Construction of minimum-norm fixed points of pseudocontractions in Hilbert spaces

Abstract

An iterative algorithm is introduced for the construction of the minimum-norm fixed point of a pseudocontraction on a Hilbert space. The algorithm is proved to be strongly convergent.

MSC:47H05, 47H10, 47H17.

1 Introduction

Construction of fixed points of nonlinear mappings is a classical and active area of nonlinear functional analysis due to the fact that many nonlinear problems can be reformulated as fixed point equations of nonlinear mappings. The research of this area dates back to Picard’s and Banach’s time. As a matter of fact, the well-known Banach contraction principle states that the Picard iterates $\{T^{n}x\}$ converge to the unique fixed point of T whenever T is a contraction of a complete metric space. However, if T is not a contraction (nonexpansive, say), then the Picard iterates $\{T^{n}x\}$ fail, in general, to converge; hence, other iterative methods are needed. In 1953, Mann [1] introduced the now called Mann’s iterative method which generates a sequence $\{x_n\}$ via the averaged algorithm

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,n\ge 0,$$

(1.1)

where $\{{\alpha }_n\}$ is a sequence in the unit interval $[0,1]$, T is a self-mapping of a closed convex subset C of a Hilbert space H, and the initial guess $x_0$ is an arbitrary (but fixed) point of C.

Mann’s algorithm (1.1) has extensively been studied [2–7], and in particular, it is known that if T is nonexpansive (i.e., $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$) and if T has a fixed point, then the sequence $\{x_n\}$ generated by Mann’s algorithm (1.1) converges weakly to a fixed point of T provided the sequence $\{{\alpha }_n\}$ satisfies the condition

$$\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }.$$

(1.2)

This algorithm, however, does not converge in the strong topology in general (see [[8], Corollary 5.2]).

Browder and Petryshyn [9] studied weak convergence of Mann’s algorithm (1.1) for the class of strict pseudocontractions (in the case of constant stepsizes ${\alpha }_n=\alpha$ for all n; see [10] for the general case of variable stepsizes). However, Mann’s algorithm fails to converge for Lipschitzian pseudocontractions (see the counterexample of Chidume and Mutangadura [11]). It is therefore an interesting question of inventing iterative algorithms which generate a sequence converging in the norm topology to a fixed point of a Lipschitzian pseudocontraction (if any). The interest of pseudocontractions lies in their connection with monotone operators; namely, T is a pseudocontraction if and only if the complement $I-T$ is a monotone operator.

We also notice that it is quite usual to seek a particular solution of a given nonlinear problem, in particular, the minimum-norm solution. For instance, given a closed convex subset C of a Hilbert space $H_1$ and a bounded linear operator $A:H_1\to H_2$, where $H_2$ is another Hilbert space. The C-constrained pseudoinverse of A, $A_C^{†}$ is then defined as the minimum-norm solution of the constrained minimization problem

$$A_C^{†}(b):=arg\underset{x\in C}{min}\parallel Ax-b\parallel$$

(1.3)

which is equivalent to the fixed point problem

$$x=P_C(x-\lambda A^{\ast }(Ax-b)),$$

(1.4)

where $P_C$ is the metric projection from $H_1$ onto C, $A^{\ast }$ is the adjoint of A, $\lambda >0$ is a constant, and $b\in H_2$ is such that $P_{\overline{A(C)}}(b)\in A(C)$.

It is therefore an interesting problem to invent iterative algorithms that can generate sequences which converge strongly to the minimum-norm solution of a given fixed point problem. The purpose of this paper is to solve such a problem for pseudocontractions. More precisely, we shall introduce an iterative algorithm for the construction of fixed points of Lipschitzian pseudocontractions and prove that our algorithm (see (3.1) in Section 3) converges in the strong topology to the minimum-norm fixed point of the mapping.

For the existing literature on iterative methods for pseudocontractions, the reader can consult [10, 12–26]; for finding minimum-norm solutions of nonlinear fixed point and variational inequality problems, see [27–29]; and for related iterative methods for nonexpansive mappings, see [2, 3, 30, 31] and the references therein.

2 Preliminaries

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. The class of nonlinear mappings which we will study is the class of pseudocontractions. Recall that a mapping $T:C\to C$ is a pseudocontraction if it satisfies the property

$$〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

(2.1)

It is not hard to find that T is a pseudocontraction if and only if T satisfies one of the following two equivalent properties:

• (a) ${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+{\parallel (I-T)x-(I-T)y\parallel }^2$ for all $x,y\in C$; or

• (b) $I-T$ is monotone on C: $〈x-y,(I-T)x-(I-T)y〉\ge 0$ for all $x,y\in C$.

