Iterated nonexpansive mappings in Hilbert spaces

In [T. Dominguez Benavides and E. Llorens-Fuster, Iterated nonexpansive mappings, J. Fixed Point Theory Appl. 20 (2018), no. 3, Paper No. 104, 18 pp.], the authors raised the question about the existence of a fixed point free continuous INEA mapping T defined on a closed convex and bounded subset (or on a weakly compact convex subset) of a Banach space with normal structure. Our main goal is to give the affirmative answer to this problem in the very special case of a Hilbert space.


Introduction and preliminaries
Let C be a weakly compact convex subset of a Banach space. In [1], the following concept of iterated nonexpansive mappings (INE in short) was stated: for all x ∈ C.
It is not surprising that an INE mapping does not have to have fixed points even if it is defined on a subset of a finite-dimensional Hilbert space (see, for instance, [2, Example 1.1]). Thus, it seems natural to raise the question of whether the same mapping must have a fixed point provided it is continuous. Clearly, according to the Klee result (see [3]), in this case we are considering a noncompact domain C, so the space must have infinite dimension (but still being a Hilbert space or a Banach one with normal structure). A Banach space is said to have normal structure if each convex subset C which contains more than one point has a point x ∈ C which is not a diametral one of C, i.e., the following condition holds: sup{ x − y : y ∈ C} < diam C (see, for instance, [4]).
The negative answer to this question was given in [2,5], where the authors presented an example of a class of fixed point free mappings which are INE and continuous on any closed convex bounded but noncompact subset of a Banach space. The same is true when the set is convex and weakly compact (and noncompact with respect to the norm topology). The common denominator of these mappings was the fact that all of them satisfy therefore, they were not asymptotically regular. Let us remind that the selfmapping T : C → C is asymptotically regular if for each x ∈ C the sequence T n x − T n+1 x tends to 0. This condition can be generalized to the case of mappings which have the so-called almost fixed point sequence. A sequence (x n ) in C is called an almost fixed point sequence (a.f.p.s. for short) for the mapping T on C whenever x n −T (x n ) → 0 . It is well known that if the selfmapping T : C → C is nonexpansive then T has an a.f.p.s. in C. Combining this fact with the normal structure of a space leads to the existence of fixed points (see, for instance, [6, Theorem 4.1] and [7, Theorem 2.7]). Iterated nonexpansive mappings which have a.f.p.s. are called INEA for short. Since the assumption of the existence of a.f.p.s. seems to play a crucial role, one may ask whether there is any fixed point free continuous INEA self-mapping of a closed convex bounded (or weakly-compact convex) subset C of a Banach space into C. Here we suppose additionally that the Banach space is a Hilbert one or Banach with normal structure.
As it was mentioned before, our main goal is to give an example of a continuous and INEA mapping T defined on a closed convex bounded subset (more precisely, on a closed unit ball) of the Hilbert space into itself for which the set of fixed points is empty. To do it, let us take the Hilbert space l 2 and let B be its closed unit ball. Further, we will apply two kinds of geometry. The first one is Cartesian geometry based on the standard base of l 2 denoted by {e n : n ∈ N}. Then let ·, · mean the inner product in l 2 . Moreover, we denote the unit sphere by S. This set will be very often considered with spherical geometry based on the spherical metric ρ, i.e.,

ρ(A, B) = arccos A, B
for a pair of two elements A, B ∈ S. By the angle between two curves c andc (c(0) =c(0)) on the sphere with respect to spherical geometry we mean the Alexandrov angle, defined by lim s,t→0 + ∠ c(0) (c(s),c(t)).
