Modern Analysis, Homework 2
Possible solutions
1. Since
{
f
m
}
is uniformly bounded on [
a, b
], there exists
C
∈
R
such that for all
m
∈
N
we have sup
x
∈
[
a,b
]

f
(
x
)
 ≤
M.
This implies that

F
n
(
x
)

=
R
x
a
f
n
(
t
)
dt
≤
R
x
a

f
n
(
t
)

dt
≤
C
(
b

a
), and the sequence
{
F
n
}
is uniformly bounded. Similarly, we have

F
n
(
x
)

F
n
(
y
)
 ≤
C

x

y

, so the sequence
{
F
n
}
is equicontinuous.
By the Arzel`aAscoli
theorem, there exists a uniformly convergent subsequence
{
F
n
k
}
.
2. Recall that for metric spaces, compactness is the same thing as sequential compactness,
so
S
is compact if and only if every sequence in
S
contains a convergent subsequence,
with limit in
S
.
(
⇐
) Let
{
f
n
}
n
∈
N
be a sequence in
S
. By the Arzel`
aAscoli theorem,
{
f
n
}
contains a
uniformly convergent subsequence
{
f
n
i
}
i
∈
N
. Suppose that
f
n
i
→
f
. Since
S
is uniformly
closed, we have
f
∈
S
. Therefore
S
is compact.
(
⇒
) Let’s prove the contrapositive. There are three cases:
(a)
S
is not uniformly closed.
This is the same as saying that
S
is not a closed
subset of the metric space
C
(
K
), but compact subsets of metric spaces are closed
(contradiction).
(b) If
S
is not pointwise bounded, then there exists a point
x
∈
K
and a sequence
{
f
n
}
n
∈
N
in
S
such that

f
n
+1
(
x
)
 ≥ 
f
n
(
x
)

+ 1 for every
n
∈
N
.
Then every
subsequence
{
f
n
i
}
i
∈
N
of
{
f
n
}
n
∈
N
satisfies

f
n
i
(
x
)
 → ∞
as
i
→ ∞
. It follows that
no subsequence of
{
f
n
}
can converge pointwise, much less uniformly. This shows
that there is a sequence in
S
with no convergent subsequence (contradiction).
(c) Suppose that
S
is not equicontinuous. Then there exists
ε
0
>
0 and a sequence
{
f
n
}
n
∈
N
, sequences
{
x
n
}
n
∈
N
,
{
y
n
}
n
∈
N
in
K
such that

f
n
(
x
n
)

f
n
(
y
n
)
 ≥
ε
0
yet

x
n

y
n

<
1
n
.
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 Summer '11
 Staff
 Taylor Series, Trigraph, Metric space, Compact space, convergent subsequence

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