since the term in brackets is strictly greater than 1 for any
𝑥 >
0. Similarly
𝑒
𝑥
𝑥
=
(
𝑒
𝑥/
(
𝑛
+1)
)
𝑛
𝑒
𝑥/
(
𝑛
+1)
(
𝑛
+ 1)
𝑛
(
𝑥
𝑛
+1
)
𝑛
=
1
(
𝑛
+ 1)
𝑛
(
𝑒
𝑥/
(
𝑛
+1)
𝑥/
(
𝑛
+ 1)
)
𝑛
𝑒
𝑥/
(
𝑛
+1)
→ ∞
2.8
Assume that
𝑆
⊆ ℜ
is compact. Then
𝑆
is bounded (Proposition 1.1), and there
exists
𝑀
such that
∣
𝑥
∣ ≤
𝑀
for every
𝑥
∈
𝑆
. For all
𝑛
≥
𝑚
≥
2
𝑀
∣
𝑓
𝑛
(
𝑥
)
−
𝑓
𝑚
(
𝑥
)
∣
=
𝑛
∑
𝑘
=
𝑚
+1
𝑥
𝑘
𝑘
!
≤
𝑥
𝑚
+1
(
𝑚
+ 1)!
𝑛
−
𝑚
∑
𝑘
=0
(
𝑥
𝑚
)
𝑘
≤
𝑀
𝑚
+1
(
𝑚
+ 1)!
𝑛
−
𝑚
∑
𝑘
=0
(
𝑀
𝑚
)
𝑘
≤
𝑀
𝑚
+1
(
𝑚
+ 1)!
(
1 +
1
2
+
1
4
+
⋅ ⋅ ⋅
+
(
1
2
)
𝑛
−
𝑚
)
≤
2
𝑀
𝑚
+1
(
𝑚
+ 1)!
≤
2
(
𝑀
𝑚
+ 1
)
𝑚
+1
≤
(
1
2
)
𝑚
by Exercise 1.206. Therefore
𝑓
𝑛
converges to
𝑓
for all
𝑥
∈
𝑆
.
2.9
This is a special case of Example 2.8. For any
𝑓, 𝑔
∈
𝐹
(
𝑋
), define
(
𝑓
+
𝑔
) =
𝑓
(
𝑥
) +
𝑔
(
𝑥
)
(
𝛼𝑓
)(
𝑥
) =
𝛼𝑓
(
𝑥
)
With these definitions
𝑓
+
𝑔
and
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 Fall '10
 Dr.DuMond
 Macroeconomics, Vector Space, All rights reserved, Metric space, scalar multiplication, Lp space

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