Projection and proximal point methods: convergence results and counterexamples

Transcription

1 Nonlinear Analysis 56 (004) Projection and proximal point methods: convergence results and counterexamples Heinz H. Bauschke a, Eva Matouskova b, Simeon Reich c; a Department of Mathematics and Statistics, University of Guelph, Guelph, Ont., Canada NG W b Mathematical Institute, Academy of Sciences of the Czech Republic, Zitna 5, Prague CZ-567, Czech Republic c Department of Mathematics, The Technion Israel Institute of Technology, Haifa 3000, Israel Received 7 June 003; accepted 7 October 003 Abstract Recently, Hundal has constructed a hyperplane H, a cone K, and a starting point y 0 in such that the sequence of alternating projections ((P KP H ) n y 0) n N converges weakly to some point in H K, but not in norm. We show how this construction results in a counterexample to norm convergence for iterates of averaged projections; hence, we give an armative answer to a question raised by Reich two decades ago. Furthermore, new counterexamples to norm convergence for iterates of rmly nonexpansive maps (a la Genel and Lindenstrauss) and for the proximal point algorithm (a la Guler) are provided. We also present a counterexample, along with some weak and norm convergence results, for the new framework of string-averaging projection methods introduced by Censor et al. Extensions to Banach spaces and the situation for the Hilbert ball are discussed as well.? 003 Elsevier Ltd. All rights reserved. MSC: 47H09; 47J5; 90C5 Keywords: Alternatingprojections; Averagedprojections; Hilbertspace; Nonexpansive; Proximalpointalgorithm; Weak convergence Corresponding author. addresses: hbauschk@uoguelph.ca (H.H. Bauschke), matouse@math.cas.cz (E. Matouskova), sreich@tx.technion.ac.il (S. Reich) X/$ - see front matter? 003 Elsevier Ltd. All rights reserved. doi:0.06/j.na

2 76 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Introduction Throughout (most of) this paper, we assume that X is a real Hilbert space with inner product ; and induced norm : () Let A and B be closed convex sets in X with corresponding projectors (also known as projection operators or nearest point mappings, see Fact.4) P A and P B, respectively. The convex feasibility problem asks to nd a point in A B; () assuming this intersection is nonempty. This problem is of considerable importance in mathematics and the physical sciences; see [8,5,0,3], and the references therein. Perhaps the oldest algorithmic approach to solve () is to generate the sequence of alternating projections, which is dened by x 0 x = P A x 0 x = P B x x 3 = P A x for some starting point x 0 X: (3) If A and B are subspaces, then the sequence (x n ) n N converges in norm to the point in the intersection that is nearest to the starting point this basic result was proved by von Neumann [5] in 933. Thirty-two years later, Bregman [5] proved that the sequence (x n ) n N converges at least weakly to some point in A B. Ever since, there has been a nagging gap between von Neumann s and Bregman s result: Is it possible that norm convergence fails? In 000, Hundal announced his armative answer to this question; full details of his ingenious construction became available two years later, see [34]. In the broader setting of xed point theory, Hundal s counterexample to norm convergence of the iterates of compositions of projectors is similar to the counterexample by Genel and Lindenstrauss [9] (see also [4, pp. 7 74]) to norm convergence of the iterates of a rmly nonexpansive map, as well as to Guler s counterexample [3] to norm convergence of the proximal point algorithm. In passing, we note that certain modications of these algorithms are able to always generate norm convergent sequences; see, for instance, [,53]. Another classical algorithmic approach to solve () is to employ (midpoint) averages of projectors, rather than compositions. This amounts to constructing a sequence (x n ) n N via ( n N) x n+ = ( P A + P B) xn ; where x 0 X: (4) In his 969 thesis, Auslender [] established weak convergence of (x n ) n N to some point in A B; this result also follows from [45, Corollary.6]. Closely related are Cimmino s method of averaged reectors for solving linear equations [4] and Merzlyakov s method of extrapolated averaged projectors for solving linear inequalities [39]. On the other hand, a more general result by Reich (see [45, Theorems.7 and.3]) implies that if A and B are subspaces, then the sequence (x n ) n N generated by (4) does converge in norm to some point in A B. (Alternatively, one can apply von Neumann s result to the subspaces A B and {(x; x):x X } in the product space X X.)

3 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Analogously to the above discussion of alternating projections, Auslender s weak convergence result for general sets and Reich s norm convergence for subspaces lead to the question originally raised by Reich at the end of [45, Section ] in 983 on whether norm convergence of a sequence generated by (4) may actually fail. The main objective in this paper is to show that Hundal s recent ingenious counterexample can also be used to provide an armative answer to Reich s question. We then explain how this construction leads to new counterexamples to norm convergence for the classical proximal point algorithm (see [38,49]) and for the string-averaging projection method recently introduced by Censor et al. []. The notion of a strongly nonexpansive map allows us to also comment on the situation in Banach spaces and for the Hilbert ball. The remainder of this paper is organized as follows. Section contains auxiliary results on projectors and reectors as well as a review of the useful properties of strongly nonexpansive maps. The classical results by von Neumann and by Bregman are reviewed in Section 3, where we also include a new elementary proof of von Neumann s result. In Section 4, we describe Hundal s counterexample and show that the composition of the two projectors is not rmly nonexpansive. Our solution to Reich s question is presented in Section 5; it also gives rise to a new counterexample a la Genel and Lindenstrauss. Section 6 contains a self-contained and somewhat more explicit proof of Moreau s result that the proximal maps form a convex set. This is used in Section 7, where it leads to a new counterexample a laguler. Convergence results on string-averaging methods as well as a counterexample are given in Section 8. In the nal Section 9, we discuss the situation in the Hilbert ball. Notation employed is standard in convex analysis: I denotes the identity map and N = {0; ; ;:::} are the nonnegative integers. Also, S = {x X :( s S) x ;s =0} (respectively, S = {x X :( s S) x ;s 6 0}, span S, cone S, conv S, int S, S )is the orthogonal complement (respectively, polar cone, closed linear span, closed convex conical hull, convex hull, interior, and indicator function) of a set S in X. The subdierential map (respectively, gradient map, Fenchel conjugate) of a function f is denoted (respectively, f, f ), and f f stands for the inmal convolution of the functions f and f.ift is a map dened on S, then its xed point set is Fix T = {x S : x = Tx}. Finally, if r R, then r denotes the largest integer less than or equal to r.. Projectors, reectors, and strongly nonexpansive maps Denition. (Nonexpansive and rmly nonexpansive). Let C be a set in X T : C X be a map. Dene a family of functions by and ( x C)( y C) x;y :[0; ] [0; + [: (( )x + Tx) (( )y + Ty) : (5) Then T is rmly nonexpansive (respectively, nonexpansive), if x;y (respectively, x;y (0) x;y ()), for all x and y in C. is decreasing

