Theoretical Mathematics & Applications, vol.2, no.3, 2012, 23-38
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2012
Hybrid descent-like halpern iteration
methods for two nonexpansive mappings
and semigroups on two sets
Nguyen Buong 1 and Nguyen Duc Lang2
Abstract
In this paper, we introduce some new iteration methods based on the
hybrid method in mathematical programming, the descent-like iterative
method and the Halpern’s method for finding a common fixed point of
two nonexpansive mappings and nonexpansive semigroups on two closed
and convex subsets in Hilbert spaces.
Mathematics Subject Classification: 47J05, 47H09, 49J30
Keywords: Metric projection, Fixed point, Nonexpansive Mappings and
Semigroups
1
2
Vietnamese Academy of Science and Technology Institute of Information Technology
Technology 18, Hoang Quoc Viet, Cau Giay, Ha Noi, Viet Nam.
Viet Nam, Thainguyen University, University of Sciences,
e-mail: nguyenduclang2002@yahoo.com
Article Info: Received : July 12, 2012. Revised : August 25, 2012
Published online : December 30, 2012
24
Hybrid descent-like halpern iteration methods...
1
Introduction
Let H be a real Hilbert space with the scalar product and the norm denoted
by the symbols h., .i and k.k, respectively, and let C be a nonempty, closed
and convex subset of H. Denote by PC x the metric projection from x ∈ H
onto C. Let T be a nonexpansive mapping on C, i.e., T : C → C and
kT x − T yk ≤ kx − yk for all x, y ∈ C. We use F (T ) to denote the set of fixed
points of T , i.e., F (T ) = {x ∈ C : x = T x}. We know that F (T ) is nonempty,
if C is bounded, for more details see [1].
Let {T (t) : t > 0} be a nonexpansive semigroup on C, that is,
(1) for each t > 0, T (t) is a nonexpansive mapping on C;
(2) T (0)x = x for all x ∈ C;
(3) T (t1 + t2 ) = T (t1 ) ◦ T (t2 ) for all t1 , t2 > 0; and
(4) for each x ∈ C, the mapping T (.)x from (0, ∞) into C is continuous.
Denote by F = ∩t>0 F (T (t)) the set of common fixed points for the semigroup {T (t) : t > 0}. We know that F is a closed convex subset in H and
F 6= ∅ if C is compact (see, [2]).
Let Ci , i = 1, 2, be two closed and convex subsets in H. Let Ti and
{Ti (t) : t > 0}, i = 1, 2, be two nonexpansive mappings and semigroups
on Ci , respectively. The problems studied in this paper is to find two elements
p ∈ F := F (T1 ) ∩ F (T2 )
(1.1)
q ∈ F1,2 := F1 ∩ F2 ,
(1.2)
and
where Fi = ∩t>0 F (Ti (t)). Assume that F and F1,2 are not empty. Some
particular cases of (1.1) and (1.2) are the following:
(i) when T1 = T2 = I, the identity mapping in H, (1.1) is the convex feasibility
problem studied in [3].
(ii) when C1 = C2 = C, problems (1.1) and (1.2) are considered in [4]-[6].
For finding a fixed point of a nonexpansive mapping T on C, in 1953, Mann
[7] proposed the following method:
x0 ∈ C
any element,
xn+1 = αn xn + (1 − αn )T xn ,
n ≥ 0,
(1.3)
25
Nguyen Buong and Nguyen Duc Lang
that converges only weakly, in general (see [8] for an example). In 1967,
Halpern [9] firstly proposed the following iteration process:
xn+1 = βn u + (1 − βn )T xn ,
n ≥ 0,
(1.4)
where u, x0 are two fixed elements in C and {βn } ⊂ (0, 1). He pointed out
P
that the conditions limn→∞ βn = 0 and ∞
n=0 βn = ∞ are necessary in the
sense that, if the iteration (1.4) converges to a fixed point of T , then these
conditions must be satisfied. Further, the iteration method was investigated
by Lions [10], Reich [11], Wittmann [12] and Song [13]. Recently, Alber [14]
proposed the following descent-like method
xn+1 = PC (xn − µn [xn − T xn ]), n ≥ 0,
(1.5)
and proved that if {µn } : µn > 0, µn → 0, as n → ∞ and {xn } is bounded,
then:
(i) there exists a weak accumulation point x̃ ∈ C of {xn };
(ii) all weak accumulation points of {xn } belong to F (T ); and
(iii) if F (T ) is a singleton, i.e., F (T ) = {x̃}, then {xn } converges weakly to x̃.
