Proof of Theorem 1.1 By the property of ρ--mixing sequence, it is easy to see that {Y
n
} is a strictly stationary ρ--mixing sequence with EY1 = 0 and . We first prove
(3.1)
Let . Obviously,
From condition (a4) in Theorem 1.1 and Lemma 2.3, we have
and
By stationarity of {Y
n
, n ≥ 1} and E |X1|2 < ∞, we know that is uniformly integrable, and from condition (a2) in Theorem 1.1, we get , so applying Lemma 2.5, we have
Notice that
so (3.1) is valid. Let f(x) be a bounded Lipschitz function and have a Radon-Nikodyn derivative h(x) bounded by Γ. From (3.1), we have
thus
(3.2)
On the other hand, note that (1.1) is equivalent to
(3.3)
from Section 2 of Peligrad and Shao [7] and Theorem 7.1 on P42 from Billingsley [8]. Hence, to prove (3.3), it suffices to show that
(3.4)
by (3.2). Let , 1 ≤ k ≤ n m we have
(3.5)
By the fact that f is bounded, we have
(3.6)
Now we estimate I2, if l > 2k, we have
and
By Lemma 2.3 and condition (a2) in Theorem 1.1, we have
(3.7)
and
By Lemma 2.6, the definition of ρ--mixing sequence and condition (a4), we have
By the inequality (cf. Zhang [[2], p. 254] or Ledoux and Talagrand [[9], p. 251]), we get
and
thus
similarly,
hence
Similarly to (3.7), we have
and
Since f is a bounded Lipschitz function, we have
where . Hence if l > 2k, we have
Thus
(3.8)
Associated with (3.5), (3.6), and (3.8), we have
(3.9)
To prove (3.4), let , where τ > 1. From (3.9), we have
Thus ∀ε > 0, we have
By Borel-Cantelli lemma, we have
Note that
For every n, there exist n
k
and nk+1satisfying n
k
< n ≤ nk+1, we have
(3.4) is completed, so the proof of Theorem 1.1 is completed.
Proof of Theorem 1.2 Let , we have
Hence (1.1) is equivalent to
(3.10)
On the other hand, to prove (1.2), it suffices to show that
(3.11)
By Lemma 2.7, for enough large i, for some we have
It is easy to know that log(1+ x) = x + O(x2) for , thus
and
Hence for arbitrary small ε > 0, there is n0 = n0(ω, ε), such that for every n > n0 and arbitrary x,
so by (3.10), we know that (3.11) is true, and (3.11) is equivalent to (1.2), thus the proof of Theorem 1.2 is complete.