1 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:

The mapping \(T:C\rightarrow C\) is INE if satisfies

$$\begin{aligned} \Vert T(Tx)-Tx\Vert \le \Vert Tx-x\Vert , \qquad \text{ for } \text{ all } x\in C. \end{aligned}$$

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 \(x \in C\) which is not a diametral one of C, i.e., the following condition holds: \(\sup \{\Vert x-y\Vert : \ y\in C\}< \text{ diam } C\) (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

$$\begin{aligned} \Vert Tx-x\Vert =\Vert T(Tx)-Tx\Vert ; \end{aligned}$$

therefore, they were not asymptotically regular. Let us remind that the self-mapping \(T:C\rightarrow C\) is asymptotically regular if for each \(x\in C\) the sequence \(\Vert T^nx-T^{n+1}x\Vert \) 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 \(\Vert x_n - T (x_n)\Vert \rightarrow 0\) . It is well known that if the self-mapping \(T :C \rightarrow C\) 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 \(\{e_n: \ n\in {\mathbb {N}}\}\). Then let \(\langle \cdot ,\cdot \rangle \) 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 \(\rho \), i.e.,

$$\begin{aligned} \rho (A,B) = \arccos \langle A,B\rangle \end{aligned}$$

for a pair of two elements \(A,B \in S\). By the angle between two curves c and \(\tilde{c}\) (\(c(0)=\tilde{c}(0)\)) on the sphere with respect to spherical geometry we mean the Alexandrov angle, defined by

$$\begin{aligned} \lim _{s,t \rightarrow 0^+} \angle _{c(0)} (c(s),\tilde{c}(t)). \end{aligned}$$

This limit always exists (see, for instance, [8, p. 16]). More details about spherical geometry can be found in [8, 9]).

2 Example

Our example may seem to be rather complicated. So, for the reader’s convenience, we divide its description into six steps.

Step 1—construction of the curve

Let us take an infinite set of points \(\{t_n: \ n\in {\mathbb {N}}\}\), where

$$\begin{aligned} t_n=\dfrac{e_n+e_{n+1}}{\sqrt{2}}, \end{aligned}$$

and first consider the convex hull of \(\{e_1,e_2,e_3\}\) with respect to spherical geometry, i.e., the set

$$\begin{aligned} \text{ conv}_S (\{e_1,e_2,e_3\}) = \left\{ x\in S: \ \langle x,e_1\rangle ,\langle x,e_2\rangle ,\langle x,e_3\rangle \ge 0, \ \langle x,e_n\rangle = 0, n>3\right\} . \end{aligned}$$

Further, we will use the same denotation \(\text{ conv}_S (A)\) for any nonempty set \(A\subset S\).

Now, still with respect to spherical geometry on \(\text{ conv}_S (\{e_1,e_2,e_3\})\), let us notice that \(a=3^{-1/2}(e_1+e_2+e_3) \in \text{ conv}_S (\{e_1,e_2,e_3\})\). 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 \(\pi /2\). 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

$$\begin{aligned} \text{ conv}_S (\{e_1,e_2,e_3\}) \cap S(a,\Vert a-t_1\Vert ), \end{aligned}$$

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 \(\pi /2\), 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).

Fig. 1
figure 1

The construction of the curve

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Now, we repeat our construction on each three-dimensional set \(\text{ conv}_S (\{e_n,e_{n+1},e_{n+2}\})\) and we obtain a smooth curve of infinite length joining all points \(t_n\), \(n=1,2,\ldots \). 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 point \(\tilde{a}=\dfrac{\sqrt{2}}{3\sqrt{3}}(e_n+e_{n+1}+e_{n+2})\) with a radius of \(r=\sqrt{\dfrac{11-4\sqrt{3}}{9}}\) (now with respect to Cartesian geometry). Then, the angle between radiuses \([\tilde{a},t_n]\) and \([\tilde{a},t_{n+1}]\) is independent of n and smaller than \(\pi \).

Let us denote this angle by \(\alpha \) and for all \(t\in [0,\infty )\) we define \(\varphi (t)\) as a point of the curve. If \(t=n \; \alpha + \tau \) then \(\varphi (t)\) is located between \(t_n\) and \(t_{n+1}\) in such a way that \(\angle _{\tilde{a}}(t_n,\varphi (t))=\tau \).

