1. No category

Zhang Journal of Inequalities and Applications 2015, 2015:2
http://www.journalofinequalitiesandapplications.com/content/2015/1/2
RESEARCH
Open Access
Some results on a viscosity splitting
algorithm in Hilbert spaces
Yunpeng Zhang*
*
Correspondence:
[email protected]
College of Electric Power, North
China University of Water Resources
and Electric Power, Zhengzhou,
450011, China
Abstract
In this paper, a viscosity splitting for common solution problems is proposed. Strong
convergence theorems are obtained in the framework of Hilbert spaces. Applications
are also provided to support the main results.
Keywords: zero point; fixed point; variational inclusion; nonexpansive mapping
1 Introduction; preliminaries
In this paper, we always assume that H is a real Hilbert space with the inner product x, y
√
and the induced norm x = x, x for x, y ∈ H. Recall that a set-valued mapping M : H ⇒
H is said to be monotone iff, for all x, y ∈ H, f ∈ Mx, and g ∈ My imply x – y, f – g ≥ .
In this paper, we use M– () to denote the zero point set of M. A monotone mapping
M : H ⇒ H is maximal iff the graph Graph(M) of M is not properly contained in the graph
of any other monotone mapping. It is well known that a monotone mapping M is maximal
if and only if, for any (x, f ) ∈ H × H, x – y, f – g ≥ , for all (y, g) ∈ Graph(M) implies
f ∈ Mx. For a maximal monotone operator M on H, and r > , we may define the singlevalued resolvent Jr : H → Dom(M), where Dom(M) denotes the domain of M. It is well
known that Jr is firmly nonexpansive, and M– () = F(Jr ).
The proximal point algorithm, which was proposed by Martinet [, ] and generalized
by Rockafellar [, ] is one of the classical methods for solving zero points of maximal
monotone operators. In this paper, we investigate the problem of finding a zero of the
sum of two monotone operators. The problem is very general in the sense that it includes,
as special cases, convexly constrained linear inverse problems, split feasibility problem,
convexly constrained minimization problems, fixed point problems, variational inequalities, Nash equilibrium problem in noncooperative games and others. Because of their
importance, splitting methods, which were proposed by Lions and Mercier [] and Passty
[], for zero problems have been studied extensively recently; see, for instance, [–] and
the references therein.
Let C be a nonempty closed and convex subset of H. Let A : C → H be a mapping. Recall
that the classical variational inequality problem is to find a point x ∈ C such that
y – x, Ax ≥ ,
∀y ∈ C.
(.)
Such a point x ∈ C is called a solution of variational inequality (.). In this paper, we use
VI(C, A) to denote the solution set of variational inequality (.). Recall that A is said to be
©2015 Zhang; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any
medium, provided the original work is properly cited.
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monotone iff
Ax – Ay, x – y ≥ ,
∀x, y ∈ C.
Recall that A is said to be inverse-strongly monotone iff there exists a constant κ >  such
that
Ax – Ay, x – y ≥ κAx – Ay ,
∀x, y ∈ C.
For such a case, we also call A is κ-inverse-strongly monotone. It is also not hard to see
that every inverse-strongly monotone mapping is monotone and continuous.
Let S : C → C be a mapping. In this paper, we use F(S) to denote the fixed point set of S.
S is said to be contractive iff there exists a constant β ∈ (, ) such that
Sx – Sy ≤ βx – y,
∀x, y ∈ C.
We also call S is β-contractive. S is said to be nonexpansive iff
Sx – Sy ≤ x – y,
∀x, y ∈ C.
It is well known if C is nonempty closed convex of H, then F(S) is not empty. S is said to
be firmly nonexpansive iff

