1. Introduction Report issue for preceding element
Let X X be a real Banach space and let B B be a nonempty, convex subset. Let u 1 , x 1 ∈ B u_{1},x_{1}\in B be two arbitrary fixed points. Let T : B → B T:B\rightarrow B be a map.
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Definition 1. The map T T is said to be asymptotically nonexpansive if there exists a sequence ( k n ) n , k n ∈ [ 1 , ∞ ) , ∀ n ∈ ℕ , lim n → ∞ k n = 1 \left(k_{n}\right)_{n},k_{n}\in[1,\infty),\forall n\in\mathbb{N},\lim_{n\rightarrow\infty}k_{n}=1 , such that
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‖ T n ​ x − T n ​ y ‖ ⩽ k n ​ ‖ x − y ‖ , ∀ x , y ∈ B , ∀ n ∈ ℕ . \left\|T^{n}x-T^{n}y\right\|\leqslant k_{n}\|x-y\|,\quad\forall x,y\in B,\forall n\in\mathbb{N}.
(1)
The following remark will be useful.
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Remark 2. An asymptotically nonexpansive map is uniformly Lipschitzian for some L ⩾ 1 L\geqslant 1 , i.e., ( ∃ L ⩾ 1 : ∥ T n x − T n y ∥ ⩽ L ∥ x − y ∥ , ∀ x , y ∈ B , ∀ n ∈ ℕ \left(\exists L\geqslant 1:\left\|T^{n}x-T^{n}y\right\|\leqslant L\|x-y\|,\forall x,y\in B,\forall n\in\mathbb{N}\right. ).
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Proof. Let L := sup n ∈ ℕ k n L:=\sup_{n\in\mathbb{N}}k_{n} . Because lim n → ∞ k n = 1 \lim_{n\rightarrow\infty}k_{n}=1 and k n ⩾ 1 , ∀ n ∈ ℕ k_{n}\geqslant 1,\forall n\in\mathbb{N} , one can deduce that L ∈ [ 1 , ∞ ) L\in[1,\infty) . From Definition 1,
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‖ T n ​ x − T n ​ y ‖ ⩽ k n ​ ‖ x − y ‖ ⩽ L ​ ‖ x − y ‖ , ∀ x , y ∈ B , ∀ n ∈ ℕ . \left\|T^{n}x-T^{n}y\right\|\leqslant k_{n}\|x-y\|\leqslant L\|x-y\|,\quad\forall x,y\in B,\forall n\in\mathbb{N}.
In [5] the following class of maps was introduced.
Definition 3. A map T T is said to be asymptotically pseudocontractive if there exists a sequence ( k n ) n , k n ∈ [ 1 , ∞ ) , ∀ n ∈ ℕ , lim n → ∞ k n = 1 \left(k_{n}\right)_{n},k_{n}\in[1,\infty),\forall n\in\mathbb{N},\lim_{n\rightarrow\infty}k_{n}=1 , and there exists j ​ ( x − y ) ∈ J ​ ( x − y ) j(x-y)\in J(x-y) such that
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⟨ T n ​ x − T n ​ y , j ​ ( x − y ) ⟩ ⩽ k n ​ ‖ x − y ‖ 2 , ∀ x , y ∈ B , ∀ n ∈ ℕ . \left\langle T^{n}x-T^{n}y,j(x-y)\right\rangle\leqslant k_{n}\|x-y\|^{2},\quad\forall x,y\in B,\forall n\in\mathbb{N}.
(2)
When n = 1 n=1 in (2) we get the usual definition of a strongly pseudocontractive map. The following remark is Remark 1 from [5].
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Remark 4 [5]. An asymptotically nonexpansive map is asymptotically pseudocontractive. The converse is not true.
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We consider the following iteration (see [3]):
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u n + 1 = ( 1 − α n ) ​ u n + α n ​ T n ​ u n . u_{n+1}=\left(1-\alpha_{n}\right)u_{n}+\alpha_{n}T^{n}u_{n}.
(3)
The sequence ( α n ) n ⊂ ( 0 , 1 ) \left(\alpha_{n}\right)_{n}\subset(0,1) is convergent, such that lim n → ∞ α n = 0 \lim_{n\rightarrow\infty}\alpha_{n}=0 and ∑ n = 1 ∞ α n = ∞ \sum_{n=1}^{\infty}\alpha_{n}=\infty . This iteration is known as Mann type iteration. We consider the following iteration, known as Ishikawa type iteration (see [1]):
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x n + 1 = ( 1 − α n ) ​ x n + α n ​ T n ​ y n \displaystyle x_{n+1}=\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}T^{n}y_{n}
y n = ( 1 − β n ) ​ x n + β n ​ T n ​ x n , n = 1 , 2 , … . \displaystyle y_{n}=\left(1-\beta_{n}\right)x_{n}+\beta_{n}T^{n}x_{n},\quad n=1,2,\ldots.
(4)
The sequences ( α n ) n , ( β n ) n ⊂ ( 0 , 1 ) \left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n}\subset(0,1) are such that
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lim n → ∞ α n = 0 , lim n → ∞ β n = 0 , ∑ n = 1 ∞ α n = ∞ . \lim_{n\rightarrow\infty}\alpha_{n}=0,\quad\lim_{n\rightarrow\infty}\beta_{n}=0,\quad\sum_{n=1}^{\infty}\alpha_{n}=\infty.
(5)
The sequence ( α n ) n \left(\alpha_{n}\right)_{n} remains the same in both iterations. For β n = 0 , ∀ n ∈ ℕ \beta_{n}=0,\forall n\in\mathbb{N} , from (5) we get (4). We denote by F ​ ( T ) = { x ∗ ∈ B : F ​ ( x ∗ ) = x ∗ } F(T)=\left\{x^{*}\in B:F\left(x^{*}\right)=x^{*}\right\} . Replacing T n T^{n} by T T in (3) and (5) gives the Mann and Ishikawa iteration, respectively.