Step 2—definition of the map on the curve

Now, we would like to define the map \(T:\varphi \rightarrow \varphi \). Let us divide the curve between points \(t_1\) and \(t_2\) into 128 equal parts and let \(\alpha _0=\alpha /128\). Then, for points of the form \(\varphi (t)\), \(t \in [0,\alpha -\alpha _0]\), we take \(T(\varphi (t))=\varphi (t+\alpha _0)\). 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 \(\alpha _1\) in such a way that

$$\begin{aligned} \dfrac{\alpha _1}{\alpha _0} = \dfrac{\sin (2 \alpha _0)}{2\alpha _0}. \end{aligned}$$

Hence, for \(t = 127 \alpha _0+\tau \), \(\tau \in (0,\alpha _0)\) we define

$$\begin{aligned} T(\varphi (t)) = \varphi \left( \alpha + \tau \cdot \dfrac{\alpha _1}{\alpha _0}\right) . \end{aligned}$$

Next we will show that so far

$$\begin{aligned} \Vert T(x)-T(T(x))\Vert \le \Vert x-T(x)\Vert \end{aligned}$$
(1)

for \(x=\varphi (t)\), \(t\le 127\alpha _0\).

Fig. 2
figure 2

The construction of the map

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Let us consider \(x=\varphi (t)\), where \(t \in \left( 126 \alpha _0,127 \alpha _0\right] \). Then \(T(x)=\varphi (t+\alpha _0)\) and

$$\begin{aligned} \Vert x-T(x)\Vert =2r \sin \dfrac{\alpha _0}{2}, \qquad r=\Vert \tilde{a}-t_2\Vert . \end{aligned}$$

Simultaneously,

$$\begin{aligned} \Vert T(x)-T(T(x))\Vert \le \Vert T(x)-t_2\Vert +\Vert t_2-T(T(x))\Vert =2r \sin \dfrac{s\alpha _0}{2}+2r\sin \dfrac{(1-s)\alpha _1}{2}, \end{aligned}$$

where \(s\in [0,1)\). To prove (1), it is sufficient to notice that

$$\begin{aligned} \sin \dfrac{\alpha _0}{2} - \sin \dfrac{s\alpha _0}{2}= & {} 2\sin \dfrac{(1-s)\alpha _0}{4}\cos \dfrac{(1+s)\alpha _0}{4}\\= & {} \sin \dfrac{(1-s)\alpha _0}{2}\cdot \dfrac{\cos \frac{(1+s) \alpha _0}{4}}{\cos \frac{(1-s)\alpha _0}{4}} \ge \sin \dfrac{(1-s)\alpha _0}{2}\cdot \cos \dfrac{\alpha _0}{2}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \sin \dfrac{(1-s)\alpha _1}{2} \le \dfrac{(1-s)\alpha _1}{2} \end{aligned}$$
(2)

and

$$\begin{aligned} \sin \dfrac{(1-s)\alpha _0}{2}\ge \dfrac{(1-s)\alpha _0}{2} \cdot \dfrac{\sin \alpha _0}{\alpha _0}. \end{aligned}$$
(3)

Therefore,

$$\begin{aligned} \sin \dfrac{(1-s)\alpha _1}{2}\le & {} \sin \dfrac{(1-s)\alpha _0}{2} \cdot \dfrac{\alpha _1}{\alpha _0}\cdot \dfrac{\alpha _0}{\sin \alpha _0} =\sin \dfrac{(1-s)\alpha _0}{2}\cdot \dfrac{\sin (2\alpha _0)}{2\alpha _0} \cdot \dfrac{\alpha _0}{\sin \alpha _0}\\= & {} \sin \dfrac{(1-s)\alpha _0}{2} \cdot \cos \alpha _0 \le \sin \dfrac{(1-s)\alpha _0}{2} \cdot \cos \dfrac{\alpha _0}{2} \end{aligned}$$

and finally

$$\begin{aligned} 2r \sin \dfrac{(1-s)\alpha _1}{2}+2r \sin \dfrac{s\alpha _0}{2} \le 2r \sin \dfrac{\alpha _0}{2}. \end{aligned}$$

Now, using the angle \(\alpha _1\) we define

$$\begin{aligned} T(\varphi (\alpha +t))=\varphi (\alpha +t+\alpha _1) \text{ as } \text{ long } \text{ as } t \le 2\alpha -\alpha _1. \end{aligned}$$

Let us take the same point x as above (see Fig. 2) and notice that

$$\begin{aligned} \Vert T(T(x))-T(T(T(x)))\Vert = 2r \sin \dfrac{\alpha _1}{2}. \end{aligned}$$

