“Lion–man” game and the fixed point property in Hilbert spaces

The main goal of the paper is to show that in the setting of a Hilbert space a closed and convex K has the fixed point property for nonexpansive mappings if and only if the lion always wins the Lion–Man game played in K.


Introduction
Multiple mathematical applications can be formulated as geometric versions of the so-called cops and robbers game.The best known version supposes that a single pursuer (the cop, hereinafter referred to as the lion) tries to get close to, and possibly capture, a single evader (the robber or, in our case, the man), where players move in a continuous space, and the lion captures the man if it moves to the exact position of the man.Assuming that the players can move and observe each other at all times, we consider this situation as so-called continuous game.Otherwise, if players move in turns, the case is called a discrete game.In both cases if the game is played in a circular arena the lion always wins.Such a result was clearly shown by Besicovitch or for instance by Alonso et al. (see [2,5]).A collection of additional assumptions under which both players can move clearly modifies the result, as one can observe, for instance, in [9,10].Moreover, for the last fifty years these problems were considered in more general spaces (see for instance [4]), but the main question was still the same: under which assumptions the lion must win?
More recently, researchers interested in the topic have focused on the case of spaces with nonpositive curvature (in the Alexandrov sense).In [1] it was proved that in the discrete game the lion must win if the space is bounded.Then, in [11] the authors showed that for achieving the same goal in CAT(0) spaces, which were also Gromov hyperbolic, it is sufficient to assume that a space is geodesically bounded.Another very simple example of spaces with nonpositive curvature are simply Hilbert spaces.Our main purpose in this paper is to show which geometrical conditions guarantee that the lion wins in Hilbert spaces.
Theorem 1 Let K be a closed and convex subset of a Hilbert space.Then the following statements are equivalent: (i) K is bounded; (ii) the lion must win the discrete "Lion-Man" game, where both players move according to the rules described in the next section.
The paper is organized as follows.In Sect. 2 we introduce some preliminaries and propose the rules of the discrete Lion-Man game.They are the same as the ones considered for instance in [1,3,11].We also present an example of a closed convex and unbounded set in which the man can escape and thus win.The last section contains the main result of our work-the proof of Theorem 1.The proof is based on the construction given in the aforementioned example.Moreover, since in the case of Hilbert spaces we know about the equivalence between the fixed point property and the boundedness of a set [13,17], our result can be read as follows: in the setting of closed and convex subsets of Hilbert spaces, the following are equivalent: the set has the fixed point property, the set is bounded and the lion always wins the discrete Lion-Man game.

Preliminaries
Let (X , d) be a metric space and x, y ∈ X .A geodesic joining x and y is an isometric embedding ı : [0, l] → X such that ı(0) = x and ı(l) = y.This immediately yields l = d(x, y).The image ı([0, l]) of a geodesic ı is called a geodesic segment.A geodesic does not have to be unique, if that is the case we denote its image by [x, y].Moreover, if ı can be extended to [0, ∞), its image is called a ray.X is called a geodesic space if each couple of points is joined by at least one geodesic.
In this paper we are considering the following Lion-Man game in a complete geodesic space X , described in [1,3,11].Let L 0 and M 0 denote the starting positions of the pursuer (the lion) and the evader (the man) respectively and let us fix a maximal length of jump Dthe same for both players.Clearly, it must be d(L 0 , M 0 ) > D. At each step the lion moves from We say that the lion wins the game in one of the following two situations: Otherwise, we claim that the man is the winner.
We begin our considerations with the following example of a block set, showing that in this case the man wins.It is worth emphasizing that a similar approach has been frequently applied elsewhere, for instance to show the almost fixed point property in subsets of Hilbert spaces (see [7] and [13]).More information on the (almost) fixed point property for nonexpansive mappings can be found, for example, in the papers [12,[14][15][16] and in the book [8].
Example 1 Let us consider the block subset of l 2 : when •, • means the scalar product in l 2 or more generally in H .We define the start positions L 0 = (0, 0, . ..), M 0 = (4, 0, . ..) and suppose that D = 1.Then L 1 = (1, 0, . ..) and let us suppose that the man moves in the direction (αx 1 , √ 1 − α 2 x 2 , 0, . ..) chosen in such a way that the lion changes its direction clockwise by π 2 6 .As we will see in a moment, such movement of the man is possible within A. Indeed, let us consider the half line γ , γ (0 . Let M 1 := P. Then L 2 lies between L 1 and M 1 and we get We repeat this situation 2 5 − 1 times.Let us notice that 6 and 2 5 • π } lie on a circle inscribed in the regular 4 • 2 5 -polygon.So the radius R 1 of the circle is equal to and so Let us notice that the points L 2 5 , L 2 5 +1 , M 2 5 lie along a straight line therefore M 2 5 , e 1 = 1 2 + R 1 .In the rest of the steps, i.e., for i ≥ 2 5 , we do not modify the first coordinate.Now let us notice that there are points M i with M i , e 1 > 1/2 + R 1 (for some i ∈ {1, 2, . . ., 2 5 }) but the finite sequence ( L i+1 − M i ) 2 5  i=0 decreases so and on account of 1/t − t < cot t < 1/t for t ∈ (0, 1) we have (2.1) In the next 2 6 steps the man moves in the plane {1/2 + R 1 , e 2 , e 3 , 0, . ..} changing the direction of his movement in such a way that the direction of the movement of the lion changes clockwise by π 2 7 .In the same way as above the points (L i + L i+1 )/2, i = 2 5 , . . ., 2 5 + 2 6 lie on a circle inscribed in the regular 4 • 2 6 -polygon and the radius of the circle is equal to 7 .This guarantees that At same time M i − L i+1 < 3 for i = 2 5 , . . ., 2 5 + 2 6 which leads to Combining it with (2.1) we get that M i ∈ A and L i ∈ A for all i = 0, . . ., 2 5 .Let us notice that for each L i (and M i ), i = 0, . . ., 2 5 + 2 6 we have In the rest of the steps we do not modify the first two coordinates.
Since the block set A lies in an infinite dimensional space, after each 2 5 + 2 6 + . . .+ 2 n movements the man can change the plane in which he moves.That also guarantees that the lion begins to move in the same plane.Moreover, again on account of 1/t − t < cot t < 1/t, t ∈ (0, 1) and the fact that , so all points L i and M i lie in A and To complete our considerations we will show that lim i→∞ L i − M i > 1.Indeed, let us consider the triangle As a direct consequence of the cosine law of the form Combining it with three inequalities: H , h ≤ 4 and h > H − 1 leads to So finally lim L i − M i > 2, which completes the proof of the fact that the man wins.

