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# Hybrid iterative algorithms for nonexpansive and nonspreading mappings in Hilbert spaces

## Abstract

Recently, Iemoto and Takahashi considered a weak convergence iterative scheme for a nonspreading mapping and a nonexpansive mapping in Hilbert spaces. In this paper, we suggest two hybrid iterative algorithms by modifying Iemoto and Takahashi’s iterative scheme for a countable family of nonspreading mappings and a nonexpansive mapping in Hilbert spaces.

MSC:47H05, 47H09.

## 1 Introduction and preliminaries

Let H be a Hilbert space and C be a nonempty closed convex subset of H. Let T be a nonlinear mapping of C into itself. We use $F(T)$ and $P C$ to denote the set of fixed points of T and the metric projection from H onto C, respectively.

Recall that T is said to be nonexpansive if

$∥Tx−Ty∥≤∥x−y∥$
(1.1)

for all $x,y∈C$.

For approximating the fixed point of a nonexpansive mapping in a Hilbert space, Mann  in 1953 introduced the famous iterative scheme as follows:

$∀ x 1 ∈C, x n + 1 =(1− α n ) x n + α n T x n ,∀n≥1,$
(1.2)

where T is a nonexpansive mapping of C into itself and ${ α n }$ is a sequence in $(0,1)$. It is well known that ${ x n }$ defined in (1.2) converges weakly to a fixed point of T.

Attempts to modify the normal Mann iteration method (1.2) for nonexpansive mappings so that strong convergence is guaranteed have recently been made; see, e.g., .

Let T be a mapping from C into itself. Then T is called nonspreading  if

$2 ∥ T x − T y ∥ 2 ≤ ∥ T x − y ∥ 2 + ∥ x − T y ∥ 2$

for all $x,y∈C$. A mapping $T:C→C$ is called quasi-nonexpansive if $F(T)≠∅$ and $∥Tx−y∥≤∥x−y∥$ for all $x∈C$ and $y∈F(T)$. If T is a nonspreading mapping from C into itself and $F(T)$ is nonempty, then T is quasi-nonexpansive. Further, we know that the set of fixed points of each quasi-nonexpansive mapping is closed and convex; see .

In , by using Moudafi’s iterative scheme , Iemoto and Takahashi considered the following weak convergence theorem.

Theorem IT ()

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a nonspreading mapping of C into itself, and let T be a nonexpansive mapping of C into itself such that $F(S)∩F(T)≠∅$. Define a sequence ${ x n }$ as follows:

${ x 1 ∈ C , x n + 1 = ( 1 − α n ) x n + α n { β n S x n + ( 1 − β n ) T x n }$
(1.3)

for all $n∈N$, where ${ α n },{ β n }⊂[0,1]$. Then the following hold:

1. (i)

If $lim inf n → ∞ α n (1− α n )>0$ and $∑ n = 1 ∞ (1− β n )<∞$, then ${ x n }$ converges weakly to $v∈F(S)$;

2. (ii)

If $∑ i = 1 ∞ α n (1− α n )=∞$ and $∑ n = 1 ∞ β n <∞$, then ${ x n }$ converges weakly to $v∈F(T)$;

3. (iii)

If $lim inf n → ∞ α n (1− α n )>0$ and $lim inf n → ∞ β n (1− β n )>0$, then ${ x n }$ converges weakly to $v∈F(S)∩F(T)$.

In this paper, we modify (1.1) by a hybrid iterative scheme and obtain the strong convergence theorems for a family of nonspreading mappings and a nonexpansive mapping in a Hilbert space.

Let E be a Banach space and K be a nonempty closed convex subset of E. Let ${ T n }:K→K$ be a family of mappings. Then ${ T n }$ is said to satisfy the AKTT-condition  if for each bounded subset B of K, one has

$∑ n = 1 ∞ sup { ∥ T n + 1 z − T n z ∥ : z ∈ B } <∞.$

The following is an important result on a family of mappings ${ T n } n = 1 ∞$ satisfying the AKTT-condition.

