Construction of fixed points of asymptotically nonexpansive mappings in uniformly convex hyperbolic spaces
aa r X i v : . [ m a t h . M G ] A ug Construction of fixed points of asymptotically nonexpansivemappings in uniformly convex hyperbolic spaces
Andrei Sipo¸s a,b a Department of Mathematics, Technische Universit¨at Darmstadt,Schlossgartenstrasse 7, 64289 Darmstadt, Germany b Simion Stoilow Institute of Mathematics of the Romanian Academy,Calea Grivit¸ei 21, 010702 Bucharest, RomaniaE-mail: [email protected]
Abstract
Kohlenbach and Leu¸stean have shown in 2010 that any asymptotically nonexpansive self-mappingof a bounded nonempty
UCW -hyperbolic space has a fixed point. In this paper, we adapt aconstruction due to Moloney in order to provide a sequence that converges strongly to such a fixedpoint.
Mathematics Subject Classification 2010 : 47H09, 47H10, 47J25.
Keywords:
Hyperbolic spaces, uniformly convex hyperbolic spaces, asymptotically nonexpansivemappings, fixed points.
In 1972, Goebel and Kirk generalized [4] the classical Browder-G¨ohde-Kirk theorem to the class of asymptotically nonexpansive mappings (also introduced in that paper), which are mappings T having theproperty that there is a ( k n ) ⊆ [0 , ∞ ) with lim n →∞ k n = 0 such that for any x , y in the domain of T and any n ∈ N , d ( T n x, T n y ) ≤ (1 + k n ) d ( x, y ) , i.e. they showed that any self-mapping of a bounded closed convex nonempty subset of a uniformlyconvex Banach space with the above property has a fixed point.In the recent decades, there has been a renewed interest in fixed point theory and convex optimizationas practiced in nonlinear generalizations of the classical structures of functional analysis. For example,there exists in the literature a number of definitions of a notion of a ‘hyperbolic space’ [2, 5, 6, 9, 19, 20],that aim to axiomatize the convexity structure of normed spaces. The kind of spaces that we shall employhere have a particularly flexible definition due to Kohlenbach [10] and are called W -hyperbolic spaces (see the next section for a definition and [11, pp. 384–387] for a detailed discussion on the relationshipbetween various definitions of hyperbolicity).Uniform convexity, a property originally due to Clarkson [3], was generalized to this hyperbolicsetting (following [6, p. 105]) by Leu¸stean in [14]. The subclass that is the most natural generalizationof uniformly convex Banach spaces has then been identified with the one having a monotone modulus ofuniform convexity. Those spaces have been called U CW -hyperbolic spaces in [15] (see also [16]), wherethe corresponding Browder-G¨ohde-Kirk result was proven for spaces of this kind which are complete andnonempty. The Goebel-Kirk extension mentioned above for asymptotically nonexpansive mappings wasobtained in the same setting by Kohlenbach and Leu¸stean in [13].Thirty years ago, Moloney showed [17, 18] (by refining an earlier result of Kaniel [8]) how to explicitlyconstruct, for any asymptotically nonexpansive self-mapping of a bounded closed convex nonempty subsetof a uniformly convex Banach space, a sequence that converges strongly to one of its fixed points. Whatwe do in this paper is to show that this construction may be adapted to also work in a bounded nonempty
U CW -hyperbolic space. 1
Facts on hyperbolic spaces
As stated in the Introduction, the following definition is due to Kohlenbach [10].
Definition 2.1. A W -hyperbolic space is a triple ( X, d, W ) where ( X, d ) is a metric space and W : X × [0 , → X such that for all x , y , z , w ∈ X and λ , µ ∈ [0 , , we have that(W1) d ( z, W ( x, y, λ )) ≤ (1 − λ ) d ( z, x ) + λd ( z, y ) ;(W2) d ( W ( x, y, λ ) , W ( x, y, µ ) = | λ − µ | d ( x, y ) ;(W3) W ( x, y, λ ) = W ( y, x, − λ ) ;(W4) d ( W ( x, z, λ ) , W ( y, w, λ )) ≤ (1 − λ ) d ( x, y ) + λd ( z, w ) . Clearly, any normed space is a W -hyperbolic space. As per [10, 14], a particular nonlinear class of W -hyperbolic spaces is the one of CAT(0) spaces, introduced by A. Aleksandrov [1] and named as suchby M. Gromov [7].A subset C of a W -hyperbolic space ( X, d, W ) is called convex if for any x , y ∈ C and λ ∈ [0 , W ( x, y, λ ) ∈ C . If ( X, d, W ) is a W -hyperbolic space, x , y ∈ X and λ ∈ [0 , W ( x, y, λ ) by (1 − λ ) x + λy . We will mainly write x + y for x + y . The following properties areimmediate consequences of the definition of a W -hyperbolic space. Proposition 2.2.
