Page 324 - Elements of Distribution Theory

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

310 Orthogonal Polynomials

It now follows, as in the proof of Theorem 6.4, that

∞

2
ˆ
lim ( f (x) − f 0 (x)) dF(x) = 0.
n
n→∞
−∞
Now return to the proof of the theorem. Write f (x) = f 0 (x) + d(x). Then, for each
j = 1, 2,...,
∞ ∞ ∞

f (x) ¯ p (x) dF(x) = f 0 (x) ¯ p (x) dF(x) + d(x) ¯ p (x) dF(x).
j
j
j
−∞ −∞ −∞
Note that, by the Cauchy-Schwarz inequality,
∞

ˆ
n j
( f (x) − f 0 (x)) ¯ p (x) dF(x)
−∞
1
∞ 1 ∞ 2

2
2
ˆ
2
≤ ( f (x) − f 0 (x)) dF(x) ¯ p (x) dF(x) .
n j
−∞ −∞
Since
∞

2
¯ p (x) dF(x) = 1, j = 1,...,
j
−∞
and

∞
ˆ
2
( f (x) − f 0 (x)) dF(x) → 0as n →∞,
n
−∞
it follows that

∞
ˆ
f 0 (x) ¯ p (x) dF(x) = lim f (x) ¯ p (x) dF(x) = α j . (10.7)
j
n
j
−∞ n→∞
Since
∞

f (x) ¯ p (x) dF(x) = α j ,
j
−∞
it follows that

∞
d(x) ¯ p (x) dF(x) = 0, j = 0, 1,...
j
−∞
so that
∞

d(x)p j (x) dF(x) = 0, j = 0, 1,....
−∞
By completeness of {p 0 , p 1 ,...}, d = 0 and, hence,
∞

ˆ
2
lim ( f (x) − f (x)) dF(x) = 0. (10.8)
n
n→∞
−∞
This verifies part (i) of the theorem.