Recall that a mapping $T:C\to C$ is nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

It is immediately clear that nonexpansive mappings are pseudocontractions.

Recall also that the nearest point (or metric) projection from H onto C is defined as follows: For each point $x\in H$, $P_{C}x$ is the unique point in C with the property

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,y\in C.$$

Note that $P_C$ is characterized by the inequality

$$P_{C}x\in C,〈x-P_{C}x,y-P_{C}x〉\le 0,y\in C.$$

(2.2)

Consequently, $P_C$ is nonexpansive.

In the sequel we shall use the following notations:

• $Fix(S)$ stands for the set of fixed points of S;

• $x_n\rightharpoonup x$ stands for the weak convergence of $(x_n)$ to x;

• $x_n\to x$ stands for the strong convergence of $(x_n)$ to x.

Below is the so-called demiclosedness principle for nonexpansive mappings.

Lemma 2.1 (cf. [32])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $S:C\to C$ be a nonexpansive mapping with fixed points. If $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup x^{\ast }$ and $(I-S)x_n\to y$, then $(I-S)x^{\ast }=y$.

We also need the following lemma whose proof can be found in literature (cf. [33]).

Lemma 2.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Assume that a mapping $F:C\to H$ is monotone and weakly continuous along segments (i.e., $F(x+ty)\to F(x)$ weakly as $t\to 0$, whenever $x+ty\in C$ for $x,y\in C$). Then the variational inequality

$$x^{\ast }\in C,〈F{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C$$

(2.3)

is equivalent to the dual variational inequality

$$x^{\ast }\in C,〈Fx,x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C.$$

(2.4)

Finally, we state the following elementary result on convergence of real sequences.

Lemma 2.3 ([30])

Let $\{a_n\}$ be a sequence of nonnegative real numbers satisfying

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\gamma }_n{\sigma }_n,n\ge 0,$$

where $\{{\gamma }_n\}\subset (0,1)$ and $\{{\sigma }_n\}$ satisfy

• (i) ${\sum }_{n=0}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (ii) either ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\sigma }_n\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}|{\gamma }_n{\sigma }_n|<\mathrm{\infty }$.

Then $\{a_n\}$ converges to 0.

3 An iterative algorithm and its convergence

Throughout this section we assume that C is a nonempty closed subset of a real Hilbert space H and $T:C\to C$ is a pseudocontraction with a nonempty fixed point set $Fix(T)$. The aim of this section is to introduce an iterative method for finding the minimum-norm fixed point of T. Towards this, we select two sequences of real numbers, $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ in the interval $(0,1)$ such that

$${\alpha }_n+{\beta }_n<1$$

(3.1)

for all n. We also take an arbitrary initial guess $x_0\in C$. We then define an iterative algorithm which generates a sequence $\{x_n\}$ via the following recursion:

$$x_{n+1}=P_C[(1-{\alpha }_n-{\beta }_n)x_n+{\beta }_{n}T{x}_n],n\ge 0.$$

(3.2)

We shall prove that this sequence strongly converges to the minimum-norm fixed point of T provided $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ satisfy certain conditions. To this end, we need the following lemma.

Lemma 3.1 Let $f:C\to H$ be a contraction with coefficient $\rho \in (0,1)$. Let $S:C\to C$ be a nonexpansive mapping with $Fix(S)\ne \mathrm{\varnothing }$. For each $t\in (0,1)$, let $x_t$ be defined as the unique solution of the fixed point equation

$$x_t=S{P}_C[tf(x_t)+(1-t)x_t].$$

(3.3)

Then, as $t\to 0^{+}$, the net $\{x_t\}$ converges strongly to a point $x^{\ast }\in Fix(S)$ which solves the following variational inequality:

$$x^{\ast }\in Fix(S),〈(I-f)x^{\ast },x-x^{\ast }〉\ge 0,x\in Fix(S).$$

In particular, if we take $f=0$, then the net $\{x_t\}$ defined via the fixed point equation

$$x_t=S{P}_C[(1-t)x_t],$$

(3.4)

converges in norm, as $t\to 0^{+}$, to the minimum-norm fixed point of S.

Proof First observe that, for each $t\in (0,1)$, $x_t$ is well defined. Indeed, if we define a mapping $S_t:C\to C$ by

$$S_{t}x=S{P}_C[tf(x)+(1-t)x],x\in C.$$

For $x,y\in C$, we have

$$\begin{array}{rcl}\parallel S_{t}x-S_{t}y\parallel & =& \parallel S{P}_C[tf(x)+(1-t)x]-S{P}_C[tf(y)+(1-t)y]\parallel \\ \le & t\parallel f(x)-f(y)\parallel +(1-t)\parallel x-y\parallel \\ \le & [1-(1-\rho )t]\parallel x-y\parallel ,\end{array}$$

which implies that $S_t$ is a self-contraction of C. Hence $S_t$ has a unique fixed point $x_t\in C$ which is the unique solution of fixed point equation (3.3).