Sx – Sy ≤ x – y – (I – S)x – (I – S)y ,
∀x, y ∈ C.
In order to prove our main results, we also need the following lemmas.
Lemma . [] Let A be a maximal monotone operator on H. For λ > , μ > , and x ∈ E,
we have Jλ x = Jμ ( μλ x + ( – μλ )Jλ x), where Jλ = (I + λA)– and Jμ = (I + μA)– .
Lemma . [] Let {xn } and {yn } be bounded sequences in H. Let {βn } be a sequence in
(, ) with  < lim infn→∞ βn ≤ lim supn→∞ βn < . Suppose that xn+ = ( – βn )yn + βn xn ,
∀n ≥  and
lim sup yn+ – yn – xn+ – xn ≤ .
n→∞
Then limn→∞ yn – xn = .
Lemma . [] Let {an } be a sequence of nonnegative numbers satisfying the condition an+ ≤ ( – tn )an + tn bn , ∀n ≥ , where {tn } is a number sequence in (, ) such that
limn→∞ tn =  and ∞
n= tn = ∞, {bn } is a number sequence such that lim supn→∞ bn ≤ .
Then limn→∞ an = .
Lemma . [] Let C be a nonempty closed convex subset of H. Let A : C → H be a
mapping and let B : H ⇒ H be a maximal monotone operator. Then x – (I + sB)– x ≤
x – (I + rB)– x for all  < s ≤ r.
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Lemma . [] Let {λn } be a real sequence that does not decreasing at infinity, in the sense
that there exists a subsequence {λnk } such that λnk ≤ λnk + for all k ≥ . For every n > n ,
define an integer sequence d(n) as d(n) = max{n ≤ k ≤ n|λnk ≤ λnk + }. Then limn→∞ d(n) =
 and for all n > n max{λd(n) , λn } ≤ λd(n)+ .
Lemma . [] Let C be a nonempty closed convex subset of H. Let S : C → C be a
nonexpansive mapping with a nonempty fixed point set. If {xn } converges weakly to x and
{xn – Txn } converges to zero. Then x ∈ F(S).
2 Main results
Now, we are in a position to state our main results.
Theorem . Let C be a nonempty closed convex subset of H. Let S : C → C be a nonexpansive mapping with fixed points and let f : C → C be a β-contractive mapping. Let
A : C → H be an α-inverse-strongly monotone mapping and let B be a maximal monotone
operator on H. Assume that Dom(B) ⊂ C and F(S) ∩ (A + B)– () is not empty. Let {αn } and
{βn } be real number sequences in (, ) and let {rn } be a positive real number sequence in
(, α). Let {xn } be a sequence generated in the following process: x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )S(I + rn B)– (yn – rn Ayn ),
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
(b)  < lim infn→∞ βn ≤ lim supn→∞ βn < ;
(c)  < a ≤ rn ≤ b < α and ∞
n= |rn – rn– | < ∞,
where a and b are two real numbers. Then {xn } converges strongly to a point x¯ ∈ F(S) ∩ (A +
B)– (), which is also a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ F(S) ∩ (A + B)– ().
Proof Note that the mapping I – rn A is nonexpansive. Indeed, we have
(I – rn A)x – (I – rn A)y
= x – y – rn x – y, Ax – Ay + rn  Ax – Ay
≤ x – y – rn (α – rn )Ax – Ay .
In light of restriction (c), one finds that I – rn A is nonexpansive. It is obvious that F((I +
rn B)– (I – rn A)) = (A + B)– (). Fix p ∈ (A + B)– () ∩ F(S). It follows that
yn – p ≤ αn f (xn ) – p + ( – αn )xn – p
≤  – αn ( – β) xn – p + αn f (p) – p.
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Putting Jrn = (I + rn B)– , we see that
xn+ – p ≤ βn xn – p + ( – βn )SJrn (yn – rn Ayn ) – p
≤ βn xn – p + ( – βn )(yn – rn Ayn ) – p
f (p) – p
.
≤  – αn ( – βn )( – β) xn – p + αn ( – βn )( – β)
–β
By mathematical induction, we find that the sequence {xn } is bounded. Note that
yn – yn– ≤ αn f (xn ) – f (xn– ) + ( – αn )xn – xn– + |αn – αn– |f (xn– ) – xn– ≤  – αn ( – β) xn – xn– + |αn – αn– |f (xn– ) – xn– .
Putting zn = yn – rn Ayn , we find from Lemma . that
rn–
rn–
Jrn zn – Jrn– zn– ≤ (z
–
z
)
+