To obtain strong convergence for (1.3), Nakajo and Takahashi [15] introduced the hybrid Mann’s iteration method:
x0 ∈ C,
yn = αn xn + (1 − αn )T xn ,
Cn = {z ∈ C : kyn − zk ≤ kxn − zk},
(1.6)
Qn = {z ∈ C : hxn − z, x0 − xn i ≥ 0},
xn+1 = PCn ∩Qn x0 ,
where {αn } ⊂ [0, a] for some a ∈ [0, 1). They showed that {xn } defined by
(1.6) converges strongly to PF (T ) x0 as n → ∞. Recently, Yanes and Xu [16]
adapted the iteration process (1.4) as follows:
x0 ∈ C
any element,
yn = βn x0 + (1 − βn )T xn ,
Cn = {z ∈ C : kyn − zk2 ≤ kxn − zk2
+ βn (kx0 k2 + 2hxn − x0 , zi)},
Qn = {z ∈ C : hxn − z, x0 − xn i ≥ 0},
xn+1 = PCn ∩Qn x0 .
(1.7)
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Hybrid descent-like halpern iteration methods...
They proved that if T is a nonexpansive mapping on a closed convex subset C
with F (T ) 6= ∅ and the sequence {βn } ⊂ (0, 1) is chosen such that limn→∞ βn =
0, then the sequence {xn } defined by (1.7) converges strongly to PF (T ) x0 as
n → ∞.
For finding an element p ∈ F, Nakajo and Takahashi [15] also introduced
an iteration procedure as follows:
x0 ∈ C
any element,
1
yn = αn xn + (1 − αn )
tn
Z
tn
T (s)xn ds,
0
Cn = {z ∈ C : kyn − zk ≤ kxn − zk},
(1.8)
Qn = {z ∈ C : hxn − z, x0 − xn i ≥ 0},
xn+1 = PCn ∩Qn x0 , n ≥ 0,
where αn ∈ [0,a] for some a ∈ [0,1) and {tn } is a positive real number divergent
sequence. Under the conditions on {αn } and {tn }, the sequence {xn } defined
by (1.8) converges strongly to PF x0 .
If C ≡ H, then Cn and Qn in (1.6)-(1.8) are two halfspaces. So, the
projection xn+1 onto Cn ∩ Qn in these methods can be found by an explicit
formula [17]. Clearly, if C is a proper subset of H, then Cn and Qn in (1.6)(1.8) are not two halfspaces. Then, the following problem is posed: how to
construct the closed convex subsets Cn and Qn and if we can express xn+1 of
(1.6)-(1.8) in a similar form as in [17]? This problem is solved very recently
in [18]-[20]. In this works, Cn and Qn are replaced by two halfspaces and yn is
the right hand side of (1.5) with a modification. In this paper, motivated by
(1.5), (1.7) and [14], [15], to solve problems (1.1) and (1.2) we introduce the
following new iteration processes:
x0 ∈ H
any element,
zn = xn − µn (xn − T1 PC1 xn ),
yn = βn x0 + (1 − βn )T2 PC2 zn ,
Hn = {z ∈ H : kyn − zk2 ≤ kxn − zk2
+ βn (kx0 k2 + 2hxn − x0 , zi)},
Wn = {z ∈ H : hxn − z, x0 − xn i ≥ 0},
xn+1 = PHn ∩Wn x0 , n ≥ 0;
(1.9)
27
Nguyen Buong and Nguyen Duc Lang
and
x0 ∈ H
any element,
Z
1 tn
z n = xn − µ n xn −
T1 (s)PC1 xn ds ,
tn 0
Z
1 tn
T2 (s)PC2 zn ds,
yn = βn x0 + (1 − βn )
tn 0
Hn = {z ∈ H : kyn − zk2 ≤ kxn − zk2
(1.10)
+ βn (kx0 k2 + 2hxn − x0 , zi)},
Wn = {z ∈ H : hxn − z, x0 − xn i ≥ 0},
xn+1 = PHn ∩Wn x0 ,
n ≥ 0.
