7.10. THE TIETZE EXTENSION THEOREM 157
Proof: Let H = f−1 ([−1,−1/3]) ,K = f−1 ([1/3,1]) . Thus H and K are disjoint closedsubsets of M. Suppose first H,K are both nonempty. Then by Lemma 7.10.3 there existsg such that g is a continuous function defined on all of X and g(H) =−1/3, g(K) = 1/3,and g(X)⊆ [−1/3,1/3] . It follows || f −g||M < 2/3. If H = /0, then f has all its values in[−1/3,1] and so letting g≡ 1/3, the desired condition is obtained. If K = /0, let g≡−1/3.This proves the lemma.
Lemma 7.10.6 Suppose M is a closed set in X where (X ,d) is a metric space and supposef : M→ [−1,1] is continuous at every point of M. Then there exists a function, g which isdefined and continuous on all of X such that g = f on M and g has its values in [−1,1] .
Proof: Let g1 be such that g1 (X) ⊆ [−1/3,1/3] and || f −g1||M ≤ 23 . Suppose that
g1, · · · ,gm have been chosen such that g j (X)⊆ [−1/3,1/3] and∣∣∣∣∣∣∣∣∣∣ f − m
∑i=1
(23
)i−1
gi
∣∣∣∣∣∣∣∣∣∣M
<
(23
)m
. (7.10.11)
Then ∣∣∣∣∣∣∣∣∣∣(
32
)m(
f −m
∑i=1
(23
)i−1
gi
)∣∣∣∣∣∣∣∣∣∣M
≤ 1
and so( 3
2
)m(
f −∑mi=1( 2
3
)i−1gi
)can play the role of f in the first step of the proof.
Therefore, there exists gm+1 defined and continuous on all of X such that its values arein [−1/3,1/3] and ∣∣∣∣∣
∣∣∣∣∣(
32
)m(
f −m
∑i=1
(23
)i−1
gi
)−gm+1
∣∣∣∣∣∣∣∣∣∣M
≤ 23.
Hence ∣∣∣∣∣∣∣∣∣∣(
f −m
∑i=1
(23
)i−1
gi
)−(
23
)m
gm+1
∣∣∣∣∣∣∣∣∣∣M
≤(
23
)m+1
.
It follows there exists a sequence, {gi} such that each has its values in [−1/3,1/3] and forevery m 7.10.11 holds. Then let
g(x)≡∞
∑i=1
(23
)i−1
gi (x) .
It follows
|g(x)| ≤
∣∣∣∣∣ ∞
∑i=1
(23
)i−1
gi (x)
∣∣∣∣∣≤ m
∑i=1
(23
)i−1 13≤ 1
and since convergence is uniform, g must be continuous. The estimate 7.10.11 impliesf = g on M.
The following is the Tietze extension theorem.