–
z
–
z
)
(J
n
n–
r
n
n–
n
r
rn
n
|rn – rn– |
Jrn zn – zn a
|rn – rn– |
Jrn zn – zn ≤ yn – yn– + |rn – rn– |Ayn– +
a
≤  – αn ( – β) xn – xn– + |αn – αn– |f (xn– ) – xn– ≤ zn – zn– +
+ |rn – rn– |Ayn– +
|rn – rn– |
Jrn zn – zn .
a
This yields
SJrn zn – SJrn– zn– ≤ Jrn zn – Jrn– zn– ≤ xn – xn– + |αn – αn– |f (xn– ) – xn– + |rn – rn– |Ayn– +
|rn – rn– |
Jrn zn – zn .
a
It follows from restrictions (a) and (c) that
lim sup SJrn zn – SJrn– zn– – xn – xn– ≤ .
n→∞
Using Lemma ., we have limn→∞ SJrn zn – xn = . It follows that
lim xn+ – xn = .
n→∞
(.)
Since yn – xn = αn f (xn ) – xn , we find that
lim yn – xn = .
n→∞
(.)
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Since ·  is convex, we find that

yn – p ≤ αn f (xn ) – p + ( – αn )xn – p .
(.)
It follows that
xn+ – p ≤ βn xn – p + ( – βn )SJrn zn – p

≤ βn xn – p + ( – βn )Jrn (I – rn A)yn – p

≤ βn xn – p + ( – βn )(I – rn A)yn – p
≤ βn xn – p + ( – βn )yn – p – rn ( – βn )(α – rn )Ayn – Ap

≤ xn – p + αn f (xn ) – p – rn ( – βn )(α – rn )Ayn – Ap .
Hence, we have
rn ( – βn )(α – rn )Ayn – Ap

≤ xn – p + xn+ – p xn+ – xn + αn f (xn ) – p .
In view of restrictions (a), (b), and (c), we find from (.) that
lim Ayn – Ap = .
n→∞
Since Jrn is firmly nonexpansive, we have
Jrn zn – p ≤ Jrn zn – p, (yn – rn Ayn ) – (p – rn Ap)
=


Jrn zn – p + (yn – rn Ayn ) – (p – rn Ap)

 – (Jr zn – p) – (yn – rn Ayn ) – (p – rn Ap) n
≤

Jrn zn – p + yn – p – Jrn zn – yn 

+ rn Ayn – ApJrn zn – yn .
This implies from (.) that
Jrn zn – p ≤ yn – p – Jrn zn – yn  + rn Ayn – ApJrn zn – yn 
≤ αn f (xn ) – p + ( – αn )xn – p – Jrn zn – yn 
+ rn Ayn – ApJrn zn – yn .
On the other hand, we have
xn+ – p ≤ βn xn – p + ( – βn )SJrn zn – p
≤ βn xn – p + ( – βn )Jrn zn – p
(.)
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
≤ xn – p + αn f (xn ) – p – ( – βn )Jrn zn – yn 
+ rn Ayn – ApJrn zn – yn .
This implies that

