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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
)
)
²
=
n
X
i
=1
³(
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
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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
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This note was uploaded on 10/03/2009 for the course ECON 6170 taught by Professor Mitra during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 MITRA
 Economics

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