We would like to show that

$$\begin{aligned} 2r \sin \dfrac{\alpha _1}{2} \le \Vert T(x)-T(T(x))\Vert . \end{aligned}$$

Let us denote \(\angle _{t_2}(T(x),T(T(x)))=\beta \). In the sequel, we will show that

$$\begin{aligned} \beta >\pi -\alpha _0. \end{aligned}$$
(4)

Let us notice that, on account of the cosine law, we get

$$\begin{aligned} \Vert T(x)-T(T(x))\Vert ^2= & {} \left( 2r\sin \dfrac{s\alpha _0}{2}\right) ^2 +\left( 2r\sin \dfrac{(1-s)\alpha _1}{2}\right) ^2\\&-2 \cdot \left( 2r\sin \dfrac{s\alpha _0}{2}\right) \cdot \left( 2r\sin \dfrac{(1-s)\alpha _1}{2}\right) \cdot \cos \beta . \end{aligned}$$

Simultaneously,

$$\begin{aligned} \Vert T(T(x))-T(T(T(x)))\Vert= & {} \left( 2r\sin \dfrac{s\alpha _1}{2}\right) ^2 +\left( 2r\sin \dfrac{(1-s)\alpha _1}{2}\right) ^2\\&- 2 \cdot \left( 2r\sin \dfrac{s\alpha _1}{2}\right) \cdot \left( 2r\sin \dfrac{(1-s)\alpha _1}{2}\right) \cdot \cos \pi . \end{aligned}$$

So, it is sufficient to prove the following inequality

$$\begin{aligned} \sin \dfrac{s\alpha _0}{2} \cdot \cos (\pi -\beta ) \ge \sin \dfrac{s\alpha _1}{2}. \end{aligned}$$

We know that (see (2) and (3))

$$\begin{aligned} \dfrac{\sin \dfrac{s\alpha _1}{2}}{\sin \dfrac{s\alpha _0}{2}} \le \dfrac{\dfrac{s\alpha _1}{2}}{\dfrac{s\alpha _0}{2}} \cdot \dfrac{\alpha _0}{\sin \alpha _0} =\dfrac{\sin (2 \alpha _0)}{2\alpha _0} \cdot \dfrac{\alpha _0}{\sin \alpha _0}=\cos \alpha _0. \end{aligned}$$

Thus, showing (4) we complete the proof of the inequality

$$\begin{aligned} \Vert T(T(x))-T(T(T(x)))\Vert \le \Vert T(x)-T(T(x))\Vert . \end{aligned}$$
Fig. 3
figure 3

Estimation of \(\beta \)

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The angle between the arcs \({\mathop {t_1t_2}\limits ^{\frown }}\) and \({\mathop {t_2t_3}\limits ^{\frown }}\) is equal to \(\pi \). 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

$$\begin{aligned} \angle _{O}(a,b) \ge \angle _{O}(a,b^\prime ). \end{aligned}$$

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 \(2r\sin \dfrac{\alpha _0}{2}\) greater than \(\Vert T(x)-t_2\Vert \) and \(\Vert t_2-T(T(x))\Vert \). See also points a, \(a^\prime \), \(b^\prime \) and \(b^{\prime \prime }\) (Fig. 3). Thus, the angle \(\angle _{t_2}(T(x),T(T(x)))\) is not smaller than \(\beta ^\prime \). To estimate \(\beta ^\prime \), let us consider the triangle of the sides of length equal to \(2r\sin \dfrac{\alpha _0}{2}\), \(2r\sin \dfrac{\alpha _0}{2}\) and \(2r\sin \dfrac{2\alpha _0}{2}\) (all vertices are located on a circle with radius r). Therefore,

$$\begin{aligned} \sin \dfrac{\beta ^\prime }{2} =\dfrac{r\sin \dfrac{2\alpha _0}{2}}{2r\sin \dfrac{\alpha _0}{2}} =\cos \dfrac{\alpha _0}{2}, \end{aligned}$$

which completes the proof of (4).

