On a Class of Generalized Nonexpansive Mappings

Abstract

In our recent work we have introduced and studied a notion of a generalized nonexpansive mapping. In the definition of this notion the norm has been replaced by a general function satisfying certain conditions. For this new class of mappings, we have established the existence of unique fixed points and the convergence of iterates. In the present paper we construct an example of a generalized nonexpansive self-mapping of a bounded, closed and convex set in a Hilbert space, which is not nonexpansive in the classical sense.

1. Introduction

The by now well-developed fixed point theory of nonexpansive operators has stemmed from Stefan Banach’s classical result, which was obtained in [1]. The main goals of this theory include establishing the existence of fixed points of nonexpansive operators and the study of the convergence of their orbits to their fixed points [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]. This theory also concerns investigations of feasibility, common fixed point problems and variational inequalities [15,16,17,18,19,20,21].

In our recent work we have introduced and studied a notion of a generalized nonexpansive mapping. In the definition of this notion the norm has been replaced by a general function satisfying certain conditions. For this new class of mappings, we have established the existence of unique fixed points and the convergence of iterates.

In the present paper, we construct an example of a generalized nonexpansive self-mapping of a bounded, closed and convex set in a Hilbert space, which is not nonexpansive in the classical sense.

To wit, in [22,23] we examine the following class of nonlinear operators.

Let $(X,\parallel ·\parallel )$ be a Banach space and let K be a bounded, closed and convex subset of X. Let $f:X\to [0,\infty )$ be a continuous function such that $f\left(0\right)=0$, the set $f(K-K)$ is bounded, and such that f enjoys the following three properties:

(i) For every positive number $ϵ$, there is a positive number $\delta$ such that for every pair of points $x,y\in K$ satisfying $f(x-y)\le \delta$, we have $\parallel x-y\parallel \le ϵ$;

(ii) For every positive number $\lambda <1$, there is a positive number $\varphi \left(\lambda \right)<1$ for which

$$f\left(\lambda \right(x-y\left)\right)\le \varphi \left(\lambda \right)f(x-y)\mathrm{for}\mathrm{every}\mathrm{pair}\mathrm{of}\mathrm{points} x,y\in K;$$

(iii) The function $(x,y)\mapsto f(x-y)$, $x,y\in K$, is uniformly continuous on $K\times K$.

Denote by $\mathcal{A}$ the set of all continuous operators $A:K\to K$ such that

$$f(Ax-Ay)\le f(x-y)\mathrm{for}\mathrm{all}x,y\in K.$$

For every pair of operators $A,B\in \mathcal{A}$, put

$$d(A,B):=sup\{\parallel Ax-Bx\parallel :x\in K\}.$$

Evidently, $(\mathcal{A},d)$ is a complete metric space.

In [22], we establish the existence of a set $\mathcal{F}$, which is a countable intersection of open and everywhere dense subsets of $\mathcal{A}$, such that each operator $C\in \mathcal{F}$ has a unique fixed point and all its iterates converge uniformly to this fixed point.

At this juncture, it is worthwhile mentioning that the classical theorem of Francesco De Blasi and Józef Myjak [24] is a special case of this result, where the function f is the norm. Clearly, the operators defined above may be considered generalized nonexpansive mappings with respect to the function f. Note that in [25,26] this approach was applied to generalized best approximation problems.

In [23] we improve the results of [22]. To wit, we introduce there a notion of a contractive mapping, show that most mappings in $\mathcal{A}$ (in the sense of Baire category) are contractive, every contractive mapping possesses a unique fixed point and that all its iterates converge to this point uniformly. We emphasize that all these results were obtained for a bounded set K.

In [27] we extend one of the main results of [23] to unbounded sets. Moreover, to establish this result, which we present in the next section, it turns out that we do not need property (ii).

2. A Convergence Result

Let $(X,\parallel ·\parallel )$ be a Banach space, and let K be a nonempty and closed subset of X. Let $f:X\to [0,\infty )$ be a continuous function with $f\left(0\right)=0$ such that f enjoys the following two properties:

(P1) for every positive number $ϵ$, there is a positive number $\delta$ such that for every pair of points $x,y\in K$ which satisfies $f(x-y)\le \delta$, we have $\parallel x-y\parallel \le ϵ$;

(P2) the function $(x,y)\mapsto f(x-y)$, $x,y\in K$, is uniformly continuous on $K\times K$ and for every point $\xi \in K$, the function $f(x-\xi )$, $x\in D$, is bounded on every bounded subset D of K.