We shall prove the strong convergence of the sequences {xn }, {yn } and {zn }
defined by (1.9) and (1.10) to some elements p and q in sections 2 and 3,
respectively.
Below, the symbols * and → denote weak and strong convergence, respectively.
2
Strong convergence to a common fixed point
of two nonexpansive mappings
We formulate the following facts needed in the proof of our results.
Lemma 2.1 [21]. Let H be a real Hilbert. There holds the following identity:
kλx + (1 − λ)yk2 = λkxk2 + (1 − λ)kyk2 − λ(1 − λ)kx − yk2 .
Lemma 2.2 [16]. Let C be a nonempty, closed and convex subset of a real
Hilbert space H. For any x ∈ H, there exists a unique z ∈ C such that
kz − xk ≤ ky − xk for all y ∈ C, and z = PC x if and only if hz − x, y − zi ≥ 0
for all y ∈ C.
Lemma 2.3. (Demiclosedness principle) [21]. If C is a nonempty, closed and
convex subset of a real Hilbert space H, T is a nonexpansive mapping on C,
{xn } is a sequence in C such that xn * x and xn − T xn → 0, then x − T x = 0.
Lemma 2.4 [22]. Every Hilbert space H has Randon-Riesz property or Kadec-
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Hybrid descent-like halpern iteration methods...
Klee property, that is, for a sequence {xn } ⊂ H with xn * x and kxn k → kxk,
then there holds xn → x.
Now, we are in a position to prove the following result.
Theorem 2.5. Let C1 and C2 be two nonempty, closed and convex subsets in
a real Hilbert space H and let T1 and T2 be two nonexpansive mappings on C1
and C2 , respectively, such that F := F (T1 ) ∩ F (T2 ) 6= ∅. Assume that {µn }
and {βn } are sequences in [0,1] such that µn ∈ (a, b) for some a, b ∈ (0, 1) and
βn → 0. Then, the sequences {xn }, {zn } and {yn }, defined by (1.9), converge
strongly to the same point u0 = PF x0 , as n → ∞.
Proof. First, note that
kyn − zk2 ≤ kxn − zk2 + βn (kx0 k2 + 2hxn − x0 , zi)
is equivalent to
βn
1
h(1 − βn )xn + βn x0 − yn , zi ≤ hxn − yn , xn i − kyn − xn k2 + kx0 k2 .
2
2
Thus, Hn is a halfspace. It is clear that F (T ) = F (T PC ) := {p ∈ H : T PC p =
p} for any mapping T from C into C. So, we have that F = F (T̃1 ) ∩ F (T̃2 )
where T̃i = Ti PCi , i = 1, 2, and T̃i , i = 1, 2, are also two nonexpansive mappings
on H. Hence, by (1.9) and Lemma 2.1, we obtain for any p ∈ F that
kzn − pk2 = k(1 − µn )(xn − p) + µn (T̃1 xn − p)k2
= (1 − µn )kxn − pk2 + µn kT̃1 xn − pk2
− (1 − µn )µn kxn − T̃1 xn k2
≤ (1 − µn )kxn − pk2 + µn kxn − pk2
− (1 − µn )µn kxn − T̃1 xn k2
≤ kxn − pk2 − (1 − µn )µn kxn − T̃1 xn k2 ≤ kxn − pk2 .
(2.1)
Nguyen Buong and Nguyen Duc Lang
29
By the similar argument and the convexity of k.k2 , we also obtain
kyn − pk2 = kβn x0 + (1 − βn )T̃2 zn − pk2
≤ βn kx0 − pk2 + (1 − βn )kT̃2 zn − T̃2 pk2
≤ βn kx0 − pk2 + (1 − βn )kzn − pk2
≤ βn kx0 − pk2 + (1 − βn )kxn − pk2
= kxn − pk2 + βn (kx0 − pk2 − kxn − pk2 )
= kxn − pk2 + βn (kx0 k2 + 2hxn − x0 , pi).
Therefore, p ∈ Hn for all n ≥ 0. It means that F (T ) ⊂ Hn for all n ≥ 0.