To define the map on \(\left\{ \varphi (t): \ t \in \left( 2\alpha -\alpha _1,2\alpha \right) \right\} \), let us choose

$$\begin{aligned} \alpha _2 = \alpha _1 \cdot \dfrac{\sin (2\alpha _0)}{2\alpha _0}. \end{aligned}$$

Since \(\alpha _1 < \alpha _0\) one may repeat considerations from the previous part to show that T is still INEA. We can also define the map on \(\left\{ \varphi (t): \ t \in \left( \alpha ,3\alpha -\alpha _2 \right) \right\} \) as a movement along the curve. Repeating all steps with

$$\begin{aligned} \alpha _{n+1}=\alpha _n \cdot \dfrac{\alpha _1}{\alpha _0}, \qquad n\in {\mathbb {N}} \end{aligned}$$

one may extend the map T on the whole curve \(\gamma \). 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

$$\begin{aligned} U=\left\{ x\in B: \; \exists t\in [0,\infty ): \Vert \varphi (t)-x\Vert \le \alpha _{m-1}, \text{ if } t\in [(m-1)\alpha ,m \alpha ]\right\} . \end{aligned}$$

We will see that this set is closed. Indeed, let us take any Cauchy sequence \((x_n)\), \(x_n\in U\). Let \(x_n \rightarrow \bar{x}\in B\). Since \(\Vert t_n-t_m\Vert \ge 1>> \alpha _0\) for \(n\ne m\), without loss of generality we may assume that there is a sequence \((\tau _n)\) such that \(\tau _n \in [m\alpha ,(m+1)\alpha ]\) and \(\Vert x_n-\varphi (\tau _n)\Vert \le \alpha _{m}\). Then, there is a convergent subsequence (denoting again by \((\tau _n)\)) such that \(\tau _n \rightarrow \bar{\tau }\). If \(\bar{\tau }\in (m\alpha ,(m+1)\alpha ]\), then the same holds for almost all \(\tau _n\), so \(\Vert \bar{x}-\varphi (\bar{\tau })\Vert \) as the limit is also not greater than \(\alpha _{m}\). If \(\bar{\tau } = m\alpha \), then

$$\begin{aligned} \Vert x_n-\varphi (\tau _n)\Vert \le \alpha _{m} \end{aligned}$$

and so

$$\begin{aligned} \Vert \bar{x}-\varphi (\bar{\tau })\Vert \le \alpha _m<\alpha _{m-1} \end{aligned}$$

and \(\bar{x}\) is also an element of U.

Step 4—definition of the hyperplanes

Now, for each point \(x\in \varphi \) there is a unique hyperplane

$$\begin{aligned} H_x=\{y \in l^2: \ \langle y-x,T(x)-x\rangle =0\}. \end{aligned}$$
(5)

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 \(x=\varphi (t_0)\) and let \(x^\prime =\varphi (t_0+\tau )\), where \(\tau \in (0,9\alpha _0)\). Let us also assume that \(t_0 \in [m\alpha ,(m+1)\alpha )\).

Claim: For all possible positions of \(x,x^\prime , T(x)\) and \(T(x^\prime )\), the angle between vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T(x^\prime )}\) is not greater than \(\tau \).

  • Case 1. First, we assume that all points are located on the curve between \(t_m\) and \(t_{m+1}\). So, the aforementioned vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T(x^\prime )}\) span one two-dimensional space and it is sufficient to consider only points on this space. Clearly, the intersection of \(H_x\) (or \(H_{x^\prime }\)) with this space is a line—see Fig. 4.

    From the equality \(\Vert x-T(x)\Vert =\Vert x^\prime -T(x^\prime )\Vert =\alpha _m\) it follows that

    $$\begin{aligned} \angle _x(p,\tilde{a})=\angle _{x^\prime }(p,\tilde{a}), \end{aligned}$$

    where p is the projection of x onto the set of common points of \(H_x\) and \(H_{x^\prime }\). Clearly, p also belongs to the same two-dimensional space. Since \(\angle _{x}(p,T(x)) =\angle _{x^\prime }(p,T(x^\prime ))=\pi /2\), we get that the angle between vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T(x^\prime )}\) is equal to the angle \( \angle _{\tilde{a}}(x,x^\prime )\), i.e., is equal to \(\tau \).

  • Case 2. Now, we assume that the three points x, \(x^\prime \), T(x) are located on the curve between \(t_m\) and \(t_{m+1}\) and \(T(x^\prime )\) is between \(t_{m+1}\) and \(t_{m+2}\)—see Fig. 5. Without loss of generality we may assume that \(\tau <\alpha _m\).

