ +t = x for all t ≥ 0
We can verify that k is feasible, and for t sufficiently large, kt = x. Then we
19
have:
Φ(k) =
∞
∑
β t u(ct ) + ϕ(u(c0 ), u(c1 ), ...)
t=0
=
=
≥
∞
∑
t=0
T
−1
∑
t=0
T
−1
∑
β t u(ct ) + ϕ(u(c), u(c), ...)
β t u(ct ) + ϕ(u(c), u(c), ...) +
∞
∑
β t u(ct )
t=T
β t u(c∗t ) + ϕ(u(c), u(c), ...)
t=0
>
∞
∑
β t u(c∗t ) + ϕ(u(c), u(c), ...) − ϵ
t=0
>
∞
∑
β t u(c∗t ) + ϕ(u(cs ), u(cs ), ...) = Φ(k∗ )
t=0
Φ(k) > Φ(k∗ ), then k∗ is NOT a solution to problem C.
Corollary 4 Assume f ′ (0) > 1. Then problem C has no solution.
Proof : It comes from Theorems 7 and 8
Theorem 9 Suppose f ′ (0) ≤ 1. Problem C has a solution which coincides with
the one to problem R .
Proof : Any feasible path (kt ) converges to zero. Let k∗ be the solution to R.
We have
Φ(k∗ ) = W (k∗ )
≥ W (k)
= W (k) + ϕ(0)
= W (k) + ϕ(u(c0 ), u(c1 ), . . .) = Φ(k)
with ct = f (kt ) − kt+1 . Thus k∗ is a solution to Problem C. The converse comes
from Theorem 7.
8.2
Criterion 2
We consider the model usually called the Rawls model:
max inf ct
t≥0
20
under the constraint ct + kt+1 ≤ f (kt ) for all t, with k0 > 0 given.
Hypothesis 1 f : R+ → R+ is strictly concave, strictly increasing, differentiable, and f (0) = 0.
Hypothesis 2 limx→∞ f ′ (x) < 1.
Define Π(k0 ) as the set of feasible paths. We know that Π(k0 ) is compact in
the product topology. Define also the reward function V : Π(k0 ) → R as:
V (k) = inf ct with ct = f (kt ) − kt+1
t≥0
Observe that if f ′ (0) ≤ 1, then for all feasible path k we have kt → 0 and so
V (k) = 0. The problem is over. Now we will consider only the case f ′ (0) > 1.
With this additional assumption, we have the lemma:
Lemma 4 supk∈Π(k0 ) V (k) > 0.
Proof : Define k as the unique solution to the equation f (k) = k. We distinguish two cases:
Case 1: 0 < k0 < k. Then k0 < f (k0 ), so the path k = (k0 , k0 , ...) is feasible,
and V (k) = f (k0 ) − k0 > 0.
Case 1: k0 ≥ k. By choosing 0 < k1 < k ≤ f (k0 ), we have that the path
k = (k0 , k1 , k1 , ...) is a feasible path, and V (k) = f (k1 ) − k1 > 0.
Theorem 10 V is upper semi-continuous for the product topology.
Proof : Suppose that the sequence of feasible paths kn → k for the product
topology.
For all ϵ > 0, there exists T such that cT < inf t≥0 ct + ϵ.
We have limn→∞ cnT = cT , then for all n sufficient large cnT < cT +ϵ ⇒ inf t≥0 cnt ≤
cnT < cT + ϵ < inf t≥0 ct + 2ϵ ⇒ lim supn→∞ V (kn ) < V (k) + 2ϵ. Since this is
true for all ϵ > 0, we get the result.
Remark V is not continuous. For example, let k ∈ Π(k0 ) with V (k) > 0.
Define kn as: ktn = kt for all 0 ≤ t ≤ n and ktn = 0 if t > n. Then limn→∞ kn =
k in the product topology, whereas for all n, V (kn ) = inf t≥0 cnt = 0.