( – βn )Jrn zn – yn  ≤ xn – p + xn+ – p xn – xn+ + αn f (xn ) – p
+ rn Ayn – ApJrn zn – yn .
In view of restrictions (a) and (b), we find from (.) and (.) that
lim Jrn zn – yn = .
(.)
n→∞
Next, we show that lim supn→∞ f (¯x) – x¯ , yn – x¯ ≤ , where x¯ = ProjF(S)∩(A+B)– () f (¯x). To
show it, we can choose a subsequence {yni } of {yn } such that
lim sup f (¯x) – x¯ , yn – x¯ = lim f (¯x) – x¯ , yni – x¯ .
i→∞
n→∞
Since {yni } is bounded, we can choose a subsequence {ynij } of {yni } which converges weakly
some point x. We may assume, without loss of generality, that yni converges weakly to x.
Now, we are in a position to show that x ∈ (A + B)– (). Set λn = Jrn (yn – rn Ayn ). It follows
n
– Ayn ∈ Bλn . Since B is monotone, we get, for any (μ, ν) ∈ B,
that ynr–λ
n
yn – λn
λn – μ,
– Ayn – ν ≥ .
rn
Replacing n by ni and letting i → ∞, we obtain from (.) that
x – μ, –Ax – ν ≥ .
This gives –Ax ∈ Bx, that is,  ∈ (A + B)(x). This proves that x ∈ (A + B)– ().
Now, we are in a position to prove that x ∈ F(S). Notice that
Sλn – yn ≤

βn
xn+ – yn +
yn – xn .
 – βn
 – βn
This implies that limn→∞ Sλn – yn = . This implies from (.) that Sλn – λn → .
Since I – S is demiclosed at zero, we find that x ∈ F(S). This complete the proof that x ∈
F(S) ∩ (A + B)– (). It follows that
lim sup f (¯x) – x¯ , yn – x¯ ≤ .
n→∞
Finally, we show that xn → x¯ . Notice that
yn – x¯  ≤ αn f (xn ) – x¯ , yn – x¯ + ( – αn )xn – x¯ yn – x¯ ≤  – αn ( – β) xn – x¯ yn – x¯ + αn f (¯x) – x¯ , yn – x¯ .
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This implies that
yn – x¯  ≤ αn f (¯x) – x¯ , yn – x¯ +  – αn ( – β) xn – x¯  .
It follows that

xn+ – x¯  ≤ βn xn – x¯  + ( – βn )SJrn (I – rn A)yn – x¯ ≤ βn xn – x¯  + ( – βn )yn – x¯ 
≤  – αn ( – βn )( – β) xn – x¯  + αn ( – βn ) f (¯x) – x¯ , yn – x¯ .
In view of restrictions (a) and (b), we find from Lemma . that xn → x¯ . This completes
the proof.
From Theorem ., we have the following results immediately.
Corollary . Let C be a nonempty closed convex subset of H. Let S : C → C be a nonexpansive mapping with fixed points and let f : C → C be a β-contractive mapping. Let
{αn } and {βn } be real number sequences in (, ). Let {xn } be a sequence generated in the
following process: x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )Syn ,
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
(b)  < lim infn→∞ βn ≤ lim supn→∞ βn < .
Then {xn } converges strongly to a point x¯ ∈ F(S), which is also a unique solution to the
following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ F(S).
Corollary . Let C be a nonempty closed convex subset of H. Let f : C → C be a βcontractive mapping. Let A : C → H be an α-inverse-strongly monotone mapping and let
B be a maximal monotone operator on H. Assume that Dom(B) ⊂ C and (A + B)– () is
not empty. Let {αn } and {βn } be real number sequences in (, ) and {rn } be a positive real
number sequence in (, α). Let {xn } be a sequence generated in the following process: x ∈ C
and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )(I + rn B)– (yn – rn Ayn ),
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
(b)  < lim infn→∞ βn ≤ lim supn→∞ βn < ;
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(c)  < a ≤ rn ≤ b < α and ∞
n= |rn – rn– | < ∞,
where a and b are two real numbers. Then {xn } converges strongly to a point x¯ ∈ (A+B)– (),
which is also a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ (A + B)– ().
Next, we give a result on the zeros of the sum of the operators A and B based on a
different method.
Theorem . Let C be a nonempty closed convex subset of H. Let f : C → C be a βcontractive mapping. Let A : C → H be an α-inverse-strongly monotone mapping and let
B be a maximal monotone operator on H. Assume that Dom(B) ⊂ C and (A + B)– () is
not empty. Let {αn } and {βn } be real number sequences in (, ) and {rn } be a positive real
number sequence in (, α). Let {xn } be a sequence generated in the following process: x ∈ C
and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )(I + rn B)– (yn – rn Ayn ),
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
¯
(b)  ≤ βn ≤ β < ;
(c)  < a ≤ rn ≤ b < α,
¯ a, and b are real numbers. Then {xn } converges strongly to a point x¯ ∈ (A + B)– (),
where β,
which is also a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ (A + B)– ().
Proof From the proof of Theorem ., we find that {xn } is bounded. Since PF(S)∩(A+B)– () f
is contractive, it has a unique fixed point. Next, we use x¯ to denote the unique fixed point.
Note that
yn – x¯  = αn f (xn ) – x¯ , yn – x¯ + ( – αn )xn – x¯ , yn – x¯ ≤  – αn ( – β) xn – x¯ yn – x¯ + αn f (¯x) – x¯ , yn – x¯
≤
( – αn ( – β) xn – x¯  + yn – x¯  + αn f (¯x) – x¯ , yn – x¯ .

