Page 322 - Elements of Distribution Theory

P1: JZP
052184472Xc10 CUNY148/Severini May 24, 2005 2:50

308 Orthogonal Polynomials

Fix > 0. By the Weierstrass approximation theorem, there exists a polynomial q n such
that
sup | f (x) − q n (x)| < .
|x|≤1
Hence,
1 1
2 2 2 2
f (x) dF(x) ≤ − q n (x) dF(x) ≤ .
−1 −1
Since is arbitrary,

1
2
f (x) dF(x) = 0,
−1
establishing completeness.
Note that this argument shows that any set of orthogonal polynomials on a bounded
interval is complete.

Completeness plays an important role in the approximation of functions by series of
orthogonal polynomials, as shown by the following theorem.

Theorem 10.6. Let {p 0 , p 1 ,...} denote orthogonal polynomials with respect to F and
define

p n (x)
, n = 0, 1,....
n ∞ 1
¯ p (x) =
2
[ p n (x) dF(x)] 2
−∞
Let f denote a function satisfying
∞

2
f (x) dF(x) < ∞
−∞
and let
∞

α n = f (x) ¯ p (x) dF(x), n = 0, 1,....
n
−∞
For n = 0, 1, 2,... define
n
ˆ
f (x) = α j ¯ p (x).
n j
j=0
If {p 0 , p 1 ...} is complete, then
∞ ˆ 2

(i) lim n→∞ [ f (x) − f (x)] dF(x) = 0
n
−∞
(ii)
n
∞ ∞
ˆ 2 2 2
[ f (x) − f (x)] dF(x) ≤ f (x) dF(x) − α , n = 1, 2,...
n
j
−∞ −∞ j=0
∞ 2 ∞ 2
(iii) α = f (x) dF(x)
j=0 j −∞