Let ( X, d, W ) be a W -hyperbolic space. Let x , y ∈ X and λ ∈ [0 , . Then we have:(i) x + 0 y = x ;(ii) x + 1 y = y ;(iii) (1 − λ ) x + λx = x ;(iv) d ( x, (1 − λ ) x + λy ) = λd ( x, y ) ;(v) d ( y, (1 − λ ) x + λy ) = (1 − λ ) d ( x, y ) . Definition 2.3. If ( X, d, W ) is a W -hyperbolic space, then a modulus of uniform convexity for ( X, d, W ) is a function η : (0 , ∞ ) × (0 , ∞ ) → (0 , such that for any r , ε > and any a , x , y ∈ X with d ( x, a ) ≤ r , d ( y, a ) ≤ r , d ( x, y ) ≥ εr we have that d (cid:18) x + y (cid:19) ≤ (1 − η ( r, ε )) r. We call the modulus monotone if for any r , s , ε > with s ≤ r , we have η ( r, ε ) ≤ η ( s, ε ) . Definition 2.4. A U CW -hyperbolic space is a quadruple ( X, d, W, η ) where ( X, d, W ) is a W -hyperbolicspace and η is a monotone modulus of uniform convexity for ( X, d, W ) . As remarked in [14, Proposition 2.6], CAT(0) spaces are
U CW -hyperbolic spaces having as a modulusof uniform convexity the function ( r, ε ) ε , quadratic in ε . Note that a closed convex nonempty subsetof a (complete) ( U C ) W -hyperbolic space is itself a (complete) nonempty ( U C ) W -hyperbolic space (incontrast to e.g. normed spaces).The following is an adaptation of a result due to Kohlenbach and Leu¸stean, namely [12, Lemma 3.2]. Proposition 2.5.
Let ( X, d, W, η ) be a U CW -hyperbolic space. Define, for any r , ε > , u ( r, ε ) := ε · η ( r, ε ) . Then, for any r , ε > and any a , x , y ∈ X with d ( x, a ) ≤ d ( y, a ) ≤ r and d ( x, y ) ≥ εr wehave that d (cid:18) x + y , a (cid:19) ≤ d ( y, a ) − u ( r, ε ) r. In addition, if there is a function η ′ which is nondecreasing in its second argument such that for all r and ε , η ( r, ε ) = εη ′ ( r, ε ) (e.g. in the case of CAT(0) spaces, as per the above remark), then one can take u to be simply η . roof. Let r , ε > a , x , y ∈ X be as required. First, note that εr ≤ d ( x, y )2 ≤ d ( x, a ) + d ( y, a )2 ≤ d ( y, a ) , so, using that η is a monotone modulus of uniform convexity, we get that d (cid:18) x + y , a (cid:19) ≤ (1 − η ( d ( y, a ) , ε )) · d ( y, a ) ≤ (1 − η ( r, ε )) · d ( y, a ) ≤ d ( y, a ) − η ( r, a ) εr d ( y, a ) − u ( r, ε ) r. The hypotheses imply that d ( y, a ) = 0, so we may write d ( x, y ) ≥ εr = εrd ( y, a ) · d ( y, a ) , and therefore, in the second case, d (cid:18) x + y , a (cid:19) ≤ (cid:18) − η (cid:18) d ( y, a ) , εrd ( y, a ) (cid:19)(cid:19) · d ( y, a ) = d ( y, a ) − d ( y, a ) · εrd ( y, a ) · η ′ (cid:18) d ( y, a ) , εrd ( y, a ) (cid:19) = d ( y, a ) − εrη ′ (cid:18) d ( y, a ) , εrd ( y, a ) (cid:19) ≤ d ( y, a ) − εrη ′ ( d ( y, a ) , ε )= d ( y, a ) − rη ( d ( y, a ) , ε ) ≤ d ( y, a ) − η ( r, ε ) r. Definition 2.6.
Let ( X, d ) be a metric space, T : X → X and ( k n ) ⊆ [0 , ∞ ) such that lim n →∞ k n = 0 .Then T is called asymptotically nonexpansive with respect to ( k n ) if for any x , y ∈ X and any n ∈ N , d ( T n x, T n y ) ≤ (1 + k n ) d ( x, y ) . For any self-mapping T (of an arbitrary set), we denote the set of its fixed points by F ix ( T ). In [13],Kohlenbach and Leu¸stean have proved that any asymptotically nonexpansive self-mapping of a boundedcomplete nonempty U CW -hyperbolic space has a fixed point.