    Let \(\tilde{a}\) be the center of the circle containing \(t_m\) and \(t_{m+1}\) while \(\tilde{b}\) is the center of the circle containing \(t_{m+1}\) and \(t_{m+2}\). Then, there is a number \(s\in (0,1)\) such that

    $$\begin{aligned} \angle _{\tilde{a}}(x^\prime ,t_{m+1})=(1-s)\alpha _m \quad \text{ and } \quad \angle _{\tilde{b}}(t_{m+1},T(x^\prime ))=s \; \alpha _{m+1}. \end{aligned}$$

    Let us choose \(T^\prime \) on the same circle as \(t_m\) and \(t_{m+1}\) (Fig. 5) for which

    $$\begin{aligned} \angle _{\tilde{a}}(x^\prime ,T^\prime )=(1-s)\alpha _m+s \; \alpha _{m+1}. \end{aligned}$$

    Since the curve is smooth and \(T(x^\prime )\) does not belong to the same two-dimensional space as the rest of the points, the following inequalities hold

    $$\begin{aligned} \Vert x^\prime -T(x^\prime )\Vert> \Vert x^\prime -T^\prime \Vert \quad \text{ and } \quad \angle _{x^\prime }(T(x),T^\prime )> \angle _{x^\prime }(T(x),T(x^\prime )). \end{aligned}$$
    (6)

    Clearly, the angle between vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T^\prime }\) is equal to the sum of angles \(\angle _{T(x)}(x,x^\prime )\) and \(\angle _{x^\prime }(T(x),T^\prime )\). Moreover, this sum is smaller than \(\tau \), because

    $$\begin{aligned} \Vert x-T(x)\Vert >\Vert x^\prime -T^\prime \Vert . \end{aligned}$$

    However, from (6) and the fact that \(T(x^\prime )\) does not belong on the same space as the rest of points it follows that the angle between vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T(x^\prime )}\) is smaller than the angle between vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T^\prime }\) and so smaller than \(\tau \).

    The proofs for the cases where \(\tau \ge \alpha _m\) or three points \(x^\prime \), T(x) and \(T(x^\prime )\) are between \(t_{m+1}\) and \(t_{m+2}\) go with the same patterns.

  • Case 3. Now we assume that x and \(x^\prime \) are located between \(t_m\) and \(t_{m+1}\) while T(x) and \(T(x^\prime )\) are between points \(t_{m+1}\) and \(t_{m+2}\)—see Fig. 6

    Let us fix T and \(T^\prime \) on the circle containing x and \(x^\prime \) in such a way that

    $$\begin{aligned} \angle _{\tilde{a}}(x,T)=\angle _{\tilde{a}}(x,t_{m+1}) +\angle _{\tilde{b}}(t_{m+1},T(x)) \end{aligned}$$

    and

    $$\begin{aligned} \angle _{\tilde{a}}(x^\prime ,T^\prime )=\angle _{\tilde{a}} (x^\prime ,t_{m+1}) + \angle _{\tilde{b}}(t_{m+1},T(x^\prime )), \end{aligned}$$

    where \(\tilde{a}\) and \(\tilde{b}\) are defined in the same way as in Case 2.

    We want to show that

    $$\begin{aligned} \angle _{T(x)}(x,x^\prime ) < \angle _{T}(x,x^\prime ). \end{aligned}$$
    (7)

    To do it, let us notice that

    $$\begin{aligned} \Vert x-T(x)\Vert > \Vert x-T\Vert \end{aligned}$$

    while

    $$\begin{aligned} \angle _{x^\prime }(x,T(x))> \angle _{x^\prime }(x,T) > \dfrac{\pi }{2}. \end{aligned}$$

    Hence, the inequality (7) follows directly from the sine law.

    In a similar way one may see that

    $$\begin{aligned} \angle _{x^\prime }(T(x),T(x^\prime )) < \angle _{x^\prime }(T,T^\prime ). \end{aligned}$$

    Moreover, since all four points do not belong to the same two-dimensional space, the angle between vectors \(\overline{x\; T(x)}\) and \(\overline{x^\prime \; T(x^\prime )}\) is smaller than the sum of \(\angle _{T(x)}(x,x^\prime )\) and \(\angle _{x^\prime }(T(x), T(x^\prime ))\). 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.

Fig. 4
figure 4

Estimation of \(\Vert p-x\Vert \)

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Fig. 5
figure 5

Estimation of angle between vectors

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Fig. 6
figure 6

Estimation of the angle between vectors

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Now, we consider the projection of \(x^\prime \) onto \(H_x\).

First, we want to estimate the angle \(\angle _{x}(x^\prime ,T(x))\). 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 \(\overline{x\;x^\prime }\) and the curve at the point x and the angle between the vector \(\overline{x\;T(x)}\) and the curve at the same point x.