4 78 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Clearly, if T is rmly nonexpansive, then it is nonexpansive, which in turn is equivalent to ( x C)( y C) Tx Ty 6 x y : (6) Lemma.. Let C be a set in X and let T : C X. Then the following properties are equivalent: (i) T is rmly nonexpansive. (ii) ( x C)( y C) Tx Ty 6 Tx Ty; x y. (iii) T I is nonexpansive. Proof. See [3, Lemma..]. Remark.3 (Extensions to Banach spaces): In Hilbert space, a rmly nonexpansive map is usually dened as in Lemma.(ii); nonetheless, we begin with Denition. because the latter is more useful in Banach spaces (see [3, Section.]). Using the duality map, the characterization in Lemma.(ii) holds true in Banach space (see [3, Lemma..]). Fact.4 (Projector and reector). Suppose that C is a nonempty closed convex set in X. Then, for every point x X, there exists a unique point P C x C such that x P C x = inf y C x y. The point P C x is the projection of x onto C; it is characterized by P C x C and ( c C) c P C x; x P C x 6 0: (7) The corresponding map P C : X C is the projector (or projection operator) onto C. It is rmly nonexpansive and, hence, the associated reector R C =P C I is nonexpansive. Proof. See [30, Chapter, 3, Propositions.3.5 and..], or [55, Lemma.]. Corollary.5. Let C be a nonempty closed convex set in X. Suppose that U : X X is unitary, i.e., a surjective linear isometry. Then U(C) is a nonempty closed convex set, and P U(C) = UP C U. Proof. This follows easily from Fact.4. Corollary.6. Let H be a hyperplane in X and suppose 0 H. Then the corresponding reector R H is unitary and it satises R H = R H = R H. Let C be a nonempty closed convex set in X and set D = R H (C). Then the following hold true: (i) D is closed and convex; (ii) P D = R H P C R H ; (iii) (P H P C ) H = (P C + P D ) H.

5 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Proof. Write H = {a}, where a =. Fix x X arbitrarily. Then P H x = x a; x a and hence R H x = x a; x a. It follows that R H is one-to-one. Also, one veries easily that R H = I and that R H is an isometry. Hence R H is unitary and (i) and (ii) follow from Corollary.5. Now pick h H. Then R H h =P H h h = h and so, using (ii), P D h = R H P C R H h = R H P C h =P H P C h P C h. Hence P C h + P D h =P H P C h, as claimed. We conclude this section with a discussion of the class of strongly nonexpansive maps, introduced by Bruck and Reich [8] in 977. Denition.7 (Strongly nonexpansive): Let C be a set in X and let T : C X be a map. Then T is strongly nonexpansive if it is nonexpansive and (x n y n ) (Tx n Ty n ) 0 whenever (x n ) n N and (y n ) n N are sequences in X such that (x n y n ) n N is bounded and x n y n Tx n Ty n 0. While the class of rmly nonexpansive maps is convex (use Lemma.(iii)), it is not closed under composition (in the Euclidean plane, consider projectors corresponding to two distinct nonorthogonal intersecting lines). This serious limitation does not occur for strongly nonexpansive maps. Fact.8. Let T ;T ;:::;T m be strongly nonexpansive maps dened on a set in X. Suppose T is given in one of the following ways: (i) T = T m T T. (ii) T = m i= it i, where { ; ;:::; m } ]0; [ and m i= i =. Then T is strongly nonexpansive. IfF = m i= Fix T i, then Fix T = F. Proof. [8, Propositions. and.3] imply that T is strongly nonexpansive. To obtain the identity for F, which we now assume to be nonempty, use [8, Lemma.] (for (i)) and [45, Lemma.4] (for (ii)). Denition.9 ((Sunny) retraction). A map T : X C isa(sunny) retraction onto C if it is continuous with Fix T = C (and Tx = c implies T(c + (x c)) = c, for all x X, c C, and 0). Note that C is necessarily closed. Remark.0. Some comments on retractions are in order. See also [3, Section.3] for further information. (i) If C is a nonempty closed convex set in X, then it not hard to show that the projector P C is a sunny nonexpansive retraction onto C [3, (3.7) on p.7]. (ii) Outside Hilbert space, projectors are still sunny retractions, but they fail to be nonexpansive. (iii) Every sunny nonexpansive retraction is rmly nonexpansive [45, Lemma.]. The class of strongly nonexpansive maps is quite rich.