Lemma 1.1 ()

Let K be a nonempty and closed subset of a Banach space E, and let ${ T n } n = 1 ∞$ be a family of mappings of K into itself which satisfies the AKTT-condition. Then, for each $x∈K$, ${ T n x}$ converges strongly to a point in K. Moreover, let the mapping $T:K→K$ be defined by

$Tx= lim n → ∞ T n x,∀x∈K.$

Then, for each bounded subset B of K,

$lim sup n → ∞ { ∥ T z − T n z ∥ : z ∈ B } =0.$

Obviously, if a family of mappings ${ T n } n = 1 ∞$ satisfies the AKTT-condition and $Tx= lim n → ∞ T n x$ for each $x∈K$, then it is unnecessary that $F(T)= ⋂ i = 1 ∞ F( T i )$. To show this, see the following example.

Example 1.1 Let $E=R$ and $K=[0,2]$. Define a family of mappings ${ T n } n = 1 ∞ :K→K$ by

$T 1 x=0, T n = 1 n (1+x),n≥2.$

Then ${ T n } n = 1 ∞$ satisfy the AKTT-condition. It is easy to see that for each $x∈K$, $lim n → ∞ T n x=0$. Define the mapping $T:K→K$ by $Tx= lim n → ∞ T n x$. That is, $Tx=0$ for all $x∈K$. But $F(T)≠ ⋂ n = 1 ∞ F( T n )$.

In this paper, we call that ${ T n ,T}$ satisfy the AKTT-condition if ${ T n } n = 1 ∞$ satisfy the AKTT-condition with $F(T)= ⋂ n = 1 ∞ F( T n )$.

Lemma 1.2 ()

Let C be a nonempty closed subset of a Hilbert space H. Then a mapping $T:C→C$ is nonspreading if and only if

$∥ T x − T y ∥ 2 ≤ ∥ x − y ∥ 2 +2〈x−Tx,y−Ty〉$

for all $x,y∈C$.

By using Lemma 1.2, we get the following simple but important result.

Lemma 1.3 Let H be a Hilbert space and C be a nonempty subset of H. Let ${ T n }$ be a family of nonspreading mappings of C into itself, and assume that $lim n → ∞ T n x$ exists for each $x∈C$. Define the mapping $T:C→C$ by $Tx= lim n → ∞ T n x$. Then the mapping T is a nonspreading mapping.

Proof In fact, for all $x,y∈C$, we have

$∥ T x − T y ∥ 2 = ∥ lim n → ∞ T n x − lim n → ∞ T n y ∥ 2 = lim n → ∞ ∥ T n x − T n y ∥ 2 ≤ lim n → ∞ [ ∥ x − y ∥ 2 + 2 〈 x − T n x , y − T n y 〉 ] = ∥ x − y ∥ 2 + 2 〈 x − lim n → ∞ T n x , y − lim n → ∞ T n y 〉 = ∥ x − y ∥ 2 + 2 〈 x − T x , y − T y 〉 .$

Lemma 1.2 shows that the mapping T is a nonspreading mapping. □

Lemma 1.4 Let C be a closed convex subset of a real Hilbert space H, and let $P C$ be the metric projection from H onto C (i.e., for $x∈H$, $P C x$ is the only point in C such that $∥x− P C x∥=inf{∥x−z∥:z∈C}$). Given $x∈H$ and $z∈C$. Then $z= P C x$ if and only if the following relation holds:

$〈x−z,y−z〉≤0,∀y∈C.$

Lemma 1.5 ()

Let H be a real Hilbert space. Then the following equation holds:

$∥ t x + ( 1 − t ) y ∥ 2 =t ∥ x ∥ 2 +(1−t) ∥ y ∥ 2 −t(1−t) ∥ x − y ∥ 2 ,∀x∈Cand∀t∈[0,1].$

## 2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of a Hilbert space H. Let $S:C→C$ be a nonexpansive mapping and ${ T i } i = 1 ∞ :C→C$ be a countable family of nonspreading mappings such that $F=F(S)∩[ ⋂ i = 1 ∞ F( T i )]≠∅$. Let ${ x n }$ be a sequence generated in the following manner:

${ x 1 = x ∈ C chosen arbitrarily , y n = ( 1 − α n ) x n + α n [ β n S x n + ∑ i = 1 n ( β i − 1 − β i ) T i x n ] , C n = { v ∈ C : ∥ y n − v ∥ ≤ ∥ x n − v ∥ } , D n = ⋂ j = 1 n C j , x n + 1 = P D n x , n ≥ 1 ,$
(2.1)

where ${ α n },{ β n }⊂[0,1]$. Assume that ${ β n }$ is strictly decreasing and $β 0 =1$. Then the following hold:

1. (i)

If $lim inf n → ∞ α n >0$ and $lim n → ∞ β n =0$, then ${ x n }$ strongly converges to $q∈ ⋂ i = 1 ∞ F( T i )$;

2. (ii)

If $lim inf n → ∞ α n (1− α n )>0$ and $lim inf n → ∞ β n >0$, then ${ x n }$ converges strongly to $q∈F$.

