Page 238 - DMTH401_REAL ANALYSIS

Real Analysis

Notes n n æ n ö
å k(k – 1)p (x) = å k(k – 1) ç ÷ x (1 – x) n – k
k
k 0 kn k 0 è k ø
=
=
n æ n - 2ö
k
= å n(n – 1) ç ÷ x (1 – x) n – k
2
k 2 è k - ø
=
n æ n - 2ö
2
= n(n – 1)x å x k – 2 (1 – x) n – k
ç
2
k 2 k - ø ÷
= è
2
æ
-
n 2 n - ö
2 i n – 2 – i 2
= n(n – 1)x å ç ÷ x (1 – x) = n(n – 1)x .
i 0 è i ø
=
Hence
n n
2
å (k – nx) p (x) = å (k – 2knx + n x )p (x)
2 2
2
=
=
k 0 kn k 0 kn
n
2 2
= å (k(k – 1) + k – 2knx + n x )p (x)
kn
=
k 0
2
2
= n(n – 1)x + nx – 2nxnx – n x
2 2
2 2
2
2
n x – nx + nx – 2n x + n x = nx(1 – x).
f
Definition 2: For any f: [0, 1]   define its Bernstein polynomials B (x) such that
n
n æ k ö
f
f
B (x) = å ç ÷ p (x).
n kn
k 0 è ø
n
=
Theorem 2: Weierstrass Approximation Theorem, special case
f
Let f be a real-valued function on [0, 1]. If f is continuous then B  f uniformly.
n
Proof: We want
f
" > 0 N Î  n  N x Î [0, 1] : | B (x) – f(x)| < .
"
"
n
Let  > 0 be given. Since f is uniformly continuous on [0, 1]
 > 0 x,y" Î [0, 1] : |x – y| <  |f(x) – f(y)| < /2.
Using this fact we can estimate
n æ k ö n
f
|B (x) – f(x)|= å ç ÷ p (x) f(x) å p (x)
-
f
n
kn
k 0 è ø kn 
n
k 0
=
=
n n
-
-
= å (f(k/n) f(x))p (x)  å (f(k/n) f(x) p (x)
kn
kn
=
k 0 k 0
=
n n
-
-
= å (f(k/n) f(x) p (x) + å (f(k/n) f(x) p (x)
kn
kn
k: k x- < k: k x- 
n n
 n n
×
< å p (x) 2 f å 1 p (x).
+
kn
kn
2 k: k x- < sup k: k x- 
n n
We used estimate
|f(k/n) – f(x)|  |f(k/n)| +|f(x)|  2||f|| .
sup
Now observe that in the second sum we have the following condition on k
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