6 70 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Fact.. Let T be a map dened on a set in X. Then T is strongly nonexpansive provided that one of the following conditions holds: (i) T is rmly nonexpansive. (ii) T =( )T + T, where T is strongly nonexpansive, T is nonexpansive, and [0; [. (iv) T : X X is a sunny nonexpansive retraction onto Fix T. Proof. For (i) and (ii), see [8, Propositions.3 and.]. Item (iii) follows from (i) and Remark.0(iii). Remark.. In any smooth Banach space, there is at most one sunny nonexpansive retraction onto a given subset [3, Lemma.3.]; consequently, if T is as in Fact.(iii), then T must coincide with the projector onto Fix T. The next result shows that the iterations of strongly nonexpansive maps are well understood. Fact.3. Let C be a closed convex nonempty set in X, and let T : C C be strongly nonexpansive. Set F = Fix T and pick x C. Then the following hold true: (i) If F =, then lim n T n x =+. (ii) If F, then (T n x) n N converges weakly to some point in F. (iii) If C = C and T is odd, then (T n x) n N converges in norm to some point in F. (iv) T n x T n+ x inf c C c Tc Proof. See [8, Corollaries.4,.3,., and Proposition.], respectively. Remark.4 (Extensions to Banach spaces): The results cited above hold true in considerably more general settings: indeed, Fact.8 is valid for a general Banach space X, while Fact. holds true when X is uniformly convex. We now provide some sucient conditions for the items of Fact.3. (i) C is boundedly weakly compact and each weakly compact convex subset of C has the xed point property for nonexpansive mappings [8, Corollary.4]. (ii) X and X are uniformly convex [45, Proposition.4]. (iii) X is uniformly convex (combine either [8, Corollary.] or [45, Proposition.5] with [3, Theorem.]). (iv) does not require any additional assumption (use [8, Proposition.], or combine [45, Proposition.5] with [43, Proposition 4.3]). Furthermore, norm convergence of (T n x T n+ x) n N to the unique element of minimum norm in the closure of the range of I T is guaranteed either when X is uniformly convex [43, Theorem 3.7(b)], or when the norm of X is Gâteaux dierentiable and the norm of its dual X is Frechet dierentiable [44, Corollary 5.3(b)]. All these conditions are satised provided both X and X are uniformly convex this holds, of course, for the classical L p and p spaces, where p +.

7 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) von Neumann s and Bregman s classical results We now present an elementary geometric proof of von Neumann s original norm convergence result. Theorem 3. (von Neumann). Let A and B be closed linear subspaces in X. Dene the sequence of alternating projections by x 0 X; ( n N) x n+ = P A x n and x n+ = P B x n+ : (8) Then (x n ) n N converges in norm to P A B x 0. Proof. Set C = A B. By Pythagoras, we have ( n N) x n = x n+ + x n x n+ : (9) In particular, ( x n ) n N is decreasing and nonnegative; hence convergent: (0) By (strong) induction on n, we now establish the related statement ( n N) ( k N)( l N) 6 k = l n x k x l 6 x k x l : () Clearly, () is true for n = 0 and also for n = (by (9)). So assume () holds true for some n, and take k; l in N such that 6 k = l (n + ). Case : n is even. Then n +=l k is odd. If l is odd, then both x k+ = P A x k and x l = P A x l belong to A, whereas x k x k+ =(I P A )x k A ; hence altogether x k x k+ ;x k+ x l =0: () If l is even, we argue similarly with A replaced by B and we derive () once again. Using (), (9), and the induction hypothesis, we now obtain x k x l = x k x k+ + x k+ x l = x k x k+ + x k+ x l 6 x k x k+ + x k+ x l = x k x l : Case : nisodd. Then n +=l k is even, which similarly to the derivation of () implies that x k x l ;x l x l =0: (3)

8 7 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Thus x k x l = x k x l + x l x l. Using this, (9), the induction hypothesis, and (0), we conclude x k x l = x k x l x l x l = x k x l x l + x l 6 x k x l x l + x l 6 x k x l : Altogether, statement () is veried. Now, by (0) and (), the sequence (x n ) n N say to is Cauchy and hence convergent, x = lim x n : (4) n N Since (x n+ ) n N lies in A and (x n+ ) n N lies in B, we conclude that x A B = C. Therefore, P C x n P C x = x : (5) Now x n N and t R, and set c =( t)p C x n + tp C x n+. Then c C = A B = Fix (P A ) Fix (P B ) and so P A c = P B c = c. Also, x n+ {P A x n ;P B x n }. Since projectors are nonexpansive (Fact.4), we obtain x n+ c 6 x n c. After squaring and simplifying, this inequality turns into ( t) P C x n+ P C x n + P C x n+ 6 P C x n : (6) Since n and t were chosen arbitrarily, we conclude ( n N) P C x n = P C x n+ : (7) Combining (4), (5) and (7), we obtain lim n N x n = P C x 0. Remark 3.. See [50 5] for classical proofs of Theorem 3., and also [7, Chapter 9] for recent information and further pointers to the literature. The convergence part in the proof given above is a modication of the proof of [8, Proposition, p. 05]. The part determining the limit is borrowed from [, Fact.(v)]; see also [4, Theorem 6..], [3, Theorem.]. Let us sketch the following dierent approach. Dene L : X A B by x 0 lim n x n. Then L is nonexpansive and Fix L = A B. Hence A B is a nonexpansive retract of A B. Therefore, using either [4, bottom of p. 6] or [45, Lemma 3.], L = P A B. (Alternatively, one can check directly that L = L, which implies x 0 Lx 0 (A B) and thus Lx 0 = P A B x 0.) Theorem 3. immediately raises the following question: what can be said about the case where the sets A and B are merely two closed convex sets with nonempty intersection? Let us recall Bregman s basic weak convergence result from 965.