Next, we show by mathematical induction that F (T ) ⊂ Hn ∩ Wn for each
n ≥ 0. For n = 0, we have W0 = H, and hence F (T ) ⊂ H0 ∩ W0 . Suppose
that xi is given and F (T ) ⊂ Hi ∩ Wi for some i > 0. There exists a unique
element xi+1 ∈ Hi ∩ Wi such that xi+1 = PHi ∩Wi x0 . Therefore, by Lemma 2.2,
hxi+1 − x0 , p − xi+1 i ≥ 0
for each p ∈ Hi ∩ Wi . Since F (T ) ⊂ Hi ∩ Wi , we get F (T ) ⊂ Wi+1 . So, we
have F (T ) ⊂ Hi+1 ∩ Wi+1 .
Further, since F (T ) is a nonempty, closed and convex subset of H, there
exists a unique element u0 ∈ F (T ) such that u0 = PF (T ) x0 . From xn+1 =
PHn ∩Wn (x0 ), we obtain
kxn+1 − x0 k ≤ kz − x0 k
for every z ∈ Hn ∩ Wn . As u0 ∈ F (T ) ⊂ Wn , we get
kxn+1 − x0 k ≤ ku0 − x0 k ∀ n ≥ 0.
(2.2)
This implies that {xn } is bounded. Now, we show that
lim kxn+1 − xn k = 0.
n→∞
(2.3)
From the definition of Wn and Lemma 2.2, we have xn = PWn x0 . As xn+1 ∈
Hn ∩ Wn , we obtain
kxn+1 − x0 k ≥ kxn − x0 k ∀ n ≥ 0.
Therefore, {kxn − x0 k} is a nondecreasing and bounded sequence. So, there
exists limn→∞ kxn − x0 k = c. On the other hand, from xn+1 ∈ Wn , it follows
that
hxn − x0 , xn+1 − xn i ≥ 0,
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Hybrid descent-like halpern iteration methods...
and hence
kxn − xn+1 k2 = kxn − x0 − (xn+1 − x0 )k2
= kxn − x0 k2 − 2hxn − x0 , xn+1 − x0 i + kxn+1 − x0 k2
≤ kxn+1 − x0 k2 − kxn − x0 k2
∀n ≥ 0.
Thus, (2.3) is proved by using the last inequality and limn→∞ kxn − x0 k = c.
Next, since xn+1 ∈ Hn we have that
kyn − xn+1 k2 ≤ kxn − xn+1 k2 + βn (kx0 k + 2hxn − x0 , zi)}.
Therefore, from (2.3), the boundness of {xn }, βn → 0 and the last inequality,
it follows that
lim kyn − xn+1 k = 0.
(2.4)
n→∞
This together with (2.3) implies that
lim kyn − xn k = 0.
n→∞
(2.5)
Noticing that T̃2 zn = yn − βn (xn − T̃2 zn ) + βn (xn − x0 ), we have
kxn − T̃2 zn k ≤ kxn − yn k + βn kxn − T̃2 zn k + βn kxn − x0 k.
From (2.2) and the last inequality, it follows that
1
kxn − yn k + βn ku0 − x0 k .
kxn − T̃2 zn k ≤
1 − βn
By βn → 0 (βn ≤ 1 − β for some β ∈ (0, 1)), (2.5) and the last inequality, we
obtain
lim kxn − T̃2 zn k = 0.
(2.6)
n→∞
Now, we shall prove that kxn − T̃1 xn k → 0 and kxn − T̃2 xn k → 0, as n → ∞.
Indeed, since {xn } is bounded, for any p ∈ F and any subsequence {T̃1 xnk −
xnk } of {T̃1 xn − xn } there exists a subsequence {xnj } ⊂ {xnk } such that
lim kxnj − pk = lim sup kxnk − pk = a.
j→∞
k→∞
By (2.6), (2.1) and the following inequalities
kxnj − pk ≤ kxnj − T̃2 znj k + kT̃2 znj − pk
≤ kxnj − T̃2 znj k + kznj − pk
≤ kxnj − T̃2 znj k + kxnj − pk,
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Nguyen Buong and Nguyen Duc Lang
we get that
lim kxnj − pk = lim kznj − pk = a.
j→∞
j→∞
Again from (2.1) and the condition on µn , it implies that
a(1 − b)kT̃1 xnj − xnj k ≤ kxnj − pk − kznj − pk.