Next we prove that $\{x_t\}$ is bounded. Take $u\in Fix(S)$. From (3.3) we have

$$\begin{array}{rcl}\parallel x_t-u\parallel & =& \parallel S{P}_C[tf(x_t)+(1-t)x_t]-S{P}_{C}u\parallel \\ \le & t\parallel f(x_t)-f(u)\parallel +t\parallel f(u)-u\parallel +(1-t)\parallel x_t-u\parallel \\ \le & [1-(1-\rho )t]\parallel x_t-u\parallel +t\parallel f(u)-u\parallel ,\end{array}$$

that is,

$$\parallel x_t-u\parallel \le \frac{\parallel f(u)-u\parallel }{1-\rho }.$$

Hence, $\{x_t\}$ is bounded and so is $\{f(x_t)\}$.

From (3.3) we have

$$\begin{array}{rcl}\parallel x_t-S{x}_t\parallel & =& \parallel S{P}_C[tf(x_t)+(1-t)x_t]-S{P}_C{x}_t\parallel \\ \le & t\parallel f(x_t)-x_t\parallel \to 0\text{as }t\to 0^{+}.\end{array}$$

(3.5)

Next we show that $\{x_t\}$ is relatively norm-compact as $t\to 0^{+}$, i.e., we show that from any sequence in $\{x_t\}$, a convergent subsequence can be extracted. Let $\{t_n\}\subset (0,1)$ be a sequence such that $t_n\to 0^{+}$ as $n\to \mathrm{\infty }$. Put $x_n:=x_{t_n}$. From (3.5) we have

$$\parallel x_n-S{x}_n\parallel \to 0.$$

(3.6)

Again from (3.3) we get

$$\begin{array}{rcl}{\parallel x_t-u\parallel }^2& =& {\parallel S{P}_C[tf(x_t)+(1-t)x_t]-S{P}_{C}u\parallel }^2\\ \le & {\parallel x_t-u+t(f(x_t)-x_t)\parallel }^2\\ =& {\parallel x_t-u\parallel }^2+2t〈f(x_t)-x_t,x_t-u〉+t^2{\parallel f(x_t)-x_t\parallel }^2\\ =& {\parallel x_t-u\parallel }^2+2t〈f(x_t)-f(u),x_t-u〉+2t〈f(u)-u,x_t-u〉\\ +2t〈u-x_t,x_t-u〉+t^2{\parallel f(x_t)-x_t\parallel }^2\\ \le & [1-2(1-\rho )t]{\parallel x_t-u\parallel }^2+2t〈f(u)-u,x_t-u〉+t^2{\parallel f(x_t)-x_t\parallel }^2.\end{array}$$

It turns out that

$${\parallel x_t-u\parallel }^2\le \frac{1}{1-\rho }〈f(u)-u,x_t-u〉+tM,$$

(3.7)

where $M>0$ is a constant such that

$$M>\frac{1}{2(1-\rho )}sup\{{\parallel f(x_t)-x_t\parallel }^2:t\in (0,1)\}.$$

In particular, we get from (3.7)

$${\parallel x_n-u\parallel }^2\le \frac{1}{1-\rho }〈f(u)-u,x_n-u〉+t_{n}M,u\in Fix(S).$$

(3.8)

Since $\{x_n\}$ is bounded, without loss of generality, we may assume that $\{x_n\}$ converges weakly to a point $x^{\ast }\in C$. Noticing (3.6) we can use Lemma 2.1 to get $x^{\ast }\in Fix(S)$. Therefore we can substitute $x^{\ast }$ for u in (3.8) to get

$${\parallel x_n-x^{\ast }\parallel }^2\le \frac{1}{1-\rho }〈f\left(x^{\ast }\right)-x^{\ast },x_n-x^{\ast }〉+t_{n}M.$$

(3.9)

However, $x_n\rightharpoonup x^{\ast }$. This together with (3.9) guarantees that $x_n\to x^{\ast }$. The net $\{x_t\}$ is therefore relatively compact, as $t\to 0^{+}$, in the norm topology.