It follows that
αn ( – β)
αn
f (¯x) – x¯ , yn – x¯ .
xn – x¯  +
yn – x¯ ≤  –
 + αn ( – β)
 + αn ( – β)

Since Jrn is firmly nonexpansive and A is inverse-strongly monotone, we find that
Jr (yn – rn Ayn ) – x¯  ≤ (yn – rn Ayn ) – (¯x – rn A¯x)
n

– (I – Jrn )(yn – rn Ayn ) – (I – Jrn )(¯x – rn A¯x)
(.)
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≤ yn – x¯  – rn (α – rn )Ayn – A¯x

– (I – Jrn )(yn – rn Ayn ) – (I – Jrn )(¯x – rn A¯x) .
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(.)
Substituting (.) into (.), we find that
Jr (yn – rn Ayn ) – x¯ 
n
αn ( – β)
αn
f (¯x) – x¯ , yn – x¯
xn – x¯  +
≤ –
 + αn ( – β)
 + αn ( – β)

 – rn (α – rn )Ayn – A¯x – (I – Jrn )(yn – rn Ayn ) – (I – Jrn )(¯x – rn A¯x) .
It follows that

xn+ – x¯  ≤ βn xn – x¯  + ( – βn )Jrn (yn – rn Ayn ) – x¯ αn ( – βn )( – β)
αn ( – βn ) xn – x¯  +
f (¯x) – x¯ , yn – x¯
≤ –
 + αn ( – β)
 + αn ( – β)
– ( – βn )rn (α – rn )Ayn – A¯x

– ( – βn )(I – Jrn )(yn – rn Ayn ) – (I – Jrn )(¯x – rn A¯x) .
(.)
Next, we consider the following possible two cases.
Case . Suppose that there exists some nonnegative integer m such that the sequence
{xn – x¯ } is eventually decreasing. Then limn→∞ xn – x¯ exists. By (.), we find that
( – βn )rn (α – rn )Ayn – A¯x
≤ xn – x¯  – xn+ – x¯  + αn f (¯x) – x¯ yn – x¯ .
By use of restrictions (b) and (c), we have limn→∞ Ayn – A¯x = . It also follows from (.)
that

