6.10 Sequences Of Polynomials, Weierstrass Approximation
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 for its
own sake in probability. It is actually an estimate for the variance of a binomial
distribution.
Lemma 6.10.1The following estimate holds for x ∈
[0,1]
and m ≥ 2.
∑m (m ) 1
(k− mx )2 xk(1− x)m−k ≤ -m
k=0 k 4
Proof: Using the binomial theorem,
m∑ (m ) 2 k m −k
k (k − mx) x (1 − x)
k=0
m∑ ( ) ∑m ( )
= m k2xk(1 − x )m −k − 2mx m kxk(1− x)m−k + m2x2
k=0 k k=0 k
m ( ) m ( )
= ∑ m k (k − 1)xk(1− x)m− k + ∑ m kxk (1 − x)m−k
k=0 k k=0 k
∑m ( )
− 2mx m kxk(1− x)m− k + m2x2
k=0 k
∑n (n ) ( k) k n− k
pn(x) ≡ k f n- x (1 − x) . (6.5)
k=0
(6.5)
Theorem 6.10.2The sequence of polynomials in 6.5converges uniformly to fon
[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)− pn(x)| ≤ n ||f k- − f (x)||xk(1− x)n−k
|x−(k∕n)|<δ k n
∑ (n ) || (k ) ||
+ ||f -- − f (x)||xk (1 − x)n− k (6.6)
|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 4.7.2 and 6.7.2 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 6.3.1
|f (x)|
for
x ∈
[0,1]
is bounded by some number M. Thus 6.6 implies that for x ∈
[0,1]
,
( )
|f (x) − p (x)| ≤ ∑ n ε xk(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 6.10.1 there is an estimate for the last sum. Using this estimate yields for all
x ∈