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Journal of Nonlinear and Convex AnalysisVolume 6, Number 1, 2005, 93–116

93

A CONJECTURE BY DE PIERRO IS TRUE FOR TRANSLATES

OF REGULAR SUBSPACES

HEINZ H. BAUSCHKE AND MCLEAN R. EDWARDS

Abstract. Suppose we are given finitely many nonempty closed convex sets ina real Hilbert space and their associated projections. For suitable arrangementsof the sets, it is known that the sequence obtained by iterating the composi-tion of the underrelaxed projections is weakly convergent. The question ariseshow these weak limits vary as the underrelaxation parameter tends to zero. In2001, De Pierro conjectured that the weak limits approach the least squares so-lution nearest to the starting point of the sequence. In fact, a result by Censor,Eggermont, and Gordon implies De Pierro’s conjecture for affine subspaces inEuclidean space.

This paper extends the result by Censor et al. from Euclidean to Hilbert space.We show that De Pierro’s conjecture is true for translates of regular subspaces andthe limits all exist with respect to the norm topology. Regularity always holds inEuclidean space. However, this condition is not automatic in infinite-dimensionalHilbert space. Two subspaces are constructed to illustrate the possible divergenceof the iterates of the composition of the underrelaxed projections. Somewhatsurprisingly, examples in the Euclidean plane demonstrate that the approach tothe least squares solution can be nonlinear.

1. Introduction

Throughout this paper, we assume that

(1.1) X is a real Hilbert space with inner product 〈·, ·〉 and induced norm ‖ · ‖,

and that

(1.2) C1, . . . , CN are finitely many nonempty closed convex subsets of X

with

(1.3) corresponding projectors PC1 , . . . , PCN.

Recall (see, e.g., [15, Chapter 5]) that the projector (or nearest point mapping)associated with a nonempty closed convex set C in X is the mapping PC : X →C : x 7→ PCx, where PCx is the unique minimizer of the optimization problem

2000 Mathematics Subject Classification. Primary 47H09, 65K05; Secondary 65F10, 65F20,90C25.

Key words and phrases. angle, Hilbert space, least squares solution, nonexpansive, projection,regularity, subspaces.

94 HEINZ H. BAUSCHKE AND MCLEAN R. EDWARDS

defining the distance of x to C,

(1.4) d(x,C) := infc∈C‖x− c‖,

i.e., PCx is the projection of x onto C, characterized by

(1.5) PCx ∈ C and supc∈C

〈c− PCx, x− PCx〉 ≤ 0.

For every λ ∈ ]0, 1], we set

(1.6) Qλ :=((1− λ) Id+λPCN

)· · ·

((1− λ) Id+λPC1

)

and we define the corresponding sets of fixed points by

(1.7) Fλ := Fix Qλ := {x ∈ X : x = Qλx}.

We aim to understand the behaviour of the sequence (Qnλx)n∈

� in terms of λ ∈ ]0, 1],for an arbitrary x ∈ X. Bruck and Reich’s seminal work on strongly nonexpansivemappings (see [9] and also [20]), specialized to our present setting, aids us greatlyin this task.

Definition 1.1 (strongly nonexpansive). Let T : X → X be nonexpansive, i.e.,

(1.8) (∀x ∈ X)(∀y ∈ X) ‖Tx− Ty‖ ≤ ‖x− y‖.Then T is said to be strongly nonexpansive, if (xn−yn)−(Txn−Tyn)→ 0 whenever(xn)n∈

� and (yn)n∈� are sequences in X such that (xn − yn)n∈

� is bounded and‖xn − yn‖ − ‖Txn − Tyn‖ → 0.

Fact 1.2 (basic properties of strongly nonexpansive mappings). The following state-ments are true for mappings from X to X.

(i) Every projector associated with a nonempty closed convex set is stronglynonexpansive.

(ii) The class of strongly nonexpansive mappings is closed under convex combi-nations. Moreover, if T1, . . . , TM are strongly nonexpansive,

{λ1, . . . , λM} ⊂ ]0, 1],∑M

i=1 λi = 1, and⋂M

i=1 FixTi 6= Ø, then

(1.9) Fix∑M

i=1 λiTi =⋂M

i=1 Fix Ti.

(iii) The class of strongly nonexpansive mappings is closed under composition.

Moreover, if T1, . . . , TM are strongly nonexpansive and⋂M

i=1 Fix Ti 6= Ø,then

(1.10) FixTM · · · T1 =⋂M

i=1 Fix Ti.

(iv) If a strongly nonexpansive mapping possesses at least one fixed point, thenevery sequence generated by iterating the mapping converges weakly to somefixed point; otherwise, the sequence has no bounded subsequence.

(v) The iterates of an odd strongly nonexpansive mapping converge strongly toa fixed point.

