On a Class of Generalized Nonexpansive Mappings

: 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 In the deﬁnition of this notion the norm has been replaced by a general function satisfying certain conditions.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.

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 In the definition of this notion the norm has been replaced by a general function satisfying certain conditions.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, · ) be a Banach space and let K be a bounded, closed and convex subset of X. Let f : X → [0, ∞) be a continuous function such that f (0) = 0, the set f (K − K) is bounded, and such that f enjoys the following three properties: (i) For every positive number , there is a positive number δ such that for every pair of points x, y ∈ K satisfying f (x − y) ≤ δ, we have x − y ≤ ; (ii) For every positive number λ < 1, there is a positive number φ(λ) < 1 for which for every pair of points x, y ∈ K; (iii) The function (x, y) → f (x − y), x, y ∈ K, is uniformly continuous on K × K. Denote by A the set of all continuous operators A : for all x, y ∈ K. Evidently, (A, d) is a complete metric space.
In [22], we establish the existence of a set F , which is a countable intersection of open and everywhere dense subsets of A, such that each operator C ∈ 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 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).

A Convergence Result
Let (X, · ) be a Banach space, and let K be a nonempty and closed subset of X. Let f : X → [0, ∞) be a continuous function with f (0) = 0 such that f enjoys the following two properties: (P1) for every positive number , there is a positive number δ such that for every pair of points Next, assume that A : K → K is a continuous mapping, ψ : [0, ∞) → [0, 1] is a decreasing function satisfying 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 ∈ K and A i x → x A as i → ∞ for all x ∈ 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 (x) = x .

An Example
Let (X, ·, · ) be a separable Hilbert space endowed with the inner product ·, · , which generates the complete norm · , and let {e i } ∞ i=1 be an orthonormal basis in X. In other words, for every natural number i, we have e i = 1, and for every pair of natural numbers j > i, we have e i , e j = 0.
Let K of the set of all points x ∈ X such that for every natural numbers i, and x, e 2i ∈ [0, i −2 ].
Evidently, K is a bounded, closed and convex subset of X. It is not difficult to see that the set is bounded, closed and convex. It is well known that there exists a projection P : X → K − K such that for every point x ∈ X, we have and such that P(y 1 ) − P(y 2 ) ≤ y 1 − y 2 for all y 1 , y 2 ∈ X.
For every point x ∈ K − K, put Furthermore, for every point Evidently, the function f : Proof. In view of (6), it suffices to show that f is uniformly continuous on K − K. Let a positive number be given. Choose an integer n 0 ≥ 1 for which and then fix a positive number Assume that z 1 , z 2 ∈ K − K and z 1 − z 2 ≤ δ.
We claim that Indeed, we may suppose without any loss of generality that It follows from (4) and (7) that there is an integer j ≥ 1 for which f (z 2 ) = max{| z 2 , e 2j−1 |, j| z 2 , e 2j |}.
There are two cases: j > n 0 and j ≤ n 0 . First, assume that j > n 0 .

Proposition 2.
For every point x ∈ K − K and every positive number λ < 1, we have Proof. Assume that x ∈ K − K and that λ ∈ (0, 1). Clearly, there is an integer j ≥ 1 such that By (7) and (15), Proposition 2 has been proved.

Proposition 3.
Let ∈ (0, 1) be given. Then there is a positive number δ such that for every x ∈ K − K satisfying f (x) ≤ δ, the inequality x ≤ is true.
Proposition 3 implies property (i) (see Section 1). Next, we define a mapping A : K → K. To this end, let x ∈ K be given. In view of (2) and (3), there is a unique point A(x) ∈ X such that for every natural number i, we have and Clearly, A(x) ∈ K for every point x ∈ K.
It is clear that and that for every natural number i ≥ 1, Hence Therefore the mapping A : K → K is not Lipschitz with respect to any norm which is equivalent to the norm induced by the inner product. Let x ∈ K be given. In view of (21) and (22), for every natural number i, Thus for every point x ∈ K. Therefore the origin is the unique fixed point of A in K, but the iterates of A do not converge to it. Let λ ∈ (0, 1). Define Clearly, By (21), (22) and (26), for all x ∈ K, A λ (x) = A(λx).
Therefore, the mapping A λ : K → K is not Lipschitz with respect to any norm which is equivalent to the norm induced by the inner product.