( – βn )(I – Jrn )(yn – rn Ayn ) – (I – Jrn )(¯x – rn A¯x)
≤ xn – x¯  – xn+ – x¯  + αn f (¯x) – x¯ yn – x¯ .
From restrictions (b) and (c), we obtain
lim (I – Jrn )(yn – rn Ayn ) – (I – Jrn )(¯x – rn A¯x) = .
n→∞
Hence, we have limn→∞ yn – Jrn (yn – rn Ayn ) = . From Lemma ., we find that yn –
Jr (yn – rAyn ) ≤ yn – Jrn (yn – rn Ayn ). This implies that limn→∞ yn – Jr (yn – rAyn ) = .
Next, we show that lim supn→∞ f (¯x) – x¯ , yn – x¯ ≤ . To show it, we can choose a subsequence {yni } of {yn } such that
lim sup f (¯x) – x¯ , yn – x¯ = lim f (¯x) – x¯ , yni – x¯ .
n→∞
i→∞
Since {yni } is bounded, we can choose a subsequence {ynij } of {yni } which converges weakly
to some point x. We may assume, without loss of generality, that yni converges weakly to x.
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Since the mapping Jr (I – rA) is nonexpansive, we find that x ∈ F(Jr (I – rA)) = (A + B)– ().
It follows that lim supn→∞ f (¯x) – x¯ , yn – x¯ ≤ . In view of (.), we find from Lemma .
that xn → x¯ .
Next, we consider another case.
Case . Suppose that the sequence {xn – x¯ } is not eventually decreasing. There exists a
subsequence {xni – x¯ } such that xni – x¯ ≤ xni + – x¯ for all i ≥ . We define an integer
sequence k(n) as in Lemma .. By use of (.), we have
xk(n)+ – x¯ 
αk(n) ( – βk(n) )( – β)
xk(n) – x¯ 
≤ –
 + αk(n) ( – β)
+
αk(n) ( – βk(n) ) f (¯x) – x¯ , yk(n) – x¯
 + αk(n) ( – β)
– ( – βk(n) )rk(n) (α – rk(n) )Ayk(n) – A¯x

– ( – βk(n) )(I – Jrk(n) )(yk(n) – rk(n) Ayk(n) ) – (I – Jrk(n) )(¯x – rk(n) A¯x) .
(.)
It follows that
lim yk(n) – Jrk(n) (yk(n) – rk(n) Ayk(n) ) = .
n→∞
(.)
Hence, we have lim supn→∞ f (¯x) – x¯ , yk(n) – x¯ ≤ . In view of (.), we find that
lim xk(n) – x¯ = .
n→∞
Note that
yk(n)+ – x¯ ≤ yk(n)+ – xk(n)+ + xk(n)+ – xk(n) + xk(n) – x¯ ≤ αk(n)+ f (xk(n)+ ) – xk(n)+ + ( – βk(n) )Jrk(n) (yk(n) – rk(n) Ayk(n) ) – xk(n) + xk(n) – x¯ ≤ αk(n)+ f (xk(n)+ ) – xk(n)+ + Jrk(n) (yk(n) – rk(n) Ayk(n) ) – yk(n) + αk(n) f (xk(n) ) – xk(n) + xk(n) – x¯ .
By use of (.), we find that limn→∞ yn – x¯ = . Since xn – x¯ ≤ αn f (xn ) – xn + yn – x¯ ,
we find that limn→∞ xn – x¯ = . This completes the proof.
Remark . Comparing Theorem . with the recent results announced in [, ] and
[], we have the following:
(i) Our proofs are different from theirs.
(ii) We remove the additional restriction ∞
n= |rn+ – rn | < ∞.
3 Applications
In this section, we investigate solutions of equilibrium problems, variational inequalities
and convex minimization problems, respectively.
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Let F be a bifunction of C × C into R, where R denotes the set of real numbers. Recall
the following equilibrium problem:
Find x ∈ C such that F(x, y) ≥ ,
∀y ∈ C.
(.)
In this paper, we use EP(F) to denote the solution set of the equilibrium problem.
To study equilibrium problems (.), we may assume that F satisfies the following conditions:
(A) F(x, x) =  for all x ∈ C;
(A) F is monotone, i.e., F(x, y) + F(y, x) ≤  for all x, y ∈ C;
(A) for each x, y, z ∈ C,
lim sup F tz + ( – t)x, y ≤ F(x, y);
t↓
(A) for each x ∈ C, y → F(x, y) is convex and weakly lower semi-continuous.
Lemma . [] Let C be a nonempty closed convex subset of a real Hilbert space H. Let
F be a bifunction from C × C to R which satisfies (A)-(A) and let B be a multivalued
mapping of H into itself defined by
Bx =
⎧
⎨{z ∈ H : F(x, y) ≥ y – x, z, ∀y ∈ C},
x ∈ C,
⎩∅,
x ∈/ C.
(.)
Then B is a maximal monotone operator with the domain D(BF ) ⊂ C, EP(F) = B– (), and
Tr x = (I + rB)– x, ∀x ∈ H, r > , where Tr is defined as

