Best approximation mappings in Hilbert spaces
BBest approximation mappings in Hilbert spaces
Heinz H. Bauschke ∗ , Hui Ouyang † , and Xianfu Wang ‡ June 2, 2020
Abstract
The notion of best approximation mapping (BAM) with respect to a closed affine subspace in finite-dimensionalspace was introduced by Behling, Bello Cruz and Santos to show the linear convergence of the block-wisecircumcentered-reflection method. The best approximation mapping possesses two critical properties of thecircumcenter mapping for linear convergence.Because the iteration sequence of BAM linearly converges, the BAM is interesting in its own right. In thispaper, we naturally extend the definition of BAM from closed affine subspace to nonempty closed convexset and from R n to general Hilbert space. We discover that the convex set associated with the BAM mustbe the fixed point set of the BAM. Hence, the iteration sequence generated by a BAM linearly converges tothe nearest fixed point of the BAM. Connections between BAMs and other mappings generating convergentiteration sequences are considered. Behling et al. proved that the finite composition of BAMs associated withclosed affine subspaces is still a BAM in R n . We generalize their result from R n to general Hilbert space andalso construct a new constant associated with the composition of BAMs. This provides a new proof of the linearconvergence of the method of alternating projections. Moreover, compositions of BAMs associated with generalconvex sets are investigated. In addition, we show that convex combinations of BAMs associated with affinesubspaces are BAMs. Last but not least, we connect BAM with circumcenter mapping in Hilbert spaces. Primary 90C25, 41A50, 65B99; Secondary 46B04, 41A65.
Keywords:
Best approximation mapping, linear convergence, fixed point set, best approximation problem, projector, cir-cumcentered isometry method, circumcentered reflection method, method of alternating projections.
Throughout this paper, we shall assume that H is a real Hilbert space,with inner product (cid:104)· , ·(cid:105) and induced norm (cid:107) · (cid:107) , N = {
0, 1, 2, . . . } and m ∈ N (cid:114) { } .In [10], Behling, Bello Cruz and Santos introduced the circumcentered Douglas-Rachford method, whichis a special instance of the circumcentered-reflection method (C-RM) and the first circumcentered isometrymethod in the literature. Then the same authors contributed [11], [12] and [13] on C-RMs. In [12], in order toprove the linear convergence of the block-wise C-RM that is the sequence of iterations of finite composition ofcircumcentered-reflection operators, they introduced the best approximation mapping (BAM) and proved thatthe finite composition of BAMs is still a BAM. Our paper is inspired by [12], and we provide the following mainresults: R1:
Proposition 3.10 states that the sequence of iterations of BAM solves the best approximation problemassociated with the fixed point set of the BAM.
R2:
Theorem 4.4 generalizes [12, Theorem 1] and shows that the finite composition of BAMs associated withclosed affine subspaces in Hilbert space is a BAM. It also provides a new constant associated with thecomposition of BAMs. In fact, we provide examples showing that our new constant is independent withthe one constructed in [12, Lemma 1]. In particular, as a corollary of the Theorem 4.4, in Corollary 5.12(i)we show the linear convergence of the method of alternating projections (MAP). ∗ Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail: [email protected] . † Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail: [email protected] . ‡ Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail: [email protected] . a r X i v : . [ m a t h . O C ] J un Theorems 5.4 and 5.10 use two different methods to show that the convex combination of finitely manyBAMs associated with affine subspaces is a BAM.
R4:
Theorems 6.26 to 6.28 show linear convergence of the iteration sequences generated from composition andconvex combination of circumcenter mappings in Hilbert spaces.The paper is organized as follows. In Section 2, we present some auxiliary results to be used in the sequel.Section 3 includes definition and properties of the BAM in Hilbert spaces. In particular, the comparisons: BAMvs convergent mapping, BAM vs Banach contraction, and BAM vs linear regular operator are provided. InSection 4, we generalize results shown in [12, Section 2] from R n to the general Hilbert space and show that thefinite composition of BAMs with closed and affine fixed point sets in Hilbert space is still a BAM. In addition,compositions of BAMs associated with general convex sets are considered in Section 4 as well. In Section 5,we use two methods to show that the convex combination of finitely many BAMs with closed and affine fixedpoint sets is a BAM. In Section 6, we review definitions and facts on circumcenter mapping and circumcen-tered isometry methods. We also provide sufficient conditions for the circumcenter mapping to be a BAM inHilbert spaces. Moreover, we show linear convergence of sequences generated from composition and convexcombination of circumcenter mappings as BAMs in Hilbert spaces.We now turn to the notation used in this paper. Let C be a nonempty subset of H . The orthogonal complement of C is the set C ⊥ : = { x ∈ H | (cid:104) x , y (cid:105) = y ∈ C } . C is an affine subspace of H if C (cid:54) = ∅ and ( ∀ ρ ∈ R ) ρ C + ( − ρ ) C = C . The smallest affine subspace of H containing C is denoted by aff C and called the affine hull of C . An affine subspace C is said to be parallel to an affine subspace M if C = M + a for some a ∈ H . Supposethat C is a nonempty closed convex subset of H . The projector (or projection operator ) onto C is the operator,denoted by P C , that maps every point in H to its unique projection onto C . R C : = C − Id is the reflector associated with C . Moreover, ( ∀ x ∈ H ) d C ( x ) : = min c ∈ C (cid:107) x − c (cid:107) = (cid:107) x − P C x (cid:107) . Let x ∈ H and ρ ∈ R ++ .Denote the ball centered at x with radius ρ as B [ x ; ρ ] .Let T : H → H be an operator. The fixed point set of the operator T is denoted by Fix T , i.e., Fix T : = { x ∈H | Tx = x } . Denote by B ( H ) : = { T : H → H : T is bounded and linear } . For every T ∈ B ( H ) , the operatornorm (cid:107) T (cid:107) of T is defined by (cid:107) T (cid:107) : = sup (cid:107) x (cid:107)≤ (cid:107) Tx (cid:107) .For other notation not explicitly defined here, we refer the reader to [3]. Projections and Friedrichs angle
Fact 2.1 [3, Proposition 3.19]
Let C be a nonempty closed convex subset of the Hilbert space H and let x ∈ H . SetD : = z + C, where z ∈ H . Then P D x = z + P C ( x − z ) . Fact 2.2 [15, Theorems 5.8]
Let M be a closed linear subspace of H . Then Id = P M + P M ⊥ . Note that the case in which M and N are linear subspaces in the following result has already been shown in[15, Lemma 9.2]. Lemma 2.3
Let M and N be closed affine subspaces of H with M ∩ N (cid:54) = ∅ . Assume M ⊆ N or N ⊆ M. Then P M P N = P N P M = P M ∩ N .Proof. Let z ∈ M ∩ N . By [18, Theorem 1.2], the parallel linear subspaces of M and N are par M = M − z andpar N = N − z respectively. By assumption, M ⊆ N or N ⊆ M , we know, par M ⊆ par N or par N ⊆ par M .Then by Fact 2.1 and [15, Lemma 9.2], for every x ∈ H , P M P N x = z + P par M ( P N ( x ) − z ) = z + P par M ( z + P par N ( x − z ) − z ) = z + P par M P par N ( x − z ) = z + P par M ∩ par N ( x − z ) = P M ∩ N x , which implies that P M P N = P M ∩ N . The proof of P N P M = P M ∩ N is similar. (cid:4) Definition 2.4 [15, Definition 9.4] The
Friedrichs angle between two linear subspaces U and V is the angle α ( U , V ) between 0 and π whose cosine, c ( U , V ) : = cos α ( U , V ) , is defined by the expression c ( U , V ) : = sup {|(cid:104) u , v (cid:105)| : u ∈ U ∩ ( U ∩ V ) ⊥ , v ∈ V ∩ ( U ∩ V ) ⊥ , (cid:107) u (cid:107) ≤ (cid:107) v (cid:107) ≤ } . Fact 2.5 [15, Theorem 9.35]
Let U and V be closed linear subspaces of H . Then the following statements are equivalent. (i) c ( U , V ) < . (ii) U + V is closed. (iii) U ⊥ + V ⊥ is closed. onexpansive operators Definition 2.6 [3, Definition 4.1] Let D be a nonempty subset of H and let T : D → H . Then T is(i) nonexpansive if it is Lipschitz continuous with constant 1, i.e., ( ∀ x ∈ D ) ( ∀ y ∈ D ) (cid:107) Tx − Ty (cid:107) ≤ (cid:107) x − y (cid:107) ;(ii) quasinonexpansive if ( ∀ x ∈ D ) ( ∀ y ∈ Fix T ) (cid:107) Tx − y (cid:107) ≤ (cid:107) x − y (cid:107) ;(iii) and strictly quasinonexpansive if ( ∀ x ∈ D (cid:114) Fix T ) ( ∀ y ∈ Fix T ) (cid:107) Tx − y (cid:107) < (cid:107) x − y (cid:107) . Definition 2.7 [3, Definition 4.33] Let D be a nonempty subset of H , let T : D → H be nonexpansive, and let α ∈ ]
0, 1 [ . Then T is averaged with constant α , or α -averaged for short, if there exists a nonexpansive operator R : D → H such that T = ( − α ) Id + α R . Lemma 2.8
Let T : H → H be affine operator with
Fix T (cid:54) = ∅ . Then T is quasinonexpansive if and only if T isnonexpansive.Proof. By Definition 2.6, T is nonexpansive implies that T is quasinonexpansive. Suppose that T is quasinonex-pansive. Because Fix T (cid:54) = ∅ , take z ∈ Fix T . Define ( ∀ x ∈ H ) F ( x ) : = T ( x + z ) − z . (2.1)Then by [9, Lemma 3.8], F is linear. Because T is quasinonexpansive, ( ∀ x ∈ H ) (cid:107) Fx (cid:107) = (cid:107) T ( x + z ) − z (cid:107) ≤ (cid:107) ( x + z ) − z (cid:107) = (cid:107) x (cid:107) ,which, by the linearity of F , implies that ( ∀ x ∈ H )( ∀ y ∈ H ) (cid:107) Fx − Fy (cid:107) ≤ (cid:107) x − y (cid:107) . (2.2)Now, for every x ∈ H and for every y ∈ H , (cid:107) Tx − Ty (cid:107) (2.1) = (cid:107) z + F ( x − z ) − ( z + F ( y − z )) (cid:107) = (cid:107) F ( x − z ) − F ( y − z ) (cid:107) (2.2) ≤ (cid:107) x − y (cid:107) ,which means that T is nonexpansive. (cid:4) The best approximation mapping with respect to a closed affine subspaces in R n was introduced by Behling,Bello-Cruz and Santos in [12]. In this section, we extend the definition of BAM from closed affine subspace tononempty closed convex set, and from R n to general Hilbert space. Moreover, we provide some examples andproperties of the generalized version of BAM. Definition of BAM
Definition 3.1
Let G : H → H , and let γ ∈ [
0, 1 [ . Then G is a best approximation mapping with constant γ (forshort γ -BAM), if(i) Fix G is a nonempty closed convex subset of H ,(ii) P Fix G G = P Fix G , and(iii) ( ∀ x ∈ H ) (cid:107) Gx − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) .In particular, if γ is unknown or not necessary to point out, we just say that G is a BAM.The following Lemma 3.2(ii) illustrates that in [12, Definition 2], the set C is uniquely determined by theoperator G , and that, moreover, C = Fix G . Hence, our Definition 3.1 is indeed a natural generalization of [12,Definition 2]. 3 emma 3.2 Let G : H → H , let C be a nonempty closed convex subset of H , and let γ ∈ [
0, 1 [ . Suppose that P C G = P C and that ( ∀ x ∈ H ) (cid:107) Gx − P C x (cid:107) ≤ γ (cid:107) x − P C x (cid:107) . (3.1) Then the following hold: (i) G P C = P C . (ii) Fix G = C. (iii) G is a γ -BAM.Proof. (i): For every y ∈ H , use the idempotent property of P C and apply (3.1) with x = P C y to obtain that (cid:107) G P C y − P C y (cid:107) = (cid:107) G P C y − P C P C y (cid:107) ≤ γ (cid:107) P C y − P C P C y (cid:107) = ( ∀ y ∈ H ) G P C y = P C y , that is, G P C = P C .(ii): Let x ∈ H . On the one hand, by (i), x ∈ C ⇒ x = P C x = G P C x = Gx ⇒ x ∈ Fix G . On the other hand, x ∈ Fix G ⇒ x = Gx ⇒ (cid:107) x − P C x (cid:107) = (cid:107) Gx − P C x (cid:107) ≤ γ (cid:107) x − P C x (cid:107) ⇒ x − P C x = ⇒ x ∈ C , where the secondand third implications are from (3.1), and γ < (cid:4) Proposition 3.3
Let γ ∈ [
0, 1 [ . Suppose that G is a γ -BAM. Then d Fix G ◦ G ≤ γ d Fix G .Proof. Let x ∈ H . By Definition 3.1(i), Fix G is a nonempty closed convex set, so d Fix G is well defined. Moreover,by Definition 3.1(ii)&(iii),d Fix G ( Gx ) = (cid:107) Gx − P Fix G Gx (cid:107) = (cid:107) Gx − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) = γ d Fix G x . (cid:4) Example 3.4
Let C be a nonempty closed convex subset of H . Then for every γ ∈ [
0, 1 [ , ( − γ ) P C + γ Id is a γ -BAM with Fix G = C . Moreover, Id is a 0-BAM with Fix Id = H . Proof.
Let γ ∈ [
0, 1 [ . Then by [3, Proposition 3.21], P C (( − γ ) P C + γ Id ) = P C . In addition, (cid:107) ( − γ ) P C x + γ x − P C x (cid:107) = γ (cid:107) x − P C x (cid:107) . The last assertion is clear from definitions. (cid:4) Remark 3.5
Let C be a nonempty closed convex subset of H and let γ ∈ R .(i) Because ( ∀ x ∈ H ) (cid:107) ( − γ ) P C x + γ x − P C x (cid:107) = | γ |(cid:107) x − P C x (cid:107) , and | γ | < ⇔ γ ∈ ] −
1, 1 [ , by Defini-tion 3.1(iii), we know that ( − γ ) P C + γ Id is a BAM implies that γ ∈ ] −
1, 1 [ .(ii) Let (cid:101) ∈ R ++ . Suppose that H = R , C : = B [
0; 1 ] and γ : = − (cid:101) . Let x : = ( + (cid:101) , 0 ) . ThenP C (( − γ ) P C + γ Id ) x = (cid:40) ( − (cid:101) , 0 ) if (cid:101) ≤ √ ( −
1, 0 ) if (cid:101) > √ C (( − γ ) P C + γ Id ) x (cid:54) = (
1, 0 ) = P C x , which yields that ( − γ ) P C + γ Id is not aBAM.Hence, using the two items above, we conclude that if ( − γ ) P C + γ Id is a BAM, then γ ∈ ] −
1, 1 [ and thatgenerally if γ ∈ ] −
1, 0 ] , then ( − γ ) P C + γ Id is not a BAM. Therefore, the assumption in Example 3.4 is tight.
Example 3.6
Suppose that H = R n . Let T : H → H be α -averaged with α ∈ ]
0, 1 [ and let T be linear. Then (cid:107) T P ( Fix T ) ⊥ (cid:107) ∈ [
0, 1 [ and T is a (cid:107) T P ( Fix T ) ⊥ (cid:107) -BAM. Proof.
