Basic Analysis – Gently Done Topological Vector Spaces

38 Basic Analysis
that g 0 (x) = − 1
3 for x ∈ A 0 and g 0
(x) = 1
3 for x ∈ B 0 . Then |(f −g 0
)(x)| ≤ 2
3 for
each x ∈ A, so that f −g0 is a continuous real-valued function on A with values
is a continuous map from A
in the interval [− 2
3
, 2
3 ]. Set f 1
= 3
2 (f −g 0 ). Then f 1
into R with values in [−1,1].
Set f 0 = f, and suppose that n ≥ 0 and that f n : A → R has been constructed
such that f n is continuous and takes its values in the interval [−1,1]. The above
argument can be applied to f n instead of f to yield a continuous map g n : X → R
with values in [− 1
3
, 1
3 ] such that f n − g n
f n+1 = 3
2 (f n −g n ), so that f n+1
Thus f n and g n are defined recursively for all n ≥ 0.
By construction,
f = f 0 = g 0 + 2
3 f 1
Put sn = g0 + 2
3g1 + ···+� 2
3
has values in [−2 3 ]. We then put
: A → R is continuous and has values in [−1,1].
� 2f2
= g0 + 2
3g1 +� 2
3
= g0 + 2
3g1 +� �
2 2g2
3 + � �
2 3f3
3
= g0 + 2
3g1 +···+� �
2 ngn
+ 3
� 2
3
, 2
3
� n+1fn+1 .
� ngn on X. Then it is clear that (s n ) is a Cauchy
sequence in C b (X,R) so that there is g such that s n → g in C b (X,R). Since
|gi (x)| ≤ 1
3 for each i ≥ 0, we see that sn for all x ∈ X. But we also have that, for any x ∈ A, � 2
3
n → ∞. It follows that, for x ∈ A,
f(x) = s n (x)+ � 2
3
→ g(x)+0
(x) ≤ 1 and so it follows that |g(x)| ≤ 1
�n+1fn+1 (x) → 0 in R as
� n+1fn+1 (x)
and so g = f on A, and g : X → R is a continuous extension of f, as required.
(ii) Suppose now that f : A → R is continuous and that f takes values in the
open interval (−1,1). We shall show that f can be extended to a continuous
function with values also in (−1,1). Certainly, f takes its values in [−1,1] and
so, as above, there is a continuous map g : X → R with values in [−1,1] such
that g = f on A. We must modify g to remove the possibility of the values ±1.
Set D = g −1 ({−1,1}). Then D is a closed subset of X, since g is continuous.
Furthermore, on A, g agrees with f which does not assume either of the values 1
or −1, so A∩D = ∅. By Urysohn’s lemma, there is a continuous map ϕ : X → R
such that ϕ takes its values in [0,1] and such that ϕ vanishes on D and is equal to
1 on A. Put h(x) = ϕ(x)g(x) for x ∈ X. Then h : X → R is continuous, since it
is the product of continuous maps. Moreover, for any x ∈ A, h(x) = ϕ(x)g(x) =
1g(x) = f(x). Finally, we note that for x ∈ D, h(x) = ϕ(x)g(x) = 0, and, for
x /∈ D, |h(x)| = |ϕ(x)||g(x)|< 1 since |ϕ(x)| ≤ 1, for all x ∈ X, and |g(x)| < 1 for
x /∈ D. Hence h has values in the interval (−1,1).
Department of Mathematics King’s College, London