We fix a complete nonempty
U CW -hyperbolic space (
X, d, W, η ) and b > k n ) ⊆ [0 , ∞ ) be such that lim n →∞ k n = 0 and T : X → X be asymptotically nonexpansivewith respect to ( k n ), so F ix ( T ) = ∅ .We shall construct a mapping S : X → X such that:(i) F ix ( T ) = F ix ( S ) (so F ix ( S ) = ∅ );(ii) for any p ∈ F ix ( S ) and x ∈ X , d ( Sx, p ) ≤ d ( x, p );(iii) for any ( x n ) ⊆ X having x ∈ X as its limit and with lim n →∞ d ( x n , Sx n ) = 0, we have x ∈ F ix ( S ).Note that if T is nonexpansive we may simply take S := T . Lemma 3.1.
Let x ∈ X . Then for any n ∈ N there is an m ∈ { n, n + 1 } such that d ( T m x, x ) ≥
12 + k d ( T x, x ) . Proof.
Let n ∈ N . Assume by way of contradiction that d ( T n x, x ) <
12 + k d ( T x, x )and d ( T n +1 x, x ) <
12 + k d ( T x, x ) . d ( T n +1 x, T x ) ≤ (1 + k ) d ( T n x, x ) < k k d ( T x, x ) , we have that d ( T x, x ) ≤ d ( T n +1 x, x ) + d ( T n +1 x, T x ) <
12 + k d ( T x, x ) + 1 + k k d ( T x, x ) = d ( T x, x ) , a contradiction.Let now x ∈ X . If T x = x , put Sx := x . If T x = x , then put n be minimal such that k n and k n +1 are both smaller or equal than min (cid:18) d ( x, T x )2 b , η (cid:18) b, d ( x, T x )2 b (cid:19)(cid:19) > n exists since lim n →∞ k n = 0). Then, making use of Lemma 3.1, put m ∈ { n, n + 1 } be minimalsuch that d ( T m x, x ) ≥
12 + k d ( T x, x )and set Sx := T m x + x . The following proposition shows that S has all the required properties. Proposition 3.2.
Let x ∈ X , p ∈ F ix ( S ) and ( x n ) ⊆ X . We have that:(i) d ( Sx, x ) ≥ k ) d ( T x, x ) ;(ii) F ix ( T ) = F ix ( S ) ;(iii) d ( Sx, p ) ≤ d ( x, p ) ;(iv) if x is the limit of ( x n ) and lim n →∞ d ( x n , Sx n ) = 0 , we have x ∈ F ix ( S ) .Proof. (i) If T x = x , there is nothing to show. If T x = x , then, by putting m to be the one from theconstruction of Sx , we have that Sx = T m x + x , so d ( Sx, x ) = d ( T m x, x ) ≥ k ) d ( T x, x ).(ii) The inclusion
F ix ( T ) ⊆ F ix ( S ) follows by the construction of S ; F ix ( S ) ⊆ F ix ( T ) follows by (i).(iii) We have that p ∈ F ix ( T ). If T x = x , there is nothing to show. Suppose, then, that T x = x , so,again by putting m to be the one from the construction of Sx , we have that Sx = T m x + x .Put c := k ) d ( T x, x ), q := d ( x, p ) and ε := c b . Since m was chosen such that k m ≤ ε , we havethat b ( k m + ε ) ≤ bε = c. Therefore, since q ≤ b , d ( T m x, x ) ≥ c ≥ b ( k m + ε ) ≥ q ( k m + ε ) , so d ( T m x, x ) − qk m ≥ qε . In addition, using the monotonicity of η and (again) the way m waschosen, η ( q, ε ) ≥ η ( b, ε ) ≥ k m / . If d ( T m x, p ) < q , then d ( Sx, p ) = d (cid:18) T m x + x , p (cid:19) ≤ d ( T m x, p ) + 12 d ( x, p ) ≤ q. Consider now the case where d ( T m x, p ) ≥ q . Then d ( T m x, p ) ≤ (1 + k m ) d ( x, p ) = (1 + k m ) q , sothere is a β ∈ [0 , k m ] with d ( T m x, p ) = (1 + β ) q . Put y := 11 + β T m x + β β p. d ( T m x, y ) = β β d ( T m x, p ) = qβ ≤ qk m , so d ( T m x, x ) ≤ d ( x, y ) + d ( T m x, y ) = d ( x, y ) + qk m and thus d ( x, y ) ≥ d ( T m x, x ) − qk m ≥ qε. On the other hand, we have that d ( x, p ) = q and d ( y, p ) = 11 + β d ( T m x, p ) = q, so, since η is a modulus of uniform convexity, d (cid:18) x + y , p (cid:19) ≤ (1 − η ( q, ε )) q. Then d ( Sx, p ) = d (cid:18) T m x + x , p (cid:19) ≤ d (cid:18) x + y , p (cid:19) + d (cid:18) x + y , x + T m x (cid:19) ≤ (1 − η ( q, ε )) q + 12 d ( T m x, y ) ≤ (cid:18) − k m (cid:19) q + qk m q. (iv) From (i), we get that lim n →∞ d ( x n , T x n ) = 0, then, by the continuity of T , we get x ∈ F ix ( T ) = F ix ( S ).Given S with these properties and x ∈ X , we will now construct a sequence converging to a fixedpoint of S .For any j ≥
1, we shall set an m j ≥ z ij ) m j i =1 , and we shall put y j := z m j j .The sequence ( y j ) will be the one we are after.Put m := 1 and z := x . Assume now that we have constructed the j th finite sequence and we areseeking the next one. We distinguish two cases. Construction case I.
