This will use the characterization of compact metric spaces to give a proof of a general version
of the Arzella Ascoli theorem. See Naylor and Sell [24] which is where I saw this general
formulation.
Definition 2.10.1Let
(X, dX)
be a compact metric space. Let
(Y,dY)
be anothercomplete metric space. Then C
(X, Y)
will denote the continuous functions which map X toY . Then ρ is a metric on C
(X,Y )
defined by
ρ(f,g) ≡ sxu∈pX dY (f (x) ,g (x )).
Theorem 2.10.2
(C(X, Y),ρ)
is a complete metric space.
Proof:It is first necessary to show that ρ is well defined. In this argument, I will just
write d rather than d_{X} or d_{Y }. To show this, note that from Lemma 2.2.5,
x → d(f (x),g(x))
is a continuous function. Here is a review why this is so. It happens because following
inequality holds.
are metricspaces. Thus A is a set of continuous functions mapping X to Y . Then A is said to beequicontinuousif for every ε > 0 there exists a δ > 0 such that if d_{X}
(x1,x2)
< δ thenfor all f ∈A, d_{Y }
(f (x1),f (x2))
< ε. (This is uniform continuity which is uniform inA.) A is said to be pointwise compactif
{f (x) : f ∈ A}
has compact closure in Y .
Here is the Ascoli Arzela theorem.
Theorem 2.10.5Let
(X, dX)
be a compact metric space and let
(Y,dY )
be acomplete metric space. Thus
(C (X, Y) ,ρ)
is a complete metric space. Let A⊆ C
(X,Y )
be pointwise compact and equicontinuous. Then Ais compact. Here the closure is takenin
(C(X, Y),ρ)
.
Proof:The more useful direction is that the two conditions imply compactness of A. I
prove this first. Since A is a closed subset of a complete space, it follows from Theorem 2.5.5,
that A will be compact if it is totally bounded. In showing this, it follows from Lemma 2.10.3
that it suffices to verify that A is totally bounded. Suppose this is not so. Then there
exists ε > 0 and a sequence of points of A,
{fn}
such that ρ
(fn,fm)
≥ ε whenever
n≠m.
By equicontinuity, there exists δ > 0 such that if d
(x,y)
< δ, then d_{Y }
(f (x),f (y))
<
-ε
8
for
all f ∈A. Let
{xi}
_{i=1}^{p} be a δ net for X. Since there are only finitely many x_{i}, it follows from
pointwise compactness that there exists a subsequence, still denoted by
{fn}
which converges
at each x_{i}. Now let x ∈ X be arbitrary. There exists N such that for each x_{i} in that δ
net,
dY (fn(xi),fm (xi)) < ε∕8 whenever n,m ≥ N
Then for m,n ≥ N,
dY (fn(x),dYm (x)) ≤ dY (fn(x),fn(xi))+ dY (fn(xi),fm(xi))+ dY (fm (xi),fm (x))
< dY (fn(x),fn(xi))+ ε∕8+ dY (fm (xi),fm (x))
Pick x_{i} such that d
(x,xi)
< δ.
{xi}
_{i=1}^{p} is a δ net and so this is surely possible. Then by
equicontinuity, the two ends are each less than ε∕8 and so for m,n ≥ N,
3ε
dY (fn(x),fm (x)) ≤--
8
Since x is arbitrary, it follows that ρ
(f ,f )
n m
≤ 3ε∕8 < ε which is a contradiction. It
follows that A and hence A is totally bounded. This proves the more important
direction.
Next suppose A is compact. Why must A be pointwise compact and equicontinuous? If it
fails to be pointwise compact, then there exists x ∈ X such that
{f (x) : f ∈ A}
is not
contained in a compact set of Y . Thus there exists ε > 0 and a sequence of functions in A
{fn}
such that d
(fn (x),fm (x))
≥ ε. But this implies ρ
(fm,fn)
≥ ε and so A fails to be
totally bounded, a contradiction. Thus A must be pointwise compact. Now why must it be
equicontinuous? If it is not, then for each n ∈ ℕ there exists ε > 0 and x_{n},y_{n}∈ X such that
d
(xn,yn)
< 1∕n but for some f_{n}∈A, d
(fn (xn),fn(yn))
≥ ε. However, by compactness,
there exists a subsequence