9 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Theorem 3.3 (Bregman). Let A and B be closed convex sets in X such that A B. Dene the sequence of alternating projections by x 0 X; ( n N) x n+ = P A x n and x n+ = P B x n+ : (8) Then (x n ) n N converges weakly to some point in A B. Proof. The original proof can be found in [5]. We include the following short proof (for another approach via Fejer monotonicity, see [5, Theorem.0(i)]). The projectors P A and P B are rmly nonexpansive (Fact.4), and hence strongly nonexpansive (Fact.(i)). By Fact.8(i), P B P A is strongly nonexpansive with Fix (P B P A )=A B. Using Fact.3(ii), we see that (x n ) n N converges weakly to some point c A B. Now x n c x n+ c x (n+) c, for all n N; consequently, x n c P A x n c 0. Since P A is strongly nonexpansive, it follows that x n x n+ =x n P A x n 0. Hence (x n+ ) n N converges weakly to c, and so does the entire sequence (x n ) n N. Remark 3.4 (Extensions to the inconsistent case): For further results on the behavior of the sequence of alternating projections in the inconsistent case (i.e., when A B= ), see [6,7,9], and the references therein. Remark 3.5 (Extensions to Banach spaces): A closer inspection of the proof of Theorem 3.3 reveals that the following generalization holds true. Suppose X is a Banach space such that both X and X are uniformly convex, and C is a closed convex nonempty set in X. Let T ;T be two strongly nonexpansive maps from C to C with F = Fix (T ) Fix (T ). Then for every x 0 C, the sequence (x n ) n N generated by ( n N) x n+ = T x n and x n+ = T x n+ (9) converges weakly to some point in F. 4. Sequential projectors and Hundal s counterexample The gap of knowledge between Theorems 3. and 3.3 was closed after nearly four decades. In 000, during the Haifa workshop on Inherently Parallel Algorithms in Feasibility and Optimization and Their Applications [9], Hein Hundal outlined his construction of two sets such that the corresponding sequence of alternating projections converges weakly, but not in norm. Moreover, since the sets are a hyperplane and a cone, his counterexample shows that there is no hope of extending Theorem 3. even to cones. Let us write = span {e ; e ;:::;}, where the nth standard unit vector e n has a one at position n, and zeros elsewhere. We now describe Hundal s construction. Fact 4. (Hundal s counterexample). Let X = = span {e ; e ;:::;} and dene v by v :[0; + [ r exp( 00r 3 )e + cos((r r )=)e r + + sin((r r )=)e r +3 : (0)

10 74 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Further, dene Hundal s hyperplane H, cone K, and starting point y 0 by H = {e } ; K = cone {v(r):r 0}; and y 0 = v(): () Then the closed convex cone K satises sup e ;K =0 and H K = {0}. Moreover, Hundal s sequence of alternating projections (y n ) n N, given by ( n N) y n+ = P K P H y n ; () converges weakly to 0, but not in norm. Proof. See [34]. Remark 4.. Some comments regarding Fact 4. are in order. (i) While lengthy, Hundal s construction is self-contained and elementary (in the sense that no external advanced results are utilized). (ii) Hundal formulated his example with the cone K dened in () and with H replaced by the halfspace {e } = {x : e ;x 6 0}. This is fully equivalent to how we stated his example because K int({e } )=. However, for our purposes, it is more convenient to work with the hyperplane H instead. (iii) For future use, we point out now that (P H y n ) n N converges weakly to 0; but not in norm: (3) Indeed, weak convergence is implied by the fact that H K = {0} and Theorem 3.3 (or, since H is a closed linear subspace, by the weak continuity of P H ). The lack of norm convergence is seen as follows: since 0 H K and projectors are nonexpansive (Fact.4), we have y n = y n 0 P H y n P H 0 = P H y n 0 P K P H y n P K 0 = y n+, for all n N. Hence 0 inf n N y n = inf n N P H y n, and thus (P H y n ) n N fails to converge to 0 in norm. We now show that neither P H P K nor P K P H is rmly nonexpansive; thus, Fact 4. does not contain an obvious counterexample to norm convergence of iterates of rmly nonexpansive maps a la Genel and Lindenstrauss [9]. However, it is conceivable but it seems unlikely that some powers of these compositions are rmly nonexpansive. Conversely, the rmly nonexpansive map in the counterexample by Genel and Lindenstrauss [9] does not appear to be the product of projectors. Lemma 4.3. Let H and K be Hundal s hyperplane and Hundal s cone, respectively. Then neither P K P H nor P H P K is rmly nonexpansive. Proof. P K P H : We require some notation and results from [34]. Let p, h, and P be as in [34], and let R satisfy [34, Eq. (4)]. For R, set x()=p()+e. For all suciently negative, we have e ;x() 6 0. Then e ;P H x() = 0 for all suciently negative and hence P H x() =P p(). By [34, Theorem 3.4], there exists ˆ ]; + [ and ˆ ]0; [ such that P K P H x()=p K P p()= ˆp(ˆ). Note that