The items (i), (ii) and (iii) in Definition 3.1 follow from [4, Lemmas 3.12 and 3.14] and [9, Proposition 2.22]respectively. (cid:4)
It is easy to see that − Id is linear and nonexpansive but not a BAM. Hence, the condition “ T is α -averaged”in Example 3.6 can not be replaced by “ T is nonexpansive”.4 roposition 3.7 Let T : H → H be a Banach contraction on H , say, there exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H )( ∀ y ∈ H ) (cid:107) Tx − Ty (cid:107) ≤ γ (cid:107) x − y (cid:107) . (3.2) Then T is a γ -BAM.Proof. By [3, Theorem 1.50(i)], Fix T is a singleton, say Fix T = { z } for some z ∈ H . Let x ∈ H . Then P Fix T Tx = z = P Fix T x , which implies that P Fix T T = P Fix T . Moreover, (cid:107) Tx − P Fix T x (cid:107) = (cid:107) Tx − z (cid:107) = (cid:107) Tx − Tz (cid:107) (3.2) ≤ γ (cid:107) x − z (cid:107) = γ (cid:107) x − P Fix T x (cid:107) . Altogether, T is a γ -BAM. (cid:4) Remark 3.8 (i) Proposition 3.7 illustrates that every Banach contraction is a BAM.(ii) Note that a contraction must be continuous. By Example 6.17 below, a BAM (even with fixed point setbeing singleton) is generally not continuous. Hence, we know that a BAM is generally not a contractionand that the converse of Proposition 3.7 fails.
Proposition 3.9
Let A ∈ R n × n be a normal matrix. Denote by ρ ( A ) the spectral radius of A, i.e., ρ ( A ) : = max {| λ | : λ is an eigenvalue of A } .(i) Suppose one of the following holds: (a) ρ ( A ) < . (b) ρ ( A ) = , where λ = is the only eigenvalue on the unit circle and semisimple.Then A is a BAM. (ii) The following are equivalent: (a) lim k → ∞ A k exists. (b) lim k → ∞ A k = P Fix A . (c) A is a BAM.Proof. (i): If ρ ( A ) <
1, then by [3, Example 2.19], A is a Banach contraction. Hence, by Proposition 3.7, A is aBAM.Suppose that ρ ( A ) = λ = A on the unit circle and semisimple. Then by theSpectral Theorem for Diagonalizable Matrices [17, page 517] and Properties of Normal Matrices [17, page 548], A = P U + λ P U + · · · + λ k P U k ,where σ ( A ) = { λ , λ , . . . , λ k } with λ = A and ( ∀ i ∈ {
1, . . . , k } ) U i : = ker ( A − λ i Id ) .Then clearly Fix A = ker ( A − Id ) = U . Moreover, by the Spectral Theorem for Diagonalizable Matrices [17,page 517] again, it is easy to see that P Fix A A = P Fix A , ( ∀ x ∈ R n ) (cid:107) Ax − P Fix A x (cid:107) ≤ | λ |(cid:107) x − P Fix A x (cid:107) ,where | λ | <
1. Therefore, A is a BAM.(ii): By the Theorem of Limits of Powers [17, Page 630], lim k → ∞ A k exists if and only if ρ ( A ) < ρ ( A ) = λ = A on the unit circle and semisimple, which implies that lim k → ∞ A k = P Fix A . Moreover, by Definition 3.1, A being a BAM implies that lim k → ∞ A k = P Fix A . Combine these results with(i) to obtain (ii). (cid:4) Properties of BAM
The following Proposition 3.10(ii) states that any sequence of iterates of a BAM must linearly converge to thebest approximation onto the fixed point set of the BAM. Therefore, we see the importance of the study of BAMs.The following Proposition 3.10 reduces to [12, Proposition 1] when H = R n and Fix G is an affine subspaceof R n . In fact, there is little difficulty to extend the space from R n to H and the related set from closed affinesubspace to nonempty closed convex set. 5 roposition 3.10 Let γ ∈ [
0, 1 [ and let G : H → H . Suppose that G is a γ -BAM. Then for every k ∈ N , (i) P Fix G G k = P Fix G , and (ii) ( ∀ x ∈ H ) (cid:107) G k x − P Fix G x (cid:107) ≤ γ k (cid:107) x − P Fix G x (cid:107) .Consequently, for every x ∈ H , ( G k x ) k ∈ N converges to P Fix G x with a linear rate γ .Proof. Because G is a γ -BAM, by Definition 3.1, we have that Fix G is a nonempty closed and convex subset of H , and that P Fix G G = P Fix G , (3.3a) ( ∀ y ∈ H ) (cid:107) Gy − P Fix G y (cid:107) ≤ γ (cid:107) y − P Fix G y (cid:107) . (3.3b)We argue by induction on k . It is trivial that (i) and (ii) hold for k =
0. Assume (i) and (ii) are true for some k ∈ N , that is, P Fix G G k = P Fix G , (3.4a) ( ∀ y ∈ H ) (cid:107) G k y − P Fix G y (cid:107) ≤ γ k (cid:107) y − P Fix G y (cid:107) . (3.4b)Let x ∈ H . Now P Fix G G k + x = P Fix G G ( G k x ) (3.3a) = P Fix G ( G k x ) (3.4a) = P Fix G .Moreover, (cid:107) G k + x − P Fix G x (cid:107) (3.4a) = (cid:107) G ( G k x ) − P Fix G ( G k x ) (cid:107) (3.3b) ≤ γ (cid:107) G k x − P Fix G ( G k x ) (cid:107) (3.4a) = γ (cid:107) G k x − P Fix G x (cid:107) (3.4b) ≤ γ k + (cid:107) x − P Fix G x (cid:107) .Hence, the proof is complete by the principle of mathematical induction. (cid:4) Proposition 3.11
Let T : H → H be quasinonexpansive with
Fix
T being a closed affine subspace of H . Let γ ∈ [
0, 1 [ .Suppose that ( ∀ x ∈ H ) (cid:107) Tx − P Fix T x (cid:107) ≤ γ (cid:107) x − P Fix T x (cid:107) . Then T is a γ -BAM.Proof. By assumptions and Definition 3.1, it remains to prove P
Fix T T = P Fix T .Let x ∈ H . By [3, Example 5.3], T is quasinonexpansive and Fix T (cid:54) = ∅ imply that ( T k x ) k ∈ N is Fej´er monotonewith respect to Fix T . This, the assumption that Fix T is a closed affine subspace, and [3, Proposition 5.9(i)] implythat ( ∀ k ∈ N ) P Fix T T k x = P Fix T x ,which yields P Fix T T = P Fix T when k = (cid:4) The following result shows further connection between BAMs and linear convergent mappings.
Corollary 3.12
Let T : H → H be quasinonexpansive with
Fix
T being a closed affine subspace of H . Let γ ∈ [
0, 1 [ .Then T is a γ -BAM if and only if ( ∀ k ∈ N ) ( ∀ x ∈ H ) (cid:107) T k x − P Fix T x (cid:107) ≤ γ k (cid:107) x − P Fix T x (cid:107) .Proof. “ ⇒ ”: This is clearly from Proposition 3.10.“ ⇐ ”: This comes from the assumptions and Proposition 3.11. (cid:4) The following result states that BAM with closed affine fixed point set is strictly quasinonexpansive. Inparticular, the inequality shown in Proposition 3.13(i) is interesting on its own.
Proposition 3.13
Let G : H → H with
Fix
G being a closed affine subspace of H . Let γ ∈ [
0, 1 [ . Suppose that G is a γ -BAM. The the following hold: (i) ( ∀ x ∈ H ) ( ∀ y ∈ Fix G ) (cid:107) Gx − y (cid:107) + ( − γ ) (cid:107) x − P Fix G ( x ) (cid:107) ≤ (cid:107) x − y (cid:107) . (ii) G is strictly quasinonexpansive. roof. (i): Because G is a γ -BAM, by Definition 3.1,P Fix G G = P Fix G , (3.5a) ( ∀ x ∈ H ) (cid:107) Gx − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) . (3.5b)Because Fix G is a closed affine subspace of H , by [7, Proposition 2.10], for every x ∈ H and y ∈ Fix G , (cid:107) Gx − y (cid:107) = (cid:107) Gx − P Fix G ( Gx ) (cid:107) + (cid:107) P Fix G ( Gx ) − y (cid:107) (3.6a) (3.5a) = (cid:107) Gx − P Fix G ( x ) (cid:107) + (cid:107) P Fix G ( x ) − y (cid:107) (3.6b) (3.5b) ≤ γ (cid:107) x − P Fix G ( x ) (cid:107) + (cid:107) P Fix G ( x ) − y (cid:107) (3.6c)and, by [7, Proposition 2.10] again, (cid:107) x − y (cid:107) = (cid:107) x − P Fix G ( x ) (cid:107) + (cid:107) P Fix G ( x ) − y (cid:107) . (3.7)Combine (3.6) with (3.7) to see that (cid:107) Gx − y (cid:107) − (cid:107) x − y (cid:107) ≤ ( γ − ) (cid:107) x − P Fix G ( x ) (cid:107) , (3.8)which yields (i).(ii): Because ( ∀ x ∈ H (cid:114) Fix G ) , (cid:107) x − P Fix G x (cid:107) > γ ∈ [
0, 1 [ , by (3.8), ( ∀ x ∈ H (cid:114) Fix G )( ∀ y ∈ Fix G ) (cid:107) Gx − y (cid:107) < (cid:107) x − y (cid:107) .Hence, by Definition 2.6(iii), we obtain that G is strictly quasinonexpansive. (cid:4) Corollary 3.14
Let G : H → H be an affine BAM. Then G is nonexpansive.Proof.
By Definition 3.1(i), G is a BAM yields that Fix G is a nonempty closed and convex subset of H . Moreover,because G is affine, ( ∀ x ∈ Fix G )( ∀ y ∈ Fix G )( ∀ α ∈ R ) G ( α x + ( − α ) y ) = α G ( x ) + ( − α ) G ( y ) = α x + ( − α ) y ,which implies that Fix G is an affine subspace. Hence, by Proposition 3.13(ii), G is strictly quasinonexpansive.Therefore, by Lemma 2.8, G is nonexpansive. (cid:4) Let T : H → H with Fix T (cid:54) = ∅ and let κ ∈ R + . We say T is linear regular with constant κ if ( ∀ x ∈ H ) d Fix T ( x ) ≤ κ (cid:107) x − Tx (cid:107) .By the following two results, we know that every BAM is linearly regular, but generally linearly regular operatoris not a BAM. Proposition 3.15
Let G : H → H and let γ ∈ [
0, 1 [ . Suppose that G is a γ -BAM. Then G is linearly regular withconstant − γ .Proof. Because G is a γ -BAM, by Definition 3.1, Fix G is a nonempty closed convex subset of H and ( ∀ x ∈ H ) (cid:107) Gx − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) . (3.9)Let x ∈ H . By the triangle inequality and (3.9), (cid:107) x − P Fix G x (cid:107) ≤ (cid:107) x − Gx (cid:107) + (cid:107) Gx − P Fix G x (cid:107) ≤ (cid:107) x − Gx (cid:107) + γ (cid:107) x − P Fix G x (cid:107) , ⇒ ( − γ ) (cid:107) x − P Fix G x (cid:107) ≤ (cid:107) x − Gx (cid:107)⇔ (cid:107) x − P Fix G x (cid:107) ≤ − γ (cid:107) x − Gx (cid:107) .Hence, ( ∀ x ∈ H ) d Fix T ( x ) ≤ − γ (cid:107) x − Gx (cid:107) , that is, G is linearly regular with constant − γ . (cid:4) xample 3.16 Suppose that H = R . Let C = B [
0; 1 ] and G = R C . Let x = (
2, 0 ) . P C R C x = (
0, 0 ) (cid:54) = (
1, 0 ) = P C x , which, by Definition 3.1, yields that R C is not a BAM. On the other hand, apply [5, Example 2.2] with λ = C = ( − ) Id + C is linearly regular with constant . Proposition 3.17
Let
I : = {
1, . . . , m } . Let ( ∀ i ∈ I ) G i : H → H be operators with
Fix G i being a closed affine subspaceof H and ( ∀ i ∈ I ) γ i ∈ [
0, 1 [ . Suppose that ( ∀ i ∈ I ) G i is a γ i -BAM and that ∩ j ∈ I Fix G j (cid:54) = ∅ . The following hold: (i) G m · · · G is strictly quasinonexpansive. (ii) Fix G m · · · G = Fix ∩ i ∈ I Fix G i . (iii) Let ( ω i ) i ∈ I be real numbers in ]
0, 1 ] such that ∑ i ∈ I ω i = . Then Fix ∑ i ∈ I ω i G i = Fix ∩ i ∈ I Fix G i .Proof. Because ( ∀ i ∈ I ) G i is a γ i -BAM with Fix G i being a closed affine subspace of H , by Proposition 3.13, ( ∀ i ∈ I ) G i is strictly quasinonexpansive. Moreover, by assumption, ∩ i ∈ I Fix G i (cid:54) = ∅ .(i)&(ii): These are from [3, Corollary 4.50].(iii): This comes from [3, Proposition 4.47]. (cid:4) Proposition 3.18
Let G : H → H with
Fix
G being a nonempty closed convex subset of H . Then G is a -BAM if andonly if G = P Fix G .Proof. “ ⇒ ”: Assume that G is a 0-BAM. By Definition 3.1, ( ∀ x ∈ H ) (cid:107) Gx − P Fix G x (cid:107) ≤ (cid:107) x − P Fix G x (cid:107) = G = P Fix G .“ ⇐ ”: Assume that G = P Fix G . Then by Example 3.4, G is a BAM with constant 0. (cid:4) Corollary 3.19
Let ( ∀ i ∈ {
1, 2 } ) G i : H → H be such that
Fix G i is a closed affine subspace of H . Suppose that ( ∀ i ∈ {
1, 2 } ) G i is a BAM and that Fix G ∩ Fix G (cid:54) = ∅ . Then G G is a -BAM if and only if G G = P Fix G ∩ Fix G .Proof. Because Fix G and Fix G are closed affine subspaces and Fix G ∩ Fix G (cid:54) = ∅ , Fix G ∩ Fix G is a closedaffine subspace.“ ⇒ ”: By Proposition 3.17(ii), Fix G G = Fix G ∩ Fix G is a closed affine subspace. Hence, by Proposi-tion 3.18, G G = P Fix G G = P Fix G ∩ Fix G .“ ⇐ ”: By Example 3.4, G G = P Fix G ∩ Fix G is a 0-BAM. (cid:4) According to the following Example 3.20 and Example 6.18 below, we know that the composition of BAMsis a projector is not sufficient to deduce that the individual BAMs are projectors. Hence, the condition “ G i is aBAM” in the Corollary 3.19 above is more general than “ G i is a projector”. Example 3.20
Let U : = R (
1, 0 ) and U : = R (
0, 1 ) . Set T : = P U and T : = P U . Then neither T nor T is aprojection. Moreover, T T = P { ( ) } . Corollary 3.21
Let C and C be closed convex subsets of H with C ∩ C (cid:54) = ∅ . Then P C P C is a -BAM if and only if P C P C = P C ∩ C .Proof. Because C ∩ C (cid:54) = ∅ , by [14, Corollary 4.5.2], Fix P C P C = C ∩ C is nonempty, closed, and convex.Therefore, the desired result follows from Proposition 3.18. (cid:4) Proposition 3.22
Let z ∈ H . Let ( ∀ i ∈ {
1, 2 } ) G i : H → H satisfy ( ∀ x ∈ H ) G x = z + G ( x − z ) . (3.10) Then the following assertions hold: (i) Fix G = Fix G − z. (ii) Suppose that
Fix G or Fix G is a nonempty closed convex subset of H . Let γ ∈ [
0, 1 [ . Then G is a γ -BAM if andonly if G is a γ -BAM.Proof. (i): Let x ∈ H . Then, x ∈ Fix G ⇔ x = G x = G ( x + z ) − z ⇔ x + z = G ( x + z ) ⇔ x + z ∈ Fix G ⇔ x ∈ Fix G − z .(ii): Clearly, by (i), Fix G is a nonempty closed convex subset of H if and only if Fix G is a nonempty closedconvex subset of H . 8ote thatP Fix G G = P Fix G ⇔ ( ∀ x ∈ H ) P Fix G G x = P Fix G x ⇔ ( ∀ x ∈ H ) P z + Fix G G x = P z + Fix G x ( by (i) ) ⇔ ( ∀ x ∈ H ) z + P Fix G ( G x − z ) = z + P Fix G ( x − z ) ( by Fact 2.1 ) (3.10) ⇔ ( ∀ x ∈ H ) P Fix G ( G ( x − z )) = P Fix G ( x − z ) ⇔ ( ∀ x ∈ H ) P Fix G ( G x ) = P Fix G x ⇔ P Fix G G = P Fix G ,and that ( ∀ x ∈ H ) (cid:107) G x − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107)⇔ ( ∀ x ∈ H ) (cid:107) G x − P z + Fix G x (cid:107) ≤ γ (cid:107) x − P z + Fix G x (cid:107) ( by (i) ) ⇔ ( ∀ x ∈ H ) (cid:107) G x − (cid:0) z + P Fix G ( x − z ) (cid:1) (cid:107) ≤ γ (cid:107) x − (cid:0) z + P Fix G ( x − z ) (cid:1) (cid:107) ( by Fact 2.1 ) (3.10) ⇔ ( ∀ x ∈ H ) (cid:107) G ( x − z ) − P Fix G ( x − z ) (cid:107) ≤ γ (cid:107) ( x − z ) − P Fix G ( x − z ) (cid:107)⇔ ( ∀ x ∈ H ) (cid:107) G x − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) .Altogether, by Definition 3.1, (ii) above is true. (cid:4) Lemma 3.23
Set
I : = {
1, . . . , m } . Let ( ∀ i ∈ I ) F i : H → H . Define ( ∀ i ∈ I )( ∀ x ∈ H ) T i x : = z + F i ( x − z ) . (3.11) Let γ ∈ [
0, 1 [ . Then the following hold: (i) Suppose that
Fix F m · · · F or Fix T m · · · T is a nonempty closed and convex subset of H . Then F m · · · F is a γ -BAM if and only if T m · · · T is a γ -BAM. (ii) Let ( ω i ) i ∈ I be in R such that ∑ i ∈ I ω i = . Suppose that Fix ∑ i ∈ I ω i F i or Fix ∑ i ∈ I ω i T i is a nonempty closed andconvex subset of H . Then ∑ i ∈ I ω i F i is a γ -BAM if and only if ∑ i ∈ I ω i T i is a γ -BAM.Proof. Let x ∈ H . By (3.11), it is easy to see that T m · · · T T x = T m · · · T ( z + F ( x − z )) = · · · = z + F m · · · F F ( x − z ) ∑ i ∈ I ω i T i x = ∑ i ∈ I ω i (cid:0) z + F i ( x − z ) (cid:1) = z + ∑ i ∈ I ω i F i ( x − z ) .Therefore, both (i) and (ii) follow from Proposition 3.22(ii). (cid:4) In this section, we study compositions of BAMs and determine whether the composition of BAMs is still a BAMor not.