Next, assume that $A:K\to K$ is a continuous mapping, $\psi :[0,\infty )\to [0,1]$ is a decreasing function satisfying

$$\psi \left(t\right)<1\mathrm{for}\mathrm{all}t>0,$$

and that

$$f(Ax-Ay)\le \psi \left(f\right(x-y\left)\right)f(x-y)\mathrm{for}\mathrm{all}x,y\in K.\left(1\right)$$

In the literature, such an operator A is said to be contractive [14].

In [27] we have established the following theorem.

Theorem 1.

The mapping A has a unique fixed point $x_A\in K$ and $A^{i}x\to x_A$ as $i\to \infty$ for all $x\in K$, uniformly on bounded subsets of K.

This theorem is also a generalization of the result of [28], which was obtained for the case where $f\left(x\right)=\parallel x\parallel$.

3. An Example

Let $(X,\langle ·,·\rangle )$ be a separable Hilbert space endowed with the inner product $\langle ·,·\rangle$, which generates the complete norm $\parallel ·\parallel$, and let ${\left\{e_i\right\}}_{i=1}^{\infty }$ be an orthonormal basis in X. In other words, for every natural number i, we have $\parallel e_i\parallel =1$, and for every pair of natural numbers $j>i$, we have $\langle e_i,e_j\rangle =0.$

Let K of the set of all points $x\in X$ such that for every natural numbers i,

$$\langle x,e_{2i-1}\rangle \in [0,i^{-1}]$$

(2)

and

$$\langle x,e_{2i}\rangle \in [0,i^{-2}].$$

(3)

Evidently, K is a bounded, closed and convex subset of X. It is not difficult to see that the set

$$K-K=\{x\in X:|\langle x,e_{2i-1}\rangle |\le i^{-1},$$

$$|\langle x,e_{2i}\rangle |\le i^{-2}\mathrm{for}\mathrm{every}\mathrm{natural}\mathrm{number}i\}$$

(4)

is bounded, closed and convex. It is well known that there exists a projection $P:X\to K-K$ such that for every point $x\in X$, we have

$$\parallel P\left(x\right)-x\parallel \le \parallel y-x\parallel \mathrm{for}\mathrm{every}\mathrm{point}y\in K-K\left(5\right)$$

and such that

$$\parallel P\left(y_1\right)-P\left(y_2\right)\parallel \le \parallel y_1-y_2\parallel \mathrm{for}\mathrm{all}{y}_1,y_2\in X.$$

(6)

For every point $x\in K-K$, put

$$f\left(x\right):=max\{sup\{|\langle x,e_{2i-1}\rangle |:i=1,2,\cdots \},sup\left\{i\right|\langle x,e_{2i}\rangle |:i=1,2,\cdots \}\}.$$

(7)

Furthermore, for every point $x\in X\backslash (K-K)$, define

$$f\left(x\right):=f\left(P\right(x\left)\right).$$

Evidently, the function $f:X\to [0,\infty )$ is well defined, $f\left(0\right)=0$, and the set $f(K-K)$ is bounded.

Proposition 1.

The function $f:X\to [0,\infty )$ is uniformly continuous.

Proof. 

In view of (6), it suffices to show that f is uniformly continuous on $K-K$. Let a positive number $ϵ$ be given. Choose an integer $n_0\ge 1$ for which

$$n_0^{-1}<ϵ/4$$

(8)

and then fix a positive number

$$\delta <ϵn_0^{-1}.$$

(9)

Assume that

$$z_1,z_2\in K-K\mathrm{and}\parallel z_1-z_2\parallel \le \delta .$$

(10)

We claim that

$$|f\left(z_1\right)-f\left(z_2\right)|\le ϵ.$$

Indeed, we may suppose without any loss of generality that

$$f\left(z_2\right)\ge f\left(z_1\right).$$

(11)

It follows from (4) and (7) that there is an integer $j\ge 1$ for which

$$f\left(z_2\right)=max\left\{\right|\langle z_2,e_{2j-1}\rangle |,j|\langle z_2,e_{2j}\rangle \left|\right\}.$$

(12)

There are two cases: $j>n_0$ and $j\le n_0$. First, assume that

$$j>n_0.$$

(13)

Then, by (4), (8), (10), (12) and (13), we have

$$f\left(z_2\right)\le j^{-1}<n_0^{-1}<ϵ/4$$

and

$$f\left(z_1\right)\ge 0\ge f\left(z_2\right)-ϵ.$$

Next, assume that

$$j\le n_0.$$

(14)

It then follows from (9), (10), (12) and (14) that

$$f\left(z_1\right)\ge max\left\{\right|\langle z_1,e_{2j-1}\rangle |,j|\langle z_1,e_{2j}\rangle \left|\right\}$$

$$\ge max\left\{\right|\langle z_2,e_{2j-1}\rangle |,j|\langle z_2,e_{2j}\rangle \left|\right\}-j\parallel z_2-z_1\parallel$$

$$\ge f\left(z_2\right)-n_0\parallel z_1-z_2\parallel \ge f\left(z_2\right)-n_0\delta \ge f\left(z_2\right)-ϵ.$$

Proposition 1 has been established. ☐

Proposition 1 implies property (iii) (see Section 1).