Now we return to (3.8) and take the limit as $n\to \mathrm{\infty }$ to get

$${\parallel x^{\ast }-u\parallel }^2\le \frac{1}{1-\rho }〈f(u)-u,x^{\ast }-u〉,u\in Fix(S).$$

In particular, $x^{\ast }$ solves the following variational inequality:

$$x^{\ast }\in Fix(S),〈(I-f)u,u-x^{\ast }〉\ge 0,u\in Fix(S).$$

By Lemma 2.2, we see that $x^{\ast }$ solves the variational inequality

$$x^{\ast }\in Fix(S),〈(I-f)x^{\ast },u-x^{\ast }〉\ge 0,u\in Fix(S).$$

(3.10)

Therefore, $x^{\ast }=(P_{Fix(S)}f)x^{\ast }$. That is, $x^{\ast }$ is the unique fixed point in $Fix(S)$ of the contraction $P_{Fix(S)}f$. Clearly this is sufficient to conclude that the entire net $\{x_t\}$ converges in norm to $x^{\ast }$ as $t\to 0^{+}$.

Finally, if we take $f=0$, then variational inequality (3.10) is reduced to

$$0\le 〈x^{\ast },u-x^{\ast }〉,u\in Fix(S).$$

Equivalently,

$${\parallel x^{\ast }\parallel }^2\le 〈x^{\ast },u〉,u\in Fix(S).$$

This clearly implies that

$$\parallel x^{\ast }\parallel \le \parallel u\parallel ,u\in Fix(S).$$

Therefore, $x^{\ast }$ is the minimum-norm fixed point of S. This completes the proof. □

We are now in a position to prove the strong convergence of algorithm (3.2).

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H, and let $T:C\to C$ be L-Lipschitzian and pseudocontractive with $Fix(T)\ne \mathrm{\varnothing }$. Suppose that the following conditions are satisfied:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_n}{{\beta }_n}={lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n^2}{{\alpha }_n}=0$;

• (iii) ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_n{\beta }_{n-1}-{\alpha }_{n-1}{\beta }_n}{{\alpha }_n^2{\beta }_{n-1}}=0$.

Then the sequence $\{x_n\}$ generated by algorithm (3.2) converges strongly to the minimum-norm fixed point of T.

Proof First we prove that the sequence $\{x_n\}$ is bounded. We will show this fact by induction. According to conditions (i) and (ii), there exists a sufficiently large positive integer m such that

$$1-2(L+1)(L+2)({\alpha }_n+2{\beta }_n+\frac{{\beta }_n^2}{{\alpha }_n})>0,n\ge m.$$

(3.11)

Fix $p\in Fix(T)$ and take a constant $M_1>0$ such that

$$max\{\parallel x_0-p\parallel ,\parallel x_1-p\parallel ,\dots ,\parallel x_m-p\parallel ,2\parallel p\parallel \}\le M_1.$$

(3.12)

Next, we show that $\parallel x_{m+1}-p\parallel \le M_1$.

Set

$$y_m=(1-{\alpha }_m-{\beta }_m)x_m+{\beta }_{m}T{x}_m;\text{thus }{x}_{m+1}=P_C[y_m].$$

Then, by using property (2.2) of the metric projection, we have

$$〈x_{m+1}-y_m,x_{m+1}-p〉\le 0.$$

(3.13)

By the fact that $I-T$ is monotone, we have

$$〈(I-T)x_{m+1}-(I-T)p,x_{m+1}-p〉\ge 0.$$

(3.14)

From (3.2), (3.13) and (3.14), we obtain

$$\begin{array}{rcl}{\parallel x_{m+1}-p\parallel }^2& =& 〈x_{m+1}-p,x_{m+1}-p〉\\ =& 〈x_{m+1}-y_m,x_{m+1}-p〉+〈y_m-p,x_{m+1}-p〉\\ \le & 〈y_m-p,x_{m+1}-p〉\\ =& 〈x_m-p,x_{m+1}-p〉-{\alpha }_m〈x_m,x_{m+1}-p〉+{\beta }_m〈T{x}_m-x_m,x_{m+1}-p〉\\ =& 〈x_m-p,x_{m+1}-p〉+{\alpha }_m〈x_{m+1}-x_m,x_{m+1}-p〉-{\alpha }_m〈p,x_{m+1}-p〉\\ -{\alpha }_m〈x_{m+1}-p,x_{m+1}-p〉+{\beta }_m〈T{x}_m-T{x}_{m+1},x_{m+1}-p〉\\ +{\beta }_m〈x_{m+1}-x_m,x_{m+1}-p〉-{\beta }_m〈x_{m+1}-T{x}_{m+1},x_{m+1}-p〉\\ \le & \parallel x_m-p\parallel \parallel x_{m+1}-p\parallel +{\alpha }_m\parallel x_{m+1}-x_m\parallel \parallel x_{m+1}-p\parallel \\ +{\alpha }_m\parallel p\parallel \parallel x_{m+1}-p\parallel -{\alpha }_m{\parallel x_{m+1}-p\parallel }^2\\ +{\beta }_m(\parallel T{x}_m-T{x}_{m+1}\parallel +\parallel x_{m+1}-x_m\parallel )\parallel x_{m+1}-p\parallel \\ \le & \parallel x_m-p\parallel \parallel x_{m+1}-p\parallel +{\alpha }_m\parallel p\parallel \parallel x_{m+1}-p\parallel -{\alpha }_m{\parallel x_{m+1}-p\parallel }^2\\ +(L+1)({\alpha }_m+{\beta }_m)\parallel x_{m+1}-x_m\parallel \parallel x_{m+1}-p\parallel .\end{array}$$