From the upper semi-continuity of V and the compactness of Π(k0 ), we know
that there exists always k∗ ∈ Π(k0 ) such that V (k∗ ) = supΠ(k0 ) V (k). The next
theorem will give us the solutions to the problem. Let x satisfy f ′ (x) = 1. Note
that x is the unique solution to problem maxx≥0 [f (x) − x].
Theorem 11 If k0 ≤ x, then the problem has a unique solution k∗ = (k0 , k0 , . . .).
If x < k0 , then the problem has an infinity of solutions.
21
Proof : Case 1 k0 ≤ x.
Denote k∗ as a solution to the problem. From the Lemma 4, V (k∗ ) ≥ V (k) =
∗
f (k0 ) − k0 with k = (k0 , k0 , ...). So, we have f (kt∗ ) − kt+1
≥ f (k0 ) − k0 for
∗
∗
all t ≥ 0. Let t = 0. We then have k1 ≤ k0 . Observe that k0 − kt+1
≥
∗
′
∗
∗
∗
f (k0 ) − f (kt ) ≥ f (k0 )(k0 − kt ) ≥ k0 − kt if kt ≤ k0 . By induction, we find that
k0 ≥ kt∗ for all t, and furthermore, the sequence (kt∗ ) is decreasing and then
converges to k̂ ≤ k0 .
From the continuity of f , we have that f (k̂) − k̂ ≥ f (k0 ) − k0 . But the function
f (x) − x is increasing in [0, x]. Thus, f (k̂) − k̂ ≤ f (k0 ) − k0 , then k̂ = k0 , and
kt∗ = k0 for all t, because k0 ≥ kt∗ ↓ k0 .
Case 2 k0 > x. First, consider the sequence k = (k0 , x, x, . . .) which is
feasible. So, if k∗ is a solution then V (k∗ ) ≥ V (k) = f (x) − x.
∗
∗
We have that for all t ≥ 0, f (kt∗ ) − kt+1
≥ f (x) − x ⇒ x − kt+1
≥ f (x) − f (kt∗ ) ≥
x − kt∗ . So, kt∗ ↓ k̂ and f (k̂) − k̂ ≥ f (x) − x ⇒ k̂ = x and V (k∗ ) = f (x) − x.
Since k0 > x, by induction, we can construct a sequence k which satisfies: for
all t, x < kt+1 < f (kt ) − f (x) + x. With this sequence, we have f (kt ) − kt+1 >
f (x) − x, and kt+1 < kt , since f (kt ) − kt < f (x) − x. So kt ↓ k̂. Using the
same argument as above, we have k̂ = x and V (k) = f (x) − x. We have an
infinity number of sequences like that. That means the problem has an infinity
of solutions.
Now we consider the case more general:
max inf F (kt , kt+1 )
t≥0
under the constraint kt+1 ∈ Γ(kt ) ∀t, k0 > 0 given.
Hypothesis 1 The correspondence Γ : X → X is continuous, with non-empty,
compact values. Moreover 0 ∈ Γ(0).
Define: GraphΓ := {(x, y) | y ∈ Γ(x)}.
Hypothesis 2 GraphΓ is closed, convex.
Hypothesis 3 F : GraphΓ → R+ is concave, increasing with respect to the first
variable and decreasing with respect to the second variable. We also assume
F (0, 0) = 0.
Hypothesis 4 The set {x | x ∈ Γ(x)} is compact.
With this hypothesis, we know that there exists x which satisfies F (x, x) =
max F (x, x).
Theorem 12 If k0 ≤ x, then there exists a unique solution which is k∗ =
(k0 , k0 , . . .).
Proof : Since k0 ≤ x, then there exists 0 ≤ λ ≤ 1 such that k0 = λx. But both
(0, 0), (x, x) belong to GraphΓ, and this set is convex. So, we have (k0 , k0 ) =
22
λ(x, x) and belongs to GraphΓ too. Thus k0 ∈ Γ(k0 ) ⇒ (k0 , k0 , . . .) is a feasible
path.
