TA: Yankun Wang
Economics 617
Problem Set 8 Solution Key
1. [Unconstrained Optimization: Sufficient Conditions for a Global Maximum]
Suppose
A
is an open convex set in
R
n
, and
f
:
A
→
R
is twice continuously
differentiable and quasiconcave on
A
.
Suppose there is
x
̄
in
A
satisfying:
(i)
∇
f
x
̄
0
, and (ii)
H
f
x
̄
is negative definite.
Show that
x
̄
is the unique point of global maximum of
f
on
A
.
Solution:
We will prove this result via two steps: first we prove that
x
̄
is a strict local
maximum; second we prove that this strict local maximum has to be the unique
global maximum given
f
is quasiconcave.
Claim1:
x
̄
is a strict local maximum of
f
on
A
.
Proof:
In the Lecture Notes, we are given that if
f
:
A
→
R
is twice continuously
differentiable and
x
̄
in
A
satisfies: (i)
∇
f
x
̄
0
, and (ii)
H
f
x
̄
is negative definite, then
x
̄
is a local maximum of
f
on
A
. However, this result is not strong enough for our
purpose. We actually need to prove that
x
̄
is a
strict
local maximum of
f
on
A
.The
main idea is to use Taylor’s Theorem. However, before using it, we need to prove
one thing: if
H
f
x
̄
is negative definite and
f
is twice continuously differentiable, then
H
f
x
is negative definite for all the
x
in a small neighborhood around
x
̄
.
We are given that
f
is twice continuously differentiable, which means
D
ij
f
:
A
→
R
is a continuous function for
i
,
j
1,.
.,
n
.
We will also utilize the fact that the
summation and multiplication of continuous functions defined on
A
is continuous. Let
F
r
:
A
→
R
,
r
1,.
.,
n
denote the corresponding leading principal minors (lpm) of
H
f
.
Note here that each
F
r
x
is a continuous function from
A
to
R
.S
ince
H
f
x
̄
is
negative definite,
lpm of degree 1;
F
1
x
̄
D
11
f
x
̄
0
For
−
F
1
x
̄
0,
∃
1
0:
whenever
d
x
,
x
̄
1
,
we have

F
1
x
̄
−
F
1
x

,
which means
D
11
f
x
0.
Since
A
is an open set,
∃
1
0:
B
x
̄
,
1
⊂
A
.
Set
1
min
1
,
1
,
so that we have
B
x
̄
,
1
⊂
A
,
with
D
11
f
x
0
for all
x
∈
B
x
̄
,
1
.
lpm of degree 2;
F
2
x
̄
D
11
f
x
̄
D
22
f
x
̄
−
D
12
f
x
̄
D
21
f
x
̄
0
For
F
2
x
̄
0,
∃
2
0:
whenever
d
x
,
x
̄
2
,
we have

F
2
x
̄
−
F
2
x

,
which means
D
11
f
x
D
22
f
x
−
D
12
f
x
D
21
f
x
0.
Since
A
is an open set,
∃
2
0:
B
x
̄
,
2
⊂
A
.
Set
2
min
2
,
2
,
so that we have
B
x
̄
,
2
⊂
A
,
with
D
11
f
x
D
22
f
x
−
D
12
f
x
D
21
f
x
0
for all
x
∈
B
x
̄
,
1
.
...