L
2
(
x
∗
,λ
∗
)=
c
−
G
(
x
∗
)
≥
0
,
(2)
λ
∗
≥
0
,
(3)
and
λ
∗
[
c
−
G
(
x
∗
)] = 0
.
(4)
Proof
Consider two possible cases, depending on whether or not the constraint is binding
at
x
∗
.
Case 1: Nonbinding constraint.
If
c>G
(
x
∗
)
,thenlet
λ
∗
=0
. Clearly, (2)-(4) are satis
f
ed, so it only remains to show
that(1)mustho
ld
.W
ith
λ
∗
, (1) holds if and only if
F
0
(
x
∗
)=0
.
(5)
We can show that (5) must hold using a proof by contradiction. Suppose that
instead of (5), it turns out that
F
0
(
x
∗
)
<
0
.
Then, by the continuity of
F
and
G
, there must exist an
ε>
0
such that
F
(
x
∗
−
ε
)
>F
(
x
∗
)
and
(
x
∗
−
ε
)
.
But this result contradicts the assumption that
x
∗
maximizes
F
(
x
)
subject to
c
≥
G
(
x
)
. Similarly, if it turns out that
F
0
(
x
∗
)
>
0
,
then by the continuity of
F
and
G
there must exist an
0
such that
F
(
x
∗
+
ε
)
(
x
∗
)
and
(
x
∗
+
ε
)
.
But, again, this result contradicts the assumption that
x
∗
maximizes
F
(
x
)
subject
to
c
≥
G
(
x
)
. This establishes that (5) must hold, completing the proof for case 1.
Case 2: Binding Constraint.
If
c
=
G
(
x
∗
)
,thenle
t
λ
∗
=
F
0
(
x
∗
)
/G
0
(
x
∗
)
. This is possible, given the assumption
that
G
0
(
x
∗
)
6
.C
lear
ly
,(1)
,(2)
,and(4)aresat
is
f
ed, so it only remains to show
that(3)mustho
λ
∗
=
F
0
(
x
∗
)
/G
0
(
x
∗
)
, (3) holds if and only if
F
0
(
x
∗
)
/G
0
(
x
∗
)
≥
0
.
(6)
We can show that (6) must hold using a proof by contradiction. Suppose that
instead of (6), it turns out that
F
0
(
x
∗
)
/G
0
(
x
∗
)
<
0
.