Mann iteration for generalized pseudocontractive maps in Hilbert spaces
Abstract
If XX is a real Hilbert space, BB is a nonempty, bounded, convex and closed subset, T:B rarr BT: B \rightarrow B is a generalized pseudocontraction; then the iteration
{:[(1)x_(1) in B","],[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Tx_(n)","],[(alpha_(n))_(n) sub(0","1)","sum_(n=1)^(oo)alpha_(n)=oo","],[sum_(n=1)^(oo)|alpha_(n+1)-alpha_(n)| < oo","lim_(n rarr oo)alpha_(n)=0","]:}\begin{align*}
x_{1} & \in B, \tag{1}\\
x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T x_{n}, \\
\left(\alpha_{n}\right)_{n} & \subset(0,1), \sum_{n=1}^{\infty} \alpha_{n}=\infty, \\
\sum_{n=1}^{\infty}\left|\alpha_{n+1}-\alpha_{n}\right| & <\infty, \lim _{n \rightarrow \infty} \alpha_{n}=0,
\end{align*}
strongly converges to the fixed point of TT.
Key words: Mann iteration, fixed points
AMS subject classifications: 47H10,47H0647 \mathrm{H} 10,47 \mathrm{H} 06
Received June 10, 2000
Accepted May 18, 2001
1. Preliminaries
In this note we study the convergence of the Mann iteration process (1) for generalized pseudocontractions. According to [8] the generalized pseudocontractions are more general than the pseudocontractions introduced by Browder.
Definition 1. [8]. Let XX be a Hilbert space, let BB be a nonempty subset. AA map T:B rarr BT: B \rightarrow B is said to be a generalized pseudocontraction if for x,y in Bx, y \in B there exists r > 0r>0 such that
{:(2)(:Tx-Ty","x-y:) <= r||x-y||^(2).:}\begin{equation*}
\langle T x-T y, x-y\rangle \leq r\|x-y\|^{2} . \tag{2}
\end{equation*}
Clearly, (2) is equivalent to
(:(I-T)x-(I-T)y,x-y:) >= (1-r)||x-y||^(2).\langle(I-T) x-(I-T) y, x-y\rangle \geq(1-r)\|x-y\|^{2} .
The map TT is a strong pseudocontraction if there exists k in(0,1)k \in(0,1) such that for all x,y in Bx, y \in B,
(:(I-T)x-(I-T)y,x-y:) >= k||x-y||^(2),\langle(I-T) x-(I-T) y, x-y\rangle \geq k\|x-y\|^{2},
see, for example [6]. Remark that both generalized pseudocontractivity and strong pseudocontractivity generalize the pseudocontractivity, but in a different manner. Iteration (1), where TT is a strong pseudocontraction in Banach spaces, was studied in [1], [2], [3], [4], [6], [9].
The following lemma can be found in [9] as Lemma4. Also, it can be found in [4] as Lemma 1.2, with another proof. A more general case is in Lemma 2 from [5]. The proof from [5] is similar to the proof of Lemma 4 from [9].
Lemma 1. [9], [4]. Let (rho_(n))_(n)\left(\rho_{n}\right)_{n} be a nonnegative real sequence satisfying
rho_(n+1) <= (1-lambda_(n))rho_(n)+sigma_(n)\rho_{n+1} \leq\left(1-\lambda_{n}\right) \rho_{n}+\sigma_{n}
where lambda_(n)in(0,1),AA n in N,sum_(n=1)^(oo)lambda_(n)=oo\lambda_{n} \in(0,1), \forall n \in N, \sum_{n=1}^{\infty} \lambda_{n}=\infty and sigma_(n)=o(lambda_(n))\sigma_{n}=o\left(\lambda_{n}\right). Then lim_(n rarr oo)rho_(n)=0\lim _{n \rightarrow \infty} \rho_{n}=0.
The normalized duality mapping JJ is the identity, when XX is a Hilbert space, see [4]. Thus Lemma 1.1 from [4] becomes:
Lemma 2. [4]. If XX is a Hilbert space, then
||x+y||^(2) <= ||x||^(2)+2(:y,(x+y):),\|x+y\|^{2} \leq\|x\|^{2}+2\langle y,(x+y)\rangle,
for all x,y in Xx, y \in X.
The following result is a corollary of Lemma 1 from [7]:
Lemma 3. [7]. If XX is a real Hilbert space, BB is a nonempty, bounded, convex and closed subset, and T:B rarr BT: B \rightarrow B is a generalized pseudocontraction, then the sequence given by (1) satisfies
lim_(n rarr oo)||x_(n+1)-x_(n)||=0\lim _{n \rightarrow \infty}\left\|x_{n+1}-x_{n}\right\|=0
In [7], the map TT is nonexpansive. If we consider the proof of Lemma 1 from [7], we see that the result is true, when our assumptions are fulfilled.