Main result
In this section, based on Example 1, we will prove the main result of the paper, i.e., Theorem 1.
To do this, we also apply the following result shown by W. O. Ray: Proposition 1 (see [13,Lemma 3] and [13, Corollary 1]) Let K be a closed, convex, linearly bounded and unbounded subset of a Hilbert space H .If 0 ∈ K , then there is an orthonormal set {e n } ⊂ H such that, for each n, n j=1 e j ∈ K and the set m=1 is decreasing sequence} is a nonempty, closed and convex subset of K .
Proof of Theorem 1.First let us notice that although the paper [1] is not correct (see for instance [3] or [11]), the proof of the "only if" part can be deduced from Theorem 8 therein.Thus, if the domain is bounded, the lion must win.We show the "if" part.Assume to the contrary that K is unbounded.Then, let us consider two separate cases: (i) K is linearly bounded; (ii) K contains a ray γ (a half line).
Clearly, in Case (ii) it is sufficient to choose L 0 = γ (0) and M 0 = γ (t) for a fixed t > D.
Then the distance between the lion and the man states constant, so lim n→∞ M n − L n > D and the man is able to escape.Next, we consider Case (i).Without loss of generality we may suppose that 0 H ∈ K .So on account of Proposition 1 there is a nonempty, closed and convex subset K 0 ⊂ K such that Clearly, K 0 is not a block set, however we will show that in K 0 the man can escape in the same way as in the set from Example 1.
Since 0 H ∈ K 0 and k 2 is perpendicular to k 1 we may define L 0 = 0 H and M 0 = 4 • k 1 .During the next 2 5 steps the man can move in the two-dimensional plane {x • k 1 + y • k 2 }.We must only prove that L i , M i ∈ K 0 .But from (2.2) we obtain so from the definition of directions k j : M i , e 1 = M i , e 2 = . . .= M i , e 625 ≥ M i , e 626 = . . .= M i , e 625+16•625 .
Clearly, the same inequalities hold for L i , i = 1, . . ., 2 5 and these guarantee that L i , M i ∈ K 0 for the same i.Further on the man can move in the same way as in Example 1 taking directions k j instead of e j .Again on account of (2.3) we have Let us notice that to prove Theorem 1 for linearly bounded sets we apply the construction given by Ray in [13].In his paper the construction was used to show the geometrical conditions that guarantee the existence of fixed points for nonexpansive mappings.Thus, using this result and our main theorem, we get the following equivalence between the result of the lion-man game and the fixed point property: Corollary 3.1 Let K be a closed and convex subset of a Hilbert space.Then the following statements are equivalent: (i) K is bounded; (ii) the lion must win the discrete "Lion-Man" game; (iii) K has the fixed point property for nonexpansive mappings.
It is also worth emphasizing that the equivalence (ii) − (iii) also holds in spaces with negative curvature in the sense of Alexandrov, although in such spaces the other equivalences are not satisfied, as one may see in [11].Moreover, our main result gives the affirmative answer to the question raised in the same paper.However, it is still an open question whether the same result can be generalized to other subclasses of spaces with nonpositive curvature, for instance the spaces considered in [6].