Tr x = z ∈ C : F(z, y) + y – z, z – x ≥ , ∀y ∈ C .
r
Theorem . Let C be a nonempty closed convex subset of H. Let S : C → C be a nonexpansive mapping with fixed points and let f : C → C be a β-contractive mapping. Let
A : C → H be an α-inverse-strongly monotone mapping and let F be a bifunction from
C × C to R which satisfies (A)-(A). Assume that F(S) ∩ EP(F) is not empty. Let {αn } and
{βn } be real number sequences in (, ) and let {rn } be a positive real number sequence in
(, α). Let {xn } be a sequence generated in the following process: x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )S(I + rn B)– (yn – rn Ayn ),
∀n ≥ ,
where B is a mapping defined as in (.). Assume that the control sequences satisfy the
following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
(b)  < lim infn→∞ βn ≤ lim supn→∞ βn < ;
(c)  < a ≤ rn ≤ b < α and ∞
n= |rn – rn– | < ∞,
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where a and b are two real numbers. Then {xn } converges strongly to a point x¯ ∈ F(S) ∩
EP(F), which is a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ F(S) ∩ EP(F).
Theorem . Let C be a nonempty closed convex subset of H. Let f : C → C be a βcontractive mapping. Let A : C → H be an α-inverse-strongly monotone mapping and Let F
be a bifunction from C × C to R which satisfies (A)-(A). Assume that EP(F) is not empty.
Let {αn } and {βn } be real number sequences in (, ) and let {rn } be a positive real number
sequence in (, α). Let {xn } be a sequence generated in the following process: x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )(I + rn B)– (yn – rn Ayn ),
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
(b)  ≤ βn ≤ β¯ < ;
(c)  < a ≤ rn ≤ b < α,
¯ a, and b are real numbers. Then {xn } converges strongly to a point x¯ ∈ (A + B)– (),
where β,
which is also a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ (A + B)– ().
Let g : H → (–∞, +∞] be a proper convex lower semi-continuous function. Then the
subdifferential ∂g of g is defined as follows:
∂g(x) = y ∈ H : g(z) ≥ g(x) + z – x, y, z ∈ H ,
∀x ∈ H.
From Rockafellar [], we know that ∂g is maximal monotone. It is easy to verify that  ∈
∂g(x) if and only if g(x) = miny∈H g(y). Let IC be the indicator function of C, i.e.,
IC (x) =
⎧
⎨,
x ∈ C,
⎩+∞, x ∈/ C.
(.)
Since IC is a proper lower semi-continuous convex function on H, we see that the subdifferential ∂IC of IC is a maximal monotone operator.
Lemma . [] Let C be a nonempty closed convex subset of H and let ProjC be the metric
projection from H onto C. Let ∂IC be the subdifferential of IC , where IC is as defined in (.).
Then y = (I + λ∂IC )– x ⇐⇒ y = ProjC x, ∀x ∈ H, y ∈ C,
Theorem . Let C be a nonempty closed convex subset of H. Let S : C → C be a nonexpansive mapping with fixed points and let f : C → C be a β-contractive mapping. Let
A : C → H be an α-inverse-strongly monotone mapping. Assume that F(S) ∩ VI(C, A) is
not empty. Let {αn } and {βn } be real number sequences in (, ) and let {rn } be a positive
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real number sequence in (, α). Let {xn } be a sequence generated in the following process:
x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )S Proj (yn – rn Ayn ),
C
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
() limn→∞ αn = , ∞
n= αn = ∞;
()  < lim infn→∞ βn ≤ lim supn→∞ βn < ;
()  < a ≤ rn ≤ b < α and ∞
