(b)
?
(10 bonus points)
Prove that
{
f
n
}
has
a
subsequence
{
f
n
k
}
converging pointwise to some
f
:
R
→
R
such that
for every closed and bounded interval [
a,b
] the sequence of restrictions
{
(
f
n
k
)

[
a,b
]
}
converges uniformly to
f

[
a,b
]
.
Solution.
For this part, one needed to assume that the sequence was uniformly bounded. I’ll give credit for
pointing this out. Even more credit for realizing that it was needed, adding it as a hypothesis, and proving the
result. I assume from now on equicontinuity and uniform boundedness. This time ArzelaAscoli only does 70% or
so of the job. I’ll merely sketch a proof, but ﬁlling in the details should be easy (in a perfect world?)
By Arzela Ascoli, the sequence has a subsequence converging uniformly to some continuous
g
1
on the interval
[

1
,
1]. This subsequence has in turn a subsequence converging uniformly to some
g
2
in [

2
,
2]. Necessarily,
the restriction of
g
2
to [

1
,
1] must be
g
1
. The subsequence of the subsequence now has in turn a subsequence
converging uniformly to some continuous
g
3
on the interval [

3
,
3
,
]. And so forth. The functions
g
m
: [

m,m
]
to
R
satisfy
g
m

[

k,k
]
=
g
k
if
k
≤
m
and this allows us to deﬁne
f
:
R
→
R
by
f
(
x
) =
g
m
(
x
) if
x
∈
[

m,m
];
f
is
then clearly (?) continuous. Returning to our subsequence of a subsequence of a subsequence.
.., the diagonal
subsequence; that is the sequence whose
n
th element is the
n
th element of the
n
th subsequence, is a subsequence
of the original subsequence converging uniformly to
g
m
; i.e., to
f
, on every interval [

m,m
]. Since every [
a,b
] is
contained in some (and many) interval [

m,m
], this diagonal subsequence converges uniformly to
f
on all closed
and bounded intervals.
5.
(20 points)
Let [
a,b
] be a closed and bounded interval and assume
f
: [
a,b
]
→
R
is continuous and such that
Z
b
a
x
3
k
f
(
x
)
dx
= 0
for
k
= 0
,
1
,
2
,...
. Prove:
f
(
x
) = 0 for all
x
∈
[
a,b
].
Solution.
It is immediate that if
p
(
x
) =
∑
n
k
=0
a
k
x
3
k
, then
Z
b
a
p
(
x
)
f
(
x
)
dx
= 0. Let
A
be the set of all polynomials of
this type, restricted to the interval [
a,b
]. That is,
g
∈ A
iﬀ
g
=
p

[
a,b
]
where
p
is a polynomials involving only powers that
3
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View Full Documentare multiples of 3. It is immediate that
A
is a subalgebra of
C
([
a,b
]). It separates points; if
x
6
=
y
, then
x
3
6
=
y
3
. Since
1
∈ A
, for every
x
∈
[
a,b
] there is
g
∈ A
with
g
(
x
)
6
= 0. Thus
A
satisﬁes all hypotheses of StoneWeierstrass, hence is dense
in
C
([
a,b
]). In particular, there is a sequence
{
g
n
}
in
C
([
a,b
]) converging uniformly to
f
. It is then very easy, immediate
perhaps, that
g
n
f
converges uniformly to
f
2
. We then have
Z
b
a
f
(
x
)
2
dx
= lim
n
→∞
Z
b
a
g
n
(
x
)
f
(
x
)
dx
= 0
.
(Certain steps that were detailed in the solution of a homework exercise were skipped). Since
f
2
≥
0 and continuous, we
conclude that
f
= 0.
4
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 Spring '11
 Speinklo
 Calculus, Topology, FN, Arzela Ascoli

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