3The topology of C(K)
II Linear Analysis
3.3 Arzel`a-Ascoli theorem
Let
K
be compact Hausdorff, and
{f
n
}
∞
n=1
be a sequence of continuous functions
f
n
:
K → R
(or
C
). When does (
f
n
) have a convergent subsequence in
C
(
K
)?
In other words, when is there a subsequence which converges uniformly?
This will be answered by the Arzel`a-Ascoli theorem. Before we get to that,
we look at some examples.
Example.
Let
K
= [0
,
1], and
f
n
(
x
) =
n
. This does not have a convergent
subsequence in C(K) since it does not even have a subsequence that converges
pointwise. This is since f
n
is unbounded.
We see that unboundedness is one “enemy” that prevents us from having a
convergent subsequence.
Example. We again let K[0, 1]), and let f be defined as follows:
1
1
n
1
We know that
f
n
does not have a convergent subsequence in
C
(
K
), since any
subsequence must converge pointwise to
f(x) =
(
0 x 6= 0
1 x = 0
,
which is not continuous.
What is happening here? For every n, fixed x and every ε, by continuity of
f
n
, there ie some
δ
such that
|x − y| < δ
implies
|f
n
(
x
)
− f
n
(
y
)
| < ε
, but this
choice of
δ
depends on
n
, and there is no universal choice that works for us. This
is another problem that leads to the lack of a limit.
The Arzel`a-Ascoli theorem tells us these are the only two “enemies” that
prevent us from extracting a convergent subsequence.
To state this theorem, we first need a definition.
Definition
(Equicontinuous)
.
Let
K
be a topological space, and
F ⊆ C
(
K
).
We say
F
is equicontinuous at
x ∈ K
if for every
ε
, there is some
U
which is an
open neighbourhood of x such that
(∀f ∈ F )(∀y ∈ U) |f(y) − f(x)| < ε.
We say F is equicontinuous if it is equicontinuous at x for all x ∈ K.
Theorem
(Arzel`a-Ascoli theorem)
.
Let
K
be a compact topological space. Then
F ⊆ C
(
K
) is pre-compact, i.e.
¯
F
is compact, if and only if
F
is bounded and
equicontinuous.
This indeed applies to the problem of extracting a uniformly convergent
subsequence, since
C
(
K
) is a metric space, and compactness is equivalent to
sequential compactness. Indeed, let (
f
n
) be a bounded and equicontinuous
sequence in
C
(
K
). Then
F
=
{f
n
:
n ∈ N} ⊆ C
(
K
) is bounded and equicon-
tinuous. So it is pre-compact, and hence (
f
n
), being a sequence in
¯
F
, has a
convergent subsequence.
To prove this, it helps to introduce some more terminology and a few lemmas
first.
Definition
(
ε
-net)
.
Let
X
be a metric space, and let
E ⊆ X
. For
ε >
0, we say
that N ⊆ X is an ε-net for E if and only if
S
x∈N
B(x, ε) ⊇ E.
Definition
(Totally bounded subset)
.
Let
X
be a metric space, and
E ⊆ X
.
We say that E is totally bounded for every ε, there is a finite ε-net N
ε
for E.
An important result about totally bounded subsets is the following:
Proposition.
Let
X
be a complete metric space. Then
E ⊆ X
is totally
bounded if and only if for every sequence
{y
i
}
∞
i=1
⊆ E
, there is a subsequence
which is Cauchy.
By completeness, we can rewrite this as
Corollary.
Let
X
be a complete metric space. Then
E ⊆ X
is totally bounded
if and only if
¯
E is compact.
We’ll prove these later. For now, we assume this corollary and we will prove
Arzel`a-Ascoli theorem.
Theorem
(Arzel`a-Ascoli theorem)
.
Let
K
be a compact topological space. Then
F ⊆ C
(
K
) is pre-compact, i.e.
¯
F
is compact, if and only if
F
is bounded and
equicontinuous.
Proof.
By the previous corollary, it suffices to prove that
F
is totally bounded if
and only if F is bounded and equicontinuous. We first do the boring direction.
(
⇒
) Suppose
F
is totally bounded. First notice that
F
is obviously bounded,
since F can be written as the finite union of ε-balls, which must be bounded.
Now we show
F
is equicontinuous. Let
ε >
0. Since
F
is totally bounded,
there exists a finite
ε
-net for
F
, i.e. there is some
{f
1
, ··· , f
n
} ⊆ F
such that
for every f ∈ F , there exists an i ∈ {1, ··· , n} such that kf − f
i
k
C(K)
< ε.
Consider a point
x ∈ K
. Since
{f
1
, ··· , f
n
}
are continuous, for each
i
, there
exists a neighbourhood U
i
of x such that |f
i
(y) − f
i
(x)| < ε for all y ∈ U
i
.
Let
U =
n
\
i=1
U
i
.
Since this is a finite intersection,
U
is open. Then for any
f ∈ F
,
y ∈ U
, we can
find some i such that kf − f
i
k
C(K)
< ε. So
|f(y) − f(x)| ≤ |f(y) − f
i
(y)| + |f
i
(y) − f
i
(x)| + |f
i
(x) −f (x)| < 3ε.
So F is equicontinuous at x. Since x was arbitrary, F is equicontinuous.
(
⇐
) Suppose
F
is bounded and equicontinuous. Let
ε >
0. By equicontinuity,
for every
x ∈ K
, there is some neighbourhood
U
x
of
x
such that
|f
(
y
)
−f
(
x
)
| < ε
for all y ∈ U
x
, f ∈ F . Obviously, we have
[
x∈K
U
x
= K.