Fig. 7
figure 7

Measure of the angle \(\angle _x{y,z}\)

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In both cases, the angles are of the largest measure if all points \(x,x^\prime \) and T(x) are located between points \(t_m\) and \(t_{m+1}\). Hence, using denotations from Fig. 7, we get

$$\begin{aligned} \angle _x(y,z)=\dfrac{\pi }{2}-\angle _x(y,\tilde{a}) =\dfrac{\angle _{\tilde{a}}(x,y)}{2}. \end{aligned}$$

Since \(\angle _{\tilde{a}}(x,T(x))\le \alpha _m\) and \(\angle _{\tilde{a}}(x,x^\prime )=\tau \le 9\alpha _0\), it follows finally that

$$\begin{aligned} \angle _x(x^\prime ,T(x)) \le 5\alpha _0. \end{aligned}$$

Now we find the projection of \(x^\prime \) onto the hyperplane \(H_x\) and denote this by \(p_x\). Since \(H_x\) is determined by the vector \(\overline{x\;T(x)}\) and

$$\begin{aligned} \overline{x\;T(x)} \ \parallel \ \overline{p_x \; x^\prime }, \end{aligned}$$

we obtain two estimations:

$$\begin{aligned} \Vert x^\prime -p_x\Vert \ge \Vert x-x^\prime \Vert \cdot \cos (5\alpha _0) \ge 2r\sin \dfrac{\tau }{2}\cdot \cos (5\alpha _0) \end{aligned}$$

and

$$\begin{aligned} \Vert x-p_x\Vert \le \Vert x-x^\prime \Vert \cdot \sin (5\alpha _0) \le \tau \cdot \sin (5\alpha _0). \end{aligned}$$

Let p be the projection of \(p_x\) onto the set \(H_x \cap H_{x^\prime }\). Clearly, this set is closed and convex, so the projection is a single, i.e., is well defined. Since vectors \(\overline{x\;T(x)}\) and \(\overline{p_x \; x^\prime }\) are parallel, we can calculate the measure of the angle \(\angle _{x^\prime }(p_x,T(x^\prime ))\) in the following way:

$$\begin{aligned} \angle _{x^\prime }(p_x,T(x^\prime ))=\pi - \gamma , \end{aligned}$$

where \(\gamma \) denotes the angle between vectors \(\overline{x\;T(x)}\) and \(\overline{x^\prime \; T(x^\prime )}\) (Fig. 8).

Fig. 8
figure 8

Measure of the angle \(\angle _x{y,z}\)

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$$\begin{aligned} \angle _{x^\prime }(T(x),p)=\dfrac{\pi }{2}, \end{aligned}$$

we can estimate \(\angle _{x^\prime }(p_x,p)\) by

$$\begin{aligned} \angle _{x^\prime }(p_x,p) = \dfrac{\pi }{2} -\gamma \ge \dfrac{\pi }{2}-\tau . \end{aligned}$$

For all points \(h\in H_x \cap H_{x^\prime }\), we have

$$\begin{aligned} \Vert h-x\Vert \ge \Vert h-p_x\Vert -\Vert p_x-x\Vert \ge \Vert p-p_x\Vert -\Vert p_x-x\Vert . \end{aligned}$$

Simultaneously,

$$\begin{aligned} \Vert p-p_x\Vert = \Vert x^\prime -p_x\Vert \cdot \tan \angle _{x^\prime }(p,p_x) =\Vert x^\prime -p_x\Vert \cdot \cot \gamma \ge \Vert x^\prime -p_x\Vert \cdot \cot \tau . \end{aligned}$$

Combining it with the earlier estimations, we obtain

$$\begin{aligned} \Vert h-x\Vert \ge 2r\sin \dfrac{\tau }{2}\cdot \cos (5\alpha _0) \cdot \cot \tau - \tau \cdot \sin (5\alpha _0). \end{aligned}$$

The minuend is equal to

$$\begin{aligned} r\cdot \dfrac{2\sin \frac{\tau }{2}}{\sin \tau }\cdot \cos (5\alpha _0) \cdot \cos \tau = r \cdot \dfrac{\cos (5\alpha _0)\;\cos \tau }{\cos \frac{\tau }{2}} \ge r \cdot \cos ^2 (9\alpha _0). \end{aligned}$$

In a similar way, we can estimate the subtrahend by

$$\begin{aligned} \tau \cdot \sin (5\alpha _0) \le 45(\alpha _0)^2. \end{aligned}$$

Our considerations finally lead to

$$\begin{aligned}&\Vert h-x\Vert \ge r \cdot \cos ^2 (9\alpha _0) - 45(\alpha _0)^2 \ge \sqrt{\dfrac{11-4\sqrt{3}}{9}} \cdot \cos ^2 (9 \cdot \pi /128) \\&\qquad - 45(\pi /128)^2\ge 0.6 \end{aligned}$$

for all points belonging to \(H_x\cap H_{x^\prime }\).

