Page 86 - Textos de Matemática Vol. 47

76 P. E. OLIVEIRA
As re (x) ! r(x), taking into account Lemma 4.1, the second term above
nF (x,h)
numbers and write
⇣2 2 b 2⌘ E (rb (x) re (x))(f (x) f (x))
⇣⌘
is easily seen to be an o 1 + 2(K,x,h) . Now, let ↵n and "n be real
n
nnn0 ⇣22b2⌘
= E (rb (x) re (x))(f (x) f (x)) I
⇣ n n n 0 {rbn (x)↵n ,|rbn (x) ren (x)|"n } ⌘
22b2 +E (rb (x) re (x))(f (x) f (x)) I
n
n
n
0 {rbn (x)↵n ,|rbn (x) ren (x)|>"n } ⇣22b2⌘
+E (rb (x) re (x))(f (x) f (x)) I =:A +B +C . n n n 0 {rbn(x)>↵n} nnn
Lemma 4.4. Assume that (SB1), (SB2) and (K) hold. If ↵n % +1, "n & 0 are such that ↵n"n ! 0 then
An =o✓ 1 + 2(K,x,h)◆. nF (x, h)
Proof. The result is obvious by noting that A  (↵ + re (x))" E (fb (x) nnnnn
f0(x))2 and taking into account Lemma 4.1. ⇤ To bound Bn, we start by remarking that
B (↵ +re (x))2⇣E(fb(x) f (x))4⌘1/2(P(|rb (x) re (x)|>" ))1/2, nnnn0nnn
and using (3.1) to obtain P(|rbn(x) ren(x)| > "n)
⇣ b ⌘ ⇣ b b P |gbn(x) Egbn(x)|>"nEfn(x) +P fn(x) Efn(x) >"n
( E fb ( x ) ) 2 ⌘ n .
◆,
4 4
E gb n ( x )
Using now Markov’s inequality it follows that P⇣|gb(x) Egb(x)|>"nEfb(x)⌘ 16Vargbn(x) = 1O✓ 1
n n 4 n " 2 ( E fb ( x ) ) 2 " 2n n F ( x , h ) nn
and analogously for the other probability term.
Lemma 4.5. Assume that (SB1), (SB2), (K) and (3.2) hold. Then
Bn =o✓ 1 + (K,x,h) + 2(K,x,h)◆. nF (x, h) n1/2 F 1/2 (x, h)