11 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) the last term is independent of and that = e ; ˆp(ˆ) =ˆ + h() 0. Thus P K P H x();x() = P K P H x();p H x() + P K P H x();p H x() = P K P H x();p H x() + ˆp(ˆ); (h()+)e = P K P H x();p H x() +(h()+): (4) The last term, (h()+), can be made arbitrarily negative by choosing suciently negative. In particular, for all suciently negative, we have P K P H x() P K P H 0 = P K P H x() P K P H x();x() = P K P H x() P K P H 0;x() 0 : (5) In view of Lemma.(ii), P K P H is not rmly nonexpansive. P H P K : Assume to the contrary that P H P K is rmly nonexpansive. On the one hand, since P H P K 0 = 0, it follows that P H P K x 6 P H P K x; x, for all x X. On the other hand, using the Moreau decomposition (see [4, Corollaire 4.b]) I = P K + P K and the fact that P H is a self-adjoint idempotent, we have for all x X : P H P K x; x = P H P K x; P K x + P K x = P H P K x + P H P K x; P K x. Altogether, P H P K x; P K x 0 and so P H (K);K 0: Now set X + = {(x n ) n N :( n N)x n 0}. Then K X + and (X + ) = X + ; hence, X + K. Thus, using (6), we see that P H (K); X + 0: Further, P H (K) P H (X + ) X + and hence P H (K) X +. Now (7) yields P H (K) 0 so that P H (K)={0}, which is the desired contradiction. (6) (7) 5. Averaged projectors We now show that Hundal s sequence can be viewed as the iterates of an average of two projectors. Consequently, since Hundal s sequence fails to converge in norm, we have obtained a counterexample to the norm convergence of iterates of averaged projectors. This provides not only an answer to a question posed by Reich (see the last paragraph in [45, Section ]) that was unresolved for two decades, but also a new counterexample a la Genel and Lindenstrauss [9]. Theorem 5. (An iteration of averaged projectors that fails to converge in norm). Let H, K, and (y n ) n N be Hundal s hyperplane, cone, and sequence, respectively (see Fact 4.). Dene L = R H (K), z 0 = P H (y 0 ), and ( n N) z n+ = ( P K + P L) zn : (8) Then (z n ) n N =(P H y n ) n N and this sequence converges weakly to 0, but not in norm.

12 76 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Proof. Suppose that, z n = P H y n, for some n N. Then z n H; hence, using () and Corollary.6(iii), we deduce y n+ = P H P K y n = P H P K z n = ( P K + P L) zn = z n+.by induction, (z n ) n N =(P H y n ) n N. The result now follows from Remark 4.(iii). Corollary 5. (A new counterexample a la Genel and Lindenstrauss). Let K, L, and z 0 be as in Theorem 5., and set T = P K + P L. Then T is rmly nonexpansive and (T n z 0 ) n N converges weakly to 0 Fix T, but not in norm. Proof. By Fact.4, the reectors R K and R L are nonexpansive, hence so is R= R K + R L. Now Lemma. implies that T = P K + P L = R + I is rmly nonexpansive. Since 0 K, it follows that 0 R H (K)=L and hence 0 = T0. The statement regarding the convergence of (z n ) n N has already been observed in Theorem 5.. Remark 5.3. In 975, Genel and Lindenstrauss provided the rst example of an iteration of a rmly nonexpansive map that fails to converge in norm; see [9] and also [4, pp. 7 74]. Corollary 5. is a new example of this kind. It is somewhat more explicit because the original construction in [9] relies upon the Kirszbraun Valentine theorem (see [35] and also [30, Theorem.4]). Remark 5.4 (Extensions to Banach spaces): The following results are drawn from [45]; their proofs also depend on the useful properties of strongly nonexpansive maps sampled in Section. Let T = m i= ir i, where m i= i =, each i 0, and each R i is a retraction onto some closed convex set C i in X (see Denition.9) such that C = m i= C i. Fix x 0 X. Then (T n x 0 ) n N converges in norm to some point in C if (i) X is a uniformly convex Banach space and each R i is a linear projection of norm one onto a subspace C i [45, Theorem.7]; or if (ii) X is smooth and uniformly convex, each C i is symmetric, and each R i is a sunny nonexpansive retraction onto C i [45, Theorem.3]. Moreover, the sequence (T n x 0 ) n N converges at least weakly, provided that (iii) X is both uniformly convex and uniformly smooth, and each R i is a sunny nonexpansive retraction onto C i [45, Theorem.5]. In particular, all these results are applicable to projectors P Ci in Hilbert space. Returning to our usual setting, we note that the sucient conditions (i) (iii) from above become the following: (i) each C i is a closed linear subspace; (ii) each set C i is symmetric; and (iii) is always satised. 6. Proximal maps We denote the proximal map (see [6,4], or [49]) of a given proper lower semicontinuous convex function f : X ] ; + ] by Prox(f): Prox(f)=(I = ( f ) : (9) For example, if C is a nonempty closed convex set in X, then Prox( C )=P C. Rockafellar proved is maximal monotone [48]. The resolvent of a maximal monotone operator A : X X is dened by (I + A). Minty characterized resolvents as rmly nonexpansive maps with full domain [40]. In particular, every proximal

13 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) map is rmly nonexpansive with full domain, but the converse is false consider, for instance, R R :(; ) 4 5 ( ; ). In this section, we show that the proximal maps form a convex set, a result rst observed by Moreau [4] (see also Remark 6. below). Consequently, if H and K denote Hundal s hyperplane and cone, respectively (see Fact 4.), then P H + P K, the midpoint average of the projectors P H and P K, is not only rmly nonexpansive (Corollary 5.), but also a proximal map. This will be exploited in the next section, where we provide a new counterexample to norm convergence for sequences generated by the proximal point algorithm. Theorem 6.. Let f and f be functions from X to ] ; + ] that are convex, lower semicontinuous, and proper. Suppose and belong to ]0; [, + =, and set f = ( ( f ) + ( f )) : (30) Then f is a proper lower semicontinuous convex function such that f = ( ( f ) + ( f )) (3) and Prox(f) = Prox(f )+ Prox(f ). If each f i is the indicator function of a nonempty closed convex set C i in X, then f(x)= inf { c c } : c + c = x and each c i belongs to C i for every x X. = d (0; ( (C x)) ( (x C ))) (3) Proof. For convenience, we set j =. It is well-known that j = j and dom j = X. For i {; }, welet h i = j f i and g i = i h i : (33) Since each f i is proper, lower semicontinuous, and convex, so is each h i = j f i ; moreover, each h i has full domain. It follows that g + g is lower semicontinuous and convex, with full domain. Thus (g + g ) is proper, which shows that f =(g + g ) j (34) is proper as well. Later, we require (g + g ) = g g ; (35)