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The aim of this paper is to prove an equivalence between the convergences of the above two iterations when T T is an asymptotically nonexpansive respective asymptotically pseudocontractive map.
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The following lemma appears in [2].
Lemma 5 [2]. Let X X be a Banach space and x , y ∈ X x,y\in X . Then ‖ x ‖ ⩽ ‖ x + r ​ y ‖ \|x\|\leqslant\|x+ry\| for all r > 0 r>0 if and only if there exists j ​ ( x ) ∈ J ​ ( x ) j(x)\in J(x) such that ⟨ y , j ​ ( x ) ⟩ ⩾ 0 \langle y,j(x)\rangle\geqslant 0 .
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Using this lemma we are able to prove the following result.
Lemma 6. Let B B be a nonempty subset of a Banach space X X and let T : B → B T:B\rightarrow B be a map. Then the following conditions are equivalent:
(i) T T is asymptotically pseudocontractive map;
(ii) There exists k n ∈ [ 1 , ∞ ) k_{n}\in[1,\infty) such that
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‖ x − y ‖ ⩽ ‖ x − y + r ​ [ ( k n ​ I − T n ) ​ x − ( k n ​ I − T n ) ​ y ] ‖ , ∀ x , y ∈ B , r > 0 . \|x-y\|\leqslant\left\|x-y+r\left[\left(k_{n}I-T^{n}\right)x-\left(k_{n}I-T^{n}\right)y\right]\right\|,\quad\forall x,y\in B,r>0.
(6)
Proof. Lemma 5 assures that relation (6) is ⟨ ( k n ​ I − T n ) ​ x − ( k n ​ I − T n ) ​ y , j ​ ( x − y ) ⟩ ⩾ 0 \left\langle\left(k_{n}I-T^{n}\right)x-\left(k_{n}I-T^{n}\right)y,j(x-y)\right\rangle\geqslant 0 , ∀ n ⩾ n 0 \forall n\geqslant n_{0} , which is equivalent with ⟨ T n ​ x − T n ​ y , j ​ ( x − y ) ⟩ ⩽ k n ​ ⟨ x − y , j ​ ( x − y ) ⟩ = k n ​ ‖ x − y ‖ 2 , ∀ x , y ∈ B \left\langle T^{n}x-T^{n}y,j(x-y)\right\rangle\leqslant k_{n}\langle x-y,j(x-y)\rangle=k_{n}\|x-y\|^{2},\forall x,y\in B , that is (2).
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The following lemma is Lemma 4 from [6].
Lemma 7 [6]. Let ( a n ) n \left(a_{n}\right)_{n} be a nonnegative sequence which satisfies the following inequality:
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a n + 1 ⩽ ( 1 − λ n ) ​ a n + σ n , a_{n+1}\leqslant\left(1-\lambda_{n}\right)a_{n}+\sigma_{n},
(7)
where λ n ∈ ( 0 , 1 ) , ∀ n ∈ ℕ , ∑ n = 1 ∞ λ n = ∞ \lambda_{n}\in(0,1),\forall n\in\mathbb{N},\sum_{n=1}^{\infty}\lambda_{n}=\infty , and σ n = o ​ ( λ n ) \sigma_{n}=o\left(\lambda_{n}\right) . Then lim n → ∞ a n = 0 \lim_{n\rightarrow\infty}a_{n}=0 .
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2. Main results Report issue for preceding element
We are now able to give the following result.
Theorem 8. Let B B be a closed convex subset of an arbitrary Banach space X X and ( x n ) n \left(x_{n}\right)_{n} and ( u n ) n \left(u_{n}\right)_{n} defined by (3) and (4) with ( α n ) n \left(\alpha_{n}\right)_{n} and ( β n ) n \left(\beta_{n}\right)_{n} satisfying (5). Let T T be an asymptotically pseudocontractive and Lipschitzian with L ⩾ 1 L\geqslant 1 self-map of B B . Let x ∗ x^{*} be the fixed point of T T . If u 0 = x 0 ∈ B u_{0}=x_{0}\in B , then the following two assertions are equivalent:
(i) Mann type iteration (3) converges to x ∗ ∈ F ​ ( T ) x^{*}\in F(T) ;
(ii) Ishikawa type iteration (4) converges to x ∗ ∈ F ​ ( T ) x^{*}\in F(T) .