Compositions of BAMs with closed and affine fixed point sets
In this subsection, we consider compositions of BAMs with with closed and affine fixed point sets.The following result is essential to the proof of Theorem 4.2 below.
Lemma 4.1
Set
I : = {
1, 2 } . Let ( ∀ i ∈ I ) G i : H → H , and let γ i ∈ [
0, 1 [ . Set ( ∀ i ∈ I ) U i : = Fix G i . Suppose that ( ∀ i ∈ I ) G i is a γ i -BAM and that U i is a closed linear subspace of H . Denote the cosine c ( U , U ) of the Friedrichs anglebetween U and U by c F . Let x ∈ H , and let x − P U ∩ U x (cid:54) = and G x − P U ∩ U x (cid:54) = . Set β : = (cid:107) P U G x − P U ∩ U x (cid:107)(cid:107) G x − P U ∩ U x (cid:107) and β : = (cid:107) P U x − P U ∩ U x (cid:107)(cid:107) x − P U ∩ U x (cid:107) . (4.1) Then the following statements hold: (cid:107) G G x − P U ∩ U x (cid:107) ≤ (cid:0) γ + ( − γ ) β (cid:1) (cid:0) γ + ( − γ ) β (cid:1) (cid:107) x − P U ∩ U x (cid:107) . (ii) β ∈ [
0, 1 ] and β ∈ [
0, 1 ] . (iii) Suppose that P U x − P U ∩ U x (cid:54) = and P U G x − P U ∩ U x (cid:54) = . Setu : = G x − P U ∩ U x (cid:107) G x − P U ∩ U x (cid:107) , v : = P U x − P U ∩ U x (cid:107) P U x − P U ∩ U x (cid:107) , and w : = P U G x − P U ∩ U x (cid:107) P U G x − P U ∩ U x (cid:107) . Then (cid:104) v , w (cid:105) ≤ c F , (4.2a) β = (cid:104) u , w (cid:105) and β ≤ (cid:104) u , v (cid:105) , (4.2b) β β ≤ + c F { β , β } ≤ (cid:114) + c F Proof.
Because G is a γ -BAM and G is a γ -BAM, by Definition 3.1 and Lemma 3.2(i), we have thatP U G = P U = G P U and P U G = P U = G P U , (4.3)and that ( ∀ y ∈ H ) (cid:107) G y − P U y (cid:107) ≤ γ (cid:107) y − P U y (cid:107) and (cid:107) G y − P U y (cid:107) ≤ γ (cid:107) y − P U y (cid:107) . (4.4)Note that by (4.3) and Fact 2.2, we have that G G x − P U G x (4.3) = G G x − P U G G x = ( Id − P U ) G G x = P U ⊥ G G x ∈ U ⊥ , (4.5a) G x − P U G x = ( Id − P U ) G x = P U ⊥ G x ∈ U ⊥ , (4.5b) G x − P U x (4.3) = G x − P U G x = ( Id − P U ) G x = P U ⊥ G x ∈ U ⊥ , (4.5c) x − P U x = ( Id − P U ) x = P U ⊥ x ∈ U ⊥ . (4.5d)Hence, by the Pythagorean theorem, we obtain (cid:107) G G x − P U G x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U ⊥ (cid:107) + (cid:107) P U G x − P U ∩ U x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U (cid:107) = (cid:107) G G x − P U ∩ U x (cid:107) , (4.6) (cid:107) G x − P U G x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U ⊥ (cid:107) + (cid:107) P U G x − P U ∩ U x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U (cid:107) = (cid:107) G x − P U ∩ U x (cid:107) , (4.7) (cid:107) G x − P U x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U ⊥ (cid:107) + (cid:107) P U x − P U ∩ U x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U (cid:107) = (cid:107) G x − P U ∩ U x (cid:107) , (4.8) (cid:107) x − P U x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U ⊥ (cid:107) + (cid:107) P U x − P U ∩ U x (cid:124) (cid:123)(cid:122) (cid:125) ∈ U (cid:107) = (cid:107) x − P U ∩ U x (cid:107) . (4.9)(i): Note that (cid:107) G G x − P U ∩ U x (cid:107) = (cid:107) G G x − P U G x (cid:107) + (cid:107) P U G x − P U ∩ U x (cid:107) ≤ γ (cid:107) G x − P U G x (cid:107) + (cid:107) P U G x − P U ∩ U x (cid:107) = γ (cid:16) (cid:107) G x − P U G x (cid:107) + (cid:107) P U G x − P U ∩ U x (cid:107) (cid:17) + ( − γ ) (cid:107) P U G x − P U ∩ U x (cid:107) = γ (cid:107) G x − P U ∩ U x (cid:107) + ( − γ ) (cid:107) P U G x − P U ∩ U x (cid:107) = γ (cid:107) G x − P U ∩ U x (cid:107) + ( − γ ) β (cid:107) G x − P U ∩ U x (cid:107) = (cid:16) γ + ( − γ ) β (cid:17) (cid:107) G x − P U ∩ U x (cid:107) = (cid:16) γ + ( − γ ) β (cid:17) (cid:16) (cid:107) G x − P U x (cid:107) + (cid:107) P U x − P U ∩ U x (cid:107) (cid:17) (4.4) ≤ (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ (cid:107) x − P U x (cid:107) + (cid:107) P U x − P U ∩ U x (cid:107) (cid:17) = (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ (cid:16) (cid:107) x − P U x (cid:107) + (cid:107) P U x − P U ∩ U x (cid:107) (cid:17) + ( − γ ) (cid:107) P U x − P U ∩ U x (cid:107) (cid:17) (4.9) = (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ (cid:107) x − P U ∩ U x (cid:107) + ( − γ ) (cid:107) P U x − P U ∩ U x (cid:107) (cid:17) (4.1) = (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ (cid:107) x − P U ∩ U x (cid:107) + ( − γ ) β (cid:107) x − P U ∩ U x (cid:107) (cid:17) = (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ + ( − γ ) β (cid:17) (cid:107) x − P U ∩ U x (cid:107) .(ii): This comes from (4.1), (4.7) and (4.9).(iii): By Lemma 2.3 and Fact 2.2, we know thatP U x − P U ∩ U x = P U x − P U ∩ U P U x = ( Id − P U ∩ U ) P U x = P ( U ∩ U ) ⊥ P U ( x ) ∈ U ∩ ( U ∩ U ) ⊥ .By Lemma 2.3, (4.3) and Fact 2.2, P U ∩ U x = P U ∩ U P U x = P U ∩ U P U G x = P U ∩ U P U G x , so by Fact 2.2,P U G x − P U ∩ U x = P U G x − P U ∩ U P U G x = P ( U ∩ U ) ⊥ P U G x ∈ U ∩ ( U ∩ U ) ⊥ .Hence, using Definition 2.4, we obtain that (cid:104) v , w (cid:105) = (cid:68) P U x − P U ∩ U x (cid:107) P U x − P U ∩ U x (cid:107) , P U G x − P U ∩ U x (cid:107) P U G x − P U ∩ U x (cid:107) (cid:69) ≤ c F ,which yields (4.2a).It is easy to see that (cid:104) G x − P U ∩ U x , P U G x − P U ∩ U x (cid:105) = (cid:104) G x − P U G x , P U G x − P U ∩ U x (cid:105) + (cid:104) P U G x − P U ∩ U x , P U G x − P U ∩ U x (cid:105) (4.5b) = (cid:107) P U G x − P U ∩ U x (cid:107) . Hence, β = (cid:107) P U G x − P U ∩ U x (cid:107)(cid:107) G x − P U ∩ U x (cid:107) = (cid:68) G x − P U ∩ U x (cid:107) G x − P U ∩ U x (cid:107) , P U G x − P U ∩ U x (cid:107) P U G x − P U ∩ U x (cid:107) (cid:69) = (cid:104) u , w (cid:105) . (4.11)Moreover, by (4.4), (cid:107) G x − P U x (cid:107) ≤ γ (cid:107) x − P U x (cid:107) ≤ (cid:107) x − P U x (cid:107) , then using (4.8) and (4.9), we know that (cid:107) G x − P U ∩ U x (cid:107) ≤ (cid:107) x − P U ∩ U x (cid:107) . Hence, β = (cid:107) P U x − P U ∩ U x (cid:107)(cid:107) x − P U ∩ U x (cid:107) ≤ (cid:107) P U x − P U ∩ U x (cid:107)(cid:107) G x − P U ∩ U x (cid:107) = (cid:68) G x − P U ∩ U x (cid:107) G x − P U ∩ U x (cid:107) , P U x − P U ∩ U x (cid:107) P U x − P U ∩ U x (cid:107) (cid:69) = (cid:104) u , v (cid:105) , (4.12)where the second equality is from (cid:104) G x − P U ∩ U x , P U x − P U ∩ U x (cid:105) = (cid:104) G x − P U x , P U x − P U ∩ U x (cid:105) + (cid:104) P U x − P U ∩ U x , P U x − P U ∩ U x (cid:105) (4.5c) = + (cid:107) P U x − P U ∩ U x (cid:107) .Hence, (4.11) and (4.12) yield (4.2b).Note that by (4.2b) and (cid:107) u (cid:107) = (cid:107) v (cid:107) = (cid:107) w (cid:107) = β + β ≤ (cid:104) u , v + w (cid:105) ≤ (cid:107) u (cid:107)(cid:107) v + w (cid:107) = (cid:113) (cid:107) v (cid:107) + (cid:104) v , w (cid:105) + (cid:107) w (cid:107) = (cid:113) ( + (cid:104) v , w (cid:105) ) (4.2a) ≤ (cid:113) ( + c F ) ,so β β ≤ ( β + β ) ≤ + c F { β , β } > (cid:113) + c F . Then β β > + c F , whichcontradicts (4.2c).Altogether, the proof is complete. (cid:4) In the following result, we extend [12, Lemma 1] from R n to H and also provide a new constant associatedwith the composition of BAMs. Although the following proof is shorter than the proof of [12, Lemma 1], themain idea of the following proof is from the proof of [12, Lemma 1]. Theorem 4.2
Set
I : = {
1, 2 } . Let ( ∀ i ∈ I ) G i : H → H , and let γ i ∈ [
0, 1 [ . Set ( ∀ i ∈ I ) U i : = Fix G i . Supposethat ( ∀ i ∈ I ) G i is a γ i -BAM and that U i is a closed linear subspace of H such that U + U is closed. Denote the cosinec ( U , U ) of the Friedrichs angle between U and U by c F . Then the following hold: (i) Fix ( G ◦ G ) = Fix G ∩ Fix G is a closed linear subspace of H and P U ∩ U G G = P U ∩ U . (ii) Let x ∈ H . If x − P U ∩ U x = or G x − P U ∩ U x = , then (cid:107) G G x − P U ∩ U x (cid:107) = . (iii) Set r : = max (cid:40)(cid:114) γ + ( − γ ) + c F (cid:114) γ + ( − γ ) + c F (cid:41) . (4.13) Then r ∈ (cid:2) max { γ , γ } , 1 (cid:2) . Moreover, ( ∀ x ∈ H ) (cid:107) G G x − P U ∩ U x (cid:107) ≤ r (cid:107) x − P U ∩ U x (cid:107) . (4.14)(iv) Set s : = (cid:114) γ + γ − γ γ + ( − γ )( − γ ) ( + c F ) Then s ∈ (cid:2) max { γ , γ , } , 1 (cid:2) . Moreover, ( ∀ x ∈ H ) (cid:107) G G x − P U ∩ U x (cid:107) ≤ s (cid:107) x − P U ∩ U x (cid:107) . (4.16)(v) G ◦ G is a min { r , s } -BAM.Proof. Because U + U is closed, by Fact 2.5, we know that c F : = c ( U , U ) ∈ [
0, 1 [ . (4.17)Because G is a γ -BAM and G is a γ -BAM, by Definition 3.1 and Lemma 3.2(i), we have thatP U G = P U = G P U and P U G = P U = G P U , (4.18)and that ( ∀ x ∈ H ) (cid:107) G x − P U x (cid:107) ≤ γ (cid:107) x − P U x (cid:107) . (4.19)(i): By assumptions and by Proposition 3.17(ii), Fix ( G ◦ G ) = Fix G ∩ Fix G = U ∩ U is a closed linearsubspace of H . Because U ∩ U ⊆ U and U ∩ U ⊆ U , by Lemma 2.3, we know thatP U ∩ U P U = P U ∩ U = P U P U ∩ U and P U ∩ U P U = P U ∩ U = P U P U ∩ U . (4.20)Moreover,P U ∩ U G G = P U ∩ U P U G G = P U ∩ U P U G = P U ∩ U P U G = P U ∩ U P U (4.20) = P U ∩ U .(ii): If x − P U ∩ U x =
0, then by (4.20) and (4.18), then G G x = G G P U ∩ U x = G G P U P U ∩ U x = G P U P U ∩ U x = G P U P U ∩ U x = P U P U ∩ U x = P U ∩ U x . Hence, (cid:107) G G x − P U ∩ U x (cid:107) = G x − P U ∩ U x =
0, then by (4.20) and (4.19), (cid:107) G G x − P U ∩ U x (cid:107) = (cid:107) G P U ∩ U x − P U P U ∩ U x (cid:107) ≤ γ (cid:107) P U ∩ U x − P U P U ∩ U x (cid:107) =
0, that is, (cid:107) G G x − P U ∩ U x (cid:107) = γ ∈ [
0, 1 [ and γ ∈ [
0, 1 [ , by (4.17), r ∈ (cid:2) max { γ , γ } , 1 (cid:2) .We shall prove (4.14) next. Let x ∈ H . By (ii), we are able to assume x − P U ∩ U x (cid:54) = G x − P U ∩ U x (cid:54) = β and β as in Lemma 4.1.Note that if P U x − P U ∩ U x =
0, then β =
0. Moreover, by Lemma 4.1(i)&(ii), (cid:107) G G x − P U ∩ U x (cid:107) ≤ (cid:0) γ + ( − γ ) β (cid:1) γ (cid:107) x − P U ∩ U x (cid:107) ≤ γ (cid:107) x − P U ∩ U x (cid:107) . If P U G x − P U ∩ U x =
0, then β = (cid:107) G G x − P U ∩ U x (cid:107) ≤ γ (cid:0) γ + ( − γ ) β (cid:1) (cid:107) x − P U ∩ U x (cid:107) ≤ γ (cid:107) x − P U ∩ U x (cid:107) . Becausemax { γ , γ } ≤ r , we know that in these two cases, (4.14) is true. So in the rest of the proof, we assume thatP U x − P U ∩ U x (cid:54) = U G x − P U ∩ U x (cid:54) = β ≤ (cid:114) + c F β ≤ (cid:114) + c F s = (cid:113) γ + γ − γ γ + ( − γ )( − γ ) ( + c F ) = (cid:113) γ + ( − γ ) (cid:0) γ + ( − γ ) ( + c F ) (cid:1) and γ ∈ [