Proposition 2.

For every point $x\in K-K$ and every positive number $\lambda <1$, we have

$$f\left(\lambda x\right)\le \lambda f\left(x\right).$$

Proof. 

Assume that $x\in K-K$ and that $\lambda \in (0,1)$. Clearly, there is an integer $j\ge 1$ such that

$$f\left(\lambda x\right)=max\left\{\lambda \right|\langle x,e_{2j-1}\rangle |,\lambda j|\langle x,e_{2j}\rangle \left|\right\}.$$

(15)

By (7) and (15),

$$f\left(\lambda x\right)=\lambda max\left\{\right|\langle x,e_{2j-1}\rangle |,j|\langle x,e_{2j}\rangle \left|\right\}\le \lambda f\left(x\right).$$

Proposition 2 has been proved. ☐

Proposition 2 implies property (ii) (see Section 1).

Proposition 3.

Let $ϵ\in (0,1)$ be given. Then there is a positive number δ such that for every $x\in K-K$ satisfying $f\left(x\right)\le \delta$, the inequality $\parallel x\parallel \le ϵ$ is true.

Proof. 

First fix an integer $n_0\ge 1$ for which

$$\sum _{i=n_0}^{\infty }{i}^{-2}<{ϵ}^2/8$$

(16)

and then fix $\delta \in (0,1)$ satisfying

$$16\delta n_0<{ϵ}^2.$$

(17)

Let

$$x\in K-K$$

(18)

satisfy

$$f\left(x\right)\le \delta .$$

(19)

By (7), (18) and (19), we have

$$|\langle x,e_i\rangle |\le \delta ,i=1,2,\cdots .$$

(20)

In view of (4), (16), (17) and (20), we also have

$${\parallel x\parallel }^2=\sum _{i=1}^{\infty }{|\langle x,e_i\rangle |}^2$$

$$\le \sum _{i=1}^{2n_0}|\langle x,e_i\rangle {|}^2+\sum _{i=2n_0+1}^{\infty }{\left|\langle x,e_i\rangle \right|}^2$$

$$\le 2n_0\delta +2\sum _{i=2n_0+1}^{\infty }{i}^{-2}<{ϵ}^2.$$

Proposition 3 is proved. ☐

Proposition 3 implies property (i) (see Section 1).

Next, we define a mapping $A:K\to K$. To this end, let $x\in K$ be given. In view of (2) and (3), there is a unique point $A\left(x\right)\in X$ such that for every natural number i, we have

$$\langle A\left(x\right),e_{2i-1}\rangle =i\langle x,e_{2i}\rangle$$

(21)

and

$$\langle A\left(x\right),e_{2i}\rangle =i^{-1}\langle x,e_{2i-1}\rangle .$$

(22)

Clearly, $A\left(x\right)\in K$ for every point $x\in K$.

Proposition 4.

For every pair of points $x,y\in K$,

$$f\left(A\right(x)-A(y\left)\right)\le f(x-y).$$

Proof. 

Let the points $x,y\in K$ be given. By (7), (21) and (22),

$$f(A\left(x\right)-A\left(y\right))=max\{sup\{|\langle A\left(x\right)-A\left(y\right),e_{2i-1}\rangle |:i=1,2,\cdots \},$$

$$sup\left\{i\right|\langle A\left(x\right)-A\left(y\right),e_{2i}\rangle |:i=1,2,\cdots \}\}$$

$$=max\{sup\{i|\langle x-y,e_{2i}\rangle |:i=1,2,\cdots \},sup\left\{\right|\langle x-y,e_{2i-1}\rangle |:i=1,2,\cdots \}\}$$

$$=f(x-y).$$

Proposition 4 has been proved. ☐

Proposition 5.

The mapping $A:K\to K$ is uniformly continuous.

Proof. 