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Proof. If the Ishikawa iteration (4) converges then setting β n = 0 , ∀ n ∈ ℕ \beta_{n}=0,\forall n\in\mathbb{N} , the convergence of Mann iteration (3). Conversely, we shall prove that (i) ⇒ \Rightarrow (ii). The proof is similar to the proof of Theorem 4 from [4]. We have
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x n = \displaystyle x_{n}=
x n + 1 + α n ​ x n − α n ​ T n ​ y n \displaystyle x_{n+1}+\alpha_{n}x_{n}-\alpha_{n}T^{n}y_{n}
= \displaystyle=
( 1 + α n 2 ) ​ x n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ x n + 1 \displaystyle\left(1+\alpha_{n}^{2}\right)x_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}
− ( 1 + k n ) ​ α n 2 ​ x n + 1 + α n ​ x n + α n ​ ( T n ​ x n + 1 − T n ​ y n ) \displaystyle-\left(1+k_{n}\right)\alpha_{n}^{2}x_{n+1}+\alpha_{n}x_{n}+\alpha_{n}\left(T^{n}x_{n+1}-T^{n}y_{n}\right)
= \displaystyle=
( 1 + α n 2 ) ​ x n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ x n + 1 \displaystyle\left(1+\alpha_{n}^{2}\right)x_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}
− ( 1 + k n ) ​ α n 2 ​ [ x n + α n ​ ( T n ​ y n − x n ) ] + α n ​ x n + α n ​ ( T n ​ x n + 1 − T n ​ y n ) \displaystyle-\left(1+k_{n}\right)\alpha_{n}^{2}\left[x_{n}+\alpha_{n}\left(T^{n}y_{n}-x_{n}\right)\right]+\alpha_{n}x_{n}+\alpha_{n}\left(T^{n}x_{n+1}-T^{n}y_{n}\right)
= \displaystyle=
( 1 + α n 2 ) ​ x n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ x n + 1 − ( 1 + k n ) ​ α n 2 ​ x n \displaystyle\left(1+\alpha_{n}^{2}\right)x_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}-\left(1+k_{n}\right)\alpha_{n}^{2}x_{n}
+ ( 1 + k n ) ​ α n 3 ​ ( x n − T n ​ y n ) + α n ​ x n + α n ​ ( T n ​ x n + 1 − T n ​ y n ) \displaystyle+\left(1+k_{n}\right)\alpha_{n}^{3}\left(x_{n}-T^{n}y_{n}\right)+\alpha_{n}x_{n}+\alpha_{n}\left(T^{n}x_{n+1}-T^{n}y_{n}\right)
= \displaystyle=
( 1 + α n 2 ) ​ x n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ x n + 1 + [ 1 − ( 1 + k n ) ​ α n ] ​ α n ​ x n \displaystyle\left(1+\alpha_{n}^{2}\right)x_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}+\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}x_{n}
+ ( 1 + k n ) ​ α n 3 ​ ( x n − T n ​ y n ) + α n ​ ( T n ​ x n + 1 − T n ​ y n ) \displaystyle+\left(1+k_{n}\right)\alpha_{n}^{3}\left(x_{n}-T^{n}y_{n}\right)+\alpha_{n}\left(T^{n}x_{n+1}-T^{n}y_{n}\right)
(8)
Also
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u n = \displaystyle u_{n}=
u n + 1 + α n ​ u n − α n ​ T n ​ u n \displaystyle u_{n+1}+\alpha_{n}u_{n}-\alpha_{n}T^{n}u_{n}
= \displaystyle=
( 1 + α n 2 ) ​ u n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ u n + 1 \displaystyle\left(1+\alpha_{n}^{2}\right)u_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}
− ( 1 + k n ) ​ α n 2 ​ u n + 1 + α n ​ u n + α n ​ ( T n ​ u n + 1 − T n ​ u n ) \displaystyle-\left(1+k_{n}\right)\alpha_{n}^{2}u_{n+1}+\alpha_{n}u_{n}+\alpha_{n}\left(T^{n}u_{n+1}-T^{n}u_{n}\right)
= \displaystyle=
( 1 + α n 2 ) ​ u n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ u n + 1 \displaystyle\left(1+\alpha_{n}^{2}\right)u_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}
− ( 1 + k n ) ​ α n 2 ​ [ u n + α n ​ ( T n ​ u n − u n ) ] + α n ​ u n + α n ​ ( T n ​ u n + 1 − T n ​ u n ) \displaystyle-\left(1+k_{n}\right)\alpha_{n}^{2}\left[u_{n}+\alpha_{n}\left(T^{n}u_{n}-u_{n}\right)\right]+\alpha_{n}u_{n}+\alpha_{n}\left(T^{n}u_{n+1}-T^{n}u_{n}\right)
= \displaystyle=
( 1 + α n 2 ) ​ u n + 1 + α n ​ ( α n ​ k n ​ I − T n ) ​ u n + 1 + ( 1 + k n ) ​ α n 3 ​ ( u n − T n ​ u n ) \displaystyle\left(1+\alpha_{n}^{2}\right)u_{n+1}+\alpha_{n}\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}+\left(1+k_{n}\right)\alpha_{n}^{3}\left(u_{n}-T^{n}u_{n}\right)
+ [ 1 − ( 1 + k n ) ​ α n ] ​ α n ​ u n + α n ​ ( T n ​ u n + 1 − T n ​ u n ) \displaystyle+\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}u_{n}+\alpha_{n}\left(T^{n}u_{n+1}-T^{n}u_{n}\right)
(9)
From (8) and (9) we get
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x n − u n = \displaystyle x_{n}-u_{n}=
( 1 + α n 2 ) ​ ( x n + 1 − u n + 1 ) + α n ​ ( ( α n ​ k n ​ I − T n ) ​ x n + 1 − ( α n ​ k n ​ I − T n ) ​ u n + 1 ) \displaystyle\left(1+\alpha_{n}^{2}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left(\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}-\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}\right)
+ [ 1 − ( 1 + k n ) ​ α n ] ​ α n ​ ( x n − u n ) + ( 1 + k n ) ​ α n 3 ​ ( x n − u n − T n ​ y n + T n ​ u n ) \displaystyle+\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}\left(x_{n}-u_{n}\right)+\left(1+k_{n}\right)\alpha_{n}^{3}\left(x_{n}-u_{n}-T^{n}y_{n}+T^{n}u_{n}\right)
+ α n ​ ( T n ​ x n + 1 − T n ​ u n + 1 − T n ​ y n + T n ​ u n ) \displaystyle+\alpha_{n}\left(T^{n}x_{n+1}-T^{n}u_{n+1}-T^{n}y_{n}+T^{n}u_{n}\right)
(10)
The norm of the sum of the first two terms on the right-hand side of (10) is equal to
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( 1 + α n 2 ) ​ ‖ ( x n + 1 − u n + 1 ) + α n 1 + α n 2 ​ ( ( α n ​ k n ​ I − T n ) ​ x n + 1 − ( α n ​ k n ​ I − T n ) ​ u n + 1 ) ‖ . \left(1+\alpha_{n}^{2}\right)\left\|\left(x_{n+1}-u_{n+1}\right)+\frac{\alpha_{n}}{1+\alpha_{n}^{2}}\left(\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}-\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}\right)\right\|.