It follows that

$$(1+{\alpha }_m)\parallel x_{m+1}-p\parallel \le \parallel x_m-p\parallel +{\alpha }_m\parallel p\parallel +(L+1)({\alpha }_m+{\beta }_m)\parallel x_{m+1}-x_m\parallel .$$

(3.15)

By (3.2), we have

$$\begin{array}{rcl}\parallel x_{m+1}-x_m\parallel & =& \parallel P_C[(1-{\alpha }_m-{\beta }_m)x_m+{\beta }_{m}T{x}_m]-P_C[x_m]\parallel \\ \le & \parallel (1-{\alpha }_m-{\beta }_m)x_m+{\beta }_{m}T{x}_m-x_m\parallel \\ \le & {\alpha }_m(\parallel p\parallel +\parallel x_m-p\parallel )+{\beta }_m(\parallel T{x}_m-p\parallel +\parallel x_m-p\parallel )\\ \le & {\alpha }_m(\parallel p\parallel +\parallel x_m-p\parallel )+(L+1){\beta }_m\parallel x_m-p\parallel \\ \le & (L+1)({\alpha }_m+{\beta }_m)\parallel x_m-p\parallel +{\alpha }_m\parallel p\parallel \\ \le & (L+2)({\alpha }_m+{\beta }_m)M_1.\end{array}$$

(3.16)

Substitute (3.16) into (3.15) to obtain

$$\begin{array}{rcl}(1+{\alpha }_m)\parallel x_{m+1}-p\parallel & \le & \parallel x_m-p\parallel +{\alpha }_m\parallel p\parallel +(L+1)(L+2){({\alpha }_m+{\beta }_m)}^2{M}_1\\ \le & (1+\frac{1}{2}{\alpha }_m)M_1+(L+1)(L+2){({\alpha }_m+{\beta }_m)}^2{M}_1,\end{array}$$

that is,

$$\begin{array}{rcl}\parallel x_{m+1}-p\parallel & \le & [1-\frac{({\alpha }_m/2)-(L+1)(L+2){({\alpha }_m+{\beta }_m)}^2}{1+{\alpha }_m}]M_1\\ =& \{1-\frac{({\alpha }_m/2)[1-2(L+1)(L+2)({\alpha }_m+2{\beta }_m+({\beta }_m^2/{\alpha }_m))]}{1+{\alpha }_m}\}{M}_1\\ \le & M_1.\end{array}$$

By induction, we get

$$\parallel x_n-p\parallel \le M_1,\mathrm{\forall }n\ge 0,$$

(3.17)

which implies that $\{x_n\}$ is bounded and so is $\{T{x}_n\}$. Now we take a constant $M_2>0$ such that

$$M_2=\underset{n}{sup}\{\parallel x_n\parallel \vee \parallel T{x}_n-x_n\parallel \}.$$

[Here $a\vee b=max\{a,b\}$ for $a,b\in \mathbb{R}$.]

Set $S={(2I-T)}^{-1}$ (i.e., S is a resolvent of the monotone operator $I-T$). We then have that S is a nonexpansive self-mapping of C and $Fix(S)=Fix(T)$ (cf. Theorem 6 of [34]).

By Lemma 3.1, we know that whenever $\{{\gamma }_n\}\subset (0,1)$ and ${\gamma }_n\to 0^{+}$, the sequence $\{z_n\}$ defined by

$$z_n=S{P}_C[(1-{\gamma }_n)z_n]$$

(3.18)

converges strongly to the minimum-norm fixed point $x^{\ast }$ of S (and of T as $Fix(S)=Fix(T)$). Without loss of generality, we may assume that $\parallel z_n\parallel \le M_2$ for all n.

