Econ 204
Problem Set 6
Due Monday, August 17th
Exercise 1
Consider
f
:
R
2
!
R
2
such that
f
2
C
3
(
R
2
)
.
Now let
F
(
x; y; w; z
) =
f
(
x; y
)
°
(
w; z
)
and suppose that
F
(
x; y; w; z
) = 0
has solutions in
R
4
:
Let
S
±
R
4
be the set of solutions to this system.
Show that there exists a set
B
such that
B
c
has measure zero and for
(
x; y; w; z
)
2
S
where
(
w; z
)
2
B
,
there is a local implicit function
h
:
W
±
R
2
!
R
2
(
W
open) such that
F
(
h
(
w; z
)
; w; z
) = 0
for all
(
w; z
)
2
W
and
h
2
C
3
(
R
2
)
:
Exercise 2
Let
f
: [0
;
1]
!
[0
;
1]
be a correspondence de°ned as
f
(
x
) =
f
0
;
1
=
(
x
+ 1)
g
for
x
6
= 0
and
f
(0) =
f
1
=
2
g
:
Does
f
have a °xed point? If yes, °nd the point(s).
Does any of the °xed point theorems you have learned apply here?
Explain.
Answer the same questions for
f
(
x
) = [
°;
1
=
(
x
+1)]
for all
x
2
[0
;
1]
where
0
< °
<
1
=
2
.
Exercise 3
We say that a relation
R
on
X
is convex if whenever
xRy
and
zRy
then
(
±x
+ (1
°
±
)
z
)
Ry
for all
±
2
(0
;
1)
. (if
x
and
y
are in
R
,
²
is an example of
such relation). Let
R
i
be a convex relation on
R
n
for
i
= 1
;
2
; :::m
, °x
x
2
R
n
and let
B
i
=
f
y
°
x
:
yR
i
x; y
2
R
n
g
.
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 Summer '08
 ANDERSON
 Trigraph, Convex set, Halfspace, local implicit function

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