Using (6) with x := x n + 1 , y := u n + 1 x:=x_{n+1},y:=u_{n+1} , we obtain
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‖ ( 1 + α n 2 ) ​ ( x n + 1 − u n + 1 ) + α n ​ ( ( α n ​ k n ​ I − T n ) ​ x n + 1 − ( α n ​ k n ​ I − T n ) ​ u n + 1 ) ‖ \displaystyle\left\|\left(1+\alpha_{n}^{2}\right)\left(x_{n+1}-u_{n+1}\right)+\alpha_{n}\left(\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}-\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}\right)\right\|
⩾ ( 1 + α n 2 ) ​ ‖ x n + 1 − u n + 1 ‖ \displaystyle\quad\geqslant\left(1+\alpha_{n}^{2}\right)\left\|x_{n+1}-u_{n+1}\right\|
(11)
From (10) it follows that
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‖ x n − u n ‖ ⩾ \displaystyle\left\|x_{n}-u_{n}\right\|\geqslant
∥ ( 1 + α n 2 ) ( x n + 1 − u n + 1 ) \displaystyle\|\left(1+\alpha_{n}^{2}\right)\left(x_{n+1}-u_{n+1}\right)
+ α n ( ( α n k n I − T n ) x n + 1 − ( α n k n I − T n ) u n + 1 ) ∥ \displaystyle+\alpha_{n}\left(\left(\alpha_{n}k_{n}I-T^{n}\right)x_{n+1}-\left(\alpha_{n}k_{n}I-T^{n}\right)u_{n+1}\right)\|
+ [ 1 − ( 1 + k n ) ​ α n ] ​ α n ​ ‖ x n − u n ‖ \displaystyle+\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}\left\|x_{n}-u_{n}\right\|
− ( 1 + k n ) ​ α n 3 ​ ‖ x n − u n − T n ​ y n + T n ​ u n ‖ \displaystyle-\left(1+k_{n}\right)\alpha_{n}^{3}\left\|x_{n}-u_{n}-T^{n}y_{n}+T^{n}u_{n}\right\|
− α n ​ ‖ T n ​ x n + 1 − T n ​ u n + 1 − T n ​ y n + T n ​ u n ‖ \displaystyle-\alpha_{n}\left\|T^{n}x_{n+1}-T^{n}u_{n+1}-T^{n}y_{n}+T^{n}u_{n}\right\|
⩾ \displaystyle\geqslant
( 1 + α n ) ​ ‖ x n + 1 − u n + 1 ‖ + [ 1 − ( 1 + k n ) ​ α n ] ​ α n ​ ‖ x n − u n ‖ \displaystyle\left(1+\alpha_{n}\right)\left\|x_{n+1}-u_{n+1}\right\|+\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}\left\|x_{n}-u_{n}\right\|
− ( 1 + k n ) ​ α n 3 ​ ‖ x n − u n − T n ​ y n + T n ​ u n ‖ \displaystyle-\left(1+k_{n}\right)\alpha_{n}^{3}\left\|x_{n}-u_{n}-T^{n}y_{n}+T^{n}u_{n}\right\|
− α n ​ ‖ T n ​ x n + 1 − T n ​ u n + 1 − T n ​ y n + T n ​ u n ‖ \displaystyle-\alpha_{n}\left\|T^{n}x_{n+1}-T^{n}u_{n+1}-T^{n}y_{n}+T^{n}u_{n}\right\|
Thus, we have
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( 1 + \displaystyle(1+
α n 2 ) ∥ x n + 1 − u n + 1 ∥ \displaystyle\left.\alpha_{n}^{2}\right)\left\|x_{n+1}-u_{n+1}\right\|
⩽ \displaystyle\leqslant
{ 1 − [ 1 − ( 1 + k n ) ​ α n ] ​ α n } ​ ‖ x n − u n ‖ + ( 1 + k n ) ​ α n 3 ​ ‖ x n − u n − T n ​ y n + T n ​ u n ‖ \displaystyle\left\{1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}\right\}\left\|x_{n}-u_{n}\right\|+\left(1+k_{n}\right)\alpha_{n}^{3}\left\|x_{n}-u_{n}-T^{n}y_{n}+T^{n}u_{n}\right\|
+ α n ​ ‖ T n ​ x n + 1 − T n ​ u n + 1 − T n ​ y n + T n ​ u n ‖ \displaystyle+\alpha_{n}\left\|T^{n}x_{n+1}-T^{n}u_{n+1}-T^{n}y_{n}+T^{n}u_{n}\right\|
⩽ \displaystyle\leqslant
{ 1 − [ 1 − ( 1 + k n ) ​ α n ] ​ α n } ​ ‖ x n − u n ‖ + ( 1 + k n ) ​ α n 3 ​ ‖ u n − T n ​ u n ‖ \displaystyle\left\{1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}\right\}\left\|x_{n}-u_{n}\right\|+\left(1+k_{n}\right)\alpha_{n}^{3}\left\|u_{n}-T^{n}u_{n}\right\|
+ ( 1 + k n ) ​ α n 3 ​ ‖ x n − T n ​ y n ‖ + α n ​ ‖ T n ​ x n + 1 − T n ​ y n ‖ + α n ​ ‖ T n ​ u n + 1 − T n ​ u n ‖ . \displaystyle+\left(1+k_{n}\right)\alpha_{n}^{3}\left\|x_{n}-T^{n}y_{n}\right\|+\alpha_{n}\left\|T^{n}x_{n+1}-T^{n}y_{n}\right\|+\alpha_{n}\left\|T^{n}u_{n+1}-T^{n}u_{n}\right\|.