0, 1 [ and γ ∈ [
0, 1 [ are symmetric in the expression of s . So, by (4.17), s ∈ (cid:2) max { γ , γ } , 1 (cid:2) . In addition,some elementary algebraic manipulations yield γ + γ − γ γ + ( − γ )( − γ ) ( + c F ) ≥ γ + γ − γ γ +( − γ )( − γ ) ≥ . Hence, s ∈ (cid:2) max { γ , γ , } , 1 (cid:2) .We prove (4.16) next. Let x ∈ H . Because s ≥ max { γ , γ } , similarly to the proof of (iii), to show (4.16), weare able to assume x − P U ∩ U x (cid:54) = G x − P U ∩ U x (cid:54) =
0, P U x − P U ∩ U x (cid:54) = U G x − P U ∩ U x (cid:54) = β and β as in Lemma 4.1.Use Lemma 4.1(ii) and (4.2c) in Lemma 4.1(iii) respectively in the following two inequalities to obtain that (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ + ( − γ ) β (cid:17) = γ γ + γ ( − γ ) β + γ ( − γ ) β + ( − γ )( − γ ) β β ≤ γ γ + γ ( − γ ) + γ ( − γ ) + ( − γ )( − γ ) β β ≤ γ + γ − γ γ + ( − γ )( − γ ) ( + c F ) = s .This and Lemma 4.1(i) yield that (cid:107) G G x − P U ∩ U x (cid:107) ≤ (cid:16) γ + ( − γ ) β (cid:17) (cid:16) γ + ( − γ ) β (cid:17) (cid:107) x − P U ∩ U x (cid:107) ≤ s (cid:107) x − P U ∩ U x (cid:107) .Hence, (iv) holds.(v): Combine Definition 3.1 with (i), (iii) and (iv) to obtain that G ◦ G is a min { r , s } -BAM. (cid:4) Lemma 4.3
Set
I : = {
1, . . . , m } . Let U , . . . , U m be closed linear subspaces of H . The following hold: (i) Let i ∈ I (cid:114) { m } . ThenU i + + ∩ ij = U j is closed ⇔ U ⊥ i + + ( ∩ ij = U j ) ⊥ is closed ⇔ U ⊥ i + + i ∑ j = U ⊥ j is closed .(ii) ( ∀ i ∈ I (cid:114) { m } ) U i + + ∩ ij = U j is closed if and only if ( ∀ i ∈ I ) ∑ ij = U ⊥ j is closedProof. (i): The two equivalences follow by Fact 2.5 and [15, Theorem 4.6(5)] respectively.(ii): Note that by [15, Theorem 4.5(1)], U ⊥ is a closed linear subspace of H , that is, U ⊥ = U ⊥ . Then theasserted result follows from (i) by the principle of strong mathematical induction on m . (cid:4) Theorem 4.4
Set
I : = {
1, . . . , m } . Let ( ∀ i ∈ I ) γ i ∈ [
0, 1 [ and let G i : H → H be a γ i -BAM such that U i : = Fix G i is a closed affine subspaces of H with ∩ i ∈ I U i (cid:54) = ∅ . Assume that ( ∀ i ∈ I ) ∑ ij = ( par U j ) ⊥ is closed. Then the followingstatements hold: ( ∀ k ∈ {
1, . . . , m } ) Fix ( G k ◦ · · · ◦ G ) = ∩ ki = Fix G i is a closed affine subspaces of H . (ii) G m ◦ · · · ◦ G ◦ G is a BAM. (iii) Suppose that m = . Denote the cosine c ( par U , par U ) of the Friedrichs angle between par U and par U by c F .Set r : = max i ∈ I (cid:114) γ i + ( − γ i ) + c F and s : = (cid:114) γ + γ − γ γ + ( − γ )( − γ ) ( + c F ) Then min { r , s } ∈ [
0, 1 [ and G ◦ G is a min { r , s } -BAM. (iv) There exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H ) (cid:107) ( G m ◦ · · · ◦ G ◦ G ) k x − P ∩ mi = U i x (cid:107) ≤ γ k (cid:107) x − P ∩ mi = U i x (cid:107) . Proof. (i): This is from Proposition 3.17(ii).(ii)&(iii): Let z ∈ ∩ i ∈ I U i . Define ( ∀ i ∈ I ) F i : H → H by ( ∀ x ∈ H ) F i ( x ) : = G i ( x + z ) − z (4.23)By the assumptions, (4.23) and Proposition 3.22, F i is a γ i -BAM with Fix F i = par U i being a closed linearsubspace of H . Hence, by (i), (4.23) and Lemma 3.23(i), we are able to assume that U , . . . , U m are closed linearsubspaces of H . Then (iii) reduces to Theorem 4.2(v).We prove (ii) next. If m =
1, then there is nothing to prove. Suppose that m ≥
2. We prove it by induction on k ∈ {
1, . . . , m } . By assumption, G is a BAM, so the base case is true. Assume G k ◦ · · · ◦ G is a BAM for some k ∈ {
1, 2, . . . , m − } . By the assumption, ( ∀ i ∈ I ) ∑ ij = U ⊥ j is closed, and by (i) and Lemma 4.3(ii), we know thatFix ( G k ◦ · · · ◦ G ) + Fix G k + = ( ∩ kj = U j ) + U k + is closed. Hence, apply Theorem 4.2(v) with G = G k ◦ · · · ◦ G and G = G k + to obtain that G k + ◦ G k ◦ · · · ◦ G is a BAM. Therefore, (ii) holds as well.(iv): This comes from (ii) and Proposition 3.10. (cid:4) The following Remark 4.5(i) and Remark 4.6(i) exhibit a case where the new constant s associated with thecomposition of BAMs presented in Theorem 4.2(v) is better than the constant r from [12]. Moreover, Remark 4.5illustrates that generally min { r , s } in Theorem 4.2 is not a sharp constant for the composition of BAMs. Remark 4.5
Let L and L be closed linear subspaces of H . Assume that L + L is closed. Denote by c F : = c ( L , L ) the Friedrichs angle between L and L . By [14, Corollary 4.5.2], Fix P L P L = L ∩ L is a closed linearsubspace of H . By Example 3.4, both P L and P L are 0-BAM. Moreover, the following hold:(i) Apply Theorem 4.2(v) with G = P L , G = P L , γ = γ = { (cid:113) + c F , + c F } = + c F and P L P L is a + c F -BAM.(ii) By [15, Lemma 9.5(7) and Theorem 9.8], ( ∀ x ∈ H ) (cid:107) P L P L x − P L ∩ L x (cid:107) ≤ c F (cid:107) x − P L ∩ L x (cid:107) ,and c F is the smallest constant satisfying the inequality above. Hence, P L P L is a BAM with sharp con-stant c F .Recall that c F : = c ( U , U ) ∈ [
0, 1 [ , so c F < + c F . Hence, we know that generally the constant associated withthe composition of BAMs provided by Theorem 4.2(v) is not sharp.The following Remark 4.6(ii) presents examples showing that the constants s and r in Theorem 4.2 are inde-pendent. Remark 4.6
Consider the constants r , s in Theorem 4.2(v) .(i) Suppose that γ = γ =
0, that is, G = P U or G = P U . Without loss of generality, let γ =
0. Then r = (cid:114) γ + ( − γ ) + c F s = (cid:114) γ + ( − γ ) ( + c F ) s ≤ r . 14ii) Suppose that γ : = γ = γ ∈ [
0, 1 [ and that c F =
0. Then r = (cid:114) γ + ( − γ )
12 and s = (cid:114) γ − γ + ( − γ )
14 .Hence s − r = ( − γ ) ( γ − ) ,which implies that s ≥ r ⇔ γ ∈ (cid:34) √
33 , 1 (cid:34) and s < r ⇔ γ ∈ (cid:34) √ (cid:34) . Compositions of BAMs with general convex fixed point sets
In this subsection, we investigate compositions of BAMs with general closed and convex fixed point sets.By Example 3.4, the projection onto a nonempty closed convex subset of H is the most common BAM. Thefollowing results show that the order of the projections does matter to determine whether the composition ofprojections is a BAM or not. The next result considers the composition of projections onto a cone and a ball. Proposition 4.7
Let K be a nonempty closed convex cone in H , and let ρ ∈ R ++ . Denote by B : = B [ ρ ] . (i) P B P K = P K ∩ B is a -BAM. (ii) Suppose that H = R , K = R + and ρ = . Then P K P B is not a BAM.Proof. (i): By [2, Corollary 7.3], P B P K = P K ∩ B , which, by Corollary 3.21, yields that P B P K is a 0-BAM.(ii): By [14, Corollary 4.5.2], Fix P K P B = K ∩ B . By [2, Example 7.5], we know thatP K ∩ B P K P B ( − ) = (cid:18) √ (cid:19) (cid:54) = (
1, 0 ) = P K ∩ B ( − ) ,which implies that P K ∩ B P K P B (cid:54) = P K ∩ B . So, by Definition 3.1, P K P B is not a BAM. (cid:4) The following example considers projections onto an affine subspace and a cone.
Example 4.8
Suppose H = R . Let U : = { ( x , x ) ∈ R : x = − x + } and K : = R + . Then the followinghold (see also Figure 1):(i) P U P K is not a BAM.(ii) P K P U is a √ -BAM. Proof.
Define the lines L : = R · (
1, 0 ) , L : = R · (
0, 1 ) , l : = { ( x , x ) ∈ R : x = x − } and l : = { ( x , x ) ∈ R : x = x + } . It is easy to see that for every ( x , x ) ∈ R ,P U ( x , x ) = (cid:18) x − x +
12 , − x + x + (cid:19) , (4.24a)P K ( x , x ) = ( x , x ) , if x ≥ x ≥ (
0, 0 ) , if x < x < ( x , 0 ) = P L ( x , x ) , if x ≥ x < ( x ) = P L ( x , x ) , if x < x ≥
0. (4.24b)By [14, Corollary 4.5.2], Fix P U P K = U ∩ K = Fix P K P U . (4.25)15i): Let ( x , x ) ∈ R (cid:114) ( K ∪ R −− ) such that x − < x < x +
1, that is, ( x , x ) is above l and below l butneither in K nor in the strictly negative orthant. Then by (4.24),P U ∩ K ( x , x ) = P U ( x , x ) (cid:54) = P U ∩ K P U P K ( x , x ) ,which, by Definition 3.1 and (4.25), implies that P U P K is not a BAM.(ii): By Definition 2.4, the cosine of Friedrichs angles between par U and L and between par U and L is c ( par U , L ) = (cid:68) ( − ) √ (
1, 0 ) (cid:69) = √ = (cid:68) ( − ) √ (
0, 1 ) (cid:69) = c ( par U , L ) (4.26)Let ( x , x ) ∈ R . If ( x , x ) ∈ { ( y , y ) ∈ R : y − ≤ y ≤ y + } , then P U ∩ K ( x , x ) = P U ( x , x ) , whichyields that P K P U ( x , x ) = P U ∩ K ( x , x ) and P U ∩ K P K P U ( x , x ) = P U ∩ K ( x , x ) .Assume that x < x −
1. ThenP K P U ( x , x ) = P L P U ( x , x ) , (4.27a)P U ∩ K P K P U ( x , x ) = (
1, 0 ) = P U ∩ K ( x , x ) = P U ∩ L ( x , x ) . (4.27b)Moreover, because U and L are closed affine subspaces with U ∩ L (cid:54) = ∅ , (cid:107) P K P U ( x , x ) − P U ∩ K ( x , x ) (cid:107) = (cid:107) P L P U ( x , x ) − P U ∩ L ( x , x ) (cid:107) ( by (4.27a) ) ≤ √ (cid:107) ( x , x ) − P U ∩ L ( x , x ) (cid:107) ( by Remark 4.5(ii) and (4.26) )= √ (cid:107) ( x , x ) − P U ∩ K ( x , x ) (cid:107) . ( by (4.27a) ) Assume that x > x +
1. Then similarly to the case that x < x −
1, we also have that (cid:107) P K P U ( x , x ) − P U ∩ K ( x , x ) (cid:107) ≤ √ (cid:107) ( x , x ) − P U ∩ K ( x , x ) (cid:107) .Altogether, for every ( x , x ) ∈ R , we have thatP U ∩ K P K P U ( x , x ) = P U ∩ K ( x , x ) , (cid:107) P K P U ( x , x ) − P U ∩ K ( x , x ) (cid:107) ≤ √ (cid:107) ( x , x ) − P U ∩ K ( x , x ) (cid:107) ,which combining with (4.25) yield that P K P U is a √ -BAM. (cid:4) Figure 1: Composition of projections onto line and cone16 emark 4.9
By Proposition 4.7 and Example 4.8, we know that in Theorem 4.4, the assumption “ ( ∀ i ∈ I ) Fix G i is closed affine subspaces” is not tight, and that the order of the operators matters.The following example examines the composition of projections onto balls and states that generally the com-position of BAMs is not a BAM again. Example 4.10
Suppose that H = R . Consider the two closed balls K = { ( x , x ) : ( x + ) + x ≤ } andlet K = { ( x , x ) : ( x − ) + x ≤ } . Then the following statements hold (see also Figure 2):(i) For every x ∈ { ( x , x ) ∈ R (cid:114) ( K ∪ K ) : x < x (cid:54) = } , P K ∩ K P K P K x = P K P K x (cid:54) = P K ∩ K x .(ii) P K P K is not a BAM. Proof.