Proof Obviously, each $C n$ is closed and convex and hence $D n$ is closed and convex. Next, we show that $F⊂ D n$ for all $n≥1$. To end this, we need to prove that $F⊂ C n$ for all $n≥1$. Indeed, for each $p∈F$, we have

$∥ y n − p ∥ ≤ ( 1 − α n ) ∥ x n − p ∥ + α n [ β n ∥ S x n − p ∥ + ∑ i = 1 n ( β i − 1 − β i ) ∥ T i x n − p ∥ ] ≤ ( 1 − α n ) ∥ x n − p ∥ + α n [ β n ∥ x n − p ∥ + ∑ i = 1 n ( β i − 1 − β i ) ∥ x n − p ∥ ] = ∥ x n − p ∥ .$
(2.2)

This implies that

Therefore, $F⊂ C n$ and hence $C n$ is nonempty for all $n≥1$. On the other hand, from the definition of $D n$, we see that $F⊂ D n = ⋂ i = 1 n C j$ for all $n≥1$.

From $x n + 1 = P D n x$, we have

$∥ x n + 1 −x∥≤∥v−x∥,∀v∈ D n ,n≥1.$

Since $P F x∈F⊂ D n$, one has

$∥ x n + 1 −x∥≤∥ P F x−x∥,n≥1.$
(2.3)

This implies that ${ x n }$ is bounded and hence ${ y n }$ is bounded.

On the other hand, since $D n + 1 ⊂ D n$ for all $n≥1$, we have

$x n + 2 = P D n + 1 x∈ D n + 1 ⊂ D n$

for all $n≥1$. From $x n + 1 = P D n x$ one has

$∥ x n + 1 −x∥≤∥ x n + 2 −x∥$
(2.4)

for all $n≥1$. It follows from (2.3) and (2.4) that the limit of ${ x n −x}$ exists.

Since $D m ⊂ D n$ and $x m + 1 = P D m x∈ D m ⊂ D n$ for all $m≥n$ and $x n + 1 = P D n x$, by Lemma 1.4 one has

$〈 x n + 1 −x, x m + 1 − x n + 1 〉≥0.$
(2.5)

It follows from (2.5) that

$∥ x m + 1 − x n + 1 ∥ 2 = ∥ x m + 1 − x − ( x n + 1 − x ) ∥ 2 = ∥ x m + 1 − x ∥ 2 + ∥ x n + 1 − x ∥ 2 − 2 〈 x n + 1 − x , x m + 1 − x 〉 = ∥ x m + 1 − x ∥ 2 + ∥ x n + 1 − x ∥ 2 − 2 〈 x n + 1 − x , x m + 1 − x n + 1 + x n + 1 − x 〉 = ∥ x m + 1 − x ∥ 2 − ∥ x n + 1 − x ∥ 2 − 2 〈 x n + 1 − x , x m + 1 − x n + 1 〉 ≤ ∥ x m + 1 − x ∥ 2 − ∥ x n + 1 − x ∥ 2 .$
(2.6)

Since the limit of $∥ x n −x∥$ exists, we get

$lim m , n → ∞ ∥ x m − x n ∥=0.$

It follows that ${ x n }$ is a Cauchy sequence. Since H is a Hilbert space and C is closed and convex, there exists $q∈C$ such that

(2.7)

By taking $m=n+1$ in (2.6), one arrives at

$lim n → ∞ ∥ x n + 2 − x n + 1 ∥=0,$

i.e.,

$lim n → ∞ ∥ x n + 1 − x n ∥=0.$
(2.8)

Noticing that $x n + 1 = P D n x∈ D n ⊂ C n$, we get

$∥ y n − x n + 1 ∥≤∥ x n − x n + 1 ∥→0,$

and hence

$∥ y n − x n ∥≤∥ y n − x n + 1 ∥+∥ x n + 1 − x n ∥→0.$
(2.9)

From (2.7) and (2.9) it follows that

$lim n → ∞ ∥ y n −p∥= lim n → ∞ ∥ x n −p∥=∥q−p∥,∀p∈F.$
(2.10)

