Lecture 3. Classical demand theory II.
Equivalence between dual and primal problems
Proposition. Suppose that preferences are continuous and locally nonsatiated. For each p ∈ R^{L}_{ + + } and each u ≥ u(0),
v(p, e(p, u)) = u.
Proof. We need to show that (a) v(p, e(p, u)) ≥ u and (b) v(p, e(p, u)) ≤ u.
For (a), take
h ∈ h(p, u). By the definition of the dual problem,
u(h) ≥ u and
p⋅h = e(p, u). This implies that
h satisfies the constraints of the primal problem
(p, e(p, u)) (i.e.,
h ∈ B(p, e(p, u)), and thus
v(p, e(p, u)) ≥ u(h) ≥ u.
For (b), take
x ∈ x(p, e(p, u)). Because the assumptions imply that Walras Law holds, we have5
p⋅x = e(p, u).
Suppose that
u(x) > u ≥ u(0) and consider replacing
x by
x_{ε} = (1 − ε)x + ε0
for some
ε > 0. Then, for each
ε > 0,
p⋅x_{ε} = (1 − ε)p⋅x < e(p, u).
Moreover, by continuity of the utility function, for sufficiently small
ε,
u(x_{ε}) > u.
But then,
x_{ε} satisfies the constraints of the dual problem and leads to lower expenditure. Contradiction.
Upper contour sets can be deduced from the expenditure function
Assume that preferences are continuous, monotonic and convex. For each
u,
{x:u(x) ≥ u} = {x:p⋅x ≥ e(p, u) for each p ∈ R^{L}_{ + }}.
Inclusion “ ⊆ ” follows form the definition of the expenditure function. For the other inclusion, we need to show that for each x = (x_{1}, ..., x_{L}), if u(x) < u, then there exists p ∈ R^{L}_{ + } such that p⋅x < e(p, u).
Because the preferences are convex, the set
{x:u(x) ≥ u} is convex. The serparating hyperplane theorem shows that there exists
p ∈ R^{L}, p ≠ 0 (but, possibly,
p_{l} < 0 for some
l) such that
Below, we show that it must be that
p_{l} ≥ 0 for each
l. Given that, notice that the righthand side of the above inequality is equal to
e(p, u). Thus,
p⋅x < e(p, u).
To finish our proof, we need to show that if (
1↑) holds, then it must be that
p_{l} ≥ 0 for each
l. On the contrary, suppose not and there is
l such that
p_{l} < 0. For each
a > 0 and
ε > 0, define a bundle
x^{a, ε} =
x + a(ε, ..., ε, 1_{lth position}, ε, ..., ε)
=
(x_{1} + aε, ..., x_{l − 1} + aε, x_{l} + a, x_{l + 1} + aε, ..., x_{L} + aε).
Then,
p⋅x^{a, ε} = p⋅x + a(p_{l} + ε^{ }⎲⎳_{l’ ≠ l}p_{l}).
Because
p_{l} < 0, we can find
ε^{*} > 0 small enough so that
p_{l} + ε∑_{l’ ≠ l}p_{l} < 0 and for each
a > 0,
Also, we can find
a^{*} > 0 large enough so that
(Indeed, choose any
y such that
u(y) ≥ u and notice that we can find
a^{*} large enough so that for each
l’,
a^{*}ε^{*} ≥ y_{l’} − x_{l’}. Then, for each
l’
x^{a*, ε*}_{l’} ≥ x_{l’} + a^{*}ε^{*} ≥ x_{l’} + y_{l’} − x_{l’} ≥ y_{l’},
in which case the monotonicity of preferences implies (
3↑). But then, (
2↑) and (
3↑) contradicts (
1↑).
Generalized Envelope Theorems

K ⊆ R^{L} is a compact set,

f:R^{L} × [ − 1, 1] → R is continuous in (x, t) and it has partial derivative f_{t}(x, t) that is continuous in (x, t)
Let x^{*}(t) be the set of maximizers of the above problem.
Theorem. (
Generalized Envelope Theorem)
For each t,
limsup_{t’ → t}(V(t’) − V(t))/(t’ − t)
= max_{x ∈ x*(t)}f_{t}(x, t),
liminf_{t’ → t}(V(t’) − V(t))/(t’ − t)
= min_{x ∈ x*(t)}f_{t}(x, t).
In particular, if x^{*}(t) is unique, then V is differentiable at t and (d)/(dt)V(t) = f_{t}(x^{*}(t), t).
Proof. First, we show that
limsup_{t’ → t}(V(t’) − V(t))/(t’ − t) ≥ max_{x ∈ x*(t)}f_{t}(x, t).
Indeed, because the derivative
f_{t}(x, t) is continuous, there exists
x_{0} ∈ argmax_{x ∈ x*(t)}f_{t}(x, t). Moreover, by the definition of the partial derivative,
f_{t}(x_{0}, t) = ^{ }lim_{t’ → t}(f(x_{0}, t’) − f(x_{0}, t))/(t’ − t).
The claim follows from the fact that
V(t’) ≥ f(x_{0}, t’) and V(t) = f(x_{0}, t).
Next, we show that
limsup_{t’ → t}(V(t’) − V(t))/(t’ − t) ≥ max_{x ∈ x*(t)}f_{t}(x, t).
If not, then, there exists a sequence
t_{n} → t and
x_{n} such that
V(t_{n}) = f(x_{n}, t_{n})
and
limsup_{n → ∞}(f(x_{n}, t_{n}) − V(t))/(t_{n} − t) > max_{x ∈ x*(t)}f_{t}(x, t).
By ocmpactness of
K, we can choose a subseqnece
x_{n} → x. By the continuuity of function
f, it must be that
x is an optimal choice for
t, i.e.,
x ∈ x^{*}(t). Because
V(t) ≥ f(x_{n}, t), we have
limsup_{n → ∞}(f(x_{n}, t_{n}) − V(t))/(t_{n} − t) ≤ limsup_{n → ∞}(f(x_{n}, t_{n}) − f(x_{n}, t))/(t_{n} − t) = limsup_{n → ∞}f_{t}(x_{n}, t_{n}’)
for some
t_{n}’ ∈ (t, t_{n}) (the last inequality follows from the Median Value Theorem). The continuity of the partial derivative implies that
limsup_{n → ∞}f_{t}(x_{n}, t_{n}’) = f_{t}(x, t).