There is an i ∈ [2 , m j −
1] such that d ( z ij , z ( i +1) j ) < d ( z ij , z ( i − j ) and d (cid:16) z ij , z ij + Sz mjj (cid:17) ≥ d ( z ij , z ( i − j ).Let i be minimal with this property. Then put m j +1 := i + 1, put for all k ≤ i , z k ( j +1) := z kj and y j +1 = z ( i +1)( j +1) := z ij + Sz mjj . Construction case II.
There is no such i .In this case, put m j +1 := m j + 1, put for all k ≤ m j , z k ( j +1) := z kj and z ( m j +1)( j +1) := z mjj + Sz mjj .It is immediate that m = 2 and m = 3. By a simple induction, it follows that for all j ≥ m j ≥ Lemma 3.3.
Let p ∈ F ix ( S ) , j ∈ N and i ∈ [1 , m j − . Then d ( z ( i +1) j , p ) ≤ d ( z ij , p ) .Proof. We prove this by induction on j . If j = 1, the property holds vacuously. Suppose now thatthe property holds for j and we want to prove it for j + 1. If i < m j +1 − z ( i +1)( j +1) = z ( i +1) j and z i ( j +1) = z ij , so we simply apply the induction hypothesis. If i = m j +1 −
1, then z i ( j +1) = z ij and z ( i +1)( j +1) = z ij + Sz mjj , so d ( z ( i +1)( j +1) , p ) = d (cid:18) z ij + Sz m j j , p (cid:19) ≤ d ( z ij , p ) + 12 d ( Sz m j j , p ) ≤ d ( z ij , p ) + 12 d ( z m j j , p )5 d ( z ij , p ) + 12 d ( z ij , p ) = d ( z ij , p ) = d ( z i ( j +1) , p ) . Lemma 3.4.
Let p ∈ F ix ( S ) , ε > , j ∈ N and i ∈ [1 , m j − . Let u be such that the property describedby Proposition 2.5 holds. Assume that d ( z ij , z ( i +1) j ) ≥ ε . Then d ( z ij , p ) − d ( z ( i +1) j , p ) ≥ u (cid:0) b, εb (cid:1) b .Proof. We prove this by induction on j . If j = 1, the property holds vacuously. Suppose now thatthe property holds for j and we want to prove it for j + 1. If i < m j +1 − z ( i +1)( j +1) = z ( i +1) j and z i ( j +1) = z ij , so we simply apply the induction hypothesis. If i = m j +1 −
1, then z i ( j +1) = z ij and z ( i +1)( j +1) = z ij + Sz mjj . Using Lemma 3.3, we have that d ( Sz m j j , p ) ≤ d ( z m j j , p ) ≤ d ( z ij , p ) ≤ b. In addition, d ( z ij , Sz m j j ) ≥ d ( z ij , Sz m j j )2 = d ( z ij , z ( i +1)( j +1) ) = d ( z i ( j +1) , z ( i +1)( j +1) ) ≥ ε = εb · b, so, applying Proposition 2.5, d ( z ( i +1)( j +1) , p ) ≤ d ( z i ( j +1) , p ) − u (cid:16) b, εb (cid:17) b. We shall now construct a sequence p < p < . . . such that for any k ≥ m p k = k (so z kp k = y p k )and for all j ≥ p k + 1, m j ≥ k + 1 and z kj = z kp k , and p k is optimal in this regard, i.e. either z k ( p k − is not defined or z k ( p k − = z kp k (which makes it uniquely determined). We shall denote, for all k ≥ x k := y p k . We will also show that for all k , d ( x k , x k +1 ) ≥ d (cid:0) x k , x k + Sx k (cid:1) , i.e. 2 d ( x k , x k +1 ) ≥ d ( x k , Sx k ).It is clear that one must have p := 1 (so x = x ). Assume that we have constructed the sequenceup to p k and we want to find the value of p k +1 .We know that m p k +1 ≥ k + 1, but since m p k +1 ≤ m p k + 1 = k + 1, m p k +1 = k + 1 = m p k + 1. Thusthe ( p k + 1)th line was obtained using Construction case II, so z ( k +1)( p k +1) = x k + Sx k . In the case where for all t ≥ p k + 1, z ( k +1) t = z ( k +1)( p k +1) , in order to simply put p k +1 := p k + 1, wemust also show that for all j ≥ p k + 2, m j ≥ k + 2. Assume that there is a j ≥ p k + 2 with m j < k + 2, i.e. m j = k + 1. Since m j − ≥ k + 1, the j th sequence must have necessarily been obtained via Constructioncase I with i = k , so d ( z k ( j − , z ( k +1)( j − ) < d ( z k ( j − , z ( k − j − ) , (1) d (cid:18) z k ( j − , z k ( j − + Sz m j − ( j − (cid:19) ≥ d ( z k ( j − , z ( k − j − ) , (2)and z ( k +1) j = z k ( j − + Sz mj − j − . Since, by our assumption, z ( k +1) j = z ( k +1)( p k +1) = z ( k +1)( j − , wehave that (2) yields d ( z k ( j − , z ( k +1)( j − ) ≥ d ( z k ( j − , z ( k − j − ) , which contradicts (1). In this case x k +1 = z ( k +1)( p k +1) = x k + Sx k , so d ( x k , x k +1 ) = d (cid:0) x k , x k + Sx k (cid:1) .Assume now that there is a t ≥ p k + 2 with z ( k +1) t = z ( k +1)( p k +1) and take it to be minimal ( aposteriori it will be unique). Then, since m t − ≥ k + 1, we have that the t th sequence must have beenobtained via Construction case I with i = k , so m t = k + 1, d ( z k ( t − , z ( k +1)( t − ) ≥ d ( z k ( t − , z ( k − t − ) (3)and d (cid:18) z k ( t − , z k ( t − + Sz m t − ( t − (cid:19) ≥ d ( z k ( t − , z ( k − t − ) . (4)6or all s ≤ k , t − ≥ p k ≥ p s , so z st = z s ( t − , and since z ( k +1) t = z k ( t − + Sz mt − t − , (4) yields d ( z kt , z ( k +1) t ) ≥ d ( z kt , z ( k − t ) . (5)We will now show that for all j ≥ t + 1, m j ≥ k + 2 and z ( k +1) j = z ( k +1) t , so we may put p k +1 := t . Startwith j := t + 1. Suppose that m t +1 < k + 2, i.e. m t +1 = k + 1. Then the ( t + 1)th sequence must havebeen obtained via Construction case I with i = k , so d ( z kt , z ( k +1) t ) < d ( z kt , z ( k − t ), which contradicts(5). Since then m t +1 ≥ k + 2 = m t + 1, the ( t + 1)th sequence must have been obtained via Constructioncase II, so z ( k +1)( t +1) = z ( k +1) t . Assume now that the property to be proven holds for the j th sequenceand we want to prove it for the next one. By the induction hypothesis, z ( k +1) j = z ( k +1) t . Suppose that m j +1 < k + 2, i.e. m j +1 = k + 1. Then the ( j + 1)th sequence must have been obtained via Constructioncase I with i = k , so d ( z kj , z ( k +1) j ) < d ( z kj , z ( k − j ) . (6)We have that for all s ≤ k , j ≥ p k ≥ p s , so z sj = z st , and since, as stated before, z ( k +1) j = z ( k +1) t , (6)yields that d ( z kt , z ( k +1) t ) < d ( z kt , z ( k − t ), which contradicts (5). Since then m j +1 ≥ k + 2, the ( j + 1)thsequence must have been obtained via Construction case I with i ≥ k + 1 or via Construction case II, so z ( k +1)( j +1) = z ( k +1) j = z ( k +1) t . In addition, by the minimality of t , we get that z ( k +1)( t − = z ( k +1)( p k +1) = x k + Sx k , so (3) yields that d (cid:18) x k , x k + Sx k (cid:19) < d ( x k , x k − ) , while (5) means that d ( x k , x k +1 ) ≥ d ( x k , x k − ) , so d ( x k , x k +1 ) > d (cid:18) x k , x k + Sx k (cid:19) . We have now finished constructing the p k ’s. Lemma 3.5.
Let p ∈ F ix ( S ) and k ∈ N . Then for all n ≥ p k , d ( y n , p ) ≤ d ( x k , p ) .Proof. Let n ≥ p k , so m n ≥ k . By Lemma 3.3, d ( z m n n , p ) ≤ d ( z kn , p ). On the other hand z m n n = y n and since n ≥ p k , z kn = z kp k = x k , so the conclusion follows. Lemma 3.6.
Let p ∈ F ix ( S ) , ε > and i ∈ N . Let u be such that the property described byProposition 2.5 holds. Assume that d ( x i , x i +1 ) ≥ ε . Then d ( x i , p ) − d ( x i +1 , p ) ≥ u (cid:0) b, εb (cid:1) b .Proof. Set j := p i +1 . We have that m j = m p i +1 = i + 1, so i ∈ [1 , m j − z ( i +1) j = z ( i +1) p i +1 = x i +1 and that, since j ≥ p i , z ij = z ip i = x i . Now apply Lemma 3.4. Lemma 3.7.