Now we prove (i). Note that

$y n = ( 1 − α n ) x n + α n [ β n S x n + ∑ i = 1 n ( β i − 1 − β i ) ( T i x n − x n ) ] + α n ( 1 − β n ) x n = ( 1 − α n β n ) x n + α n β n S x n + α n ∑ i = 1 n ( β i − 1 − β i ) ( T i x n − x n ) .$

Hence,

$α n ∑ i = 1 n ( β i − 1 − β i )( T i x n − x n )=(1− α n β n )( y n − x n )+ α n β n ( y n −S x n ).$
(2.11)

On the other hand, for any $p∈F$, from Lemma 1.2 we have

$∥ x n − p ∥ 2 = 2 〈 x n − T i x n , p − T i p 〉 + ∥ x n − p ∥ 2 ≥ ∥ T i x n − T i p ∥ 2 = ∥ T i x n − p ∥ 2 = ∥ T i x n − x n + ( x n − p ) ∥ 2 = ∥ T i x n − x n ∥ 2 + ∥ x n − p ∥ 2 + 2 〈 T i x n − x n , x n − p 〉 ,$

and hence

$∥ T i x n − x n ∥ 2 ≤2〈 x n − T i x n , x n −p〉,∀i∈N.$
(2.12)

Note that ${ β n }$ is strictly decreasing. Hence from (2.11) and (2.12) we get

$∥ T i x n − x n ∥ 2 ≤ 1 2 α n ( β i − 1 − β i ) [ ( 1 − α n β n ) 〈 y n − x n , x n − T i p 〉 + α n β n 〈 y n − S x n , x n − p 〉 ] , i ≥ 1 .$
(2.13)

Since $lim inf n → ∞ α n >0$ and $lim n → ∞ β n =0$, from (2.9) and (2.13) it follows that

$lim n → ∞ ∥ T i x n − x n ∥=0,∀i∈N.$
(2.14)

Since each $T i$ is a nonspreading mapping, by Lemma 1.2, (2.7) and (2.10), we have

$∥ T i q − T i x n ∥ 2 ≤ ∥ x n − q ∥ 2 +2〈q− T i q, x n − T i x n 〉→0,∀i∈N.$
(2.15)

Further, one has

$∥q− T i q∥≤∥q− x n ∥+∥ x n − T i x n ∥+∥ T i x n − T i q∥→0,∀i∈N.$
(2.16)

So, we have $q∈ ⋂ i = 1 ∞ F( T i )$.

To prove (ii), first we show that $lim n → ∞ ∥ x n −S x n ∥=0$. For any $p∈F$, we have

$∥ y n − p ∥ 2 = ∥ β n [ ( 1 − α n ) x n + α n S x n − p ] + ∑ i = 1 n ( β i − 1 − β i ) [ ( 1 − α n ) x n + α n T i x n − p ] ∥ 2 ≤ β n ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 + ∑ i = 1 n ( β i − 1 − β i ) ∥ ( 1 − α n ) x n + α n T i x n − p ∥ 2 ≤ β n ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 + ∑ i = 1 n ( β i − 1 − β i ) [ ( 1 − α n ) ∥ x n − p ∥ 2 + α n ∥ T i x n − p ∥ 2 ] ≤ β n ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 + ∑ i = 1 n ( β i − 1 − β i ) ∥ x n − p ∥ 2 = β n ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 + ( 1 − β n ) ∥ x n − p ∥ 2 ≤ ∥ x n − p ∥ 2 ,$

and hence by (2.10) we get

$0 ≤ ∥ x n − p ∥ 2 − β n ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 − ( 1 − β n ) ∥ x n − p ∥ 2 = β n [ ∥ x n − p ∥ 2 − ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 ] ≤ ∥ x n − p ∥ 2 − ∥ y n − p ∥ 2 → 0 .$
(2.17)

Since $lim inf n → ∞ β n >0$, it follows from (2.17) that

$lim n → ∞ ( ∥ x n − p ∥ 2 − ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 ) =0.$
(2.18)

From (2.18) and

$∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 =(1− α n ) ∥ x n − p ∥ 2 + α n ∥ S x n − p ∥ 2 − α n (1− α n ) ∥ x n − S x n ∥ 2 ,$

we get

$α n ( 1 − α n ) ∥ x n − S x n ∥ 2 = ( ∥ x n − p ∥ 2 − ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 ) − α n ∥ x n − p ∥ 2 + α n ∥ S x n − p ∥ 2 ≤ ( ∥ x n − p ∥ 2 − ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 ) − α n ∥ x n − p ∥ 2 + α n ∥ x n − p ∥ 2 = ∥ x n − p ∥ 2 − ∥ ( 1 − α n ) x n + α n S x n − p ∥ 2 → 0 .$