∗ )≥
Denote the solution to problem as k∗ . We have that for all t ≥ 0, F (kt∗ , kt+1
F (k0 , k0 ). We know that k1∗ ≤ k0 . By induction, we will prove that kt∗ is
decreasing. First, we prove that F (x, x) is increasing in [0, x]. If it is not the
case, there exist 0 ≤ x < y ≤ x which satisfy F (x, x) > F (y, y). There exists
0 < λ ≤ 1 such that y = (1 − λ)x + λx. And we have:
F (x, x) > F (y, y) ≥ (1 − λ)F (x, x) + λF (x, x) ≥ F (x, x)
which is a contradiction.
Suppose that kt∗ ≤ k0 . We have:
∗
∗
F (k0 , kt+1
) ≥ F (kt∗ , kt+1
) ≥ F (k0 , k0 )
∗
⇒ kt+1
≤ k0 .
Then we have:
∗
F (kt∗ , kt+1
) ≥ F (k0 , k0 ) ≥ F (kt∗ , kt∗ )
∗
⇒ for all t kt+1
≤ kt∗ and limt→∞ kt∗ =: k̂. By the continuity of F we have
F (k̂, k̂) ≥ F (k0 , k0 ). So, k̂ = k0 and then kt∗ = k0 for all t, because k0 ≥ kt∗ ↓ k0 .
Theorem 13 If k0 > x, then we have an infinity of solutions, and the value of
the problem is F (x, x).
Proof : We recall that k = (x, x, . . .) is a feasible path. Then if k∗ is a solution,
∗ ) ≥ F (x, x) ≥ F (k ∗ , k ∗ ) ⇒ for all t, k ∗ ≥ k ∗
we have that for all t, F (kt∗ , kt+1
t
t
t
t+1
and then kt∗ ↓ k̂. We have F (k̂, k̂) ≥ F (x, x) ⇒ k̂ = x and the value of problem
is F (x, x).
Conversely, consider a feasible path k which satisfies kt ↓ x. Then, for all t,
F (kt , kt+1 ) ≥ F (x, kt+1 ) ≥ F (x, x) and limt→∞ F (kt , kt+1 ) = F (x, x). Then k
is a solution to problem. In other words, we have an infinity of solutions.
9
Rawls model as the limit of a sequence of Ramsey
models
With each σ > 0, we associate the Ramsey problem:
max
∞
∑
t=0
β
t u(ct )
1− σ1
1−
23
1
σ
under the constraint ct + kt+1 ≤ f (kt ) for all t, and k0 > 0 given.
From now on, we consider the model in the case f ′ (0) > 1.
We will prove that when σ → 0, the solutions of the Ramsey problem converge to a solution of the Rawls model. Moreover, the optimal consumption
paths of the Ramsey models converge to a constant consumption path.
∞ , we have:
Lemma 5 For each a ∈ l+
[∞
]1
θ
∑
t θ
lim
β at
= sup at
θ→∞
t≥0
t=0
Proof : Fix any ϵ > 0. There exists T such that aT > supt at − ϵ. Then we
have:
∞
∞
∑
∑
sup aθt
βt ≥
β t aθt ≥ β T aθT .
t≥0
t=0
t=0
That implies
1
(1 − β)
1
θ
sup at ≥
]1
[∞
∑
t≥0
θ
T
≥ β θ (sup at − ϵ).
β t aθt
t≥0
t=0
Let θ converge to infinity, we have supt at ≥ limθ→∞
for all ϵ > 0.
Let θ :=
1−σ
σ .
[∑∞
t θ
t=0 β at
]1
θ
≥ supt at − ϵ
Since σ → 0, we have θ → +∞ and our problem becomes
∞
∑
max
k∈Π(k0 )
βt
t=0
u−θ (ct )
.