(12)
But
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‖ x n − T n ​ y n ‖ \displaystyle\left\|x_{n}-T^{n}y_{n}\right\|
⩽ ‖ x n − u n ‖ + ‖ u n − T n ​ u n ‖ + ‖ T n ​ u n − T n ​ y n ‖ \displaystyle\leqslant\left\|x_{n}-u_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|+\left\|T^{n}u_{n}-T^{n}y_{n}\right\|
⩽ ‖ x n − u n ‖ + ‖ u n − T n ​ u n ‖ + L ​ ‖ u n − y n ‖ \displaystyle\leqslant\left\|x_{n}-u_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|+L\left\|u_{n}-y_{n}\right\|
(13)
and
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‖ u n − y n ‖ \displaystyle\left\|u_{n}-y_{n}\right\|
= ‖ ( 1 − β n ) ​ ( u n − x n ) + β n ​ ( u n − T n ​ x n ) ‖ \displaystyle=\left\|\left(1-\beta_{n}\right)\left(u_{n}-x_{n}\right)+\beta_{n}\left(u_{n}-T^{n}x_{n}\right)\right\|
⩽ ( 1 − β n ) ​ ‖ u n − x n ‖ + β n ​ ‖ u n − T n ​ x n ‖ \displaystyle\leqslant\left(1-\beta_{n}\right)\left\|u_{n}-x_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}x_{n}\right\|
⩽ ( 1 − β n ) ​ ‖ u n − x n ‖ + β n ​ [ ‖ T n ​ u n − T n ​ x n ‖ + ‖ u n − T n ​ u n ‖ ] \displaystyle\leqslant\left(1-\beta_{n}\right)\left\|u_{n}-x_{n}\right\|+\beta_{n}\left[\left\|T^{n}u_{n}-T^{n}x_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|\right]
⩽ ( 1 − β n ) ​ ‖ u n − x n ‖ + β n ​ L ​ ‖ u n − x n ‖ + β n ​ ‖ u n − T n ​ u n ‖ \displaystyle\leqslant\left(1-\beta_{n}\right)\left\|u_{n}-x_{n}\right\|+\beta_{n}L\left\|u_{n}-x_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}u_{n}\right\|
= ( 1 − β n + β n ​ L ) ​ ‖ u n − x n ‖ + β n ​ ‖ u n − T n ​ u n ‖ \displaystyle=\left(1-\beta_{n}+\beta_{n}L\right)\left\|u_{n}-x_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}u_{n}\right\|
⩽ L ​ ‖ u n − x n ‖ + β n ​ ‖ u n − T n ​ u n ‖ \displaystyle\leqslant L\left\|u_{n}-x_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}u_{n}\right\|
(14)
because 1 ⩽ L ⇒ 1 − β n + β n ​ L ⩽ L 1\leqslant L\Rightarrow 1-\beta_{n}+\beta_{n}L\leqslant L .
Substituting (14) into (13) we obtain
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‖ x n − T n ​ y n ‖ \displaystyle\left\|x_{n}-T^{n}y_{n}\right\|
⩽ ‖ u n − x n ‖ + ‖ u n − T n ​ u n ‖ + L ​ ( L ​ ‖ u n − x n ‖ + β n ​ ‖ u n − T n ​ u n ‖ ) \displaystyle\leqslant\left\|u_{n}-x_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|+L\left(L\left\|u_{n}-x_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}u_{n}\right\|\right)
⩽ ( 1 + L 2 ) ​ ‖ u n − x n ‖ + ( 1 + L ​ β n ) ​ ‖ u n − T n ​ u n ‖ \displaystyle\leqslant\left(1+L^{2}\right)\left\|u_{n}-x_{n}\right\|+\left(1+L\beta_{n}\right)\left\|u_{n}-T^{n}u_{n}\right\|
(15)
‖ T n ​ x n + 1 − T n ​ y n ‖ \displaystyle\left\|T^{n}x_{n+1}-T^{n}y_{n}\right\|
⩽ L ​ ‖ x n + 1 − y n ‖ = L ​ ‖ ( 1 − α n ) ​ x n + α n ​ T n ​ y n − y n ‖ \displaystyle\leqslant L\left\|x_{n+1}-y_{n}\right\|=L\left\|\left(1-\alpha_{n}\right)x_{n}+\alpha_{n}T^{n}y_{n}-y_{n}\right\|
⩽ L ​ [ ( 1 − α n ) ​ ‖ x n − y n ‖ + α n ​ ‖ T n ​ y n − y n ‖ ] \displaystyle\leqslant L\left[\left(1-\alpha_{n}\right)\left\|x_{n}-y_{n}\right\|+\alpha_{n}\left\|T^{n}y_{n}-y_{n}\right\|\right]
(16)
Using (14),
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‖ T n ​ y n − y n ‖ \displaystyle\left\|T^{n}y_{n}-y_{n}\right\|
⩽ ‖ T n ​ y n − T n ​ u n ‖ + ‖ T n ​ u n − u n ‖ + ‖ y n − u n ‖ \displaystyle\leqslant\left\|T^{n}y_{n}-T^{n}u_{n}\right\|+\left\|T^{n}u_{n}-u_{n}\right\|+\left\|y_{n}-u_{n}\right\|
⩽ L ​ ‖ y n − u n ‖ + ‖ T n ​ u n − u n ‖ + ‖ y n − u n ‖ \displaystyle\leqslant L\left\|y_{n}-u_{n}\right\|+\left\|T^{n}u_{n}-u_{n}\right\|+\left\|y_{n}-u_{n}\right\|