Since $lim inf n → ∞ α n (1− α n )>0$, we get

$lim n → ∞ ∥ x n −S x n ∥=0.$
(2.19)

Now, using (2.19), (2.7) and

$∥q−Sq∥≤∥q− x n ∥+∥ x n −S x n ∥+∥S x n −Sq∥≤2∥q− x n ∥+∥ x n −S x n ∥→0,$

which implies that $q∈F(S)$.

Note that (2.9) and (2.19) imply that $lim n → ∞ ∥ y n −S x n ∥=0$. Then, repeating (2.11) to (2.16), we get $q∈ ⋂ i = 1 ∞ F( T i )$. So, $q∈F$. This completes the proof. □

Theorem 2.2 Let C be a nonempty closed convex subset of a Hilbert space H. Let $S:C→C$ be a nonexpansive mapping and ${ T i } i = 1 ∞ :C→C$ be a countable family of nonspreading mappings such that $F=F(S)∩[ ⋂ i = 1 ∞ F( T i )]≠∅$. Let ${ x n }$ be a sequence generated in the following manner:

${ x 1 = x ∈ C chosen arbitrarily , y n = ( 1 − α n ) x n + α n [ β n S x n + ( 1 − β n ) T n x n ] , C n = { v ∈ C : ∥ y n − v ∥ ≤ ∥ x n − v ∥ } , D n = ⋂ j = 1 n C j , x n + 1 = P D n x , n ≥ 1 ,$
(2.20)

where ${ α n },{ β n }⊂[0,1]$. Assume that ${ T n ,T}$ satisfies the AKTT-condition. Then the following hold:

1. (i)

If $lim inf n → ∞ α n >0$ and $lim n → ∞ β n =0$, then ${ x n }$ strongly converges to $v∈ ⋂ i = 1 ∞ F( T i )$;

2. (ii)

If $lim inf n → ∞ α n (1− α n )>0$ and $lim inf n → ∞ β n >0$, then ${ x n }$ converges strongly to $z∈F$.

Proof By a process similar to the proof of Theorem 2.1, we can conclude that ${ x n }$ converges strongly to some $q∈C$ and

$x n − y n →0.$

We first prove (i). From (2.20) we have

$T n x n − x n = 1 α n ( 1 − β n ) ( y n − x n )− β n 1 − β n (S x n − x n ),$

and hence

$∥ T n x n − x n ∥≤ 1 α n ( 1 − β n ) ∥ y n − x n ∥+ β n 1 − β n ∥S x n − x n ∥.$

Since $lim inf n → ∞ α n >0$ and $lim n → ∞ β n =0$, we get

$lim n → ∞ ∥ T n x n − x n ∥=0.$
(2.21)

Further, by Lemma 1.1 and (2.21), we have

$∥ x n − T x n ∥ ≤ ∥ x n − T n x n ∥ + ∥ T n x n − T x n ∥ ≤ ∥ x n − T n x n ∥ + sup { ∥ T n z − T z ∥ : z ∈ { x n } } → 0 .$
(2.22)

Since each $T n$ is a nonspreading mapping, Lemma 1.3 shows that T is a nonspreading mapping. Further, by using Lemma 1.2, we have

$∥ T q − T x n ∥ 2 ≤ ∥ x n − q ∥ 2 +2〈q−Tq, x n −T x n 〉→0,∀i∈N.$
(2.23)

From (2.21) and (2.23) it follows that

$∥q−Tq∥≤∥q− x n ∥+∥ x n −T x n ∥+∥T x n −Tq∥→0.$
(2.24)

It follows that $q∈F(T)$. Since $({ T n },T)$ satisfies the AKTT-condition, one has $q∈ ⋂ i = 1 ∞ F( T i )=F(T)$. This completes (i).