By Example 3.4 and Proposition 3.17(ii), Fix P K P K = K ∩ K . The proof follows by Definition 3.1, theformula shown in [3, Example 3.18] and some elementary algebraic manipulations. (cid:4) Figure 2: Composition of BAMs may not be a BAM
In this section, we consider combinations of finitely many BAMs. In the following results, by reviewing Re-mark 3.5, we obtain constraints for the coefficients constructing the combinations.
Remark 5.1
Let C be a nonempty closed convex subset of H and let γ ∈ R . Note that by Example 3.4, P C andId = P H are BAMs.(i) Let γ ∈ R . By Remark 3.5(i) and Definition 3.1, if γ Id +( − γ ) P C is a BAM, then γ ∈ ] −
1, 1 [ .(ii) In addition, suppose that H = R and C : = B [
0; 1 ] . Then by Remark 3.5(ii), γ Id +( − γ ) P C is a BAMimplies that γ ∈ [
0, 1 [ .The following results are similar to Example 4.10. Example 5.2
Suppose that H = R . Consider the two closed balls K : = { ( x , x ) : ( x + ) + x ≤ } and let K : = { ( x , x ) : ( x − ) + x ≤ } . Let α ∈ ]
0, 1 [ . Then the following hold:(i) For every x ∈ { ( x , x ) ∈ R (cid:114) ( K ∪ K ) : x < x (cid:54) = } , P K ∩ K ( α P K +( − α ) P K ) x (cid:54) = P K ∩ K x .(ii) P K P K is not a BAM. Proof.
Note that by Example 3.4 and Proposition 3.17(iii), Fix ( α P K +( − α ) P K ) = K ∩ K . The remainingpart of the proof is similar to the proof of Example 4.10. (cid:4) In the remaining part of this section, we consider convex combinations of BAMs with closed and affine fixedpoint sets. 17 emma 5.3
Set
I : = {
1, . . . , m } . Let ( ∀ i ∈ I ) G i be a BAM such that Fix G i is a closed affine subspace of H , and ∩ i ∈ I Fix G i (cid:54) = ∅ . Let ( ∀ i ∈ I ) ω i ∈ ]
0, 1 [ such that ∑ i ∈ I ω i = . Set G : = ∑ i ∈ I ω i G i . Then (i) Fix G = ∩ i ∈ I Fix G i is a closed affine subspace of H . (ii) P Fix G G = P Fix G .Proof. (i): By Proposition 3.17(iii) and the assumptions, Fix G = Fix ( ∑ i ∈ I ω i G i ) = ∩ i ∈ I Fix G i is a closed affinesubspace of H .(ii): By (i) and [3, Proposition 29.14(i)], we know that P Fix G is affine. Note that ( ∀ j ∈ I ) Fix G = ∩ i ∈ I Fix G i ⊆ Fix G j and that both ∩ i ∈ I Fix G i and Fix G j are closed affine subspaces. Moreover, by Lemma 2.3 and Defini-tion 3.1(i), we have that ( ∀ j ∈ I ) P Fix G G j = P ∩ i ∈ I Fix G i G j = P ∩ i ∈ I Fix G i P Fix G j G j = P ∩ i ∈ I Fix G i P Fix G j = P ∩ i ∈ I Fix G i = P Fix G (5.1)Therefore, P Fix G G = P Fix G (cid:32) ∑ i ∈ I ω i G i (cid:33) = ∑ i ∈ I ω i P Fix G G i (5.1) = ∑ i ∈ I ω i P Fix G = P Fix G . (cid:4) Convex combination of BAMs with closed and affine fixed point sets
Theorem 5.4
Set
I : = {
1, 2 } . Let ( ∀ i ∈ I ) γ i ∈ [
0, 1 [ and let G i : H → H be a γ i -BAM. Suppose that ( ∀ i ∈ I ) Fix G i is a closed linear subspace of H . Suppose that Fix G + Fix G is closed. Let α ∈ ]
0, 1 [ . Set c F : = c ( Fix G , Fix G ) and γ : = max (cid:40) α (cid:114) γ + ( − γ ) + c F + ( − α ) , α + ( − α ) (cid:114) γ + ( − γ ) + c F (cid:41) . (5.2) Then max (cid:8) α (cid:113) ( + γ ) + ( − α ) , α + ( − α ) (cid:113) ( + γ ) (cid:9) ≤ γ < and α G + ( − α ) G is a γ -BAM.Proof. Set ( ∀ i ∈ I ) U i : = Fix G i . Because U + U is closed, by Fact 2.5, we know that c F : = c ( U , U ) ∈ [
0, 1 [ ,which yields that γ < ( ∀ i ∈ I ) γ i + ( − γ i ) + c F ≥ ( + γ i ) . Hence, γ ≥ (cid:8) α (cid:113) ( + γ ) + ( − α ) , α + ( − α ) (cid:113) ( + γ ) (cid:9) .Let x ∈ H . By Lemma 5.3 and Definition 3.1, it suffices to show that (cid:107) α G x + ( − α ) G x − P U ∩ U x (cid:107) ≤ γ (cid:107) x − P U ∩ U x (cid:107) . (5.3)Because ( ∀ i ∈ I ) G i is a γ i -BAM, by Definition 3.1 and Lemma 3.2(i), we have that ( ∀ i ∈ I ) P U i G i = P U i = G i P U i , (5.4)and that ( ∀ y ∈ H ) (cid:107) G i y − P U i y (cid:107) ≤ γ i (cid:107) y − P U i y (cid:107) . (5.5)If x = P U ∩ U x , then x ∈ U ∩ U and α G x + ( − α ) G x − P U ∩ U x = α G P U x + ( − α ) G P U x − x (5.4) = α P U x + ( − α ) P U x − x = x − x =
0, from which we deduce that (5.3) holds. Therefore, in the rest of theproof, we assume that x (cid:54) = P U ∩ U x . Set β : = (cid:107) P U x − P U ∩ U x (cid:107)(cid:107) x − P U ∩ U x (cid:107) and β : = (cid:107) P U x − P U ∩ U x (cid:107)(cid:107) x − P U ∩ U x (cid:107) . (5.6)By the triangle inequality, (cid:107) α G x + ( − α ) G x − P U ∩ U x (cid:107) (5.7a) ≤ α (cid:107) G x − P U ∩ U x (cid:107) + ( − α ) (cid:107) G x − P U ∩ U x (cid:107) + α ( − α ) (cid:107) G x − P U ∩ U x (cid:107)(cid:107) G x − P U ∩ U x (cid:107) . (5.7b)18ote that ( ∀ i ∈ I ) , G i x − P U i x (5.4) = G i x − P U i G i x = P U ⊥ i G i x ∈ U ⊥ i , (5.8a)P U i x − P U ∩ U x = P U i x − P U ∩ U P U i x = P ( U ∩ U ) ⊥ P U i x ∈ U i ∩ ( U ∩ U ) ⊥ . (5.8b)Now, using (5.8a) and (5.8b) in the following (5.9a) and (5.9d), we know that ( ∀ i ∈ I ) , (cid:107) G i x − P U ∩ U x (cid:107) = (cid:107) G i x − P U i x (cid:107) + (cid:107) P U i x − P U ∩ U x (cid:107) (5.9a) (5.5) ≤ γ i (cid:107) x − P U i x (cid:107) + (cid:107) P U i x − P U ∩ U x (cid:107) (5.9b) = γ i (cid:107) x − P U i x (cid:107) + γ i (cid:107) P U i x − P U ∩ U x (cid:107) + ( − γ i ) (cid:107) P U i x − P U ∩ U x (cid:107) (5.9c) = γ i (cid:107) x − P U ∩ U x (cid:107) + ( − γ i ) (cid:107) P U i x − P U ∩ U x (cid:107) (5.9d) (5.6) = γ i (cid:107) x − P U ∩ U x (cid:107) + ( − γ i ) β i (cid:107) x − P U ∩ U x (cid:107) (5.9e) = (cid:0) γ i + ( − γ i ) β i (cid:1) (cid:107) x − P U ∩ U x (cid:107) . (5.9f)Set ( ∀ i ∈ I ) η i : = (cid:113) γ i + ( − γ i ) β i . (5.10)Combine (5.7) with (5.9) to obtain that (cid:107) α G x + ( − α ) G x − P U ∩ U x (cid:107) ≤ (cid:16) α η + ( − α ) η + α ( − α ) η η (cid:17) (cid:107) x − P U ∩ U x (cid:107) (5.11a) = ( αη + ( − α ) η ) (cid:107) x − P U ∩ U x (cid:107) . (5.11b)Combining (5.3), (5.2), (5.10) and (5.11), we know that it remains to show thatmin { β , β } ≤ (cid:114) + c F i ∈ I such that P U i x − P U ∩ U x =
0, then β i = ( ∀ i ∈ I ) P U i x − P U ∩ U x (cid:54) = i ∈ I. Because P U i x − P U ∩ U x ∈ U i and x − P U i x = P U ⊥ i x ∈ U ⊥ i , we have (cid:104) P U i x − P U ∩ U x , x − P U i x (cid:105) =
0. Hence (cid:104) P U i x − P U ∩ U x , x − P U ∩ U x (cid:105) = (cid:104) P U i x − P U ∩ U x , x − P U i x (cid:105) + (cid:104) P U i x − P U ∩ U x , P U i x − P U ∩ U x (cid:105) = (cid:107) P U i x − P U ∩ U x (cid:107) and thus β i = (cid:107) P U i x − P U ∩ U x (cid:107)(cid:107) x − P U ∩ U x (cid:107) = (cid:68) P U i x − P U ∩ U x (cid:107) P U i x − P U ∩ U x (cid:107) , x − P U ∩ U x (cid:107) x − P U ∩ U x (cid:107) (cid:69) . (5.13)Set u : = P U x − P U ∩ U x (cid:107) P U x − P U ∩ U x (cid:107) , v : = P U x − P U ∩ U x (cid:107) P U x − P U ∩ U x (cid:107) and w : = x − P U ∩ U x (cid:107) x − P U ∩ U x (cid:107) . By (5.8b), P U x − P U ∩ U x ∈ U ∩ ( U ∩ U ) ⊥ and P U x − P U ∩ U x ∈ U ∩ ( U ∩ U ) ⊥ . Hence, by Definition 2.4, (cid:104) u , v (cid:105) ≤ c F . (5.14)Using (5.13), the Cauchy-Schwarz inequality, and (cid:107) u (cid:107) = (cid:107) v (cid:107) = (cid:107) w (cid:107) =
1, we deduce that β + β = (cid:104) u + v , w (cid:105) ≤ (cid:107) u + v (cid:107) = (cid:113) (cid:107) u (cid:107) + (cid:104) u , v (cid:105) + (cid:107) v (cid:107) = (cid:113) ( + (cid:104) u , v (cid:105) ) (5.14) ≤ (cid:113) ( + c F ) . (5.15)Suppose to the contrary that (5.12) is not true, that is, β > (cid:113) + c F and β > (cid:113) + c F . Then β + β > (cid:114) + c F = (cid:113) ( + c F ) ,which contradicts with (5.15). Altogether, the proof is complete. (cid:4) Example 5.5
Let U be a closed linear subspace of H . Let α ∈ ]
0, 1 [ . Then the following hold:(i) α P U +( − α ) P U ⊥ is a BAM with constant max { α √ + ( − α ) , ( − α ) √ + α } , by Theorem 5.4.(ii) α P U +( − α ) P U ⊥ is a BAM with sharp constant max { α , 1 − α } .(iii) max { α √ + ( − α ) , ( − α ) √ + α } > max { α , 1 − α } . Proof. (i): By Example 3.4, both P U and P U ⊥ are 0-BAMs. Moreover, by Definition 2.4, c F = c ( U , U ⊥ ) = α P U +( − α ) P U ⊥ is a BAM with constant max { √ α + ( − α ) , α + √ ( − α ) } .(ii): Denote by G : = α P U +( − α ) P U ⊥ . Let x ∈ H and γ ∈ [
0, 1 [ . Using (cid:104) P U x , P U ⊥ x (cid:105) =
0, Fact 2.2, andP
Fix G x = P { } x =
0, we obtain that (cid:107) Gx − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) (5.16a) ⇔ (cid:107) α P U x + ( − α ) P U ⊥ x (cid:107) ≤ γ (cid:107) x (cid:107) (5.16b) ⇔ α (cid:107) P U x (cid:107) + ( − α ) (cid:107) P U ⊥ x (cid:107) ≤ γ ( (cid:107) P U x (cid:107) + (cid:107) P U ⊥ x (cid:107) ) (5.16c) ⇔ ≤ ( γ − α ) (cid:107) P U x (cid:107) + ( γ − ( − α ) ) (cid:107) P U ⊥ x (cid:107) , (5.16d)which implies that γ ≥ max { α , 1 − α } , since x ∈ H is arbitrary. Therefore, the required result follows fromLemma 5.3, (5.16) and Definition 3.1.(iii): This is trivial from α ∈ ]
0, 1 [ and √ ∈ ]
0, 1 [ . (cid:4) Theorem 5.6
Set
I : = {
1, . . . , m } . Let ( ∀ i ∈ I ) ω i ∈ ]
0, 1 [ . Suppose that m ≥ and that ( ∀ i ∈ I ) G i is a BAM withU i : = Fix G i being a closed affine subspace of H such that ∩ i ∈ I Fix G i (cid:54) = ∅ . Suppose that ( ∀ i ∈ I ) ∑ ij = ( par U j ) ⊥ isclosed. Then ∑ i ∈ I ω i G i is a BAM.Proof. Let z ∈ ∩ i ∈ I U i . Define ( ∀ i ∈ I ) F i : H → H by ( ∀ x ∈ H ) F i ( x ) : = G i ( x + z ) − z (5.17)By the assumptions, (5.17) and Proposition 3.22, F i is a BAM with Fix F i = par U i being a closed linear subspaceof H . By Proposition 3.17(iii) and by assumptions, Fix ( ∑ i ∈ I ω i G i ) = ∩ mi = U i is a closed affine subspace. Hence,by (5.17) and Lemma 3.23(ii), to show ∑ i ∈ I ω i G i is a BAM, we are able to assume that U , . . . , U m are closedlinear subspaces of H .We prove it by induction on m . By Lemma 4.3(ii) and Theorem 5.4, we know that the base case in which m = m ≥ m −
1, that is, for any { α , . . . , α m − } ⊆ ]
0, 1 [ we have that if ( ∀ i ∈ {
1, . . . , m − } ) ∑ ij = U ⊥ j is closed, then ∑ m − i = α i G i is a BAM. Note that m ∑ i = ω i G i = (cid:0) m − ∑ j = ω j (cid:1) (cid:32) m − ∑ i = ω i ∑ m − t = ω t G i (cid:33) + ω m G m + .Because we have the assumption, ( ∀ i ∈ {