It suffices to prove that $\parallel x_{n+1}-z_n\parallel \to 0$ as $n\to \mathrm{\infty }$ (for some ${\gamma }_n\to 0^{+}$). To this end, we rewrite (3.18) as

$$(2I-T)z_n=P_C[(1-{\gamma }_n)z_n],n\ge 0.$$

By using the property of metric projection (2.2), we have

$$\begin{array}{r}〈(1-{\gamma }_n)z_n-(2z_n-T{z}_n),x_{n+1}-(2z_n-T{z}_n)〉\le 0\\ \Rightarrow 〈-{\gamma }_n{z}_n,x_{n+1}-z_n-(z_n-T{z}_n)〉+〈T{z}_n-z_n,x_{n+1}-z_n-(z_n-T{z}_n)〉\le 0\\ \Rightarrow 〈-{\gamma }_n{z}_n+T{z}_n-z_n,x_{n+1}-z_n〉+{\parallel z_n-T{z}_n\parallel }^2\le 〈{\gamma }_n{z}_n,T{z}_n-z_n〉\\ \Rightarrow 〈-{\gamma }_n{z}_n+T{z}_n-z_n,x_{n+1}-z_n〉\le {\gamma }_n\parallel z_n\parallel \parallel T{z}_n-z_n\parallel \\ \Rightarrow 〈-z_n+\frac{T{z}_n-z_n}{{\gamma }_n},x_{n+1}-z_n〉\le \parallel z_n\parallel \parallel T{z}_n-z_n\parallel .\end{array}$$

Note that

$$\begin{array}{rcl}\parallel z_n-T{z}_n\parallel & =& \parallel P_C[(1-{\gamma }_n)z_n]-z_n\parallel \\ \le & \parallel (1-{\gamma }_n)z_n-z_n\parallel \\ =& {\gamma }_n\parallel z_n\parallel .\end{array}$$

Hence, we get

$$〈-z_n+\frac{T{z}_n-z_n}{{\gamma }_n},x_{n+1}-z_n〉\le {\gamma }_n{\parallel z_n\parallel }^2.$$

(3.19)

From (3.18) we have

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel S{P}_C[(1-{\gamma }_{n+1})z_{n+1}]-S{P}_C[(1-{\gamma }_n)z_n]\parallel \\ \le & \parallel (1-{\gamma }_{n+1})z_{n+1}-(1-{\gamma }_n)z_n\parallel \\ =& \parallel (1-{\gamma }_{n+1})(z_{n+1}-z_n)+({\gamma }_n-{\gamma }_{n+1})z_n\parallel \\ \le & (1-{\gamma }_{n+1})\parallel z_{n+1}-z_n\parallel +|{\gamma }_{n+1}-{\gamma }_n|\parallel z_n\parallel .\end{array}$$

It follows that

$$\parallel z_{n+1}-z_n\parallel \le \frac{|{\gamma }_{n+1}-{\gamma }_n|}{{\gamma }_{n+1}}\parallel z_n\parallel .$$

(3.20)

Set

$${\gamma }_n:=\frac{{\alpha }_n}{{\beta }_n}.$$

By condition (ii), ${\gamma }_n\to 0^{+}$ and ${\gamma }_n\in (0,1)$ for n large enough. Hence, by (3.19) and (3.20) we have

$$〈-z_n+\frac{{\beta }_n(T{z}_n-z_n)}{{\alpha }_n},x_{n+1}-z_n〉\le \frac{{\alpha }_n}{{\beta }_n}{\parallel z_n\parallel }^2\le \frac{{\alpha }_n}{{\beta }_n}{M}_2^2$$

(3.21)

and

$$\parallel z_n-z_{n-1}\parallel \le \frac{{\alpha }_n{\beta }_{n-1}-{\alpha }_{n-1}{\beta }_n}{{\alpha }_n{\beta }_{n-1}}{M}_2.$$

(3.22)

By (3.2) we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel P_C[(1-{\alpha }_n-{\beta }_n)x_n+{\beta }_{n}T{x}_n]-P_C{x}_n\parallel \\ \le & {\alpha }_n\parallel x_n\parallel +{\beta }_n\parallel T{x}_n-x_n\parallel \\ \le & ({\alpha }_n+{\beta }_n)M_2.\end{array}$$

(3.23)

Next, we estimate $\parallel x_{n+1}-z_{n+1}\parallel$. Since $x_{n+1}=P_C[y_n]$, $〈x_{n+1}-y_n,x_{n+1}-z_n〉\le 0$. Using (3.21) and by the fact that T is L-Lipschitzian and pseudocontractive, we infer that