Next we show (ii). By a process similar to the proof of Theorem 2.1 and from (2.22) to (2.24), we can get that

$lim n → ∞ ∥ x n − S x n ∥ = 0 , lim n → ∞ ∥ x n − T n x n ∥ = 0 , lim n → ∞ ∥ T x n − T q ∥ = 0 and lim n → ∞ ∥ x n − T x n ∥ = 0 .$

Finally, by

$∥q−Sq∥≤∥q− x n ∥+∥ x n −S x n ∥+∥S x n −Sq∥≤2∥ x n −q∥+∥ x n −S x n ∥→0$

and

$∥q−Tq∥≤∥q− x n ∥+∥ x n −T x n ∥+∥T x n −Tq∥→0,$

we get $q∈F(S)∩F(T)$. Since $({ T n },T)$ satisfies the AKTT-condition, we conclude that $q∈F$. This completes (ii). □

Letting $T i =T$ for all $i∈N$ in Theorem 2.1 and Theorem 2.2, we get the following corollary.

Corollary 2.1 Let C be a nonempty closed convex subset of a Hilbert space H. Let $S:C→C$ be a nonexpansive mapping and $T:C→C$ be a nonspreading mapping such that $F(S)∩F(T)≠∅$. Let ${ x n }$ be a sequence generated in the following manner:

${ x 1 = x ∈ C chosen arbitrarily , y n = ( 1 − α n ) x n + α n [ β n S x n + ( 1 − β n ) T x n ] , C n = { v ∈ C : ∥ y n − v ∥ ≤ ∥ x n − v ∥ } , D n = ⋂ j = 1 n C j , x n + 1 = P D n x , n ≥ 1 ,$

where ${ α n },{ β n }⊂[0,1]$. Then the following hold:

1. (i)

If $lim inf n → ∞ α n >0$ and $lim n → ∞ β n =0$, then ${ x n }$ strongly converges to $x ′ ∈F(T)$;

2. (ii)

If $lim inf n → ∞ α n (1− α n )>0$ and $lim inf n → ∞ β n >0$, then ${ x n }$ converges strongly to $q∈F(S)∩F(T)$ with $q= P F x$.

Letting $S=I$ in Theorems 2.1 and 2.2, we get the following corollary.

Corollary 2.2 Let C be a nonempty closed convex subset of a Hilbert space H. Let ${ T i } i = 1 ∞ :C→C$ be a countable family of nonspreading mappings such that $⋂ i = 1 ∞ F( T i )≠∅$. Let ${ x n }$ be a sequence generated in the following manner:

${ x 1 = x ∈ C chosen arbitrarily , y n = ( 1 − α n ( 1 − β n ) ) x n + α n ∑ i = 1 n ( β i − 1 − β i ) T i x n , C n = { v ∈ C : ∥ y n − v ∥ ≤ ∥ x n − v ∥ } , D n = ⋂ j = 1 n C j , x n + 1 = P D n x , n ≥ 1 ,$

where ${ α n },{ β n }⊂[0,1]$. Assume that ${ β n }$ is strictly decreasing and $β 0 =1$. Then if $lim inf n → ∞ α n (1− α n )>0$, then ${ x n }$ strongly converges to $q∈ ⋂ i = 1 ∞ F( T i )$.

Corollary 2.3 Let C be a nonempty closed convex subset of a Hilbert space H. Let ${ T i } i = 1 ∞ :C→C$ be a countable family of nonspreading mappings such that $⋂ i = 1 ∞ F( T i )≠∅$. Let ${ x n }$ be a sequence generated in the following manner:

${ x 1 = x ∈ C chosen arbitrarily , y n = ( 1 − α n ( 1 − β n ) ) x n + α n ( 1 − β n ) T n x n , C n = { v ∈ C : ∥ y n − v ∥ ≤ ∥ x n − v ∥ } , D n = ⋂ j = 1 n C j , x n + 1 = P D n x , n ≥ 1 ,$

where ${ γ n }⊂[0,1]$. Assume that $({ T n ,T})$ satisfies the AKTT-condition. Then if $lim inf n → ∞ α n (1− α n )>0$, then ${ x n }$ strongly converges to $q∈ ⋂ i = 1 ∞ F( T i )$.

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## Acknowledgements

This work is supported by the Fundamental Research Funds for the Central Universities (Grant Number: 13MS109).

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Correspondence to Shenghua Wang.

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Wang, S. Hybrid iterative algorithms for nonexpansive and nonspreading mappings in Hilbert spaces. Fixed Point Theory Appl 2013, 314 (2013). https://doi.org/10.1186/1687-1812-2013-314

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### Keywords

• nonexpansive mappings
• nonspreading mappings
• AKTT-condition
• hybrid algorithms 