So, kT̃1 xnj − xnj k → 0 and hence kT̃1 xn − xn k → 0, as n → ∞. Further, since
kT̃2 xn − xn k ≤ kT̃2 xn − T̃2 zn k + kT̃2 zn − xn k
≤ kxn − zn k + kT̃2 zn − xn k,
lim kzn − xn k = lim µn kT̃1 xn − xn k = 0,
n→∞
n→∞
(2.7)
by (2.6) and kT̃1 xn −xn k → 0, we also obtain that kT̃2 xn −xn k → 0. Since {xn }
is bounded, there exists a subsequence {xni } of {xn } that converges weakly to
some element p ∈ H as i → ∞. By Lemmas 2.3 and kT̃1 xn −xn k, kT̃2 xn −xn k →
0, we have that p ∈ F .
Now, from (2.2) and the weak lower semicontinuity of the norm it implies
that
kx0 − u0 k ≤ kx0 − pk ≤ lim inf kx0 − xnj k ≤ lim sup kx0 − xnj k ≤ kx0 − u0 k.
j→∞
j→∞
Thus, we obtain limj→∞ kx0 − xnj k = kx0 − u0 k = kx0 − pk. This implies
xkj → p = u0 by Lemma 2.4. By the uniqueness of the projection u0 = PF x0 ,
we have that xn → u0 . Consequently, from (2.7) it follows that zn → u0 . From
(2.5), we also get that yn → u0 . This completes the proof.
We have the following corollaries.
Corollary 2.6. Let Ci , i = 1, 2, be two nonempty, closed and convex subsets
in a real Hilbert space H. Let Ti , i = 1, 2, be two nonexpansive mappings on
Ci such that F (T1 ) ∩ F (T2 ) 6= ∅. Assume that {µn } is a sequence such that
0 < a ≤ µn ≤ b < 1. Then, the sequences {xn } and {yn }, defined by
x0 ∈ H
any element,
yn = T2 PC2 (xn − µn (xn − T1 PC1 xn )),
Hn = {z ∈ H : kyn − zk ≤ kxn − zk},
Wn = {z ∈ H : hxn − z, x0 − xn i ≥ 0},
xn+1 = PHn ∩Wn (x0 ), n ≥ 0,
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Hybrid descent-like halpern iteration methods...
converge strongly to the same point u0 = PF (T ) x0 , as n → ∞.
Proof. By putting βn ≡ 0 in Theorem 2.5, we obtain the conclusion.
Corollary 2.7. Let Ci , i = 1, 2, be two nonempty, closed and convex subsets
in a real Hilbert space H such that C := C1 ∩ C2 6= ∅. Assume that {µn } and
{βn } are sequences in [0,1] such that µn ∈ (a, b) for some a, b ∈ (0, 1) and
βn → 0. Then, the sequences {xn }, {zn } and {yn }, defined by
x0 ∈ H
any element,
zn = xn − µn (xn − PC1 xn ),
yn = βn x0 + (1 − βn )PC2 zn ,
Hn = {z ∈ H : kyn − zk2 ≤ kxn − zk2
+ βn (kx0 k + 2hxn − x0 , zi)},
Wn = {z ∈ H : hxn − z, x0 − xn i ≥ 0},
xn+1 = PHn ∩Wn x0 , n ≥ 0,
converge strongly to the same point u0 = PC x0 , as n → ∞.
Proof. By putting T1 = T2 = I in Theorem 2.5, we obtain the conclusion.
3
Strong convergence to a common fixed point
of two nonexpansive semigroups
We need the following Lemma in the proof of our result.
Lemma 3.1 [23]. Let C be a nonempty bounded closed convex subset in a real
Hilbert space H and let {T (t) : t > 0} be a nonexpansive semigroup on C.
Then, for any h > 0
Z t
Z
1
1 t
lim sup sup T (h)
T (s)yds −
T (s)yds = 0.
t 0
t 0
t→∞ y∈C
Now, we prove the following result.
Theorem 3.2. Let C1 and C2 be two nonempty closed convex subsets in a real
Hilbert space H and let {T1 (t) : t > 0} and {T2 (t) : t > 0} be two nonexpansive
semigroups on C1 and C2 , respectively, such that F = F1 ∩ F2 6= ∅ where
33
Nguyen Buong and Nguyen Duc Lang
Fi = ∩t>0 F (Ti (t)), i = 1, 2. Assume that {µn } and {βn } are sequences in [0,1]
such that µn ∈ (a, b) for some a, b ∈ (0, 1) and βn → 0 and {tn } is a positive
real divergent sequence. Then, the sequences {xn }, {zn } and {yn }, defined by
(1.10), converge strongly to the same point u0 = PF x0 , as n → ∞.