2. Main result
We are now able to give the following result:
Theorem 1. If XX is a real Hilbert space, BB is a nonempty, bounded, convex and closed subset, and T:B rarr BT: B \rightarrow B is a generalized pseudocontraction, then the iteration (1) :
{:[x_(1) in B","],[x_(n+1)=(1-alpha_(n))x_(n)+alpha_(n)Tx_(n)],[(alpha_(n))_(n) sub(0","1)","sum_(n=1)^(oo)alpha_(n)=oo","sum_(n=1)^(oo)|alpha_(n+1)-alpha_(n)| < oo","],[lim_(n rarr oo)alpha_(n)=0.]:}\begin{aligned}
x_{1} & \in B, \\
x_{n+1} & =\left(1-\alpha_{n}\right) x_{n}+\alpha_{n} T x_{n} \\
\left(\alpha_{n}\right)_{n} & \subset(0,1), \sum_{n=1}^{\infty} \alpha_{n}=\infty, \sum_{n=1}^{\infty}\left|\alpha_{n+1}-\alpha_{n}\right|<\infty, \\
\lim _{n \rightarrow \infty} \alpha_{n} & =0 .
\end{aligned}
strongly converges to the fixed point of TT.
Proof. Theorem 2.1 from [8] gives us the existence and the uniqueness of the fixed point of TT. Let us denote this fixed point by qq. Using Lemma 3 and (2), we have
{:[||x_(n+1)-q||^(2)=||(1-alpha_(n))(x_(n)-q)+alpha_(n)(Tx_(n)-q)||^(2)],[ <= (1-alpha_(n))^(2)||x_(n)-q||^(2)+2alpha_(n)(:Tx_(n)-q,x_(n+1)-q:)],[=(1-alpha_(n))^(2)||x_(n)-q||^(2)+2alpha_(n)(:Tx_(n)-q,x_(n)-q:)+],[+2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):)],[ <= (1-alpha_(n))^(2)||x_(n)-q||^(2)+2alpha_(n)r||x_(n)-q||^(2)],[+2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):)],[ <= [1-alpha_(n)(2(1-r)-alpha_(n))]||x_(n)-q||^(2)],[+2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):).]:}\begin{aligned}
\left\|x_{n+1}-q\right\|^{2}= & \left\|\left(1-\alpha_{n}\right)\left(x_{n}-q\right)+\alpha_{n}\left(T x_{n}-q\right)\right\|^{2} \\
\leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-q\right\|^{2}+2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-q\right\rangle \\
= & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-q\right\|^{2}+2 \alpha_{n}\left\langle T x_{n}-q, x_{n}-q\right\rangle+ \\
& +2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle \\
\leq & \left(1-\alpha_{n}\right)^{2}\left\|x_{n}-q\right\|^{2}+2 \alpha_{n} r\left\|x_{n}-q\right\|^{2} \\
& +2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle \\
\leq & {\left[1-\alpha_{n}\left(2(1-r)-\alpha_{n}\right)\right]\left\|x_{n}-q\right\|^{2} } \\
& +2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle .
\end{aligned}
Let us denote
{:[A_(n):=(:Tx_(n)-q,x_(n+1)-x_(n):)],[lambda_(n):=alpha_(n)(2(1-r)-alpha_(n))],[rho_(n):=||x_(n)-q||^(2)],[sigma_(n):=2alpha_(n)A_(n).]:}\begin{aligned}
A_{n}: & =\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle \\
\lambda_{n}: & =\alpha_{n}\left(2(1-r)-\alpha_{n}\right) \\
\rho_{n}: & =\left\|x_{n}-q\right\|^{2} \\
\sigma_{n}: & =2 \alpha_{n} A_{n} .
\end{aligned}
Thus, we have
rho_(n+1) <= (1-lambda_(n))rho_(n)+sigma_(n)\rho_{n+1} \leq\left(1-\lambda_{n}\right) \rho_{n}+\sigma_{n}
We observe that
{:[lim_(n rarr oo)(sigma_(n))/(lambda_(n))=lim_(n rarr oo)(2alpha_(n)(:Tx_(n)-q,x_(n+1)-x_(n):))/(alpha_(n)(2(1-r)-alpha_(n)))],[=2lim_(n rarr oo)((:Tx_(n)-q,x_(n+1)-x_(n):))/((2(1-r)-alpha_(n)))=0]:}\begin{aligned}
\lim _{n \rightarrow \infty} \frac{\sigma_{n}}{\lambda_{n}} & =\lim _{n \rightarrow \infty} \frac{2 \alpha_{n}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle}{\alpha_{n}\left(2(1-r)-\alpha_{n}\right)} \\
& =2 \lim _{n \rightarrow \infty} \frac{\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle}{\left(2(1-r)-\alpha_{n}\right)}=0
\end{aligned}
the last equality is true. From Lemma 4, we have lim_(n rarr oo)||x_(n+1)-x_(n)||=0\lim _{n \rightarrow \infty}\left\|x_{n+1}-x_{n}\right\|=0. The sequence (||Tx_(n)-q||)_(n)\left(\left\|T x_{n}-q\right\|\right)_{n} is bounded, being in the bounded set BB. Hence we have lim_(n rarr oo)(:Tx_(n)-q,x_(n+1)-x_(n):)=0\lim _{n \rightarrow \infty}\left\langle T x_{n}-q, x_{n+1}-x_{n}\right\rangle=0. The assumptions from Lemma 2 are fulfilled. Hence rho_(n)rarr0\rho_{n} \rightarrow 0 as n rarr oon \rightarrow \infty. Thus x_(n)rarr qx_{n} \rightarrow q as n rarr oon \rightarrow \infty.
A prototype for (alpha_(n))_(n)\left(\alpha_{n}\right)_{n} is (1//sqrtn)_(n >= 1)(1 / \sqrt{n})_{n \geq 1}.
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