1, . . . , m } ) ∑ ij = U ⊥ j is closed, by the inductive hypothesis, ∑ m − i = ω i ∑ m − j = ω j G i is a BAM. By the assumption, ( ∀ i ∈ I ) ∑ ij = U ⊥ j is closed, by Proposition 3.17(iii) and Lemma 4.3(ii), weknow that Fix (cid:18) ∑ m − i = ω i ∑ m − t = ω t G i (cid:19) + Fix G m = ( ∩ m − i = U i ) + U m is closed. Hence, apply Theorem 5.4 with G = ∑ m − i = ω i ∑ m − j = ω j G i , G = G m , α = ∑ m − j = ω j to obtain that ∑ mi = ω i G i is a BAM. (cid:4) ew method using the Cartesian product space reformulation The main result Theorem 5.10 in this subsection is almost the same with the Theorem 5.6 proved in the previoussubsection, however, in this subsection, we use a Cartesian product space reformulation.In the whole subsection, set I : = {
1, . . . , m } . Let ( ω i ) i ∈ I be real numbers in ]
0, 1 ] such that ∑ i ∈ I ω i =
1. Let H m be the real Hilbert space obtained by endowing the Cartesian product × i ∈ I H with the usual vector spacestructure and with the weighted inner product ( ∀ x = ( x i ) i ∈ I ∈ H m )( ∀ y = ( y i ) i ∈ I ∈ H m ) (cid:104) x , y (cid:105) = ∑ i ∈ I ω i (cid:104) x i , y i (cid:105) .Clearly, ( ∀ x = ( x i ) i ∈ I ∈ H m ) (cid:107) x (cid:107) = (cid:104) x , x (cid:105) = ∑ i ∈ I ω i (cid:104) x i , x i (cid:105) = ∑ i ∈ I ω i (cid:107) x i (cid:107) . (5.18)Denote by D : = { ( x ) i ∈ I ∈ H m : x ∈ H} .The following well-known fact is critical in proofs in this subsection. Fact 5.7
Let x = ( x i ) i ∈ I ∈ H m . The following hold: (i) P D x = ( ∑ j ∈ I ω j x j ) i ∈ I . (ii) Let ( ∀ i ∈ I ) C i be nonempty closed and convex subset of H . Then P × i ∈ I C i x = ( P C i x i ) i ∈ I .Proof. (i): This is from [3, Poposition 29.16].(ii): This is similar to [3, Proposition 29.3]. Because the definition of inner product is different, we show theproof next. Clearly, ( P C i x i ) i ∈ I ∈ × i ∈ I C i . Moreover, by [3, Theorem 3.16], ( ∀ ( c i ) i ∈ I ∈ × i ∈ I C i ) (cid:68) ( x i ) i ∈ I − ( P C i x i ) i ∈ I , ( c i ) i ∈ I − ( P C i x i ) i ∈ I (cid:69) = ∑ i ∈ I ω i (cid:104) x i − P C i x i , c i − P C i x i (cid:105) ≤ × i ∈ I C i x = ( P C i x i ) i ∈ I . (cid:4) In the remaining part of this subsection, let ( ∀ i ∈ I ) G i : H → H . Define F : H m → H m , and G : H m → H m respectively by ( ∀ x = ( x i ) i ∈ I ∈ H m ) F ( x ) = ( G i x i ) i ∈ I , (5.19) ( ∀ x = ( x i ) i ∈ I ∈ H m ) G ( x ) = (cid:0) ∑ j ∈ I ω j G j x j (cid:1) i ∈ I . (5.20) Proposition 5.8 (i) Fix F = × i ∈ I Fix G i . (ii) Let ( ∀ i ∈ I ) γ i ∈ [
0, 1 [ . Suppose that ( ∀ i ∈ I ) G i is a γ i -BAM. Then F is a ( max i ∈ I { γ i } ) -BAM.Proof. Let x = ( x i ) i ∈ I ∈ H m .(i): Now x ∈ Fix F (5.19) ⇔ ( x i ) i ∈ I = ( G i x i ) i ∈ I ⇔ ( ∀ i ∈ I ) x i = G i x i ⇔ ( ∀ i ∈ I ) x i ∈ Fix G i ⇔ x ∈ × i ∈ I Fix G i .(ii): Because ( ∀ i ∈ I ) G i is a BAM, we know that ( ∀ i ∈ I ) Fix G i is a nonempty closed and convex subsets of H and, by Definition 3.1, that for every i ∈ I,P
Fix G i G i = P Fix G i (5.21) ( ∀ x ∈ H ) (cid:107) G i x − P Fix G i x (cid:107) ≤ γ i (cid:107) x − P Fix G i x (cid:107) . (5.22)By (i) and Fact 5.7(ii), Fix F = × i ∈ I Fix G i is a nonempty closed convex subset of H m andP Fix F ( x ) = P × i ∈ I Fix G i ( x ) = ( P Fix G i x i ) i ∈ I . (5.23)21ow P Fix F F ( x ) (5.19) = P Fix F (( G i x i ) i ∈ I ) (5.23) = ( P Fix G i G i x i ) i ∈ I (5.21) = ( P Fix G i x i ) i ∈ I (5.23) = P Fix F ( x ) . (5.24)Note that by (5.19) and (5.23), (cid:107) F ( x ) − P Fix F x (cid:107) = (cid:107) ( G i x i ) i ∈ I − ( P Fix G i x i ) i ∈ I (cid:107) (5.25a) (5.18) = ∑ i ∈ I ω i (cid:107) G i x i − P Fix G i x i (cid:107) (5.25b) (5.22) ≤ ∑ i ∈ I ω i γ i (cid:107) x i − P Fix G i x i (cid:107) (5.25c) ≤ max j ∈ I { γ j } ∑ i ∈ I ω i (cid:107) x i − P Fix G i x i (cid:107) (5.25d) (5.18) = max j ∈ I { γ j } (cid:13)(cid:13)(cid:13) ( x i ) i ∈ I − ( P Fix G i x i ) i ∈ I (cid:13)(cid:13)(cid:13) (5.25e) (5.23) = max j ∈ I { γ j }(cid:107) x − P Fix F x (cid:107) . (5.25f)Therefore, combine (5.24) and (5.25) with Definition 3.1 to obtain the asserted result. (cid:4) Proposition 5.9
Let γ ∈ [
0, 1 [ . Then the following hold: (i) If F is a γ -BAM, then ( ∀ i ∈ I ) G i is a γ -BAM. (ii) F is a BAM if and only if ( ∀ i ∈ I ) G i is a BAM.Proof. (i): Because F is a γ -BAM, using Definition 3.1 and Proposition 5.8(i), we know that Fix F = × i ∈ I Fix G i isa nonempty closed and convex subset of H m , and there exists γ ∈ [
0, 1 [ such thatP × i ∈ I Fix G i F = P × i ∈ I Fix G i (5.26) ( ∀ x ∈ H m ) (cid:107) Fx − P × i ∈ I Fix G i x (cid:107) ≤ γ (cid:107) x − P × i ∈ I Fix G i x (cid:107) . (5.27)Hence, ( ∀ i ∈ I ) Fix G i is a nonempty closed and convex subset of H . Let x ∈ H .Set y : = ( x ) i ∈ I ∈ H m . By Fact 5.7(ii), (5.19), and (5.26), (cid:0) P Fix G i ( G i x ) (cid:1) i ∈ I = P × i ∈ I Fix G i (cid:0) ( G i x ) i ∈ I (cid:1) = P × i ∈ I Fix G i Fy = P × i ∈ I Fix G i y = (cid:0) P Fix G i ( x ) (cid:1) i ∈ I ,which yields ( ∀ i ∈ I ) P Fix G i G i = P Fix G i .Let j ∈ I. Set x : = ( x i ) i ∈ I ∈ H m such that x j = x and ( ∀ i ∈ I (cid:114) { j } ) x i ∈ Fix G i . Then ( ∀ i ∈ I (cid:114) { j } ) x i = G i x i = P Fix G i x i . Hence, by (5.27), we have that (cid:107) Fx − P × i ∈ I Fix G i x (cid:107) ≤ γ (cid:107) x − P × i ∈ I Fix G i x (cid:107) ⇔ (cid:107) ( G i x i ) i ∈ I − ( P Fix G i x i ) i ∈ I (cid:107) ≤ γ (cid:107) ( x i ) i ∈ I − ( P Fix G i x i ) i ∈ I (cid:107) ⇔ ∑ i ∈ I ω i (cid:107) G i x i − P Fix G i x i (cid:107) ≤ γ ∑ i ∈ I ω i (cid:107) x i − P Fix G i x i (cid:107) ⇔ ω j (cid:107) G j x − P Fix G j x (cid:107) ≤ γ ω j (cid:107) x − P Fix G j x (cid:107) ⇔ (cid:107) G j x − P Fix G j x (cid:107) ≤ γ (cid:107) x − P Fix G j x (cid:107) .Hence, by Definition 3.1, we know that ( ∀ i ∈ I ) G i is a γ -BAM.(ii): The equivalence comes from (i) above and Proposition 5.8(ii). (cid:4) The following result is inspired by [8, Proposition 5.25]. With consideration of Proposition 3.10 and Exam-ple 3.4, we note that Theorem 5.10 is a refinement of [8, Proposition 5.25].22 heorem 5.10
Let ( ∀ i ∈ I ) γ i ∈ [
0, 1 [ . Suppose that ( ∀ i ∈ I ) G i is a γ i -BAM and that Fix G i is a closed affine subspaceof H with ∩ j ∈ I Fix G j (cid:54) = ∅ . Set c F : = c (cid:0) D , × j ∈ I ( par Fix G j ) (cid:1) . Suppose that ∑ j ∈ I ( par Fix G j ) ⊥ is closed. Denote by µ : = max j ∈ I { γ j } and γ : = (cid:113) µ + ( − µ ) ( + c F ) . Then Fix ∑ j ∈ I ω j G j = ∩ j ∈ I Fix G j is a closed affine subspace of H and ∑ j ∈ I ω j G j is a γ -BAM. Moreover, ( ∀ x ∈ H ) (cid:107) ( ∑ i ∈ I ω i G i ) k x − P ∩ i ∈ I Fix G i x (cid:107) ≤ γ k (cid:107) x − P ∩ i ∈ I Fix G i x (cid:107) . Proof.
By the assumptions and Lemma 5.3, Fix ∑ i ∈ I ω i G i = ∩ i ∈ I Fix G i is a closed affine subspace of H andP ∩ i ∈ I Fix G i ( ∑ i ∈ I ω i G i ) = P ∩ i ∈ I Fix G i . To show ∑ j ∈ I ω j G j is a γ -BAM, by Definition 3.1, it suffices to show that ( ∀ x ∈ H ) (cid:13)(cid:13)(cid:13) ∑ i ∈ I ω i G i x − P ∩ i ∈ I Fix G i x (cid:13)(cid:13)(cid:13) ≤ γ (cid:107) x − P ∩ i ∈ I Fix G i x (cid:107) . (5.28)By Proposition 5.8(i)&(ii), Fix F = × j ∈ I Fix G j is a closed affine subspace of H m and F is a µ -BAM. By Exam-ple 3.4, P D is a 0-BAM. By Fact 2.5, D ⊥ + (cid:0) par (cid:0) × j ∈ I Fix G j (cid:1)(cid:1) ⊥ is closed if and only if D + (cid:0) par (cid:0) × j ∈ I Fix G j (cid:1)(cid:1) is closed. Moreover, by [1, Lemma 5.18], D + (cid:0) par (cid:0) × j ∈ I Fix G j (cid:1)(cid:1) is closed if and only if ∑ j ∈ I ( par Fix G j ) ⊥ isclosed, which is our assumption. Hence, we obtain that D ⊥ + (cid:0) par (cid:0) × j ∈ I Fix G j (cid:1)(cid:1) ⊥ is closed. Then apply The-orem 4.4(iii) with H = H m , G = F and G = P D to obtain that P D F is a γ -BAM. Note that, by (5.19) andFact 5.7(i), ( ∀ y = ( y i ) i ∈ I ∈ H m ) P D F ( y ) = (cid:0) ∑ j ∈ I ω j G j y j (cid:1) i ∈ I = G ( y ) ,that is, P D F = G . By [14, Corollary 4.5.2],Fix G = Fix ( P D F ) = D ∩ (cid:0) × i ∈ I ( ∩ j ∈ I Fix G j ) (cid:1) .Let x ∈ H and set x = ( x ) i ∈ I ∈ H m . Similarly with the proof of Fact 5.7(ii), by [3, Theorem 3.16], we have thatP Fix G x = ( P ∩ j ∈ I Fix G j x ) i ∈ I . (5.29)Because G = P D F is a γ -BAM, by Definition 3.1(iii), (cid:107) Gx − P Fix G x (cid:107) ≤ γ (cid:107) x − P Fix G x (cid:107) (5.29) ⇔ (cid:13)(cid:13)(cid:13) ( ∑ j ∈ I ω j G j x ) i ∈ I − ( P ∩ j ∈ I Fix G j x ) i ∈ I (cid:13)(cid:13)(cid:13) ≤ γ (cid:13)(cid:13)(cid:13) ( x ) i ∈ I − ( P ∩ j ∈ I Fix G j x ) i ∈ I (cid:13)(cid:13)(cid:13) ⇔ ∑ i ∈ I ω i (cid:13)(cid:13)(cid:13) ∑ j ∈ I ω j G j x − P ∩ j ∈ I Fix G j x (cid:13)(cid:13)(cid:13) ≤ γ ∑ i ∈ I ω i (cid:107) x − P ∩ j ∈ I Fix G j x (cid:107) ⇔ (cid:13)(cid:13)(cid:13) ∑ j ∈ I ω j G j x − P ∩ j ∈ I Fix G j x (cid:13)(cid:13)(cid:13) ≤ γ (cid:107) x − P ∩ j ∈ I Fix G j x (cid:107) ,which yields (5.28). Hence, the proof is complete. (cid:4) Remark 5.11
Consider Theorems 5.6 and 5.10. Although the results from these two theorems are the same, butthere are different assumptions: “ ( ∀ i ∈ I ) ∑ ij = ( par U j ) ⊥ is closed” and “ ∑ i ∈ I ( par U i ) ⊥ is closed” respectively.Suppose that m =
3, that ( par U ) ⊥ + ( par U ) ⊥ is not closed, and that ( par U ) ⊥ = H , say, G = P { } .Then clearly, ∑ i = ( par U i ) ⊥ = H is closed. Hence, “ ∑ i ∈ I ( par U i ) ⊥ is closed” (cid:54)⇒ “ ( ∀ i ∈ I (cid:114) { m } ) ( par U i + ) ⊥ + ∑ ij = ( par U j ) ⊥ is closed”.Therefore, we know that the assumptions in Theorem 5.6 are more restrictive than the assumptions in Theo-rem 5.10. However, comparing the constant γ in Theorem 5.4 and in Theorem 5.10 for m =
2, we know that theconstants associated with the convex combination of two BAMs are independent in these two theorems. Hence,we keep Theorems 5.6 and 5.10 together.The following Corollary 5.12(i) is a weak version of [15, Theorem 9.33] which shows clearly the convergencerate of the method of alternating projections. 23 orollary 5.12
Let U , . . . , U m be closed affine subspaces of H with ∩ mi = U i (cid:54) = ∅ . Then the following statements hold: (i) Assume that ( ∀ i ∈ I ) ∑ ij = ( par U j ) ⊥ is closed. Then P U m · · · P U P U is a BAM; moreover, there exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H ) (cid:107) ( P U m · · · P U P U ) k x − P ∩ mi = U i x (cid:107) ≤ γ k (cid:107) x − P ∩ mi = U i x (cid:107) .(ii) Suppose that ∑ i ∈ I ( par U i ) ⊥ is closed. Let ( ω i ) ≤ i ≤ m be real numbers in ]
0, 1 ] such that ∑ mi = ω i = . Then ∑ mi = ω i P U i is a BAM. Moreover, there exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H ) (cid:13)(cid:13)(cid:13) ( m ∑ i = ω i P U i ) k x − P ∩ mi = U i x (cid:13)(cid:13)(cid:13) ≤ γ k (cid:107) x − P ∩ mi = U i x (cid:107) . Proof.