⩽ ( 1 + L ) ​ ‖ y n − u n ‖ + ‖ T n ​ u n − u n ‖ \displaystyle\leqslant(1+L)\left\|y_{n}-u_{n}\right\|+\left\|T^{n}u_{n}-u_{n}\right\|
⩽ ( 1 + L ) ​ [ L ​ ‖ x n − u n ‖ + β n ​ ‖ T n ​ u n − u n ‖ ] + ‖ T n ​ u n − u n ‖ \displaystyle\leqslant(1+L)\left[L\left\|x_{n}-u_{n}\right\|+\beta_{n}\left\|T^{n}u_{n}-u_{n}\right\|\right]+\left\|T^{n}u_{n}-u_{n}\right\|
= ( 1 + L ) ​ L ​ ‖ x n − u n ‖ + [ ( 1 + L ) ​ β n + 1 ] ​ ‖ T n ​ u n − u n ‖ \displaystyle=(1+L)L\left\|x_{n}-u_{n}\right\|+\left[(1+L)\beta_{n}+1\right]\left\|T^{n}u_{n}-u_{n}\right\|
(17)
From (4) we have
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‖ x n − y n ‖ \displaystyle\left\|x_{n}-y_{n}\right\|
= ‖ x n − ( 1 − β n ) ​ x n − β n ​ T n ​ x n ‖ = β n ​ ‖ x n − T n ​ x n ‖ \displaystyle=\left\|x_{n}-\left(1-\beta_{n}\right)x_{n}-\beta_{n}T^{n}x_{n}\right\|=\beta_{n}\left\|x_{n}-T^{n}x_{n}\right\|
⩽ β n ​ [ ‖ x n − u n ‖ + ‖ T n ​ u n − u n ‖ + ‖ T n ​ u n − T n ​ x n ‖ ] \displaystyle\leqslant\beta_{n}\left[\left\|x_{n}-u_{n}\right\|+\left\|T^{n}u_{n}-u_{n}\right\|+\left\|T^{n}u_{n}-T^{n}x_{n}\right\|\right]
⩽ β n ​ [ ( 1 + L ) ​ ‖ x n − u n ‖ + ‖ T n ​ u n − u n ‖ ] . \displaystyle\leqslant\beta_{n}\left[(1+L)\left\|x_{n}-u_{n}\right\|+\left\|T^{n}u_{n}-u_{n}\right\|\right].
(18)
Substituting (18) and (17) into (16), we obtain
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‖ T n ​ x n + 1 − T n ​ y n ‖ \displaystyle\left\|T^{n}x_{n+1}-T^{n}y_{n}\right\|
⩽ L ​ [ ( 1 − α n ) ​ ‖ x n − y n ‖ + α n ​ ‖ T n ​ y n − y n ‖ ] \displaystyle\leqslant L\left[\left(1-\alpha_{n}\right)\left\|x_{n}-y_{n}\right\|+\alpha_{n}\left\|T^{n}y_{n}-y_{n}\right\|\right]
= L { ( 1 − α n ) β n [ ( 1 + L ) ∥ x n − u n ∥ + ∥ T n u n − u n ∥ ] \displaystyle=L\left\{\left(1-\alpha_{n}\right)\beta_{n}\left[(1+L)\left\|x_{n}-u_{n}\right\|+\left\|T^{n}u_{n}-u_{n}\right\|\right]\right.
+ α n ( 1 + L ) [ L ∥ x n − u n ∥ + β n ∥ u n − T n u n ∥ ] + α n ∥ u n − T n u n ∥ } \displaystyle\left.+\alpha_{n}(1+L)\left[L\left\|x_{n}-u_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}u_{n}\right\|\right]+\alpha_{n}\left\|u_{n}-T^{n}u_{n}\right\|\right\}
= { L ​ ( 1 − α n ) ​ β n ​ ( 1 + L ) + α n ​ ( 1 + L ) ​ L 2 } ​ ‖ x n − u n ‖ \displaystyle=\left\{L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\right\}\left\|x_{n}-u_{n}\right\|
+ { β n ​ L ​ ( 1 − α n ) + α n ​ L ​ [ ( 1 + L ) ​ β n + 1 ] } ​ ‖ T n ​ u n − u n ‖ . \displaystyle+\left\{\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right\}\left\|T^{n}u_{n}-u_{n}\right\|.
(19)
Using (14) we have
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‖ x n − T n ​ y n ‖ \displaystyle\left\|x_{n}-T^{n}y_{n}\right\|
⩽ ‖ x n − u n ‖ + ‖ u n − T n ​ u n ‖ + ‖ T n ​ y n − T n ​ u n ‖ \displaystyle\leqslant\left\|x_{n}-u_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|+\left\|T^{n}y_{n}-T^{n}u_{n}\right\|
⩽ ‖ x n − u n ‖ + ‖ u n − T n ​ u n ‖ + L ​ ‖ y n − u n ‖ \displaystyle\leqslant\left\|x_{n}-u_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|+L\left\|y_{n}-u_{n}\right\|
⩽ ‖ x n − u n ‖ + ‖ u n − T n ​ u n ‖ + L ​ [ L ​ ‖ x n − u n ‖ + β n ​ ‖ u n − T n ​ u n ‖ ] \displaystyle\leqslant\left\|x_{n}-u_{n}\right\|+\left\|u_{n}-T^{n}u_{n}\right\|+L\left[L\left\|x_{n}-u_{n}\right\|+\beta_{n}\left\|u_{n}-T^{n}u_{n}\right\|\right]
= ( 1 + L 2 ) ​ ‖ x n − u n ‖ + ( 1 + β n ​ L ) ​ ‖ u n − T n ​ u n ‖ \displaystyle=\left(1+L^{2}\right)\left\|x_{n}-u_{n}\right\|+\left(1+\beta_{n}L\right)\left\|u_{n}-T^{n}u_{n}\right\|
(20)