14 78 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) an identity proven by Attouch and Brezis []. Now x x X. Then f (x ) = sup (g + g ) (x + z ) j(z ) z = sup (g + g )(x + z ) j(z ) z = sup (g + g )(y ) j(y x ) y = sup (j f )(y )+ (j f )(y ) (j(y )+j(x ) x ;y ) y = sup (j(y ) (j f )(y )) + (j(y ) (j f )(y )) y (j(y )+j(x ) y ;x ) = j(x ) + sup x ;y ( (j f )+ (j f ))(y ) y =( (j f )+ (j f )) (x ) j(x ); (36) where in the st equality, we used the fact that (g +g ) is proper (since g +g is), j=j, dom j =X, and [33, Theorem.]; the nd equality is obtained by noting that g + g is proper, lower semicontinuous and convex, so that the Biconjugate Theorem [54, Theorem.3.3] holds; we changed variables in the 3rd equality; the 4th equality follows from the denitions; Moreau s result that (w j)+(w j)=j for every proper, lower semicontinuous, and convex function w (see [4, Eq. (9.)] or prove it directly via Fenchel duality) implies the 5th equality; the two remaining equalities follow from the denitions. Since x was chosen arbitrarily, we have proved the announced identity for f, namely (( (j f )+ (j f )) j) =( (j f )+ (j f )) j: (37) Now this identity holds true for any two proper lower semicontinuous convex functions f ;f from X to ] ; + ]. Applying (37) tof ;f, we obtain (( (j f )+ (j f )) j) =( (j f )+ (j f )) j: (38) On the one hand, the left side of (38) is a conjugate function, hence it is convex and lower semicontinuous. On the other hand, the right side of (38) is j ust f. Altogether, f is convex, lower semicontinuous, and (as observed in (34)) proper. It remains to establish the identity regarding the proximal map of f. The denitions and a further dose of convex calculus now yield f =(g + g ) j j + f =(g + g ) (j + f) = g + g j f = g + g j f = h + h (39)

15 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Therefore, Prox(f)= (j f )= ( h + h )= (h )+ (h )= (j f )+ (j f )= Prox(f )+ Prox(f ). Now suppose f i = Ci, for each i {; }, and x x X. Using (35), we obtain f(x)=(g g )(x) j(x) = inf x g +x =x (x )+g(x ) j(x) ( ( ) = inf x ) x x +x =x + C + ( x = inf x +x =x and each x i= i C i ( ) ) x + C x + x x + x x + x = inf c + c =x and each c i C i c + c c + c = inf c + c =x and each c i C i c c ; (40) which establishes the rst equality in (3). If each c i belongs to C i and x = c + c, then c (x C )=, c (x C )=, and c c = x c = = x c =. This implies that f(x)= d (x; C ((x C )= )) = = = d (0; (C x) ((x x C )= )) d (0; (C x) (( = )(x C ))) d (0; ( (C x)) ( (x C ))): (4) Therefore, the second equality of (3) is veried. Remark 6.. Theorem 6. implies that the proximal maps form a convex set, an observation originally due to Moreau [4, Proposition 9.d]. The present proof appears to be more straightforward and it also provides an explicit formula of the function corresponding to the convex combination of the proximal maps. This will aid us in the derivation of the function f in Corollary 7. below.

16 730 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Proximal point algorithm We are now in a position to derive a new counterexample to the norm convergence of the proximal point algorithm a la Guler. Corollary 7. (A new proximal point iteration that fails to converge in norm). Let K, L, and z 0 be as in Theorem 5., and set f(x) = d (0; (K x) (x L)), for every x X. Then f is convex, lower semicontinuous, proper, and its proximal map is Prox(f)= P K + P L. Moreover, the sequence ((Prox(f)) n z 0 ) n N converges weakly to 0, but not in norm. Proof. Combine Corollary 5. and Theorem 6. (where = = ). Remark 7. (Proximal point algorithm): Let f : X ] ; + ] be convex, lower semicontinuous, and proper. Recall that (see the classical papers by Martinet [38] and by Rockafellar [49]) given a sequence of strictly positive parameters ( n ) n N and a starting point x 0 X, the inductive update rule of the proximal point algorithm for minimizing f is ( x n+ = argmin f(x)+ ) x x n x X n =(I + (x n ) ( = ( + n f ) ) (x n ) = (Prox( n f))(x n ) (4) for all n N. Brezis and Lions [7] show that if f has minimizers and n N n=+, then the sequence generated by (4) converges weakly to a minimizer of f. The question whether norm convergence always holds remained open until 99, when Guler [3] constructed a whole family of counterexamples for all strictly positive parameter sequences ( n ) n N such that n n =+. In [0], Fact 4. has been interpreted as a simple counterexample to norm convergence of the proximal point algorithm with n 0. We now realize that Corollary 7. provides another counterexample of this type, for the case when n. 8. String-averaging projection methods Let C, C, and C 3 be closed convex sets in X with corresponding projectors P, P, and P 3, respectively. Suppose further that C = C C C 3 and dene T = P P + 4 P + 4 P 3: (43) Note that T is neither a composition nor an average of the given projectors. However, the iterates of the operator T can be analyzed within the very exible string-averaging algorithmic structure proposed recently by Censor et al. [] (see also [0,,6]).