Proof. For each p ∈ F, we have for each s > 0 that
p = PCi p = T̃i (s)p,
i = 1, 2,
where T̃i (s) = Ti (s)PCi , and hence from (1.10) and Lemma 2.1, we obtain that
2
Z tn
1
2
T̃1 (s)xn )ds − p
kzn − pk = (1 − µn )(xn − p) + µn
tn 0
2
Z tn
1
= (1 − µn )(xn − p) + µn
[T̃1 (s)xn − T̃1 (s)p]ds
tn 0
Z
2
1 tn
2
T̃1 (s)xn − T̃1 (s)pds
= (1 − µn )kxn − pk + µn
tn 0
Z tn
2
1
(3.1)
T̃1 (s)xn ds
− (1 − µn )µn xn −
tn 0
Z
2
1 tn
2
T̃1 (s)xn ds
≤ kxn − pk − (1 − µn )µn xn −
tn 0
≤ kxn − pk2 .
By the similar argument and the convexity of k.k2 , we also obtain
2
Z tn
1
2
T̃2 (s)zn ds − p
kyn − pk = βn (x0 − p) + (1 − βn )
tn 0
Z
1 tn
2
[T̃2 (s)zn − T̃2 (s)p]ds
≤ βn kx0 − pk + (1 − βn )
tn 0
2
≤ βn kx0 − pk2 + (1 − βn )kzn − pk2
≤ βn kx0 − pk2 + (1 − βn )kxn − pk2
= kxn − pk2 + βn (kx0 − pk2 − kxn − pk2 )
= kxn − pk2 + βn (kx0 k2 + 2hxn − x0 , pi).
Therefore, p ∈ Hn for n ≥ 0. It means that F ⊂ Hn for n ≥ 0. As in the proof
of Theorem 2.5, we can obtain the following properties:
(i) F ⊂ Hn ∩ Wn ,
kxn+1 − x0 k ≤ ku0 − x0 k, u0 = PF x0
(3.2)
34
Hybrid descent-like halpern iteration methods...
for n ≥ 0. This implies that {xn } is bounded.
(ii)
lim kxn+1 − xn k = 0.
(3.3)
lim kyn − xn+1 k = 0.
(3.4)
lim kyn − xn k = 0.
(3.5)
n→∞
n→∞
n→∞
Noticing that
Z
Z
1 tn
1 tn
T̃2 (s)zn ds = yn − βn xn −
T̃2 (s)zn ds +βn (xn − x0 ),
tn 0
tn 0
we have
1
kxn −
tn
Z
tn
T̃2 (s)zn dsk ≤ kxn − yn k
Z
1 tn
T̃2 (s)zn ds +βn kxn − x0 k.
+βn xn −
tn 0
0
From (3.2) and the last inequality, it follows that
Z
1 tn
1
kxn − yn k + βn ku0 − x0 k .
T̃2 (s)zn ds ≤
xn −
tn 0
1 − βn
By βn → 0 (βn ≤ 1 − β for some β ∈ (0, 1)), (3.5) and the last inequality, we
obtain
Z
1 tn
lim xn −
(3.6)
T̃2 (s)zn ds = 0.
n→∞
tn 0
As in the proof of Theorem 2.5, by using (3.6) we can obtain that
Z
1 tn
T̃i (s)xn ds = 0, i = 1, 2,
lim xn −
n→∞
tn 0
(3.7)
and
lim kxn − zn k = 0.
n→∞
Since
1
tn
Z
tn
T̃i (s)xn ds ∈ Ci , i = 1, 2,
0
we have that
1
PCi xn −
tn
Z
tn
0
Z
1 tn
T̃i (s)xn ds = PCi xn − PCi
T̃i (s)xn ds
tn 0
Z
1 tn
≤ xn −
T̃i (s)xn ds ,
tn 0
(3.8)
Nguyen Buong and Nguyen Duc Lang
and hence from (3.7) it implies that
Z
1 tn
T̃i (s)xn ds = 0, i = 1, 2.