n= |rn – rn– | < ∞,
where a and b are two real numbers. Then {xn } converges strongly to a point x¯ ∈ F(S) ∩
VI(C, A), which is a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ F(S) ∩ VI(C, A).
Proof Putting Bx = ∂IC , we find from Theorem . and Lemma . the desired conclusion
immediately.
Theorem . Let C be a nonempty closed convex subset of H. Let f : C → C be a βcontractive mapping and let A : C → H be an α-inverse-strongly monotone mapping. Assume that FVI(C, A) is not empty. Let {αn } and {βn } be real number sequences in (, ) and
let {rn } be a positive real number sequence in (, α). Let {xn } be a sequence generated in
the following process: x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn ) Proj (yn – rn Ayn ),
C
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
(b)  ≤ βn ≤ β¯ < ;
(c)  < a ≤ rn ≤ b < α,
¯ a, and b are real numbers. Then {xn } converges strongly to a point x¯ ∈ (A + B)– (),
where β,
which is also a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ (A + B)– ().
Proof Putting B = ∂IC , we find from Theorem . and Lemma . the desired conclusion
immediately.
Let W : H → R be a convex and differentiable function and M : H → R is a convex
function. Consider the convex minimization problem minx∈H (W (x) + M(x)). From [],
we know if ∇W is L -Lipschitz continuous, then it is L-inverse-strongly monotone. Hence,
we have the following results.
Theorem . Let W : H → R be a convex and differentiable function such that ∇W is L Lipschitz continuous and let M : H → R be a convex and lower semi-continuous function
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such that (∇W + ∂M)– () is not empty. Let f be a β-contractive mapping on H. Let {αn }
and {βn } be real number sequences in (, ) and let {rn } be a positive real number sequence
in (, α). Let {xn } be a sequence generated in the following process: x ∈ C and
⎧
⎨y = α f (x ) + ( – α )x ,
n
n
n
n n
⎩xn+ = βn xn + ( – βn )(I + rn M)– (yn – rn ∇Wyn ),
∀n ≥ .
Assume that the control sequences satisfy the following restrictions:
(a) limn→∞ αn = , ∞
n= αn = ∞;
¯
(b)  ≤ βn ≤ β < ;
(c)  < a ≤ rn ≤ b < α,
¯ a, and b are real numbers. Then {xn } converges strongly to a point x¯ ∈ (∇W +
where β,
∂M)– , which is also a unique solution to the following variational inequality:
f (¯x) – x¯ , p – x¯ ≤ ,
∀p ∈ (∇W + ∂M)– .
Proof Putting A = ∇W and B = ∂M, we find from Theorem . the desired conclusion
immediately.
4 Conclusions
In this paper, we study a convex feasibility problem via two monotone mappings and a
nonexpansive mapping. The common solution is also a unique solution of another variational inequality. The restrictions imposed on the sequence {rn } are mild. The results
presented in this paper mainly improve the corresponding results in [] and [].
Competing interests
The author declares that they have no competing interests.
Acknowledgements
The author is very grateful to the editor and anonymous reviewers’ suggestions which improved the contents of the
article.
Received: 25 August 2014 Accepted: 5 December 2014 Published: 06 Jan 2015
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10.1186/1029-242X-2015-2
Cite this article as: Zhang: Some results on a viscosity splitting algorithm in Hilbert spaces. Journal of Inequalities and
Applications 2015, 2015:2
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