showing that f is continuous. Thus the sequence of functions converges uniformly to a function f ∈ C_{0}
(X)
which is what it means to be complete. Every Cauchy sequence converges. Indeed 2.19 says that
∥f − f ∥
n
_{∞}< ε for f ∈ C_{0}
(X)
. ■
The above refers to functions which have values in ℂ but the same proof works for functions which have
values in any complete normed linear space.
In the case where the functions in C_{0}
(X )
all have real values, I will denote the resulting space by
C_{0}
(X;ℝ )
with similar meanings in other cases.
The following has to do with a trick which will enable a result valid on sets which are only closed rather
than compact. In general, you consider a locally compact Hausdorff space instead of a closed subset of ℝ^{n}.
Consider the unit sphere in ℝ^{n+1}, centered at the point
(0,⋅⋅⋅,0,1)
≡
(⃗ )
0,1
.
{ ∑n }
Sn ≡ ⃗x ∈ ℝn+1 : (xn+1 − 1)2 + x2k = 1
k=1
Define a map from ℝ^{n} which is identified with ℝ^{n}×
{0}
to the surface of this sphere as follows. Extend a
line from the point,
⃗p
in ℝ^{n} to the point
(⃗0,2)
on the top of this sphere and let θ
(p)
denote the point of
this sphere which the line intersects.
PICT
This map θ is one to one onto S^{n}∖
{( )}
⃗0,2
. More precisely, if you have
(⃗a,an+1)
on S^{n}∖
( )
⃗0,2
to
get θ^{−1}
(⃗a,an+1)
, you form the line from
( )
⃗0,2
through this point and see where it hits ℝ^{n}. The line is
( )
⃗0,2
+ t
( ( ) )
(⃗a,a )− ⃗0,2
n+1
and it hits ℝ^{n} when 2 + t
(a − 2)
n+1
= 0 which is when t =
---2--
2−an+1
. Thus
θ^{−1}
(⃗a,an+1)
=
( )
2−2a⃗an+1,0
. From this formula, it is clear that θ^{−1} is continuous and one to one. It is also
onto because if
⃗x
∈ ℝ^{n} you can take the line from
(⃗ )
0,2
to
(⃗x,0)
and where it intersects S^{n} is the point
which is wanted. It is also easy to see from this that θ is continuous. Indeed, suppose
⃗x
_{k}→
⃗x
in ℝ^{n}. Does it
follow that θ
(⃗xk)
→ θ
(⃗x)
? We know that
{⃗xk }
is bounded since it converges. Therefore, there is an open
ball, B
( ( ) )
⃗0,2 ,r
such that θ
(⃗xk)
∈ S^{n}∖ B
(( ) )
⃗0,2 ,r
≡ K a compact set. If θ
(⃗xk)
fails to
converge to θ
(⃗x)
, then there is a subsequence, still denoted as θ
(⃗xk)
such that θ
(⃗xk)
→ y ∈ K
where y≠θ
(⃗x)
. But then, the continuity of θ^{−1} implies x_{k}→ θ^{−1}
(y)
and so θ^{−1}
(y)
= x which
implies y = θ
(x )
, a contradiction. Thus both θ and θ^{−1} are continuous, one to one and onto
mappings.
Note that S^{n} is a compact subset of ℝ^{n+1} so the earlier theorem applies to S^{n}. Consider θ
(X )
. If you
have a sequence θx_{n}→ y ∈ S^{n}∖
{(⃗ )}
0,2
, then by continuity of θ^{−1}, you obtain x_{n}→ θ^{−1}y. Since X is
closed, θ^{−1}y ∈ X. Therefore, y ∈ θ
(X )
and so θ
(X )
is a closed subset of the compact set S^{n} and is
therefore, compact.
Theorem 2.8.11Let A be an algebra of functions in C_{0}
(X; ℝ)
where X is a closed subset of ℝ^{n}which separates the points and annihilates no point. Then A is dense in C_{0}
(X; ℝ)
.
Proof:
A^
denote all finite linear combinations of the form
{ }
∑n ^
cifi + c0 : f ∈ A, ci ∈ ℝ
i=1
where for f ∈ C_{0}
(X;ℝ)
,
{ f (θ−1(x)) if x ∈ θ(X )
^f (x) ≡ ( ) .
0 if x = ⃗0,2
Then
A^
is obviously an algebra of functions in C
(Sn;ℝ )
. It separates points because this is true of A.
Similarly, it annihilates no point because of the inclusion of c_{0} an arbitrary element of ℝ in the definition of
= 0, you
can take each c_{0}^{n} = 0 and so this has shown the existence of a sequence of functions in A
such that it converges uniformly to f. The points of X are of the form θ^{−1}