We have that lim n →∞ d ( x n , x n +1 ) = 0 .Proof. Here is where we use that
F ix ( S ) = ∅ . Let p ∈ F ix ( S ). Assume that our conclusion is false,i.e. there is an ε > N there is an n > N such that d ( x n , x n +1 ) > ε . Denote, for all n , a n := d ( x n , x n +1 ) and c n := d ( x n , p ) − d ( x n +1 , p ). Put d := 1 and for all n ≥
0, put d n +1 > d n such that a d n +1 > ε . In particular, we have that for all i ≥ a d i > ε . Let u be such that the propertydescribed by Proposition 2.5 holds and set k := (cid:24) b +1 u ( b, εb ) b (cid:25) . Applying Lemma 3.6, we get that for all i ≥ c d i ≥ u (cid:0) b, εb (cid:1) b . We may thus write b ≥ d ( x , p ) ≥ d ( x , p ) − d ( x d k +1 , p ) = d k X n =1 c n ≥ k X i =1 c d i ≥ k · u (cid:16) b, εb (cid:17) b ≥ b + 1 , a contradiction. 7 emma 3.8. Let n be such that x n = x n +1 = x n +2 . Then:(i) x n = Sx n ;(ii) for all j ≥ p n +2 and all q ∈ [ n, m j ] , z qj = x n ;(iii) for all j ≥ p n +2 , y j = x n ;(iv) for all q ≥ n , x q = x n .Proof. (i) Put j := p n +2 . Then j > p n +1 , so j − ≥ p n +1 > p n . Then we get that z ( n +1)( j − = z ( n +1) p n +1 = x n +1 and z n ( j − = z np n = x n , so, since x n = x n +1 , d ( z ( n +1)( j − , z n ( j − ) = 0.Assume that the j th sequence was obtained using Construction case I. Then we must have, since m j = n + 2, d ( z ( n +1)( j − , z ( n +2)( j − ) < d ( z ( n +1)( j − , z n ( j − ) = 0, a contradiction. Thus, it wasobtained using Construction case II, so x n +2 = z ( n +2) j = z ( n +1)( j − + Sz ( n +1)( j − x n +1 + Sx n +1 . However, x n +1 = x n +2 , so 0 = d ( x n +1 , x n +2 ) = d ( x n +1 , Sx n +1 ) /
2. Thus, x n +1 = Sx n +1 , i.e. x n = Sx n .(ii) It is clear that this holds for j = p n +2 . Assume that it holds for a j ≥ p n +2 – since then j ≥ p n +1 +1, m j ≥ n + 2 – and we want to prove it for j + 1 – since j + 1 ≥ p n +2 + 1, m j +1 ≥ p n +3 . Thus, the( j +1)th sequence was obtained either using Construction case I with i ≥ n +2 or using Constructioncase II. Let q ∈ [ n, m j ]. Then either z q ( j +1) = z qj = x n (by the induction hypothesis) or z qj = z ( q − j + Sz m j j , but this can happen only if q − ≥ n + 2, so (by the induction hypothesis) z ( q − j = x n and z m j j = x n . Since in addition we know that x n = Sx n , we have that z qj = x n + Sx n x n . (iii) Let j ≥ p n +2 and put q := m j in the above.(iv) Let q ≥ n + 3. Then p q ≥ p n +2 and x q = y p q = x n . Lemma 3.9.
Let n ≥ be such that d ( x n , x n +1 ) < d ( x n , x n − ) . Then for all u ≥ p n +1 , d ( x n , x n − ) ≥ d ( x n , Sy u ) .Proof. Let u ≥ p n +1 , so for all i ∈ [ n − , n + 1], z iu = x i . Assume that 2 d ( x n , x n − ) < d ( x n , Sy u ), so d ( z nu , z ( n − u ) ≤ d ( z nu , Sz m u u ) = d (cid:18) z nu , z nu + Sz m u u (cid:19) . We also know that d ( z nu , z ( n +1) u ) < d ( z nu , z ( n − u ), so the ( u +1)th sequence is obtained by Constructioncase I with i ≤ n , so m u +1 ≤ n + 1. On the other hand u + 1 ≥ p n +1 + 1, so m u +1 ≥ n + 2, acontradiction. Lemma 3.10.
Let n ≥ be such that d ( x n , x n +1 ) < d ( x n , x n − ) . Then for all q ≥ n , d ( x n , x n − ) ≥ d ( x n , x q ) .Proof. Clearly, the conclusion holds for q = n and q = n + 1. Let q ≥ n + 1 and assume that theconclusion holds for q . We want to prove that it also holds for q + 1. Put u := p q +1 −
1. Since u ≥ p q , z qu = x q , so x q +1 = z qu + Sy u x q + Sy u .
8y the induction hypothesis, we have that 2 d ( x n , x n − ) ≥ d ( x n , x q ) and since u ≥ p q ≥ p n +1 , byLemma 3.9 we have that 2 d ( x n , x n − ) ≥ d ( x n , Sy u ), so d ( x n , x q +1 ) ≤ d ( x n , x q ) + 12 d ( x n , Sy u ) ≤ d ( x n , x n − ) . Proposition 3.11.
