16.2. FUNCTIONS OF MANY VARIABLES 445
=m
∑k=1
m(m−1)!(m− k)!(k−1)!
xk−1 (1− x)m−k f(
km
)
−m−1
∑k=0
(mk
)xk (m− k)(1− x)m−1−k f
(km
)
=m−1
∑k=0
m(m−1)!(m−1− k)!k!
xk (1− x)m−1−k f(
k+1m
)−
m−1
∑k=0
m(m−1)!(m−1− k)!k!
xk (1− x)m−1−k f(
km
)
=m−1
∑k=0
m(m−1)!(m−1− k)!k!
xk (1− x)m−1−k(
f(
k+1m
)− f
(km
))
=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]. Letδ > 0 be such that 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. ■