$$\begin{array}{rcl}{\parallel x_{n+1}-z_n\parallel }^2& =& 〈x_{n+1}-z_n,x_{n+1}-z_n〉\\ =& 〈x_{n+1}-y_n,x_{n+1}-z_n〉+〈y_n-z_n,x_{n+1}-z_n〉\\ \le & 〈y_n-z_n,x_{n+1}-z_n〉\\ =& 〈[(1-{\alpha }_n-{\beta }_n)x_n+{\beta }_{n}T{x}_n]-z_n,x_{n+1}-z_n〉\\ =& (1-{\alpha }_n-{\beta }_n)〈x_n-z_n,x_{n+1}-z_n〉+{\beta }_n〈T{x}_n-T{x}_{n+1},x_{n+1}-z_n〉\\ +{\beta }_n〈T{x}_{n+1}-T{z}_n,x_{n+1}-z_n〉+〈-{\alpha }_n{z}_n+{\beta }_n(T{z}_n-z_n),x_{n+1}-z_n〉,\end{array}$$

which leads to

$$\begin{array}{rcl}{\parallel x_{n+1}-z_n\parallel }^2& \le & (1-{\alpha }_n-{\beta }_n)\parallel x_n-z_n\parallel \parallel x_{n+1}-z_n\parallel +{\beta }_{n}L\parallel x_n-x_{n+1}\parallel \parallel x_{n+1}-z_n\parallel \\ +{\beta }_n{\parallel x_{n+1}-z_n\parallel }^2+{\alpha }_n〈-z_n+\frac{{\beta }_n}{{\alpha }_n}(T{z}_n-z_n),x_{n+1}-z_n〉\\ \le & \frac{1-{\alpha }_n-{\beta }_n}{2}({\parallel x_n-z_n\parallel }^2+{\parallel x_{n+1}-z_n\parallel }^2)+\frac{{\beta }_n^2}{2}{\parallel x_{n+1}-z_n\parallel }^2\\ +\frac{L^2}{2}{\parallel x_n-x_{n+1}\parallel }^2+{\beta }_n{\parallel x_{n+1}-z_n\parallel }^2+\frac{{\alpha }_n^2}{{\beta }_n}{\parallel z_n\parallel }^2.\end{array}$$

It follows that, using (3.21), (3.22) and (3.23), we get

$$\begin{array}{rcl}{\parallel x_{n+1}-z_n\parallel }^2& \le & \frac{1-{\alpha }_n-{\beta }_n}{1+{\alpha }_n-{\beta }_n}{\parallel x_n-z_n\parallel }^2+\frac{L^2}{1+{\alpha }_n-{\beta }_n}{\parallel x_{n+1}-x_n\parallel }^2\\ +\frac{2{\alpha }_n^2}{(1+{\alpha }_n-{\beta }_n){\beta }_n}{\parallel z_n\parallel }^2+\frac{{\beta }_n^2}{1+{\alpha }_n-{\beta }_n}{\parallel x_{n+1}-z_n\parallel }^2\\ \le & (1-\frac{2{\alpha }_n}{1+{\alpha }_n-{\beta }_n}){\parallel x_n-z_n\parallel }^2+\frac{{({\alpha }_n+{\beta }_n)}^2}{1+{\alpha }_n-{\beta }_n}{L}^2{M}_2^2\\ +\frac{2{\alpha }_n^2}{(1+{\alpha }_n-{\beta }_n){\beta }_n}{M}_2^2+\frac{{\beta }_n^2}{1+{\alpha }_n-{\beta }_n}4M_2^2\\ \le & (1-\frac{2{\alpha }_n}{1+{\alpha }_n-{\beta }_n}){(\parallel x_n-z_{n-1}\parallel +\parallel z_n-z_{n-1}\parallel )}^2\\ +\{\frac{{({\alpha }_n+{\beta }_n)}^2}{1+{\alpha }_n-{\beta }_n}+\frac{2{\alpha }_n^2}{(1+{\alpha }_n-{\beta }_n){\beta }_n}+\frac{{\beta }_n^2}{1+{\alpha }_n-{\beta }_n}\}M\\ \le & (1-\frac{2{\alpha }_n}{1+{\alpha }_n-{\beta }_n}){\parallel x_n-z_{n-1}\parallel }^2\\ +\frac{1}{1+{\alpha }_n-{\beta }_n}\parallel z_n-z_{n-1}\parallel (2\parallel x_n-z_{n-1}\parallel +\parallel z_n-z_{n-1}\parallel )\\ +\{\frac{{({\alpha }_n+{\beta }_n)}^2}{1+{\alpha }_n-{\beta }_n}+\frac{2{\alpha }_n^2}{(1+{\alpha }_n-{\beta }_n){\beta }_n}+\frac{{\beta }_n^2}{1+{\alpha }_n-{\beta }_n}\}M\\ \le & (1-\frac{2{\alpha }_n}{1+{\alpha }_n-{\beta }_n}){\parallel x_n-z_{n-1}\parallel }^2+\frac{1}{1+{\alpha }_n-{\beta }_n}\frac{{\alpha }_n{\beta }_{n-1}-{\alpha }_{n-1}{\beta }_n}{{\alpha }_n{\beta }_{n-1}}M\\ +\{\frac{{({\alpha }_n+{\beta }_n)}^2}{1+{\alpha }_n-{\beta }_n}+\frac{2{\alpha }_n^2}{(1+{\alpha }_n-{\beta }_n){\beta }_n}+\frac{{\beta }_n^2}{1+{\alpha }_n-{\beta }_n}\}M,\end{array}$$