17 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Roughly speaking, a string corresponds to a point obtained through the sequential application of operators to the current point, and the update step consists of averaging the resulting strings. This framework is clearly inspired by parallel computing architectures, where each processor can work independently on its string. Thus, the application of the operator T can be viewed as a string-averaging algorithm, where P P ;P ;P 3 are three strings that are averaged by the weights, 4, 4, respectively. We now provide a prototypical convergence result for the iterates of the operator T. Theorem 8.. Let T be dened by (43) and x x 0 X. Then the sequence (T n x 0 ) n N converges weakly to some point in C. If each set C i is symmetric (i.e., C i = C i ), then (T n x 0 ) n N converges in norm. Proof. Each projector is rmly nonexpansive hence strongly nonexpansive (Fact.(i)). Repeated use of Fact.8 shows that the map T is strongly nonexpansive as well, and its xed point set is C. Fact.3(ii) now implies that (T n x 0 ) n N converges weakly to a point in C. Assume that each set C i is symmetric. Then C is symmetric as well. Also, [45, Lemma.] implies that each P i is odd (P i ( x)= P i x, for all x X ). Thus T is odd. The norm convergence of (T n x 0 ) n N now follows from Fact.3(iii). Remark 8.. It is clear that the convergence proof of Theorem 8. will also work for iterations of various other maps as long as they are assembled from projectors by averaging and by composition. (See [8, Examples.4 and.0] for another example of a string-averaging method.) For convergence results in nite-dimensional spaces, see [0,6]. Remark 8.3 (Extensions to Banach spaces): A closer inspection of its proof reveals that Theorem 8. holds true under the following more general conditions: X is a Banach space such that both X and its dual X are uniformly convex, and each C i is the xed point set of a strongly nonexpansive map P i. The norm convergence requires additionally that each P i be odd. We now combine Theorem 8. with Theorem 5. to obtain a counterexample to norm convergence for string-averaging projection methods. Corollary 8.4. Let H, K, y 0 be Hundal s hyperplane, cone, and starting point, respectively. Further, dene L = R H (K) and T = P H P K + 4 P K + 4 P L: (44) Then the sequence (T n P H y 0 ) n N converges weakly to 0, but not in norm. Proof. By Fact 4., H K L = {0}. Hence, Theorem 8. implies that (T n P H y 0 ) n N converges weakly to 0. Suppose x H. By Corollary.6(iii), P H P K x = P Kx + P Lx and thus Tx = P H P K. Using induction, it follows that (T n P H y 0 ) n N is equal to the sequence (z n ) n N of Theorem 5.; consequently, it does not converge in norm.

18 73 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Remark 8.5. In fact, a second glance at the proof of Corollary 8.4 reveals that the conclusion of this result holds whenever ( T =! (P H P K )+! (P H P L )+! 3 P K + P L) ; (45) where! ;! ;! 3 belong to [0; ] and! +! +! 3 =. 9. The Hilbert ball Throughout this last section, we let Y be a complex Hilbert space with inner product ; and induced norm. Suppose that X = {x Y : x } is the open unit ball of Y and dene ( x X )( y X ) (x; y)= ( x )( y ) x; y and (x; y) = arctanh (x; y): (46) Then (X; ) is a complete metric space, commonly referred to as the Hilbert ball (see [3, p. 9 and Section.5]; for further background material on this and on what follows, see [3, Chapter ]. Given x and y in X, and [0; ], the results in [3, Section.7] show that there exists a unique point z X such that (x; z)=(x; y) and (z; y)=( )(x; y); this point is denoted by ( )x y (47) and it is a -convex combination of x and y. A set S in X is called -convex (respectively, -closed), if it contains all its -convex combinations (respectively, if it is closed in (X; )). Given T : X X, let us dene ( x X )( y X ) x;y :[0; ] [0; + [: (( )x Tx; ( )y Ty): (48) Analogously to (5), one says that T is rmly -nonexpansive of the rst kind (respectively, -nonexpansive), if x;y is decreasing (respectively, x;y (0) x;y ()) for all x and y in C (see [3, pp. 73, 4]). As in the Hilbert space case (6), T : X X is -nonexpansive if and only if ( x X )( y X ) (Tx; Ty) 6 (x; y); (49) in which case Fix T is -closed and -convex [3, Theorem.3.]. Fact 9. (Projectors in the Hilbert ball). Suppose C is a -closed and -convex set in X. Then for every x X, there exists a unique point in C, denoted P C x, such that (x; P C x) = inf y C (x; y). The induced map P C : X C is rmly -nonexpansive of the rst kind.

19 Proof. See [3, Section.9]. H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Given y X and s 0, the set {x X : (x; y) s} is the open -ball of radius s centered at y (see also [3, Section 5] for characterizations). A set S in X is -bounded, if it is contained in some open -ball. A sequence (x n ) n N in X is -bounded, if its orbit {x n : n N} is. A map T : X X is called para-strongly -nonexpansive (see [47], where this was called strongly nonexpansive for brevity), if T is -nonexpansive, Fix T, and for every -bounded sequence (x n ) n N and every y Fix T, the condition (x n ;y) (Tx n ;y) 0 implies (x n ;Tx n ) 0. Analogously to Denition.9, a map T : X C X is called a -retraction onto C, ift is continuous in (X; ) and Fix T = C. The following basic results on para-strongly -nonexpansive maps are due to Reich [47]. Fact 9.. Let T; T ;T be maps from X to X with xed point sets F; F ;F, respectively. (i) If T is rmly -nonexpansive of the rst kind and F, then T is para-strongly -nonexpansive. (ii) If T is para-strongly -nonexpansive, then the map X F : x weak lim n T n x (50) is a well-dened -nonexpansive -retraction onto F. (iii) If each T i is para-strongly -nonexpansive and F F, then T T is para-strongly -nonexpansive and F = F F. Proof. (i): [47, Lemma ]. (ii): [47, Lemmata 5 and 6]. (iii): [47, Lemmata 3 and 4]. Corollary 9.3. Suppose A and B are -closed -convex sets in X such that A B. Then for every x X, the sequence (( P B P A ) n x) n N converges weakly to some point in A B. Proof. (See also [47].) The projectors P A and P B are rmly -nonexpansive of the rst kind (Fact 9.), hence para-strongly -nonexpansive (Fact 9.(i)). By Fact 9.(iii), T = P B P A is para-strongly -nonexpansive and Fix T = A B. The weak convergence statement now follows from Fact 9.(ii). Remark 9.4. Reich asks at the end of [47] whether the convergence in Corollary 9.3 is actually strong. While Hundal settled the corresponding question in Hilbert space (Fact 4.), the problem posed by Reich remains open in the Hilbert ball. The next result exhibits two analogues of Fact.8(ii).

