Problem Set 9 Solutions
Economics 6170
TA: Christopher Handy
December 5, 2008
Problem 1
(a)
Since ¯
x
∈
C
is a point of local maximum of
F
, there is some
β >
0 such that for all
x
∈
C
satisfying
d
(
x,
¯
x
)
< β
, we have
F
(
x
)
≤
F
(¯
x
). And since
C
is open, there is
some
γ >
0 such that for all
x
∈
R
n
satisfying
d
(
x,
¯
x
)
< γ
, we have
x
∈
C
. Now, let
r
=
1
2
min
{
β, γ
}
, so that
B
≡
B
(¯
x,
2
r
) =
{
x
∈
R
n

d
(
x,
¯
x
)
<
2
r
} ⊂
C
. Then for all
x
∈
B
, we have that
F
(¯
x
)
≥
F
(
x
).
(b)
For any
k
∈ {
1
, . . . , n
}
, define
a
(
k
) = ¯
x

re
k
,
b
(
k
) = ¯
x
+
re
k
Now,
a
(
k
)
∈
B
and
b
(
k
)
∈
B
because
d
(
a
(
k
)
,
¯
x
) =
d
(
b
(
k
)
,
¯
x
) =
d
(
re
k
,
0) =
r <
2
r
.
Since the open ball
B
is convex, we have that
tb
(
k
) + (1

t
)
a
(
k
)
∈
B
for all
t
∈
I
=
[0
,
1]. Define
f
:
I
→
R
by
f
(
t
) =
F
(
tb
(
k
) + (1

t
)
a
(
k
)
)
for all
t
∈
I
Since
tb
(
k
) + (1

t
)
a
(
k
)
∈
B
⊂
C
for all
t
∈
I
and
F
is defined on
C
, the function
f
is welldefined. Let
A
= (0
,
1). We showed in problem 1 of Problem Set 8 that
f
is
continuous on [0
,
1], so it follows that
f
is continuous on
A
. Since
F
is continuously
differentiable on the open set
C
, we can use the Chain Rule to differentiate
f
on
A
.
For all
t
∈
A
, we have
f
0
(
t
) =
n
X
i
=1
D
i
F
(
tb
(
k
) + (1

t
)
a
(
k
)
)
∂
∂t
(
tb
i
(
k
) + (1

t
)
a
i
(
k
)
)
=
n
X
i
=1
(
b
i
(
k
)

a
i
(
k
)
)
D
i
F
(
tb
(
k
) + (1

t
)
a
(
k
)
)
=
(
b
(
k
)

a
(
k
)
)
∇
F
(
tb
(
k
) + (1

t
)
a
(
k
)
)
= (2
re
k
)
∇
F
(
¯
x
+ (2
t

1)
re
k
)
= 2
rD
k
F
(
¯
x
+ (2
t

1)
re
k
)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Since
F
is continuously differentiable on
C
and
tb
(
k
)+(1

t
)
a
(
k
) = ¯
x
+(2
t

1)
re
k
∈
C
whenever
t
∈
A
, we have that
f
0
(
t
) is continuous on
A
.
So
f
is continuously
differentiable on
A
.
(c)
Since
tb
(
k
) + (1

t
)
a
(
k
) = ¯
x
+ (2
t

1)
re
k
, we have
f
(
1
2
)
=
F
(¯
x
).
And since
tb
(
k
) + (1

t
)
a
(
k
)
∈
B
for all
t
∈
I
and
F
(¯
x
)
≥
F
(
x
) for all
x
∈
B
, we have by the
definition of
f
that
f
(
1
2
)
≥
f
(
t
) for all
t
∈
I
, and thus for all
t
∈
A
.
(d)
Using the result in part (c), we have by the stated Theorem that
f
0
(
1
2
)
= 0. And
from part (b), we have that
f
0
(
1
2
)
= 2
rD
k
F
(¯
x
).
Since
r >
0, these two equations
imply that
D
k
F
(¯
x
) = 0.
Because
k
∈ {
1
, . . . , n
}
was chosen arbitrarily, we have
∇
F
(¯
x
) = 0. This proves the Corollary.
Problem 2
We will prove this by verifying a series of claims.
First, we will show that the Hessian
of
f
is negative definite in some neighborhood around ¯
x
.
Second, we will use Taylor’s
Theorem to show that ¯
x
is a point of strict local maximum of
f
. Finally, we will use the
quasiconcavity of
f
on
A
to show that ¯
x
is the unique point of global maximum of
f
on
A
.
Claim 1: There is some
δ >
0 such that for all
x
∈
B
(¯
x, δ
) =
{
x
∈
R
n

d
(
x,
¯
x
)
< δ
}
, we
have that
x
∈
A
and
H
f
(
x
) is negative definite.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 MITRA
 Economics, Derivative, Optimization, Continuous function, X1, Convex function

Click to edit the document details