lim PCi xn −
n→∞
tn 0
35
(3.9)
Since {xn } is bounded, there exists a subsequence {xnj } of {xn } that converges
weakly to some element q ∈ H as j → ∞. From (3.7) and (3.9), we also obtain
that uinj := PCi xnj → q as j → ∞. It means that q ∈ C1 ∩ C2 . Then, for each
h > 0, we have that
Z tn
1
i
i
i
i
Ti (s)un ds
kTi (h)un − un k ≤ Ti (h)un − Ti (h)
tn 0
Z tn
Z
1 tn
1
i
+ Ti (h)
Ti (s)un ds −
Ti (s)uin ds
tn 0
tn 0
Z
1 tn
(3.10)
Ti (s)uin ds − uin
+
tn 0
Z
1 tn
≤2
Ti (s)uin ds − uin
tn 0
Z tn
Z
1 tn
1
i
+ T (h)
Ti (s)un ds −
Ti (s)uin ds .
tn 0
tn 0
Let C0i = {z ∈ Ci : kz − u0 k ≤ 2kx0 − u0 k}. Since u0 = PF x0 ∈ Ci , we have
that
kuinj − u0 k = kPCi xnj − PCi u0 k ≤ kxnj − u0 k ≤ 2kx0 − uo k.
So, C0i is a nonempty bounded closed convex subset. It is easy to verify that
{Ti (t) : t > 0} is a nonexpansive semigroup on C0i . By Lemma 3.1, we get
Z tn
Z
1
1 tn
i
lim Ti (h)
T (s)un ds −
T (s)uin ds = 0
n→∞
tn 0
tn 0
for every fixed h > 0 and hence by (3.9)-(3.10) we obtain that
lim kTi (h)uinj − uinj k = 0
j→∞
for each h > 0. By Lemma 2.3, q ∈ F (Ti (h)) for all h > 0. It means that
q ∈ F. As in the proof of Theorem 2.5, by using (3.2), (3.5) and (3.8), we
also obtain that the sequences {xn }, {yn } and {zn }, defined by (1.10), converge
strongly to u0 as n → ∞. This completes the proof.
Corollary 3.3. Let C be a nonempty closed convex subset in a real Hilbert
36
Hybrid descent-like halpern iteration methods...
space H and let {T (t) : t > 0} be a nonexpansive semigroup on C such that
F = ∩t>0 F (T (t)) 6= ∅. Assume that {βn } is a sequence in [0,1] such that
βn → 0. Then, the sequences {xn } and {yn }, defined by
x0 ∈ H
any element,
1
yn = βn x0 + (1 − βn )
tn
Z
tn
T (s)PC (xn )ds,
0
Hn = {z ∈ H : kyn − zk2 ≤ kxn − zk2
+ βn (kx0 k + 2hxn − x0 , zi)},
Wn = {z ∈ H : hxn − z, x0 − xn i ≥ 0},
xn+1 = PHn ∩Wn (x0 ), n ≥ 0,
converge strongly to the same point u0 = PF x0 , as n → ∞.
Proof. By putting T1 (s) = I for all s > 0, C1 = H, C2 = C and T2 (s) = T (s)
in Theorem 3.2, we obtain the conclusion.
Corollary 3.4. Let C be a nonempty closed convex subset in a real Hilbert
space H and let {T (t) : t > 0} be a nonexpansive semigroup on C such that
F = ∩t>0 F (T (t)) 6= ∅. Assume that {αn } is a sequence in [0,1] such that
αn → 1. Then, the sequences {xn } and {yn }, defined by
x0 ∈ H
any element,
Z
1 tn
1 tn
T (s)PC xn − µn xn −
T (s)PC xn ds ds ,
yn =
tn 0
tn 0
Z
Hn = {z ∈ H : kyn − zk ≤ kxn − zk},
Wn = {z ∈ H : hxn − z, x0 − xn i ≥ 0},
xn+1 = PHn ∩Wn (x0 ), n ≥ 0,
converge strongly to the same point u0 = PF x0 , as n → ∞.
Proof. By putting βn ≡ 0, C2 = H, C1 = C, T2 (s) = I and T1 (s) = T (s) for all
s > 0 in Theorem 3.2, we obtain the conclusion.
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