Substituting (19) and (20) into (12), and using the facts that ( 1 + α n 2 ) − 1 ⩽ 1 \left(1+\alpha_{n}^{2}\right)^{-1}\leqslant 1 , we get
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( 1 + α n 2 ) ​ ‖ x n + 1 − u n + 1 ‖ \displaystyle\left(1+\alpha_{n}^{2}\right)\left\|x_{n+1}-u_{n+1}\right\|
⩽ ⩽ { 1 − [ 1 − ( 1 + k n ) α n ] α n } ∥ x n − u n ∥ \displaystyle\leqslant\leqslant\left\{1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}\right\}\left\|x_{n}-u_{n}\right\|
+ ( 1 + k n ) ​ α n 3 ​ { ( 1 + L 2 ) ​ ‖ x n − u n ‖ + ( 1 + β n ​ L ) ​ ‖ u n − T n ​ u n ‖ } \displaystyle\quad+\left(1+k_{n}\right)\alpha_{n}^{3}\left\{\left(1+L^{2}\right)\left\|x_{n}-u_{n}\right\|+\left(1+\beta_{n}L\right)\left\|u_{n}-T^{n}u_{n}\right\|\right\}
+ ( 1 + k n ) ​ α n 3 ​ ‖ u n − T n ​ u n ‖ + α n ​ ‖ T n ​ u n + 1 − T n ​ u n ‖ \displaystyle\quad+\left(1+k_{n}\right)\alpha_{n}^{3}\left\|u_{n}-T^{n}u_{n}\right\|+\alpha_{n}\left\|T^{n}u_{n+1}-T^{n}u_{n}\right\|
+ α n ​ { L ​ ( 1 − α n ) ​ β n ​ ( 1 + L ) + α n ​ ( 1 + L ) ​ L 2 } ​ ‖ x n − u n ‖ \displaystyle\quad+\alpha_{n}\left\{L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\right\}\left\|x_{n}-u_{n}\right\|
+ α n ​ { β n ​ L ​ ( 1 − α n ) + α n ​ L ​ [ ( 1 + L ) ​ β n + 1 ] } ​ ‖ u n − T n ​ u n ‖ \displaystyle\quad+\alpha_{n}\left\{\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right\}\left\|u_{n}-T^{n}u_{n}\right\|
(21)
∥ x n + 1 − u n + 1 ∥ ⩽ { 1 − [ 1 − ( 1 + k n ) α n ] α n + ( 1 + k n ) α n 3 ( 1 + L 2 ) \displaystyle\left\|x_{n+1}-u_{n+1}\right\|\leqslant\left\{1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}+\left(1+k_{n}\right)\alpha_{n}^{3}\left(1+L^{2}\right)\right.
+ α n { L ( 1 − α n ) β n ( 1 + L ) + α n ( 1 + L ) L 2 } } ∥ x n − u n ∥ \displaystyle\left.\quad+\alpha_{n}\left\{L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\right\}\right\}\left\|x_{n}-u_{n}\right\|
+ { ( 1 + k n ) α n 3 ( 2 + β n L ) \displaystyle\quad+\left\{\left(1+k_{n}\right)\alpha_{n}^{3}\left(2+\beta_{n}L\right)\right.
+ α n { β n L ( 1 − α n ) + α n L [ ( 1 + L ) β n + 1 ] } } ∥ u n − T n u n ∥ \displaystyle\left.\quad+\alpha_{n}\left\{\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right\}\right\}\left\|u_{n}-T^{n}u_{n}\right\|
+ α n ​ ‖ T n ​ u n + 1 − T n ​ u n ‖ \displaystyle\quad+\alpha_{n}\left\|T^{n}u_{n+1}-T^{n}u_{n}\right\|
(22)
We may write
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a n + 1 ⩽ γ n ​ a n + σ n , a_{n+1}\leqslant\gamma_{n}a_{n}+\sigma_{n},
(23)
where
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a n := \displaystyle a_{n}:=
‖ x n − u n ‖ \displaystyle\left\|x_{n}-u_{n}\right\|
γ n := \displaystyle\gamma_{n}:=
{ 1 − [ 1 − ( 1 + k n ) α n ] α n + ( 1 + k n ) α n 3 ( 1 + L 2 ) \displaystyle\left\{1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}+\left(1+k_{n}\right)\alpha_{n}^{3}\left(1+L^{2}\right)\right.
+ α n { L ( 1 − α n ) β n ( 1 + L ) + α n ( 1 + L ) L 2 } } , \displaystyle\left.+\alpha_{n}\left\{L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\right\}\right\},
σ n := \displaystyle\sigma_{n}:=
{ ( 1 + k n ) α n 3 ( 2 + β n L ) \displaystyle\left\{\left(1+k_{n}\right)\alpha_{n}^{3}\left(2+\beta_{n}L\right)\right.
+ α n { β n L ( 1 − α n ) + α n L [ ( 1 + L ) β n + 1 ] } } ∥ u n − T n u n ∥ \displaystyle\left.+\alpha_{n}\left\{\beta_{n}L\left(1-\alpha_{n}\right)+\alpha_{n}L\left[(1+L)\beta_{n}+1\right]\right\}\right\}\left\|u_{n}-T^{n}u_{n}\right\|
+ α n ​ ‖ T n ​ u n + 1 − T n ​ u n ‖ . \displaystyle+\alpha_{n}\left\|T^{n}u_{n+1}-T^{n}u_{n}\right\|.
