Under criterion (II'), Ceng et al. [12] considered another type, RPPA:
(3.1)
and proved the weak convergence of (3.1) under the assumptions:
We note that the choice of (ρ
n
) excludes the case whenever ρ
n
∈ (1,2), the overrelaxation. The overrelaxation, however, may indeed speed up the convergence of the algorithm (see [13]). Below, we shall improve their conditions on the relaxation factor from 0 < δ ≤ ρ
n
≤ 1 to 0 < δ ≤ ρ
n
≤ 2 - δ.
Theorem 6. Assume that the following conditions hold:
(a)
(b) 0 < δ ≤ ρ
n
≤ 2 - δ;
(c)
Then the sequence generated by (3.1) converges weakly to a point in S.
Proof. The key point of our proof is to show lim
n
s
n
= 0, where To see this, let z ∈ S be fixed. Since is firmly nonexpansive and applying Lemma 1 yields This together with (3.1) enables us to get
Using the basic inequality 2ab ≤ a2 / ε + εb2 (a,b ∈ ℝ, ε > 0), we arrive at
where is a summable sequence. Substituting this into above yields
Since by Lemma 5 the limit of ||x
n
- z ||2 exists and lim inf
n
ρ
n
(2 - ρ
n
-4η
n
) ≥ δ (2 - δ), this implies that On the other hand, we note that for all n ∈ ℕ
therefore, lim
n
s
n
= 0. The rest proof is similar to that of [12, Theorem 3.1].
We now turn to the RPPA (1.2). Under the criterion (I), the assumptions on relaxation factors can be relaxed to Σρ
n
(2 - ρ
n
) = ∞ (see [3, Theorem 3.3]). Since the proof there is very technical, we wang to restate this result with a simple proof.
Theorem 7. Assume that the following conditions hold:
(a) Σ
n
||e
n
|| < ∞;
(b) Σ
n
ρn(2 - ρ
n
) = ∞;
(c)
(d) Σ
n
|c
n
+1- c
n
| < ∞.
Then the sequence generated by (1.2) converges weakly to a point in S.
Proof. The key step is to show lim
n
s
n
= 0, where It has been shown that Σ
n
ρ
n
(2 - ρ
n
)s
n
< ∞ (see [3, Lemma 3.2]). Therefore, it remains to show that lim
n
s
n
exists. By letting T
n
= 2J
n
- I, we rewrite (2) as
In view of Lemma 4 and condition (c),
where M > 0 is a suitable number. Consequently,
Using s
n
= || x
n
- T
n
x
n
||/2, we therefore arrive at
where σ
n
= 2M |c
n
+1- c
n
| + 4||e
n
|| satisfying Σ
n
σ
n
< ∞ (due to (a) and (d)). By Lemma 5, we finally conclude that lim
n
s
n
= 0.