4.10. POLYNOMIALS AND CONTINUOUS FUNCTIONS 111

Lemma 4.10.1 The following estimate holds for x ∈ [0,1] and n ≥ 2.

n

∑k=0

(nk

)(k−nx)2 xk (1− x)n−k ≤ 1

4n

Proof: Here are some observations. ∑nk=0(n

k

)kxk (1− x)n−k =

nxn

∑k=1

(n−1)!(k−1)!((n−1)− (k−1))!

xk−1 (1− x)(n−1)−(k−1)

= nxn−1

∑k=0

(n−1

k

)xk (1− x)n−1−k = nx

n

∑k=0

(nk

)k (k−1)xk (1− x)n−k

= n(n−1)x2n

∑k=2

(n−2)!(k−2)!(n−2− (k−2))!

xk−2 (1− x)(n−2)−(k−2)

= n(n−1)x2n−2

∑k=0

(n−2

k

)xk (1− x)(n−2)−k = n(n−1)x2

Now (k−nx)2 = k2−2knx+n2x2 = k (k−1)+k (1−2nx)+n2x2. From the above and thebinomial theorem, ∑

nk=0(n

k

)(k−nx)2 xk (1− x)n−k =

n

∑k=0

(nk

)k (k−1)xk (1− x)n−k +(1−2nx)

n

∑k=0

(nk

)kxk (1− x)n−k

+n2x2n

∑k=0

(nk

)xk (1− x)n−k = n(n−1)x2 +(1−2nx)nx+n2x2

= nx(1− x)≤ n14

Now let f be a continuous function defined on [0,1] . Let pn be the polynomial definedby

pn (x)≡n

∑k=0

(nk

)f(

kn

)xk (1− x)n−k . (4.4)

Theorem 4.10.2 The sequence of polynomials in 4.4 converges uniformly to f on[0,1].

Proof: By the binomial theorem,

f (x) = f (x)n

∑k=0

(nk

)xk (1− x)n−k =

n

∑k=0

(nk

)f (x)xk (1− x)n−k

and so by the triangle inequality

| f (x)− pn (x)| ≤n

∑k=0

(nk

)∣∣∣∣ f ( kn

)− f (x)

∣∣∣∣xk (1− x)n−k .