(24)
We have
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L ​ ( 1 − α n ) ​ β n ​ ( 1 + L ) + α n ​ ( 1 + L ) ​ L 2 ⩽ L ​ ( 1 + L ) ​ [ ( 1 − α n ) ​ β n + α n ​ L ] \displaystyle L\left(1-\alpha_{n}\right)\beta_{n}(1+L)+\alpha_{n}(1+L)L^{2}\leqslant L(1+L)\left[\left(1-\alpha_{n}\right)\beta_{n}+\alpha_{n}L\right]
⩽ L ​ ( 1 + L ) ​ [ L ​ β n + α n ​ L ] = L 2 ​ ( 1 + L ) ​ ( α n + β n ) \displaystyle\quad\leqslant L(1+L)\left[L\beta_{n}+\alpha_{n}L\right]=L^{2}(1+L)\left(\alpha_{n}+\beta_{n}\right)
The last inequality is true because L ⩾ 1 L\geqslant 1 . From (5) it follows that for all n n sufficiently large we have
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α n ⩽ 1 5 ​ sup ( ( 1 1 + L 2 ) , ( 1 1 + k n ) ) , α n + β n ⩽ 1 5 ​ ( 1 ( 1 + L ) ​ L 2 ) ; \alpha_{n}\leqslant\frac{1}{5}\sup\left(\left(\frac{1}{1+L^{2}}\right),\left(\frac{1}{1+k_{n}}\right)\right),\quad\alpha_{n}+\beta_{n}\leqslant\frac{1}{5}\left(\frac{1}{(1+L)L^{2}}\right);
thus
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γ n \displaystyle\gamma_{n}
⩽ 1 − [ 1 − ( 1 + k n ) ​ α n ] ​ α n + ( 1 + k n ) ​ α n 3 ​ ( 1 + L 2 ) + α n ​ L 2 ​ ( 1 + L ) ​ ( α n + β n ) \displaystyle\leqslant 1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}+\left(1+k_{n}\right)\alpha_{n}^{3}\left(1+L^{2}\right)+\alpha_{n}L^{2}(1+L)\left(\alpha_{n}+\beta_{n}\right)
⩽ 1 − [ 1 − ( 1 + k n ) ​ α n ] ​ α n + 1 25 ​ α n + 1 5 ​ α n \displaystyle\leqslant 1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}+\frac{1}{25}\alpha_{n}+\frac{1}{5}\alpha_{n}
⩽ 1 − [ 1 − ( 1 + k n ) ​ α n ] ​ α n + 2 5 ​ α n \displaystyle\leqslant 1-\left[1-\left(1+k_{n}\right)\alpha_{n}\right]\alpha_{n}+\frac{2}{5}\alpha_{n}
⩽ 1 − 4 5 ​ α n + 2 5 ​ α n = 1 − 2 5 ​ α n \displaystyle\leqslant 1-\frac{4}{5}\alpha_{n}+\frac{2}{5}\alpha_{n}=1-\frac{2}{5}\alpha_{n}
(25)
Thus γ n ⩽ 1 − ( 2 / 5 ) ​ α n \gamma_{n}\leqslant 1-(2/5)\alpha_{n} for all n n sufficiently large, from which we obtain relation (7),
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a n + 1 ⩽ ( 1 − λ n ) ​ a n + σ n a_{n+1}\leqslant\left(1-\lambda_{n}\right)a_{n}+\sigma_{n}
(26)
The fact that Mann iteration (3) converges, i.e., lim n → ∞ u n = x ∗ \lim_{n\rightarrow\infty}u_{n}=x^{*} (more precisely using lim n → ∞ ‖ u n + 1 − u n ‖ = 0 \lim_{n\rightarrow\infty}\left\|u_{n+1}-u_{n}\right\|=0 ), it is easy to see that σ n = o ​ ( λ n ) \sigma_{n}=o\left(\lambda_{n}\right) . All the assumptions from Lemma 2 are now satisfied, so lim n → ∞ a n = 0 \lim_{n\rightarrow\infty}a_{n}=0 . Hence,
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lim n → ∞ ‖ x n − u n ‖ = 0 \lim_{n\rightarrow\infty}\left\|x_{n}-u_{n}\right\|=0
(27)
Since lim n → ∞ u n = x ∗ \lim_{n\rightarrow\infty}u_{n}=x^{*} , (27) and the inequality
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‖ x n − x ∗ ‖ ⩽ ‖ x n − u n ‖ + ‖ u n − x ∗ ‖ → 0 ( n → ∞ ) \left\|x_{n}-x^{*}\right\|\leqslant\left\|x_{n}-u_{n}\right\|+\left\|u_{n}-x^{*}\right\|\rightarrow 0\quad(n\rightarrow\infty)
(28)
lead to lim n → ∞ x n = x ∗ \lim_{n\rightarrow\infty}x_{n}=x^{*} .
Because an asymptotically nonexpansive map is asymptotically pseudocontractive and Lipschitzian (see Remarks 2 and 4), from Theorem 8 we obtain the following result.
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Corollary 9. Let B B be a closed convex subset of an arbitrary Banach space X X and ( x n ) n \left(x_{n}\right)_{n} and ( u n ) n \left(u_{n}\right)_{n} defined by (3) and (4) with ( α n ) n \left(\alpha_{n}\right)_{n} and ( β n ) n \left(\beta_{n}\right)_{n} satisfying (5). Let T T be an asymptotically nonexpansive self-map of B B . Let x ∗ x^{*} be the fixed point of T T . If u 0 = x 0 ∈ B u_{0}=x_{0}\in B , then the following two assertions are equivalent:
(i) Mann type iteration (3) converges to x ∗ ∈ F ​ ( T ) x^{*}\in F(T) ;
(ii) Ishikawa type iteration (4) converges to x ∗ ∈ F ​ ( T ) x^{*}\in F(T) .
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The following result is from [4].
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Theorem 10 [4]. Let B B be a closed convex subset of an arbitrary Banach space X X and let T T be a Lipschitzian strongly pseudocontractive self-map of B B . Let x 1 = u 1 x_{1}=u_{1} and let ( x n ) n \left(x_{n}\right)_{n} and ( u n ) n \left(u_{n}\right)_{n} be the Mann and Ishikawa iterations (that is (3) and (4) without " n n " at the exponent of T T ), with ( α n ) n , ( β n ) n \left(\alpha_{n}\right)_{n},\left(\beta_{n}\right)_{n} satisfying (5). Then the following are equivalent:
(i) The Mann iteration converges strongly to x ∗ x^{*} ;
(ii) The Ishikawa iteration converges strongly to x ∗ x^{*} .
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Theorem 8 is the analog of Theorem 10 for asymptotically pseudocontractive operators. Our theorems are also true for set-valued mappings, if such maps admit appropriate single-valued selections.
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