By Example 3.4, we know that ( ∀ i ∈ {
1, . . . , m } ) P U i is a 0-BAM and Fix P U i = U i is a closed affinesubspace.(i): This comes from Theorem 4.4(ii)&(iv) with G = P U , . . . , G m = P U m .(ii): This follows by Theorem 5.10. (cid:4) In this section, we present BAMs which are not projections in Hilbert spaces. In particular, we connect thecircumcenter mapping with BAM.
Definitions and facts on circumcentered isometry methods
Before we turn to the relationship between best approximation mapping and circumcenter mapping, we needthe background and facts on the circumcenter mapping and the circumcentered method in this section.By [6, Proposition 3.3], we know that the following definition is well defined.
Definition 6.1 (circumcenter operator) [6, Definition 3.4] Let P ( H ) be the set of all nonempty subsets of H containing finitely many elements. The circumcenter operator is CC : P ( H ) → H ∪ { ∅ } : K (cid:55)→ (cid:40) p , if p ∈ aff ( K ) and {(cid:107) p − y (cid:107) : y ∈ K } is a singleton; ∅ , otherwise.In particular, when CC ( K ) ∈ H , that is, CC ( K ) (cid:54) = ∅ , we say that the circumcenter of K exists and we call CC ( K ) the circumcenter of K . Definition 6.2 (circumcenter mapping) [7, Definition 3.1] Let F , . . . , F m be operators from H to H such that ∩ mj = Fix F j (cid:54) = ∅ . Set S : = { F , . . . , F m } and ( ∀ x ∈ H ) S ( x ) : = { F x , . . . , F m x } . The circumcenter mapping inducedby S is CC S : H → H ∪ { ∅ } : x (cid:55)→ CC ( S ( x )) ,that is, for every x ∈ H , if the circumcenter of the set S ( x ) defined in Definition 6.1 does not exist, then CC S x = ∅ . Otherwise, CC S x is the unique point satisfying the two conditions below:(i) CC S x ∈ aff ( S ( x )) = aff { F ( x ) , . . . , F m ( x ) } , and(ii) (cid:107) CC S x − F ( x ) (cid:107) = · · · = (cid:107) CC S x − F m ( x ) (cid:107) .In particular, if for every x ∈ H , CC S x ∈ H , then we say the circumcenter mapping CC S induced by S is proper .Otherwise, we call CC S improper . Fact 6.3 [7, Proposition 3.7(ii)]
Let F , . . . , F m be operators from H to H with ∩ mj = Fix F j (cid:54) = ∅ . Set S : = { F , . . . , F m } .Assume that CC S is proper and that Id ∈ S . Then Fix CC S = ∩ mi = Fix F i . act 6.4 [7, Proposition 3.3] Let F , F be operators from H to H and set S : = { F , F } . Then ( ∀ x ∈ H ) CC S x = F x + F x x ∈ H and assume that CC S is proper. The circumcenter method induced by S is x : = x , and x k : = CC S ( x k − ) = CC k S x , where k =
1, 2, . . . . (6.1)
Definition 6.5 [16, Definition 1.6-1] A mapping T : H → H is said to be isometric or an isometry if ( ∀ x ∈ H )( ∀ y ∈ H ) (cid:107) Tx − Ty (cid:107) = (cid:107) x − y (cid:107) . (6.2) Fact 6.6 [9, Proposition 3.3 and Corollary 3.4]
Let T : H → H be isometric. Then T is affine. Moreover, if
Fix
T isnonempty, then
Fix
T is a closed affine subspace.
Note that by Fact 6.6, every isometry must be affine. In the rest of this section, without otherwise statement, (cid:0) ∀ i ∈ {
1, . . . , m } (cid:1) T i : H → H is affine isometry with m (cid:92) j = Fix T j (cid:54) = ∅ .Denote by S : = { T , . . . , T m − , T m } .The associated set-valued operator S : H → P ( H ) is defined by ( ∀ x ∈ H ) S ( x ) : = { T x , . . . , T m − x , T m x } .The following Fact 6.7(i) makes the circumcentered method induced by S defined in (6.1) well-defined. Sinceevery element of S is isometry, we call the circumcentered method induced by the S circumcentered isometrymethod (CIM). Fact 6.7 [8, Theorem 3.3 and Proposition 4.2]
Let x ∈ H . Then the following statements hold: (i)
The circumcenter mapping CC S : H → H induced by S is proper; moreover, CC S x is the unique point satisfyingthe two conditions below: (a) CC S x ∈ aff ( S ( x )) , and (b) {(cid:107) CC S x − Tx (cid:107) : T ∈ S } is a singleton. (ii) Let W be nonempty closed affine subspace of ∩ mi = Fix T i . Then ( ∀ k ∈ N ) P W CC k S = P W = CC k S P W . Fact 6.8 [8, Theorem 4.15(i)]
Let W be a nonempty closed affine subspace of ∩ mi = Fix T i . Assume that there existF : H → H and γ ∈ [
0, 1 [ such that ( ∀ x ∈ H ) F ( x ) ∈ aff ( S ( x )) and ( ∀ x ∈ H ) (cid:107) Fx − P W x (cid:107) ≤ γ (cid:107) x − P W x (cid:107) . Then ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k S x − P W x (cid:107) ≤ γ k (cid:107) x − P W x (cid:107) .In fact, it is easy to show that W = Fix CC S from the last inequality with k = Fact 6.9 [8, Theorem 4.16(ii)]
Suppose that H = R n . Let T S ∈ aff S satisfy Fix T S ⊆ ∩ T ∈S Fix
T. Assume that T S islinear and α -averaged with α ∈ ]
0, 1 [ . Then (cid:107) T S P ( ∩ T ∈S Fix T ) ⊥ (cid:107) ∈ [
0, 1 [ . Moreover, ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k S x − P ∩ T ∈S Fix T x (cid:107) ≤ (cid:107) T S P ( ∩ T ∈S Fix T ) ⊥ (cid:107) k (cid:107) x − P ∩ T ∈S Fix T x (cid:107) .25 ircumcenter mappings that are BAMs Theorem 6.10
Let W be a nonempty closed affine subspace of ∩ mi = Fix T i . Assume that Id ∈ S and that there existsF : H → H and γ ∈ [
0, 1 [ such that ( ∀ x ∈ H ) Fx ∈ aff ( S ( x )) and ( ∀ x ∈ H ) (cid:107) Fx − P W x (cid:107) ≤ γ (cid:107) x − P W x (cid:107) . ThenCC S is a γ -BAM and Fix CC S = ∩ mi = Fix T i .Proof. By Fact 6.7(i), CC S is proper. Then by Facts 6.3 and 6.6, Fix CC S = ∩ mi = Fix T i is a nonempty closed affinesubspace of H . Apply Fact 6.7(ii) with W = ∩ mi = Fix T i to obtain that P Fix CC S CC S = P Fix CC S . Moreover, by theassumptions, Fact 6.8 and Definition 3.1, we know that CC S is a γ -BAM. (cid:4) The following result states that in order to study whether the circumcenter mapping CC S is a BAM or not,we are free to assume the related isometries are linear. Proposition 6.11
Let z ∈ ∩ mi = Fix T i . Define ( ∀ i ∈ {
1, . . . , m } ) ( ∀ x ∈ H ) F i x : = T i ( x + z ) − z. Set S F : = { F , . . . , F m } . Then the following statements hold: (i) S F is a set of linear isometries. (ii) Let γ ∈ [
0, 1 [ . Assume that Id ∈ S . Then CC S is a γ -BAM if and only if CC S F is a γ -BAM.Proof. (i): Because z ∈ ∩ mi = Fix T i , by [9, Lemma 3.8], F , . . . , F m are linear isometries.(ii): Because both S and S F are sets of isometries, by Fact 6.7(i), both CC S and CC S F are proper. Clearly, Id ∈ S implies that Id ∈ S F as well. So, by Fact 6.3 , Fix CC S = ∩ mi = Fix T i and Fix CC S F = ∩ mi = Fix F i . In addition, by[9, Lemma 4.8], ( ∀ x ∈ H ) CC S x = z + CC S F ( x − z ) . Hence, the desired result comes from Proposition 3.22 andDefinition 3.1. (cid:4) Theorem 6.12
Suppose that H = R n . Let T S ∈ aff S satisfy that Fix T S ⊆ ∩ T ∈S Fix
T. Assume that T S is linear and α -averaged with α ∈ ]
0, 1 [ . Then γ : = (cid:107) T S P ( ∩ mi = Fix T i ) ⊥ (cid:107) ∈ [
0, 1 [ and CC S is a γ -BAM.Proof. By Facts 6.3 and 6.6, and assumptions, Fix CC S = ∩ mi = Fix T i is a closed affine subspace. Apply Fact 6.7(ii)with W = ∩ mi = Fix T i to obtain that P Fix CC S CC S = P Fix CC S . Moreover, by Fact 6.9, γ ∈ [
0, 1 [ and ( ∀ x ∈ H ) (cid:107) CC S x − P ∩ mi = Fix T i x (cid:107) ≤ γ (cid:107) x − P ∩ mi = Fix T i x (cid:107) . (cid:4) Let t ∈ N (cid:114) { } and let ( ∀ i ∈ {
1, . . . , t } ) F i : H → H . From now on, to facilitate the statements later, wedenote Ω ( F , . . . , F t ) : = (cid:110) F i r · · · F i F i (cid:12)(cid:12)(cid:12) r ∈ N , and i , . . . , i r ∈ {
1, . . . , t } (cid:111) (6.3)which is the set consisting of all finite composition of operators from { F , . . . , F t } . We use the empty productconvention: F i · · · F i = Id.
Fact 6.13 [9, Theorem 5.4]
Suppose that H = R n . Let F , F , . . . , F t be linear isometries on H . Assume that (cid:101) S is afinite subset of Ω ( F , . . . , F t ) , where Ω ( F , . . . , F t ) is defined in (6.3) . Assume that { Id, F , F , . . . , F t } ⊆ (cid:101) S . Let ( ω i ) i ∈ I be real numbers in ]
0, 1 ] such that ∑ i ∈ I ω i = and let ( α i ) i ∈ I be real numbers in ]
0, 1 [ . Denote A : = ∑ ti = ω i A i where ( ∀ i ∈ {
1, . . . , t } ) A i : = ( − α i ) Id + α i F i . Then the following statements hold: (i) Fix CC (cid:101) S = ∩ T ∈ (cid:101) S Fix T = ∩ ti = Fix F i = Fix A. (ii) (cid:107) A P ( ∩ ti = Fix F i ) ⊥ (cid:107) < . Moreover, ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k (cid:101) S x − P ∩ ti = Fix F i x (cid:107) ≤ (cid:107) A P ( ∩ ti = Fix F i ) ⊥ (cid:107) k (cid:107) x − P ∩ ti = Fix F i x (cid:107) . Fact 6.14 [9, Theorem 5.6]
Suppose that H = R n . Let F , F , . . . , F t be linear isometries. Assume that (cid:101) S is a finite subsetof Ω ( F , . . . , F t ) , where Ω ( F , . . . , F t ) is defined in (6.3) . Assume that { Id, F , F F , . . . , F t · · · F F } ⊆ (cid:101) S . Let ( ω i ) i ∈ I bereal numbers in ]
0, 1 ] such that ∑ i ∈ I ω i = and let ( α i ) i ∈ I and ( λ i ) i ∈ I be real numbers in ]
0, 1 [ . Set A : = ∑ i ∈ I ω i A i where A : = ( − α ) Id + α F and ( ∀ i ∈ I (cid:114) { } ) A i : = ( − α i ) Id + α i (( − λ i ) Id + λ i F i ) F i − · · · F . Then the following assertions hold: CC (cid:101) S = ∩ T ∈ (cid:101) S Fix T = ∩ ti = Fix F i = Fix A. (ii) (cid:107) A P ( ∩ ti = Fix F i ) ⊥ (cid:107) ∈ [
0, 1 [ . Moreover, ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k (cid:101) S x − P ∩ ti = Fix F i x (cid:107) ≤ (cid:107) A P ( ∩ ti = Fix F i ) ⊥ (cid:107) k (cid:107) x − P ∩ ti = Fix F i x (cid:107) . Proposition 6.15
Suppose that H = R n . Let F , . . . , F t be linear isometries from H to H . Let ˜ S be a finite subset of Ω ( F , . . . , F t ) . (i) If { Id, F , F , . . . , F t } ⊆ ˜ S , then Fix CC ˜ S = ∩ ti = Fix F i and CC ˜ S is a BAM (ii) If { Id, F , F F , . . . , F t · · · F F } ⊆ ˜ S , then Fix CC ˜ S = ∩ ti = Fix F i and CC ˜ S is a BAM.Proof. (i): By Fact 6.13(i), Fix CC ˜ S = ∩ T ∈ (cid:101) S Fix T = ∩ ti = Fix F i is a nonempty closed linear subspace of H . ApplyFact 6.7(ii) with W = Fix CC ˜ S = ∩ T ∈ (cid:101) S Fix T yields P Fix CC ˜ S CC ˜ S = P Fix CC ˜ S . Apply Fact 6.13(ii) to obtain thatthere exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H ) (cid:107) CC ˜ S x − P Fix CC ˜ S x (cid:107) ≤ γ (cid:107) x − P Fix CC ˜ S Fix Tx (cid:107) .Hence, by Definition 3.1, CC ˜ S is a BAM.(ii): The proof is similar to the proof of (i), however, this time we use Fact 6.14 instead of Fact 6.13. (cid:4) Theorem 6.16
Assume that H = R n and that F , . . . , F t are affine isometries from H to H with ∩ ti = Fix F i (cid:54) = ∅ . Assumethat ˜ S is a finite subset of Ω ( F , . . . , F t ) defined in (6.3) such that { Id, F , F , . . . , F t } ⊆ ˜ S or { Id, F , F F , . . . , F t · · · F F } ⊆ ˜ S . Then Fix CC ˜ S = ∩ ti = Fix F i and CC ˜ S is a BAM.Proof. This is from Proposition 6.15(i)&(ii) and Proposition 6.11(ii). (cid:4)
The following example shows that BAM is generally neither continuous nor linear.
