134 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES
=m−1
∑k=0
(m−1
k
)xk (1− x)m−1−k
(f( k+1
m
)− f
( km
)1/m
)
By the mean value theorem,f( k+1
m )− f( km )
1/m = f ′(xk,m), xk,m ∈
( km ,
k+1m
). Now the desired
result follows as before from the uniform continuity of f ′ on [0,1]. Let δ > 0 be suchthat if |x− y| < δ , then | f ′ (x)− f ′ (y)| < ε and let m be so large that 1/m < δ/2. Then if∣∣x− k
m
∣∣< δ/2, it follows that∣∣x− xk,m
∣∣< δ and so
∣∣ f ′ (x)− f ′(xk,m)∣∣= ∣∣∣∣∣ f ′ (x)− f
( k+1m
)− f
( km
)1/m
∣∣∣∣∣< ε.
Now as before, letting M ≥ | f ′ (x)| for all x,
∣∣p′m (x)− f ′ (x)∣∣≤ m−1
∑k=0
(m−1
k
)xk (1− x)m−1−k ∣∣ f ′ (xk,m
)− f ′ (x)
∣∣
≤ ∑{x:|x− k
m |< δ2
}(
m−1k
)xk (1− x)m−1−k
ε
+Mm−1
∑k=0
(m−1
k
)4(k−mx)2
m2δ2 xk (1− x)m−1−k
≤ ε +4M14
m1
m2δ2 = ε +M
1
mδ2 < 2ε
whenever m is large enough. Thus this proves uniform convergence. ■There is a more general version of the Weierstrass theorem which is easy to get. It
depends on the Tietze extension theorem, a wonderful little result which is interesting forits own sake.
5.8 A GeneralizationThis is an interesting theorem which holds in arbitrary normal topological spaces. In par-ticular it holds in metric space and this is the context in which it will be discussed. First,review Lemma 3.12.1.
Lemma 5.8.1 Let H,K be two nonempty disjoint closed subsets of X . Then there existsa 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.Then hk → x and since H is closed, x ∈ H.) Similarly, if dist(x,K) = 0, then x ∈ K andso the denominator is never zero as claimed. Hence f is continuous and from its definition,f = 0 on H and f = 1 on K. Now let g(x) ≡ 2
3
(f (x)− 1
2
). Then g has the desired
properties. ■