60 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

As to the other terms, 0≤ f (x)≤ 1/2 and so∣∣∣∣∣m−1

∑k=0

(34

)k(f(

4k (x+δ m))− f

(4kx))∣∣∣∣∣≤ m−1

∑k=0

(34

)k

=1− (3/4)m

1/4= 4−4

(34

)m

Thus∣∣∣ g(x+δ m)−g(x)

δ m

∣∣∣ ≥ 3m−(

4−4( 3

4

)m)≥ 3m− 4. Since δ m→ 0 as m→ ∞, g′ (x) does

not exist because the difference quotients are not bounded. ■This proves the following theorem.

Theorem 2.5.43 There exists a function defined on R which is continuous andbounded but fails to have a derivative at any point.

Proof: It only remains to verify that the function just constructed is bounded. However,0≤ g(x)≤ 1

2 ∑∞k=0( 3

4

)k= 2. ■

Note that you could consider (ε/2)g(x) to get a function which is continuous, hasvalues between 0 and ε which has no derivative.

2.6 Tietze Extension TheoremThis is an interesting theorem which holds in arbitrary normal topological spaces. However,I am specializing to a metric space X to keep the emphasis on that which is most familiar.The presentation depends on Lemma 2.4.8.

Lemma 2.6.1 Let H,K be two nonempty disjoint closed subsets of a metric space X .Then there exists a continuous function, g : X → [−1/3,1/3] such that g(H) = −1/3,g(K) = 1/3,g(X)⊆ [−1/3,1/3] .

Proof: Let f (x)≡ dist(x,H)dist(x,H)+dist(x,K) . The denominator is never equal to zero because if

dist(x,H) = 0, then x∈H because H is closed. (To see this, pick hk ∈ B(x,1/k)∩H. Thenhk → x and since H is closed, x ∈ H.) Similarly, if dist(x,K) = 0, then x ∈ K and so thedenominator is never zero as claimed because it is not possible for a point to be in both Hand K. Hence f is continuous and from its definition, f = 0 on H and f = 1 on K. Now letg(x)≡ 2

3

(f (x)− 1

2

). Then g has the desired properties. ■

Definition 2.6.2 For f : M ⊆ X → R, define ∥ f∥M as

sup{| f (x)| : x ∈M} .

Lemma 2.6.3 Suppose M is a closed set in X and suppose f : M→ [−1,1] is continuousat every point of M. Then there exists a function g which is defined and continuous on allof Rp such that ∥ f −g∥M < 2

3 , g(X)⊆ [−1/3,1/3] .

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 2.6.1 there exists gsuch that g is a continuous function defined on all of X and g(H) = −1/3, g(K) = 1/3,and g(Rp)⊆ [−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.If both H,K = /0, there isn’t much to show. Just let g(x) = 0 for all x. ■