This limit always exists (see, for instance, [8, p. 16]). More details about spherical geometry can be found in [8,9]). Further, we will use the same denotation conv S (A) for any nonempty set A ⊂ S. Now, still with respect to spherical geometry on conv S ({e 1 , e 2 , e 3 }), let us notice that a = 3 −1/2 (e 1 + e 2 + e 3 ) ∈ conv S ({e 1 , e 2 , e 3 }). Moreover, if we join points a and t 1 and points t 1 and e i , i = 1, 2, with the geodesic segments, then the angle between the segments is equal to π/2. Notice that the same is true for t 2 and e i , i = 2, 3; hence, we can join points t 1 and t 2 with the curve

Example
where S(a, r) is a sphere in l 2 with the radius with respect to the norm. The angle between the curve and the segments [a, t 1 ] or [a, t 2 ] is also equal to π/2, so the curve is tangent to [e 1 , e 2 ] and [e 2 , e 3 ] at the points t 1 and t 2 , respectively ( Fig. 1). Now, we repeat our construction on each three-dimensional set conv S ({e n , e n+1 , e n+2 }) and we obtain a smooth curve of infinite length joining all points t n , n = 1, 2, . . .. Let us notice that points of the curve between t n and t n+1 can be treated as points of the circle centered at a pointã = √ 2 3 √ 3 (e n + e n+1 + e n+2 ) with a radius of r = 11 − 4 √ 3 9 (now with respect to Cartesian geometry). Then, the angle between radiuses [ã, t n ] and [ã, t n+1 ] is independent of n and smaller than π. Let us denote this angle by α and for all t ∈ [0, ∞) we define ϕ(t) as a point of the curve. If t = n α + τ then ϕ(t) is located between t n and t n+1 in such a way that ∠ã(t n , ϕ(t)) = τ .
Step 2-definition of the map on the curve Figure 2. The construction of the map Now, we would like to define the map T : ϕ → ϕ. Let us divide the curve between points t 1 and t 2 into 128 equal parts and let α 0 = α/128. Then, for points of the form ϕ(t), t ∈ [0, α − α 0 ], we take T (ϕ(t)) = ϕ(t + α 0 ). So far, the map T is an isometry.
To extend the map T on the whole curve, we need to make some calculations because our map must be INEA. To do it, first, let us choose α 1 in such a way that Next we will show that so far Simultaneously, where s ∈ [0, 1). To prove (1), it is sufficient to notice that On the other hand, Therefore, Now, using the angle α 1 we define Let us take the same point x as above (see Fig. 2) and notice that We would like to show that 2r sin Let us denote ∠ t2 (T (x), T (T (x))) = β. In the sequel, we will show that Let us notice that, on account of the cosine law, we get Simultaneously, So, it is sufficient to prove the following inequality We know that (see (2) and (3)) sin The angle between the arcs t 1 t 2 and t 2 t 3 is equal to π. Since these two arcs are the subsets of two different two-dimensional spaces, the angle between segments [T (x), t 2 ] and [t 2 , T (T (x))] (with respect to Cartesian geometry) is greater than in the case when all points are located on one two-dimensional space. The same situation can be observed in Fig. 3, where Next, if all points are on one two-dimensional space, then the angle between the metric segments is greater than the angle between two metric segments joining points on the same circle and having the same length 2r sin α 0 2 greater than T (x) − t 2 and t 2 − T (T (x)) . See also points a, a , b and b (Fig. 3). Thus, the angle ∠ t2 (T (x), T (T (x))) is not smaller than β . To estimate β , let us consider the triangle of the sides of length equal to 2r sin α 0 2 , 2r sin α 0 2 and 2r sin 2α 0 2 (all vertices are located on a circle with radius r). Therefore, which completes the proof of (4).
To define the map on {ϕ(t) : t ∈ (2α − α 1 , 2α)}, let us choose Since α 1 < α 0 one may repeat considerations from the previous part to show that T is still INEA. We can also define the map on {ϕ(t) : t ∈ (α, 3α − α 2 )} as a movement along the curve. Repeating all steps with α n+1 = α n · α 1 α 0 , n∈ N one may extend the map T on the whole curve γ. Let us notice that T is also continuous.
Step 3-neighborhood of the curve In this step, we consider the neighborhood of the curve. Mainly, let us consider the set We will see that this set is closed. Indeed, let us take any Cauchy sequence (x n ), x n ∈ U . Let x n →x ∈ B. Since t n − t m ≥ 1 >> α 0 for n = m, without loss of generality we may assume that there is a sequence (τ n ) such that τ n ∈ [mα, (m +1)α] and x n − ϕ(τ n ) ≤ α m . Then, there is a convergent subsequence (denoting again by (τ n )) such that τ n →τ . Ifτ ∈ (mα, (m+1)α], then the same holds for almost all τ n , so x − ϕ(τ ) as the limit is also not greater than α m . Ifτ = mα, then andx is also an element of U .