The sequence ( x n ) is Cauchy.Proof. Let ε >
0. We want an M such that for all n , m ≥ M , d ( x n , x m ) ≤ ε . Put δ := ε . By Lemma 3.7,there is an N such that for all k ≥ N , d ( x k , x k +1 ) ≤ δ . Case I.
We have that x N = x N +1 = x N +2 .Then, by Lemma 3.8, for all q ≥ N , x q = x N , so we may take M := N . Case II.
There is a k ∈ { N, N + 1 } such that x k = x k +1 .Put ρ := d ( x k , x k +1 ) >
0. Again, by Lemma 3.7, there is a p ≥ k + 1 such that d ( x p , x p +1 ) ≤ ρ , sothere is an M ∈ [ k + 1 , p ] such that d ( x M , x M +1 ) < d ( x M , x M − ). Thus, by Lemma 3.10, for all q ≥ M ,2 d ( x M , x M − ) ≥ d ( x M , x q ).On the other hand, since M ≥ k + 1, M − ≥ k ≥ N , so d ( x M , x M − ) ≤ δ and for all q ≥ M , d ( x M , x q ) ≤ δ .Let n , m ≥ M . Then d ( x n , x m ) ≤ d ( x M , x n ) + d ( x M , x m ) ≤ δ = ε .Since X is complete, ( x n ) is convergent. We denote its limit by p . Lemma 3.12.
Let n ≥ be such that d ( x n , x n +1 ) < d ( x n , x n − ) . Then for all q ≥ n + 1 , d ( x n , x n − ) ≥ d ( x n , Sx q ) .Proof. Let q ≥ n +1 and put u := p q ≥ p n +1 . Then x q = y u and the conclusion follows by Lemma 3.9. Lemma 3.13.
We have that lim n →∞ d ( x n , Sx n ) = 0 . Thus, by the establishing properties of S , p ∈ F ix ( S ) = F ix ( T ). Proof of Lemma 3.13.
It follows immediately from Lemma 3.7 and the fact that for all k , 2 d ( x k , x k +1 ) ≥ d ( x k , Sx k ).If we do not want to use the fact that for all k , 2 d ( x k , x k +1 ) ≥ d ( x k , Sx k ), there is also the followingmore complicated argument in the style of the proof of Proposition 3.11. Second proof of Lemma 3.13.
Let ε >
0. We want an M such that for all n ≥ M , d ( x n , Sx n ) ≤ ε . Put δ := ε . By Lemma 3.7, there is an N such that for all k ≥ N , d ( x k , x k +1 ) ≤ δ . Case I.
We have that x N = x N +1 = x N +2 .Then, by Lemma 3.8, for all q ≥ N , x q = x N and Sx q = Sx N = x N , so we may take M := N . Case II.
There is a k ∈ { N, N + 1 } such that x k = x k +1 .Put ρ := d ( x k , x k +1 ) >
0. Again, by Lemma 3.7, there is a p ≥ k + 1 such that d ( x p , x p +1 ) ≤ ρ ,so there is an s ∈ [ k + 1 , p ] such that d ( x s , x s +1 ) < d ( x s , x s − ). Thus, by Lemma 3.10, for all q ≥ s ,2 d ( x s , x s − ) ≥ d ( x s , x q ). In addition, by Lemma 3.12, for all q ≥ s + 1, 2 d ( x s , x s − ) ≥ d ( x s , Sx q ).On the other hand, since s ≥ k + 1, s − ≥ k ≥ N , so d ( x s , x s − ) ≤ δ and for all q ≥ s + 1, d ( x s , x q ) + d ( x s , Sx q ) ≤ δ .Put M := s + 1 and let n ≥ M . Then d ( x n , Sx n ) ≤ d ( x s , x n ) + d ( x s , Sx n ) ≤ δ = ε .9 roposition 3.14. The sequence ( y n ) converges to p .Proof. Let ε >
0. We want an N such that for all n ≥ N , d ( y n , p ) ≤ ε . Let k be such that d ( x k , p ) ≤ ε and put N := p k . Then the conclusion follows by Lemma 3.5.We may also show that ( y n ) is Cauchy without referring to its limit, by first proving the followinganalogue of Lemma 3.10. Lemma 3.15.