−θ
Obviously, this problem is equivalent to
∞
∑
min
k∈Π(k0 )
or to :
min
t=0
[∞
∑
k∈Π(k0 )
β t u−θ (ct ).
]1
t −θ
βu
θ
(ct )
.
t=0
For each k ∈ Π(k0 ), define
V (k, θ) :=
[∞
∑
]1
t −θ
βu
θ
(ct )
.
t=0
Define
G(k0 , θ) = {k ∈ Π(k0 ) |V (k, θ) < +∞}.
24
Lemma 6 The set G(k0 , θ) is non empty.
Proof : Since f ′ (0) > 1, by Theorem 8.6, there exists a solution k of the
Rawls model which satisfies ct = c0 for all t ≥ 2. We can verify easily that
V (k, θ) < +∞.
Lemma 7 V (·, θ) is lower semi-continuous in product topology.
Proof : Take any k in Π(k0 ), and the sequence kn converges to k for the product
topology. Fix any N ∈ N. We have:
n
θ
V (k , θ) =
∞
∑
t −θ
βu
(cnt )
t=0
≥
N
∑
β t u−θ (cnt ).
t=0
Let n converges to infinity, we have lim inf V (kn , θ)θ ≥
all N . Let N converges to infinity, we have the conclusion.
∑N
t=0 β
t u−θ (c
t ),
for
Lemma 8 There exists minΠ(k0 ) V (k, θ).
Proof : We use the lower semi-continuity of V (·, θ) and the compactness of
Π(k0 ) for the product topology.
Denote by k(θ) as the solution of minΠ(k0 ) V (k, θ).
Lemma 9 For all t, we have 0 < kt+1 (θ) < f (kt (θ)). And we have also the
Euler’s equalities:
u−θ−1 (ct (θ))u′ (ct (θ)) = βu−θ−1 (ct+1 (θ))u′ (ct+1 (θ))f ′ (kt+1 (θ)).
Proof : We prove the first claim. Indeed, V (k(θ), θ) < +∞. That implies
ct (θ) > 0 for all t. Also we have f (kt (θ)) > ct (θ) > 0. This implies kt (θ) > 0
for all t.
By the very definition of k(θ), we have for all t, kt+1 (θ) is the solution of
min
0≤y≤f (kt (θ))
u−θ (f (kt (θ)) − y) + βu−θ (f (y) − kt+2 (θ)).
Since 0 < kt+1 (θ)) < f (kt (θ)), by the first order condition, we have:
−θu−θ−1 (f (kt (θ)) − kt+1 (θ))u′ (f (kt (θ)) − kt+1 (θ))
= −θβu−θ−1 (f (kt+1 (θ)) − kt+2 (θ))u′ (f (kt+1 (θ)) − kt+2 (θ))f ′ (kt+1 (θ)).
That implies
u−θ−1 (ct (θ))u′ (ct (θ)) = βu−θ−1 (ct+1 (θ))u′ (ct+1 (θ))f ′ (kt+1 (θ)).
25
Theorem 14 Denote by k one solution of the Rawls model. Define ct = f (k t )−
k t+1 . We have that every limit point of {k(θ)} is a solution of the Rawls model
and
1
lim V (k(θ), θ) = sup u−1 (ct ) =
.
θ→∞
inf t≥0 u(ct )
t≥0
Proof : Without loss of generality, we suppose that for the product topology,
k(θ) converges to k. For each t, define ct = f (kt ) − kt+1 and ct (θ) = f (kt (θ)) −
kt+1 (θ).
Fix ϵ > 0 such that ϵ < supt u−1 (ct ). There exists T > 0 such that
u−1 (cT ) > supt≥0 u−1 (ct ) − ϵ/2. For θ big enough, we have
u−1 (cT (θ)) > u−1 (cT ) − ϵ/2 > sup u−1 (ct ) − ϵ.
t≥0
Observe that
T
θ
−1
V (k(θ), θ) ≥ β u
(cT (θ)) > β
T
θ
[
−1
sup u
t≥0
]
(ct ) − ϵ .