Let $ϵ$ be a positive number. There is an integer $n_0>4$ for which

$$16\sum _{i=n_0}^{\infty }{i}^{-2}<{ϵ}^2.$$

(23)

Fix $\delta \in (0,1)$ for which

$$4n_0^3\delta <{ϵ}^2$$

(24)

and assume that

$$x,y\in K\mathrm{and}\parallel x-y\parallel \le \delta .$$

By Equations (2), (3), (21)–(24) and the above relation, we now have

$${\parallel A\left(x\right)-A\left(y\right)\parallel }^2=\sum _{i=1}^{\infty }{\langle e_i,A\left(x\right)-A\left(y\right)\rangle }^2$$

$$=\sum _{i=1}^{\infty }{\langle e_{2i-1},A\left(x\right)-A\left(y\right)\rangle }^2+\sum _{i=1}^{\infty }{\langle e_{2i},A\left(x\right)-A\left(y\right)\rangle }^2$$

$$\le \sum _{i=1}^{n_0}{\langle e_{2i-1},A\left(x\right)-A\left(y\right)\rangle }^2+\sum _{i=1}^{n_0}{\langle e_{2i},A\left(x\right)-A\left(y\right)\rangle }^2+8\sum _{i=n_0+1}^{\infty }{i}^{-2}$$

$$\le \sum _{i=1}^{n_0}{i}^2{\langle e_{2i},x-y\rangle }^2+\sum _{i=1}^{n_0}{\langle e_{2i-1},x-y\rangle }^2+{ϵ}^2/2$$

$$\le 2n_0^3{\parallel x-y\parallel }^2+{ϵ}^2/2\le 2n_0^3{\delta }^2+{ϵ}^2/2<{ϵ}^2.$$

Proposition 5 has been established. ☐

It is clear that

$$A\left(0\right)=0,$$

for each $\lambda \in (0,1]$,

$$\lambda (\sum _{i=1}^{\infty }{i}^{-1}{e}_{2i-1}+\sum _{i=1}^{\infty }{i}^{-2}{e}_{2i})\in K,$$

$$A(\lambda (\sum _{i=1}^{\infty }{i}^{-1}{e}_{2i-1}+\sum _{i=1}^{\infty }{i}^{-2}{e}_{2i}))$$

$$=\lambda (\sum _{i=1}^{\infty }{i}^{-1}{e}_{2i-1}+\sum _{i=1}^{\infty }{i}^{-2}{e}_{2i}),$$

and that for every natural number $i\ge 1$,

$$i^{-1}{e}_{2i-1},i^{-2}{e}_{2i}\in K,A\left(i^{-2}{e}_{2i}\right)=i^{-1}{e}_{2i-1}.$$

(25)

Hence

$$\parallel A\left(i^{-2}{e}_{2i}\right)\parallel (\parallel i^{-2}{e}_{2i}{\parallel )}^{-1}\to \infty \mathrm{as}i\to \infty .$$

Therefore the mapping $A:K\to K$ is not Lipschitz with respect to any norm which is equivalent to the norm induced by the inner product.

Let $x\in K$ be given. In view of (21) and (22), for every natural number i,

$$\langle A^2\left(x\right),e_{2i-1}\rangle =i\langle A\left(x\right),e_{2i}\rangle =i\left(i^{-1}\langle x,e_{2i-1}\rangle \right)=\langle x,e_{2i-1}\rangle$$

and

$$\langle A^2\left(x\right),e_{2i}\rangle =i^{-1}\langle A\left(x\right),e_{2i-1}\rangle =i^{-1}\left(i\langle x,e_{2i}\rangle \right)=\langle x,e_{2i}\rangle .$$

Thus

$$A^2\left(x\right)=x$$

for every point $x\in K$. Therefore the origin is the unique fixed point of A in K, but the iterates of A do not converge to it.

Let $\lambda \in (0,1)$. Define

$$A_{\lambda }\left(x\right):=\lambda A\left(x\right),x\in K.$$

(26)

Clearly,

$$A_{\lambda }\left(K\right)\subset K.$$

By (21), (22) and (26), for all $x\in K$,

$$A_{\lambda }\left(x\right)=A\left(\lambda x\right).$$

(27)

Proposition 4 and (27) imply that

$$f(A_{\lambda }\left(x\right)-A_{\lambda }\left(y\right))=f(A\left(\lambda x\right)-A\left(\lambda y\right))\le f(\lambda x-\lambda y)=\lambda f(x-y).$$

Evidently, (1) holds with $\psi \left(t\right)=\lambda$ for every nonnegative number t. Theorem 1 implies that the origin is the unique fixed point of $A_{\lambda }:K\to K$ and that all the iterates of $A_{\lambda }$ converge to it, uniformly on K.

It follows from (25) and (26) that for every natural number i, we have

$$A_{\lambda }\left(i^{-2}{e}_{2i}\right)=\lambda A\left(i^{-2}{e}_{2i}\right)=\lambda i^{-1}{e}_{2i-1}$$

and

$$\parallel A_{\lambda }\left(i^{-2}{e}_{2i}\right)\parallel (\parallel i^{-2}{e}_{2i}{\parallel )}^{-1}=\lambda i^{-1}/\left(i^{-2}\right)\to \infty \mathrm{as} i\to \infty .$$

Therefore, the mapping $A_{\lambda }:K\to K$ is not Lipschitz with respect to any norm which is equivalent to the norm induced by the inner product.