Let n ≥ be such that d ( x n , x n +1 ) < d ( x n , x n − ) . Then for all u ≥ p n +1 , d ( x n , x n − ) ≥ d ( x n , y u ) .Proof. If u = p n +1 , y u = x n +1 , so the conclusion holds by our hypothesis. Take an u ≥ p n +1 andassume the conclusion holds for all l ∈ [ p n +1 , u ]. We want to prove it for u + 1. Since u + 1 ≥ p n +1 + 1, m u +1 ≥ n + 2. Set i := m u +1 − ≥ n + 1. Then y u +1 = z iu + Sy u . Set s := min { k ≥ | m u − k ≤ i } . Since m p n +1 = n + 1 ≤ i , we have that s ≤ u − p n +1 . We also have that m u − s ≤ i and m u − s +1 > i (noting, for the edge case s = 0, that m u +1 > i ). Since m u − s ≥ m u − s +1 − > i −
1, we have that m u − s = i . By the minimality of s , for all k ∈ [0 , s ), m u − k > i , i.e. for all k ∈ (0 , s ], m u − s + k > i , sofor all k ∈ [0 , s ], z i ( u − s + k ) = z i ( u − s ) = z m u − s ( u − s ) = y u − s . In particular, for k := s , z iu = y u − s . Put l := u − s . Then, since s ∈ [0 , u − p n +1 ], l ∈ [ p n +1 , u ]. By the induction hypothesis, we have that2 d ( x n , x n − ) ≥ d ( x n , y l ) = d ( x n , z iu ) . In addition, by Lemma 3.9, we have that2 d ( x n , x n − ) ≥ d ( x n , Sy u ) , so d ( x n , y u +1 ) ≤ d ( x n , z iu ) + 12 d ( x n , Sy u ) ≤ d ( x n , x n − ) . Proposition 3.16.
The sequence ( y n ) is Cauchy.Proof. Let ε >
0. We want an M such that for all n , m ≥ M , d ( y n , y m ) ≤ ε . Put δ := ε . By Lemma 3.7,there is an N such that for all k ≥ N , d ( x k , x k +1 ) ≤ δ . Case I.
We have that x N = x N +1 = x N +2 .Then, by Lemma 3.8, for all j ≥ p N +2 , y j = x N , so we may take M := p N +2 . Case II.
There is a k ∈ { N, N + 1 } such that x k = x k +1 .Put ρ := d ( x k , x k +1 ) >
0. Again, by Lemma 3.7, there is a p ≥ k + 1 such that d ( x p , x p +1 ) ≤ ρ , sothere is an s ∈ [ k + 1 , p ] such that d ( x s , x s +1 ) < d ( x s , x s − ). Thus, by Lemma 3.15, for all u ≥ p s +1 ,2 d ( x s , x s − ) ≥ d ( x s , y u ).On the other hand, since s ≥ k + 1, s − ≥ k ≥ N , so d ( x s , x s − ) ≤ δ and for all u ≥ p s +1 , d ( x s , y u ) ≤ δ .Put M := p s +1 . Let n , m ≥ M . Then d ( y n , y m ) ≤ d ( x s , y n ) + d ( x s , y m ) ≤ δ = ε .Since X is complete, ( y n ) is convergent and we again denote its limit by p . Since ( x n ) is a subsequenceof ( y n ), ( x n ) also converges to p , and so, as before, p ∈ F ix ( S ) = F ix ( T ). If we do not want to use thisdetour via the convergence of ( x n ), we must prove the following analogue of Lemma 3.13.10 emma 3.17. We have that lim n →∞ d ( y n , Sy n ) = 0 .Proof. Let ε >
0. We want an M such that for all n ≥ M , d ( y n , Sy n ) ≤ ε . Put δ := ε . By Lemma 3.7,there is an N such that for all k ≥ N , d ( x k , x k +1 ) ≤ δ . Case I.
We have that x N = x N +1 = x N +2 .Then, by Lemma 3.8, for all j ≥ p N +2 , y j = x N and Sy j = Sx N = x N = y j , so we may take M := p N +2 . Case II.
There is a k ∈ { N, N + 1 } such that x k = x k +1 .Put ρ := d ( x k , x k +1 ) >
0. Again, by Lemma 3.7, there is a p ≥ k + 1 such that d ( x p , x p +1 ) ≤ ρ , sothere is an s ∈ [ k + 1 , p ] such that d ( x s , x s +1 ) < d ( x s , x s − ). Thus, by Lemma 3.15, for all u ≥ p s +1 ,2 d ( x s , x s − ) ≥ d ( x s , y u ). In addition, by Lemma 3.9, for all u ≥ p s +1 , 2 d ( x s , x s − ) ≥ d ( x s , Sy u ).On the other hand, since s ≥ k + 1, s − ≥ k ≥ N , so d ( x s , x s − ) ≤ δ and for all u ≥ p s +1 , d ( x s , y u ) + d ( x s , Sy u ) ≤ δ .Put M := p s +1 and let n ≥ M . Then d ( y n , Sy n ) ≤ d ( x s , y u ) + d ( x s , Sy u ) ≤ δ = ε . This work has been supported by the German Science Foundation (DFG Project KO 1737/6-1).
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