This implies
lim inf V (k(θ), θ) ≥ sup u−1 (ct ) − ϵ.
θ→∞
t≥0
Since this inequality is true for all ϵ > 0, hence lim inf θ V (k(θ), θ) ≥ supt≥0 u−1 (ct ).
Observe that for all θ, k ∈ G(k0 , θ) and we have V (k, θ) ≥ V (k(θ), θ). This
implies:
lim V (k, θ) ≥ lim sup V (k(θ), θ) ≥ lim inf V (k(θ), θ) ≥ sup u−1 (ct ).
θ→∞
θ→∞
θ→∞
(6)
t≥0
By Lemma 5, we have limθ V (k, θ) = supt≥0 u−1 (ct ). Hence supt u−1 (ct ) ≥
supt≥0 u−1 (ct ). This implies inf t u(ct ) ≤ inf t u(ct ). Since k is a solution of the
Rawls model, we have inf t u(ct ) = inf t u(ct ) and hence the inequalities in (6)
are equalities. This implies that k is also a solution of the Rawls model and
lim V (k(θ), θ) = sup u−1 (ct ) =
θ→∞
t≥0
1
.
inf t≥0 u(ct )
Theorem 15 If k0 ≤ x, then limθ ct (θ) = f (k0 ) − k0 for all t.
If k0 > x, then limθ ct (θ) = f (x) − x for all t.
Proof : When k0 ≤ x, the Rawls problem has a unique solution (k0 , k0 , . . . ).
So, k(θ) converges to (k0 , k0 , . . . ).
26
Now consider the case k0 ≥ x. Without loss of generality, suppose that for
the product topology, k(θ) converges to k solution to the Rawls problem. From
the Euler equation we have for all t:
[
u(ct (θ))
u (ct (θ)) = β
u(ct+1 (θ))
′
]1+θ
u′ (ct+1 (θ))f ′ (kt+1 (θ)).
If ct > ct+1 , then u(ct ) > u(ct+1 ), then we have:
′
′
′
[
′
u (ct ) = lim u (ct (θ)) = u (ct+1 )f (k t+1 ) lim
θ→∞
θ→∞
u(ct (θ))
u(ct+1 (θ))
]1+θ
=∞
which is a contradiction.
If ct < ct+1 , using the similar argument, we find that u′ (ct ) = 0, also a contradiction.
Hence for every t ≥ 0, ct = ct+1 . There exists only one solution of the Rawls
model which satisfies this property. That is the solution defined by:
k t+1 = f (k t ) − f (x) + x for all t.
The consumption path is constant, and ct = f (x) − x for all t.
References
[1] Basu, K. and T. Mitra, Aggregating infinite utility streams with intergenerational equity: The impossibility of being paretian, Econometrica, Vol.
71, No. 5 (2003), 1557–1563.
[2] Brock, W.A., On the existence of weakly maximal programmes in a multisector economy, Review of Economic Studies, Vol. 37 (1970), 275–280.
[3] Chichilnisky, G., An axiomatic approach to sustainable development, Social Choice and Welfare, Vol. 13, No. 2 (1996), 219–248.
[4] Chichilnisky, G., What is sustainable development?, Land Economics, Vol.
7, No.4 (1997), 467–491.
[5] Dana, R.A. and C. Le Van, On the Bellman equation of the overtaking
criterion, Journal of Optimization Theory and Applications, Vol. 78, No.3
(1990), 605–612.
[6] Peter A. Diamond, The evaluation of infinite utility streams, Econometrica,
Vol. 33, No. 1 (1965).
27
[7] Gale, D., On optimal development in a multi-sector economy, Review of
Economic Studies, Vol. 34, No.97 (1967), 1–18.
[8] Rawls, J., A Theory of Justice, Oxford, England: Clarendon (1971).
28