20 734 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Theorem 9.5. Let T ;T be para-strongly -nonexpansive maps from X to X such that Fix (T ) Fix (T ). Suppose ; are in ]0; [ and + =. Now dene T in one of the following two ways. (i) T : X X : x T x T x; (ii) T : X X : x T x + T x. Then T is para-strongly -nonexpansive with Fix T = Fix (T ) Fix (T ). Proof. (i). Fact.8(ii) states that the corresponding Hilbert (and even Banach) space statement is true. It turns out that the corresponding proofs of [47, Lemmata 3 and 4] carry over if we utilize [3, Lemma.7.] at the appropriate places. (ii). Clearly, Fix (T ) Fix (T ) Fix T. Conversely, x y Fix (T ) Fix (T ) and pick x Fix T. By[36, Lemma 3.3(i)], (x; y)=(tx; Ty)=(Tx; y) 6 max{(t x; y); (T x; y)}. Without loss of generality, assume that max{(t x; y);(t x; y)} = (T x; y). Since T is para-strongly -nonexpansive, we obtain (x; y) 6 (T x; y) 6 (x; y) and hence T x = x. Since x = Tx, this implies T x = x. We conclude that Fix T Fix (T ) Fix (T ). Next, pick a -bounded sequence (x n ) n N such that (x n ;y) (Tx n ;y) 0: (5) We have to show that (x n ;Tx n ) 0. Assume to the contrary this were false. Using [36, Lemma 3.3(i)] again, and after repeatedly passing to subsequences and relabelling if necessary, we obtain 0 and M 0 such that ( n N) 6 (x n ;Tx n ); max{(tx n ;y);(t x n ;y)} 6 (T x n ;y); and (x n ;y) M: (5) Hence 0 (Tx n ;y) (x n ;y) 6 (T x n ;y) (x n ;y) 6 0 and thus (T x n ;y) (x n ;y) 0: (53) Since T is para-strongly -nonexpansive, it follows that (T x n ;x n ) 0: (54) Moreover, (5) (53) imply that (T x n ;y) M; lim sup (T x n ;y) 6 M and (Tx n ;y) M: (55) Now let (u n ) n N and (v n ) n N be the two sequences in X that are uniquely dened by the following properties: { un conv{t x n ;Tx n }; v n conv{tx n ;T x n } and ( n N) (56) u n Tx n = v n Tx n =min{ T x n Tx n ; T x n Tx n }: Then, by construction, ( n N) u n + v n = Tx n and T x n T x n = u n v n min{ ; } : (57)

21 H.H. Bauschke et al. / Nonlinear Analysis 56 (004) Using [36, Lemma 3.3(i)] and (5), we obtain (u n ;y) 6 max{(t x n ;y);(tx n ;y)}= (T x n ;y) and (v n ;y) 6 max{(tx n ;y);(t x n ;y)} 6 (T x n ;y), for every n N. Combining these inequalities with (55) and (57), we see that lim sup (u n ;y) 6 M; lim sup (v n ;y) 6 M and lim ( u n + v n;y ) = M: (58) Now (58) and [46, Lemma 4] result in u n v n 0; equivalently (see (57)), T x n T x n 0. By (55), the sequences (T x n ) n N and (T x n ) n N are both -bounded; therefore, using [36, Theorem 3.4], we conclude that (T x n ;T x n ) 0. The triangle inequality and (54) yield (T x n ;x n ) 0: (59) Finally, (5), (54), (59), and [36, Lemma 3.3(i)] imply the contradiction 6 (Tx n ;x n ) 6 max{(t x n ;x n );(T x n ;x n )} 0. Corollary 9.6. Suppose A and B are -closed -convex sets in X such that A B. Then for every x X, the sequences ( ( P A P ) n ( ( B x and )n N P A + P ) n B x (60) )n N both converge weakly to some point in A B. Proof. Combine Theorem 9.5 with Fact 9.(ii). Remark 9.7. Whether the convergence in Corollary 9.6 is actually strong is an interesting open problem. Theorem 5. illustrates the failure of norm convergence in the corresponding Hilbert space setting. Remark 9.8. A map T : X X is -averaged (respectively, -averaged of the second kind), if it is of the form T =( )I T (respectively, T =( )I + T ), where T is -nonexpansive and [0; [; see [46] if furthermore =, we say that the map is -midpoint averaged. It is known that if T is -averaged of either kind with Fix T and x X, then (T n x) n N converges weakly to some point in Fix T (see [46, Theorems 3 and 5]). Remark 9.9. Kuczumow and Stachura [37] (see also [36, Example 0.6 p. 475]) constructed -midpoint averaged maps of both types such that some sequence of iterates fails to converge in norm these counterexamples are similar to (and their construction is based upon) the corresponding counterexample by Genel and Lindenstrauss in Hilbert space [9]. We do not know whether the Kuczumow Stachura maps can be expressed as either P A P B or P A + P B, for some -closed -convex sets A; B in X. We conclude with the following striking dierence between Hilbert space and the Hilbert ball. If C is a closed convex nonempty set in a Hilbert space, then (Fact.4) R C is nonexpansive and thus P C = I + R C is (even midpoint) averaged. In the Hilbert ball, the corresponding statement is known to be false; see [3, Example..].

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