Example 6.17
Suppose that H = R , set U : = R · (
1, 0 ) , and U : = R · (
1, 1 ) . Suppose that S = { Id, R U , R U } or that S = { Id, R U , R U R U } . Then the following statements hold.(i) CC S is a BAM and Fix CC S = { (
0, 0 ) } .(ii) CC S is neither continuous nor linear. Proof. (i): Because R U and R U are linear isometries and Fix R U ∩ Fix R U = U ∩ U = { (
0, 0 ) } , by Theo-rem 6.16, CC S is a BAM.(ii): This is from [7, Examples 4.19 and 4.20]. (cid:4) The following example illustrates that the composition of three BAMs is a projector does not imply that theindividual BAMs are projectors.
Example 6.18
Suppose that H = R , set U : = R · (
1, 0 ) , U : = R · (
1, 1 ) and U : = R · (
0, 1 ) . Denote by S : = { Id, R U , R U } and S : = { Id, R U , R U } . Then the following statements hold:(i) All of CC S , CC S and CC S CC S are BAMs. Moreover, Fix CC S = { (
0, 0 ) } , Fix CC S = { (
0, 0 ) } , andFix ( CC S CC S ) = { (
0, 0 ) } .(ii) None of the CC S , CC S or CC S CC S is a projector.(iii) CC S CC S CC S = P { ( ) } . Proof.
By Fact 6.4, it is easy to see that ( ∀ x ∈ H ) CC S x = P U x , if x ∈ U ;P U x , if x ∈ U ;0, otherwise. and CC S x = P U x , if x ∈ U ;P U x , if x ∈ U ;0, otherwise. (6.4)27ence, CC S CC S x = P U P U x , if x ∈ U ;0, if x ∈ U ;0, otherwise. (6.5)(i): Because R U , R U and R U are linear isometries, Fix R U ∩ Fix R U = { (
0, 0 ) } , and Fix R U ∩ Fix R U = U ∩ U = { (
0, 0 ) } , by Theorem 6.16, CC S and CC S are BAMs and Fix CC S = Fix CC S = { (
0, 0 ) } . Hence, byTheorem 4.4(ii), CC S CC S is BAM and Fix ( CC S CC S ) = { (
0, 0 ) } .(ii): Because U is not orthogonal with U , and by (6.4), the range of CC S equals U ∪ U , but CC S (cid:54) = P U ∪ U ,we know that CC S is not a projector. Similarly, neither CC S nor CC S CC S is a projector.(iii): This is clear from the definitions of CC S and CC S CC S presented in (6.4) and (6.5) respectively. (cid:4) Circumcenter and best approximation mappings in Hilbert space
Because reflectors associated with closed affine subspaces are isometries, we call the circumcenter method in-duced by a set of reflectors the circumcentered reflection method (CRM). Clearly, all facts on CIM are applicable toCRM.In this subsection, we assume that U , . . . , U m are closed linear subspaces in the real Hilbert space H , (6.6a) Ω : = Ω ( R U , . . . , R U m ) : = (cid:110) R U ir · · · R U i R U i (cid:12)(cid:12)(cid:12) r ∈ N , and i , . . . , i r ∈ {
1, . . . , m } (cid:111) , (6.6b) Ψ : = (cid:110) R U ir · · · R U i R U i (cid:12)(cid:12)(cid:12) r , i , i , . . . , i r ∈ {
0, 1, . . . , m } and 0 < i < · · · < i r (cid:111) . (6.6c)We also assume that Ψ ⊆ S ⊆ Ω and S consists of finitely many elements. (6.7)For every nonempty closed affine subset C of H , R C R C = ( C − Id )( C − Id ) = C − C − C + Id = Id. So, if m =
1, then Ω = Ψ = { Id, R U } . Hence, by the assumption, S = { Id, R U } , and, by Fact 6.4, CC S = ( Id + R U ) = P U . By Example 3.4, CC S = P U is a 0-BAM. Therefore, m = m ≥ Lemma 6.19
Fix CC S = ∩ T ∈S Fix T = ∩ mi = U i .Proof. By construction of Ψ with r = r =
1, we know that { Id, R U , . . . , R U m } ⊆ Ψ ⊆ S , so by Fact 6.3,Fix CC S = ∩ T ∈S Fix T ⊆ ∩ mi = Fix R U i = ∩ mi = U i . On the other hand, because S ⊆ Ω and ( ∀ T ∈ Ω ) ∩ mi = U i ⊆ Fix T , we know that ∩ mi = U i ⊆ ∩ T ∈S Fix T = Fix CC S . Altogether, Fix CC S = ∩ T ∈S Fix T = ∩ mi = U i . (cid:4) Fact 6.20 [9, Theorem 6.6]
Set γ : = (cid:107) P U m P U m − · · · P U P ( ∩ mi = U i ) ⊥ (cid:107) . Assume that m ≥ and that U ⊥ + · · · + U ⊥ m isclosed. Then γ ∈ [
0, 1 [ and ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k S x − P ∩ mi = U i x (cid:107) ≤ γ k (cid:107) x − P ∩ mi = U i x (cid:107) . Theorem 6.21
Set γ : = (cid:107) P U m P U m − · · · P U P ( ∩ mi = U i ) ⊥ (cid:107) . Assume that m ≥ and that U ⊥ + · · · + U ⊥ m is closed. Then γ ∈ [
0, 1 [ , Fix CC S = ∩ mi = U i , and CC S is a γ -BAM.Proof. Because Ψ ⊆ S ⊆ Ω , by Lemma 6.19, Fix CC S = ∩ T ∈S Fix T = ∩ mi = U i is a closed linear subspace. ApplyFact 6.7(ii) with W = ∩ T ∈S Fix T to obtain that P Fix CC S CC S = P Fix CC S . In addition, by Fact 6.20, γ ∈ [
0, 1 [ and ( ∀ x ∈ H ) (cid:107) CC S x − P Fix CC S x (cid:107) ≤ γ (cid:107) x − P Fix CC S x (cid:107) . Hence, by Definition 3.1, CC S is a γ -BAM. (cid:4) Corollary 6.22
Assume that m = in (6.6) , that S = { Id, R U , R U , R U R U } , and that U + U is closed. Set γ : = (cid:107) P U P U P ( U ∩ U ) ⊥ (cid:107) . Then γ ∈ [
0, 1 [ , Fix CC S = U ∩ U , and CC S is a γ -BAM. roof. By Fact 2.5, U + U is closed if and only if U ⊥ + U ⊥ is closed. Note that m = Ψ = { Id, R U , R U , R U R U } = S . Hence, the desired result is from Theorem 6.21 with m = (cid:4) Theorem 6.23
Let n ∈ N (cid:114) { } . Assume that m = n − and that U , . . . , U n are closed linear subspaces of H withU ⊥ + · · · + U ⊥ n being closed. Set ( ∀ i ∈ {
1, . . . , n − } ) U n + i : = U n − i . Denote γ : = (cid:107) P U n P U n − · · · P U P ( ∩ ni = U i ) ⊥ (cid:107) .Then γ ∈ [
0, 1 [ , Fix CC S = ∩ mi = U i , and CC S is a γ -BAM.Proof. Because Ψ ⊆ S ⊆ Ω , by Lemma 6.19, Fix CC S = ∩ T ∈S Fix T = ∩ ni = U i is a closed linear subspace.Apply Fact 6.7(ii) with W = ∩ T ∈S Fix T to obtain that P Fix CC S CC S = P Fix CC S . In addition, by [9, Theorem 6.7], γ ∈ [
0, 1 [ and ( ∀ x ∈ H ) (cid:107) CC S x − P Fix CC S x (cid:107) ≤ γ (cid:107) x − P Fix CC S x (cid:107) . Hence, by Definition 3.1, CC S is a γ -BAM. (cid:4) Corollary 6.24
Assume that m = in (6.6) , that S : = { Id, R U , R U , R U R U , R U R U , R U R U R U } , and thatU + U is closed. Set γ : = (cid:107) P U P U P ( U ∩ U ) ⊥ (cid:107) . Then γ ∈ [
0, 1 [ , Fix CC S = U ∩ U , and CC S is a γ -BAM.Proof. Let U = U in (6.6) with m = Ψ = { Id, R U , R U , R U R U , R U R U , R U R U R U } = S .Hence, the required result comes from Theorem 6.23 with n = U = U . (cid:4) Remark 6.25 [9, Theorem 6.8] shows that the sequence of iterations of the CC S in Theorem 6.23 attains theconvergence rate of the accelerated method of alternative projections which is no larger than the γ presentedin Theorem 6.23. Hence, by [9, Theorem 6.8], using the similar proof of Theorem 6.23, one can show that theconstant associated with the BAM, the CC S in Theorem 6.23, is no larger than the convergence rate of theaccelerated method of alternative projections. Compositions and convex combinations of circumcenter mapping
The following Theorems 6.26 and 6.27 with condition (i) are generalizations of [12, Theorem 2] from one class ofcircumcenter mapping induced by finite set of reflections to two classes of more general circumcenter mappingsinduced by finite set of isometries. Recall that T , . . . , T m are affine isometries from H to H with ∩ mi = Fix T i (cid:54) = ∅ . Theorem 6.26
Suppose that H = R n . Set S : = { Id, T q + , T q + , . . . , T q } , S : = { Id, T q + , T q + , . . . , T q } , . . . , S t : = { Id, T q t − + , T q t − + , . . . , T q t } , with q = q t = m and ( ∀ i ∈ {
1, . . . , t } ) q i − q i − ≥ . Suppose that one of thefollowing holds: (i) CC S = CC S t ◦ CC S t − ◦ · · · ◦ CC S . (ii) CC S = ∑ ti = ω i CC S i , where { ω i } ≤ i ≤ t ⊆ ]
0, 1 ] such that ∑ ti = ω i = .Then Fix CC S = ∩ mi = Fix T i and CC S is a BAM. Moreover, there exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k S x − P ∩ mi = Fix T I x (cid:107) ≤ γ k (cid:107) x − P ∩ mi = Fix T I x (cid:107) . Proof.
By Theorem 6.16, ( ∀ i ∈ {
1, . . . , t } ) CC S i is a BAM with Fix CC S i = ∩ q i − j = q i − Fix T j + . Using Facts 6.3and 6.7 and Proposition 3.17(ii)&(iii), we know that Fix CC S = ∩ mi = Fix T i . Note that every finite-dimensionallinear subspace must be closed. Hence, by Theorem 4.4(ii) and Theorem 5.10, we obtain that CC S is a BAM. Thelast inequality comes from Proposition 3.10. (cid:4) Theorem 6.27
Suppose that H = R n . Set I : = {
1, . . . , t } and ( ∀ i ∈ I ) S i : = { Id, T q i − + , T q i − + T q i − + , . . . , T q i · · · T q i − + T q i − + } , with q = q t = m and ( ∀ i ∈ I ) q i − q i − ≥ . Suppose that one of the following holds: (i) CC S = CC S t ◦ CC S t − ◦ · · · ◦ CC S . (ii) CC S = ∑ ti = ω i CC S i , where { ω i } ≤ i ≤ t ⊆ ]
0, 1 ] such that ∑ ti = ω i = . hen Fix CC S = ∩ mi = Fix T i and CC S is a BAM. Moreover, there exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k S x − P ∩ mi = Fix T i x (cid:107) ≤ γ k (cid:107) x − P ∩ mi = Fix T i x (cid:107) . Proof.
The proof is similar to that of Theorem 6.26. (cid:4)
We conclude this section by presenting BAMs from finite composition or convex combination of circumcentermappings, which is not projections, in Hilbert spaces. In fact, using Theorems 6.21 and 6.23, one may constructmore similar BAMs in Hilbert space.
Theorem 6.28
Let U , . . . , U m be closed affine subspaces of H with ∩ mi = U i (cid:54) = ∅ . Set I : = {
1, . . . , m } . Assume that ( ∀ i ∈ I ) par U i − + par U i is closed. Set S : = { Id, R U q + , R U q + , R U q + R U q + } , . . . , S m : = { Id, R U qm − , R U qm , R U qm R U qm − } , with ( ∀ i ∈ { } ∪ I ) q i = i. Suppose that one of the following holds: (i) ( ∀ i ∈ I ) ∑ ij = ( par U j ) ⊥ is closed, and CC S = CC S t ◦ CC S t − ◦ · · · ◦ CC S . (ii) ∑ mj = ( par U j ) ⊥ is closed, and CC S = ∑ mi = ω i CC S i , where { ω i } ≤ i ≤ m ⊆ ]
0, 1 ] such that ∑ mi = ω i = .Then Fix CC S = ∩ mi = Fix T i and CC S is a BAM. Moreover, there exists γ ∈ [
0, 1 [ such that ( ∀ x ∈ H )( ∀ k ∈ N ) (cid:107) CC k S x − P ∩ mi = U i x (cid:107) ≤ γ k (cid:107) x − P ∩ mi = U i x (cid:107) . Proof.
By Proposition 6.11, we are able to assume that ( ∀ i ∈ {
1, . . . , 2 m } ) U i is closed linear subspace of H . Forevery ( i ∈ I ) , because U i − + U i is closed, by Corollary 6.22, CC S i is a BAM with Fix CC S i = U i − ∩ U i andby Fact 2.5, U ⊥ i − + U ⊥ i = U ⊥ i − + U ⊥ i . Hence, for every i ∈ I, i ∑ j = ( par Fix CC S j ) ⊥ = i ∑ j = ( U j − ∩ U j ) ⊥ = i ∑ j = U ⊥ i − + U ⊥ i = i ∑ j = ( par U j ) ⊥ .Therefore, the asserted results follow by Theorem 4.4(ii) and Theorem 5.10. (cid:4) We discovered that the iteration sequence of BAM linearly converges to the best approximation onto the fixedpoint set of the BAM. We compared BAMs with linear convergent mappings, Banach contractions, and linearregular operators. We also generalized the result proved by Behling, Bello-Cruz and Santos that the finitecomposition of BAMs with closed affine fixed point sets in R n is still a BAM from R n to the general Hilbertspace. We constructed new constant associated with the composition of BAMs. Moreover, we proved thatconvex combinations of BAMs with closed affine fixed point sets is still a BAM. In addition, we connectedBAMs with circumcenter mappings.Although Theorem 4.4 states that the finite composition of BAMs with closed affine fixed point sets is stilla BAM, Example 4.10 shows that the composition of BAMs associated with closed Euclidean balls is generallynot a BAM. Moreover, Proposition 4.7 and Examples 4.8 and 4.10 illustrate that to determine whether the com-position of BAMs is a BAM or not, the order of the BAMs does matter. In addition, although Theorems 5.6and 5.10 state that the convex combination of BAMs with closed affine fixed point sets is a BAM, we have a littleknowledge for affine combinations of BAMs with general convex fixed point sets. It would be interesting tocharacterize the sufficient conditions for the finite composition of or affine combination of BAMs with generalconvex fixed point sets. By Remark 4.5, the constant associated with the composition of BAMs in Theorem 4.2(v)is not sharp. Using Example 5.5, we know that the constant associated with the convex combination of BAMspresented Theorem 5.4 is not sharp as well. Hence, we will also try to find better upper bound for the constantassociated with the composition of or the convex combination of BAMs. As we mentioned in Remark 5.11,although the assumption of Theorem 5.6 is more restrictive than that of Theorem 5.10, the constants in theseresults are independent. We will investigate the relation between the constants associated with the convex com-bination of BAMs in Theorems 5.6 and 5.10. Last but not least, we will try to find more BAMs with generalconvex fixed point sets and more applications of those BAMs.30 cknowledgements HHB and XW were partially supported by NSERC Discovery Grants.
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