Step 4-definition of the hyperplanes Now, for each point x ∈ ϕ there is a unique hyperplane We will show that two hyperplanes do not intersect inside U as long as they are determined by points which are not located too far from each other. Let us fix a point x = ϕ(t 0 ) and let x = ϕ(t 0 + τ ), where τ ∈ (0, 9α 0 ). Let us also assume that t 0 ∈ [mα, (m + 1)α). Claim: For all possible positions of x, x , T (x) and T (x ), the angle between vectors x T (x) and x T (x ) is not greater than τ .
• Case 1. First, we assume that all points are located on the curve between t m and t m+1 . So, the aforementioned vectors x T (x) and x T (x ) span one two-dimensional space and it is sufficient to consider only points on this space. Clearly, the intersection of H x (or H x ) with this space is a line-see Fig. 4.
where p is the projection of x onto the set of common points of H x and H x . Clearly, p also belongs to the same two-dimensional space.
Since ∠ x (p, T (x)) = ∠ x (p, T (x )) = π/2, we get that the angle between vectors x T (x) and x T (x ) is equal to the angle ∠ã(x, x ), i.e., is equal to τ . • Case 2. Now, we assume that the three points x, x , T (x) are located on the curve between t m and t m+1 and T (x ) is between t m+1 and t m+2see Fig. 5. Without loss of generality we may assume that τ < α m . Letã be the center of the circle containing t m and t m+1 whileb is the center of the circle containing t m+1 and t m+2 . Then, there is a number s ∈ (0, 1) such that ∠ã(x , t m+1 ) = (1 − s)α m and ∠b(t m+1 , T (x )) = s α m+1 .
Let us choose T on the same circle as t m and t m+1 (Fig. 5) for which Since the curve is smooth and T (x ) does not belong to the same twodimensional space as the rest of the points, the following inequalities hold Clearly, the angle between vectors x T (x) and x T is equal to the sum of angles ∠ T (x) (x, x ) and ∠ x (T (x), T ). Moreover, this sum is smaller than τ , because However, from (6) and the fact that T (x ) does not belong on the same space as the rest of points it follows that the angle between vectors x T (x) and x T (x ) is smaller than the angle between vectors x T (x) and x T and so smaller than τ . The proofs for the cases where τ ≥ α m or three points x , T (x) and T (x ) are between t m+1 and t m+2 go with the same patterns. • Case 3. Now we assume that x and x are located between t m and t m+1 while T (x) and T (x ) are between points t m+1 and t m+2 -see Fig. 6 Let us fix T and T on the circle containing x and x in such a way that and whereã andb are defined in the same way as in Case 2.
We want to show that To do it, let us notice that Hence, the inequality (7) follows directly from the sine law.
In a similar way one may see that  Moreover, since all four points do not belong to the same two-dimensional space, the angle between vectors x T (x) and x T (x ) is smaller than the sum of ∠ T (x) (x, x ) and ∠ x (T (x), T (x )). This completes the proof for Case 3. The case where two points x and T (x) are between t m and t m+1 is slightly easier. Now, we consider the projection of x onto H x . First, we want to estimate the angle ∠ x (x , T (x)). We may consider three cases as it was done in the previous part when we studied the angle between vectors but here we do not need to make the estimation so precise; therefore, only notice that this angle is smaller than the sum of the angle between the vector x x and the curve at the point x and the angle between the vector x T (x) and the curve at the same point x.
In both cases, the angles are of the largest measure if all points x, x and T (x) are located between points t m and t m+1 . Hence, using denotations from Fig. 7, we get Since ∠ã(x, T (x)) ≤ α m and ∠ã(x, x ) = τ ≤ 9α 0 , it follows finally that Now we find the projection of x onto the hyperplane H x and denote this by p x . Since H x is determined by the vector x T (x) and we obtain two estimations: Let p be the projection of p x onto the set H x ∩ H x . Clearly, this set is closed and convex, so the projection is a single, i.e., is well defined. Since vectors x T (x) and p x x are parallel, we can calculate the measure of the angle ∠ x (p x , T (x )) in the following way: where γ denotes the angle between vectors x T (x) and x T (x ) (Fig. 8).
Next, we will show that almost all points of the set U (more precisely, all points from V ) satisfy the following condition: Let us consider the Cauchy sequence of points (x n ) such that x n ∈ U and all of them satisfy this property. As it was shown the limit point x 0 = lim x n belongs to U and we will prove that x 0 also satisfies (P ).
Since (x n ) is a Cauchy sequence, we may take a subsequence with x n − x m ≤ α 0 . Then there must be