It turns out that if f is a continuous real valued function defined on an interval,

[a,b]

then there exists a sequence of polynomials,

{pn}

such that the sequence converges
uniformly to f on

[a,b]

. I will first show this is true for the interval

[0,1]

and then verify
it is true on any closed and bounded interval. First here is a little lemma which is
interesting in probability. It is actually an estimate for the variance of a binomial
distribution.

Lemma 10.8.1The following estimate holds for x ∈

[0,1]

and m ≥ 2.

( )
∑m m 2 k m− k 1
k (k− mx ) x (1− x) ≤ 4 m
k=0

Proof: First of all, from the binomial theorem

( ) ( )
m∑ m ( t(k− mx)) k m −k −tmx m∑ m ( tk) k m− k
k e x (1− x) = e k e x (1− x)
k=0 k=0

( )
= e−tmx 1− x + xetm = e−tmxg(t)m , g(0) = 1,g′(0) = g′′(0) = x

Take a partial derivative with respect to t twice.

( )
∑m m 2 t(k−mx) k m −k
k (k− mx ) e x (1− x)
k=0

= (mx )2e−tmxg (t)m + 2(− mx )e−tmxmg (t)m −1g′(t)
−tmx [ m −2 ′ 2 m −1 ′′ ]
+e m (m − 1)g(t) g (t) + mg (t) g (t)

Now let t = 0 and note that the right side is m(x − x^{2}) ≤ m∕4 for x ∈

[0,1]

.
Thus

( )
∑m m 2 k m −k 2
k (k− mx ) x (1 − x) = mx − mx ≤ m∕4 ■
k=0

Now let f be a continuous function defined on

[0,1]

. Let p_{n} be the polynomial
defined by

( ) ( )
∑n n k- k n−k
pn(x) ≡ k f n x (1− x) . (10.11)
k=0

(10.11)

Theorem 10.8.2The sequence of polynomials in 10.11convergesuniformly to f on

[0,1]

. These polynomials are called the Bernstein polynomials.

Proof: By the binomial theorem,

n ( ) n ( )
f (x) = f (x)∑ n xk(1− x)n−k = ∑ n f (x)xk(1 − x )n−k
k=0 k k=0 k

and so by the triangle inequality

( ) | ( ) |
∑n n || k- || k n−k
|f (x)− pn (x )| ≤ k |f n − f (x)|x (1− x)
k=0

At this point you break the sum into two pieces, those values of k such that k∕n
is close to x and those values for k such that k∕n is not so close to x. Thus

( ) | ( ) |
|f (x)− p (x)| ≤ ∑ n ||f k- − f (x)||xk(1− x)n−k
n |x−(k∕n)|< δ k | n |
∑ ( ) || ( ) ||
+ n ||f k- − f (x)||xk(1− x)n−k(10.12)
|x−(k∕n)|≥δ k n

where δ is a positive number chosen in an auspicious manner about to be described. Since
f is continuous on

[0,1]

, it follows from Theorems 3.5.2 and 4.7.2 that f is uniformly
continuous. Therefore, letting ε > 0, there exists δ > 0 such that if

|x − y|

< δ, then

|f (x) − f (y)|

< ε∕2. This is the auspicious choice for δ. Also, by Lemma 4.3.1

|f (x)|

for
x ∈

[0,1]

is bounded by some number M. Thus 10.12 implies that for x ∈

[0,1]

,

( )
|f (x)− p (x)| ≤ ∑ n εxk(1− x)n−k
n |x−(k∕n)|<δ k 2
∑ ( )
+2M n xk (1 − x)n− k
|nx−k|≥nδ k
∑ ( ) 2
≤ ε+ 2M n (k-−-nx)-xk(1− x)n−k
2 |nx−k|≥nδ k n2δ2
∑n ( )
≤ ε+ -2M- n (k− nx )2xk (1− x)n−k
2 n2δ2 k=0 k

Now by Lemma 10.8.1 there is an estimate for the last sum. Using this estimate yields for
all x ∈