By the compactness of K, there are some {x
1
, ··· , x
n
} such that
n
[
i=1
U
x
i
⊇ K.
Consider the restriction of functions in
F
to these points. This can be viewed
as a bounded subset of
`
n
∞
, the
n
-dimensional normed vector space with the
supremum norm. Since this is finite-dimensional, boundedness implies total
boundedness (due to, say, the compactness of the closed unit ball). In other
words, there is a finite
ε
-net
{f
1
, ··· , f
n
}
such that for every
f ∈ F
, there is a
j ∈ {1, ··· , m} such that
max
i
|f(x
i
) −f
j
(x
i
)| < ε.
Then for every f ∈ F , pick an f
j
such that the above holds. Then
kf − f
j
k
C(K)
= sup
y
|f(y) − f
j
(y)|
Since {U
x
i
} covers K, we can write this as
= max
i
sup
y∈U
x
i
|f(y) − f
j
(y)|
≤ max
i
sup
y∈U
x
i
|f(y) − f(x
i
)| + |f (x
i
) −f
j
(x
i
)| + |f
j
(x
i
) −f
j
(y)|
< ε + ε + ε = 3ε.
So done.
Now we return to prove the proposition we just used to prove Arzel`a-Ascoli.
Proposition.
Let
X
be a (complete) metric space. Then
E ⊆ X
is totally
bounded if and only if for every sequence
{y
i
}
∞
i=1
⊆ E
, there is a subsequence
which is Cauchy.
Proof.
(
⇒
) Let
E ⊆ X
be totally bounded,
{y
i
} ∈ E
. For every
j ∈ N
, there
exists a finite
1
j
-net, call it N
j
.
Now since
N
1
is finite, there is some
x
1
such that there are infinitely many
y
i
’s in B(x
1
, 1). Pick the first y
i
in B(x
1
, 1) and call it y
i
1
.
Now there is some
x
2
∈ N
2
such that there are infinitely many
y
i
’s in
B
(
x
1
,
1)
∩ B
(
x
2
,
1
2
). Pick the one with smallest value of
i > i
1
, and call this
y
i
2
.
Continue till infinity.
This procedure gives a sequence x
i
∈ N
i
and a subsequence {y
i
k
}, and also
y
i
n
∈
n
\
j=1
B
x
j
,
1
j
.
It is easy to see that {y
i
n
} is Cauchy since if m > n, then d(y
i
m
, y
i
n
) <
2
n
.
(
⇐
) Suppose
E
is not totally bounded. So there is no finite
ε
-net. Pick any
y
1
. Pick y
2
such that d(y
1
, y
2
) ≥ ε. This exists because there is no finite ε-net.
Now given
y
1
, ··· , y
n
such that
d
(
y
i
, y
j
)
≥ ε
for all
i, j
= 1
, ··· , n
,
i 6
=
j
,
we pick
y
n+1
such that
d
(
y
n+1
, y
j
)
≥ ε
for all
j
= 1
, ··· , n
. Again, this exists
because there is no finite
ε
-net. Then clearly any subsequence of
{y
n
}
is not
Cauchy.
Note that the first part is similar to the proof of Bolzano-Weierstrass in
R
n
by repeated bisection.
Recall that at the beginning of the chapter, we have seen that boundedness
and equicontinuity assumptions are necessary. The compactness of
K
is also
important. Let
X
=
R
, which is not compact, and let
φ ∈ C
∞
c
, an infinitely
differentiable function with compact support, say, the bump function.
x
We now let
f
n
(
x
) =
φ
(
x − n
), i.e. we shift our bump function to the right by
n
units. This sequence is clearly bounded and equicontinuous, but this has no
convergent subsequence —
f
n
converges pointwise to the zero function, but the
convergence is not uniform, and this is true for arbitrary subsequences as well.
We are going to look at some applications of the theorem:
Example.
Let
K ⊆ R
be a compact space,
{f
n
}
∞
n=1
be a sequence of continu-
ously differentiable functions in C(K), such that
sup
x
sup
n
(|f
n
(x)| + |f
0
n
(x)|) < C
for some
C
. Then there is a convergent subsequence. We clearly have uniform
boundedness. To obtain equicontinuity, since the derivative is bounded, by the
mean value theorem, we have
|f
n
(x) −f
n
(y)|
|x −y|
= |f
0
n
(z)| ≤ C
So
|f
n
(x) −f
n
(y)| ≤ C|x −y|.
Consider the ordinary differential equation
x
0
=
f
(
x
) with the boundary
conditions
x
(0) =
x
0
∈ R
n
. Recall from IB Analysis II that Picard-Lindel¨of
theorem says that if
f
is a Lipschitz function, then there exists some
ε >
0 such
that the ODE has a unique solution in (−ε, ε).
What if f is not Lipschitz? If so, we can get the following
Theorem
(Peano*)
.
Given
f
continuous, then there is some
ε >
0 such that
x
0
= f(x) with boundary condition x(0) = x
0
∈ R has a solution in (−ε, ε).
Note that uniqueness is false. For example, if
x
0
=
p
|x|
,
x
(0) = 0, then
x(t) = 0 and x(t) = t
2
are both solutions.
Proof.
(sketch) We approximate
f
by a sequence of continuously differentiable
functions
f
n
such that
kf − f
n
k
C(K)
→
0 for some
K ⊆ R
. We use Picard-
Lindel¨of to get a solution for all
n
. Then we use the ODE to get estimates for
the solution. Finally, we can use Arzel`a-Ascoli to extract a limit as
n → ∞
. We
can then show it is indeed a solution.