3.142
Clearly, if
𝑓
is convex, it is locally convex at every
x
∈
𝑆
, where
𝑆
is the required
neighborhood. To prove the converse, assume to the contrary that
𝑓
is locally convex
at every
x
∈
𝑆
but it is not globally convex. That is, there exists
x
1
,
x
2
∈
𝑆
such that
𝑓
(
𝛼
x
1
+ (1
−
𝛼
)
x
2
)
> 𝛼𝑓
(
x
1
) + (1
−
𝛼
)
𝑓
(
x
2
)
Let
ℎ
(
𝑡
) =
𝑓
(
𝑡
x
1
+ (1
−
𝑡
)
x
2
)
Local convexity implies that
𝑓
is continuous at every
x
∈
𝑆
(Corollary 3.8.1), and
therefore continuous on
𝑆
.
Therefore,
ℎ
is continuous on [0
,
1].
By the continuous
maximum theorem (Theorem 2.3),
𝑇
= arg
max
x
∈
[
x
1
,
x
2
]
ℎ
(
𝑡
)
is nonempty and compact. Let
𝑡
0
= max
𝑇
. For every
𝜖 >
0,
ℎ
(
𝑡
0
−
𝜖
)
≤
ℎ
(
𝑡
0
) and
ℎ
(
𝑡
0
+
𝜖
)
< ℎ
(
𝑡
0
)
Let
x
0
=
𝑡
0
x
1
+ (1
−
𝑡
0
)
x
2
and
x
𝜖
= (
𝑡
0
+
𝜖
)
x
1
+ (1
−
𝑡
0
−
𝜖
)
x
2
Every neighborhood
𝑉
of
x
0
contains
x
−
𝜖
,
x
𝜖
∈
[
x
1
,
x
2
] with
1
2
𝑓
(
x
−
𝜖
) +
1
2
𝑓
(
x
𝜖
) =
1
2
ℎ
(
𝑡
0
−
𝜖
) +
1
2
ℎ
(
𝑡
0
+
𝜖
)
< ℎ
(
𝑡
0
) =
𝑓
(
x
0
) =
𝑓
(
1
2
x
−
𝜖
+
1
2
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 Fall '10
 Dr.DuMond
 Macroeconomics, X1, Convex function, Michael Carter, Foundations of Mathematical Economics

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