(3.24)

where the finite constant $M>0$ is given by

$$M:=max\{L^2{M}_2^2,4M_2^2,M_2\underset{n}{sup}(2\parallel x_n-z_{n-1}\parallel +\parallel z_n-z_{n-1}\parallel )\}.$$

Let

$${\delta }_n=\frac{2{\alpha }_n}{1+{\alpha }_n-{\beta }_n}\approx 2{\alpha }_n(\text{as }n\to \mathrm{\infty })$$

and note that by (3.1) it follows that $\{{\delta }_n\}\subset (0,1)$. Moreover, set

$${\theta }_n=\{\frac{{\alpha }_n{\beta }_{n-1}-{\alpha }_{n-1}{\beta }_n}{2{\alpha }_n^2{\beta }_{n-1}}+\frac{1}{2}({\alpha }_n+2{\beta }_n+\frac{{\beta }_n^2}{{\alpha }_n})+\frac{{\alpha }_n}{{\beta }_n}+\frac{{\beta }_n^2}{2{\alpha }_n}\}M.$$

Then relation (3.24) is rewritten as

$${\parallel x_{n+1}-z_n\parallel }^2\le (1-{\delta }_n){\parallel x_n-z_{n-1}\parallel }^2+{\delta }_n{\theta }_n.$$

(3.25)

By conditions (i), (ii) and (iii), it is easily found that

$$\underset{n\to \mathrm{\infty }}{lim}{\delta }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\delta }_n=\mathrm{\infty },\underset{n\to \mathrm{\infty }}{lim}{\theta }_n=0.$$

We can therefore apply Lemma 2.3 to (3.25) and conclude that ${\parallel x_{n+1}-z_n\parallel }^2\to 0$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 3.3 Choose the sequences $({\alpha }_n)$ and $({\beta }_n)$ such that

$${\alpha }_n=\frac{1}{{(n+1)}^a}\text{and}{\beta }_n=\frac{1}{{(n+1)}^b},n\ge 0,$$

where $0<b<a<2b<1$. It is clear that conditions (i) and (ii) of Theorem 3.2 are satisfied. To verify condition (iii), we compute

$$\begin{array}{rcl}\left|\frac{{\alpha }_n{\beta }_{n-1}-{\alpha }_{n-1}{\beta }_n}{{\alpha }_n^2{\beta }_{n-1}}\right|& =& \frac{1}{{\alpha }_n}|1-\frac{{\alpha }_{n-1}{\beta }_n}{{\alpha }_n{\beta }_{n-1}}|\\ =& {(n+1)}^a|1-\frac{{(n+1)}^{a-b}}{n^{a-b}}|\\ =& {(n+1)}^a[{(1+\frac{1}{n})}^{a-b}-1]\\ \approx & \frac{a-b}{n}{(n+1)}^a\to 0.\end{array}$$

Therefore, $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ satisfy all three conditions (i)-(iii) in Theorem 3.2.

4 Application

To show an application of our results, we deal with the following problem.

Problem 4.1 Let $0<x_0<1$ and define the sequence $\{x_n\}$ by the recursion

$$x_{n+1}=(1-n^{-1/2}-n^{-1/3})x_n+n^{-1/3}\frac{x_n^2}{1+x_n}.$$

(4.1)

At which value does $\{x_n\}$ approach as n goes to infinity?

We claim that ${lim}_{n\to \mathrm{\infty }}{x}_n=0$ and it can be easily derived by applying Theorem 3.2.

Proof In order to apply our result, let $H=\mathbb{R}$, $C=[0,1]$ and define $T:C\to C$ by

$$Tx:=\frac{x^2}{1+x}.$$

Observe that T is Lipschitzian, pseudocontractive and that $Fix(T)=\{0\}$. Moreover, if we set ${\alpha }_n=n^{-1/2}$ and ${\beta }_n=n^{-1/3}$, then

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_n}{{\beta }_n}={lim}_{n\to \mathrm{\infty }}\frac{{\beta }_n^2}{{\alpha }_n}=0$;

• (iii) ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_n{\beta }_{n-1}-{\alpha }_{n-1}{\beta }_n}{{\alpha }_n^2{\beta }_{n-1}}=0$.

Then Theorem 3.2 ensures that

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n=0.$$

 □