On the other hand, by Exercise 3.16,
𝑓
attains its maximum at an extreme point of
𝑆
.
That is, there exists
x
1
∈
ˆ
𝑆
such that
𝑓
(
x
1
)
≥
𝑓
(
x
)
for every
x
∈
𝑆
In particular
𝑓
(
x
1
)
≥
𝑓
(
x
0
)
since
x
0
∈
ˆ
𝑆
⊂
𝑆
. This contradicts (3.64) since
x
1
∈
ˆ
𝑆
.
Thus our assumption that
𝑆
⊊
ˆ
𝑆
yields a contradiction. We conclude that
𝑆
=
ˆ
𝑆
3.210
1.
(a)
𝑃
is compact and convex, since it is the product of compact, convex
sets (Proposition 1.2, Exercise 1.165).
(b) Since
x
∈
∑
𝑛
𝑖
=1
conv
𝑆
𝑖
, there exist
x
𝑖
∈
conv
𝑆
𝑖
such that
x
=
∑
𝑛
𝑖
=1
x
𝑖
.
(
x
1
,
x
2
, . . . ,
x
𝑛
)
∈
𝑃
(
x
) so that
𝑃
(
x
)
∕
=
∅
.
(c) By the KreinMillman theorem (or Exercise 3.207),
𝑃
(
x
) has an extreme
point
z
= (
z
1
,
z
2
, . . . ,
z
𝑛
) such that
∙
z
𝑖
∈
conv
𝑆
𝑖
for every
𝑖
∙
∑
𝑛
𝑖
=1
z
𝑖
=
x
.
since
z
∈
𝑃
(
x
).
2.
(a) Exercise 1.176
(b) Since
𝑙 > 𝑚
= dim
𝑋
, the vectors
y
1
,
y
2
, . . . ,
y
𝑙
are linearly dependent
(Exercise 1.143). Consequently, there exists numbers
𝛼
′
1
, 𝛼
′
2
, . . . , 𝛼
′
𝑙
, not all
zero, such that
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 Fall '10
 Dr.DuMond
 Macroeconomics, Existence, All rights reserved, Michael Carter, extreme point, Foundations of Mathematical Economics, conv ????????

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