In this way, we have shown that the hyperplane \(H_x\) and \(H_{x^\prime }\) intersect outside the set U as long as \(\tau \le 9 \alpha _0\). Furthermore, each point of the closed ball \(\bar{B}(x,\alpha _m)\) can belong to at most one hyperplane \(H_y\) with \(|\gamma ^{-1}(y)-\gamma ^{-1}(x)|\le 9 \alpha _0\).

Next, we will show that almost all points of the set U (more precisely, all points from V) satisfy the following condition:

$$\begin{aligned} x \text{ satisfies } (P) \text{ if }\qquad \quad x \in U \wedge \left( \exists t\in [0,\infty ):\ \Vert x-\varphi (t)\Vert \le 4 \alpha _0 \wedge x \in H_{\varphi (t)}\right) . \end{aligned}$$

Let us consider the Cauchy sequence of points \((x_n)\) such that \(x_n \in U\) 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 \(\Vert x_n-x_m\Vert \le \alpha _0\). Then there must be

$$\begin{aligned} \Vert \varphi (s_n)-\varphi (s_m)\Vert \le 9\alpha _0, \end{aligned}$$

where \(x_n \in H_{\varphi (s_n)}\).

So, without loss of generality we may assume that all \(s_n\) belong to \([m\alpha ,(m+1)\alpha ]\) and \(x_0 \not \in \varphi \). Otherwise, \(x_0\in H_{x_0}\) and the proof of the claim is complete. Then there is at least one accumulation point \(s_A\). Let us assume that \(s_{l(n)}\rightarrow s_A\). Since \(x_n \not \rightarrow \varphi (s_A)\) and \(s_{l(n)}\rightarrow s_A\), it must be

$$\begin{aligned} \cos _{\varphi (s_A)}(x_{l(n)},T(\varphi (s_A)))\rightarrow 0. \end{aligned}$$

This leads to

$$\begin{aligned} \cos _{\varphi (s_A)}(x_0,T(\varphi (s_A)))= 0 \end{aligned}$$

and completes the proof of our claim.

However, we will apply our considerations to prove an additional point. Namely, we will prove that the whole sequence \((s_n)\) must tend to \(s_A\). To see it, let us repeat our pattern for \(s_B\)—another accumulation point of \((s_n)\). Let us assume that \(s_{k(n)}\rightarrow s_B\). And again if \(x_0\not \in \varphi \), there must be

$$\begin{aligned} \cos _{\varphi (s_B)}(x_{k(n)},T(\varphi (s_B)))\rightarrow 0, \end{aligned}$$

which yields

$$\begin{aligned} \cos _{\varphi (s_B)}(x_0,T(\varphi (s_B)))= 0, \end{aligned}$$

but \(x_0\) cannot belong to both hyperplanes \(H_{\varphi (s_A)}\) and \(H_{\varphi (s_B)}\), because it is too close to \(\varphi (s_A)\). The proof of uniqueness of accumulation point for \(x_0\in \varphi \) is obvious.

Let us consider a subset V of U containing points u for which

$$\begin{aligned} \langle t_0-T(t_0),u-T(t_0)\rangle \ge 0 \text{ as } \text{ long } \text{ as } \Vert u-t_0\Vert \le 3\alpha _0. \end{aligned}$$

Clearly, V is also closed. We will show that each point \(u\in V\) satisfies the property (P).

Let us fix \(u\in V\). From the inclusion \(u\in U\) it follows that there is a positive number \(\tau \) such that \(\Vert u-\varphi (\tau )\Vert \le \alpha _m\), when \(\tau \in [m\alpha ,(m+1)\alpha ]\). Let us denote \(x=\varphi (\tau )\). Let us assume that \(\langle x-u,x-T(x)\rangle \ne 0\) is a negative number and we consider the line k containing u and parallel to the vector \(\overline{x \; T(x)}\). Let \(y_1=k\cap H_x\).

Taking \(x^\prime =\varphi (\tau -3\alpha _m)\), we obtain that u belongs to the metric segment \([y_0,y_1]\), where \(y_0=k\cap H_{x^\prime }\). Otherwise, the hyperplanes \(H_x\) and \(H_{x^\prime }\) intersect too close to x. For each \(t\in [\tau -3\alpha _m,\tau ]\) one can find a point \(u_t = k\cap H_{\varphi (t)}\). Let \(t_0=\sup \{t\in [\tau -3\alpha _m,\tau ]:\; u \in [u_t,y_1]\}\). It is sufficient to prove that \(u=u_{t_0}\). Indeed, for all \(t>t_0\) there is \(u_t\in [u,y_1]\), which means that \(\langle \varphi (t)-u,\varphi (t)-T(\varphi (t))\rangle \le 0\). And from the continuity of T it follows that \( \langle \varphi (t_0)-u,\varphi (t_0)-T(\varphi (t_0))\rangle \le \), i.e., \(u_{t_0}\in [u,y_1]\). Therefore \(u=u_{t_0}\), i.e., \(u\in H_{\varphi (t_0)}\). Moreover,

$$\begin{aligned} \Vert u-\varphi (t_0)\Vert \le \Vert u-x\Vert +\Vert x-\varphi (t_0)\Vert \le \alpha _m + 3\alpha _m \le 4\alpha _0 \end{aligned}$$

and finally u satisfies the property (P).

Step 5—definition of the map on U

Now, for each \(y\in V\) there is precisely one point x on the curve such that \(y\in H_{x}\), i.e.,

$$\begin{aligned} \langle x-y,x-T(x) \rangle =0 \qquad \text{ and } \qquad \Vert x-y\Vert \le 4\alpha _0. \end{aligned}$$

Hence, since T is INEA on the curve, we obtain

$$\begin{aligned} \Vert y-T(x)\Vert \ge \Vert x-T(x)\Vert \ge \Vert T(x)-T(T(x))\Vert . \end{aligned}$$

So, one can set \(T(y)=T(x)\). Then T is INEA on the whole set V.

Simultaneously, applying the same denotations as in the proof of the closedness of V, if the sequence \((x_n)\) tends to \(x_0\) (i.e., \(x_n \in H_{\varphi (s_n)}\)), then \(x_0\in H_{\varphi (s_0)}\), where \(s_n\rightarrow s_0\). So, \(T(x_n)=T(\varphi (s_n))\) tends to \(T(\varphi (s_0))=T(x_0)\) and T is also continuous on V. Now, we must only consider points from \(\bar{B}(t_0,3\alpha _0)\). Let u be such a point. If

$$\begin{aligned} \langle t_0-u,t_0-T(t_0)\rangle \ge 0, \end{aligned}$$

then T(u) has already been defined. Otherwise, we set \(T(u)=T(t_0)\). Let us notice that T is still continuous and INEA also on the set U.

Step 6—definition of the map on the whole set B

In the previous step, we defined the mapping T on the whole set U. Since this is a closed subset of B and the curve \(\varphi \) is isomorphic to the set \([0,\infty )\), applying the Tietze extension theorem the mapping T can be extended to the whole set B. Therefore, we must show only that the continuous extension \(\tilde{T}\) is also a fixed point free INEA mapping. The fact that \(\tilde{T}\) is fixed point free is obvious. So let us take \(x_0\in B\setminus U\). Then \(\tilde{T}(x_0) \in \varphi \). Since \(\tilde{T}(x_0) = \varphi (t_0)\), one may assume that \(t_0\in [m\alpha ,(m+1)\alpha ]\). Since \(x_0\) does not belong to U, the distance between \(x_0\) and \(\varphi (t_0)\) is bigger than \(\alpha _m\). Simultaneously, \(\Vert T(\varphi (t_0))-\varphi (t_0)\Vert <\alpha _m\). Hence,

$$\begin{aligned} \Vert x_0-\tilde{T}(x_0)\Vert >\Vert \tilde{T}(x_0)-T(\tilde{T}(x_0))\Vert = \Vert \tilde{T}(x_0)-\tilde{T}(\tilde{T}(x_0))\Vert \end{aligned}$$

and \(\tilde{T}\) is INE. Moreover, since \(\tilde{T}|_\varphi =T|_\varphi \), the extension is asymptotically regular and so is INEA. This fact completes the proof.