A CONJECTURE BY DE PIERRO 95

Proof. (i): It is well-known that every projector is firmly nonexpansive (see, e.g.,[15, Chapter 5]). The conclusion now follows from [9, Proposition 2.1], which statesthat every firmly nonexpansive mapping is strongly nonexpansive. (ii): See [9,Proposition 1.3] and [20, Lemma 1.3 and Lemma 1.4]. (iii): [9, Proposition 1.1 andLemma 2.1]. (iv): [9, Corollary 1.3 and Corollary 1.4]. (v): [9, Corollary 1.2]. �

Given a nonempty closed convex set C ⊂ X and λ ∈ ]0, 1], it will be convenientto set

(1.11) Rλ,C := (1− λ) Id+λPC .

Clearly, Id = PX is strongly nonexpansive and so is Rλ,C by Fact 1.2(i)&(ii).

Corollary 1.3. Let λ ∈ ]0, 1] and x ∈ X. Then:

(i) The mapping Qλ is strongly nonexpansive.

(ii) If⋂N

i=1 Ci 6= Ø, then Fλ =⋂N

i=1 Ci.(iii) If Fλ 6= Ø, then (Qn

λx)n∈� converges weakly to some point in Fλ.

(iv) If Fλ = Ø, then limn→+∞ ‖Qnλx‖ = +∞.

Proof. Our observation following (1.11) and Fact 1.2(ii) show that

(1.12) each Rλ,Ciis strongly nonexpansive and Fix Rλ,Ci

= Ci.

(i)&(ii): Combine (1.12) with Fact 1.2(iii). (iii)&(iv): This follows from (i) andFact 1.2(iv). �

If Fλ 6= Ø, then an alternative proof of Corollary 1.3(iii) can be based uponconvergence results on averaged mappings. See [12, Section 7] for further details.

The following notion of a least squares solution is required throughout the paper.

Definition 1.4 (least squares solutions). A point x ∈ X is a least square solution(associated with the given sets C1, . . . , CN ), if and only if

(1.13)∑N

i=1‖x− PCix‖2 = inf

y∈X

∑Ni=1‖y − PCi

y‖2;

the set of all such points is denoted by L.

Note that L coincides with⋂N

i=1 Ci provided the intersection is nonempty. Theset L serves as a tremendously useful “generalized intersection” and numerous algo-rithms approximate a point in it; see, e.g., [3], [11], [14], and the references therein.Moreover, L can be characterized as follows.

Fact 1.5. L = Fix∑N

i=11N

PCi.

Proof. See, e.g., [3, Section 6]. �

We are now in a position to formulate De Pierro’s conjecture.

Conjecture 1.6 (De Pierro). (See [13, Section 3, Conjecture II].) Suppose thatFλ 6= Ø, for every λ ∈ ]0, 1]. In view of Corollary 1.3(iii), all weak limits

(1.14) (∀x ∈ X)(∀λ ∈ ]0, 1]) xλ := weak limn→+∞

Qnλx

are well defined. Then the conjecture of De Pierro states that (xλ)λ∈]0,1] approaches

PLx as λ→ 0+.


The reader is referred to the nice article [13] for further information on thisconjecture as well as its relevance to applications.

Remark 1.7. Several comments on Conjecture 1.6 are in order.

(i) De Pierro points out in [13, Section 1] that Conjecture 1.6 is true for affinesubspaces in Euclidean space; in fact, this is a consequence of a result byCensor, Eggermont, and Gordon [10, Theorem 1]. Only the case when eachaffine subspaces is a hyperplane is covered directly by [10, Theorem 1];to deal with general affine subspaces, one has to utilize the reformulationtechniques outlined at the end of [10, Section 2] and on [16, page 41].

(ii) We implemented interactive JAVA code for the visualization of De Pierro’sconjecture in the Euclidean plane, for various types of convex sets. Numer-ous experiments were performed, all of which strongly support De Pierro’sconjecture.

(iii) Little is known about De Pierro’s conjecture for general nonempty closedconvex sets; this setting remains a challenging topic for future research.

(iv) Suppose the sequence (xn)n∈� is generated by

(1.15) x0 = x, (∀n ∈ N) xn+1 := (1− λn)xn + λnQλnxn,

where

(1.16) (λn)n∈� is a sequence satisfying

∑n∈

� λn = +∞ and 0← λn ∈ ]0, 1].

Iteration (1.15) can be interpreted as a “diagonalization” of Conjecture 1.6.De Pierro also conjectures (see [13, Section 3, Conjecture I]) that L 6= Ø ifand only if each curve (xλ)λ∈]0,1] (where xλ is given by (1.14)) approachesPLx and each sequence (xn)n∈

� generated by (1.15) approaches some pointin L. Note that in the general convex case, it is possible for the sequence(xn)n∈

� to converge to a point in L\{PLx}; see De Pierro’s example at theend of [13, Section 3]. In the case of translates of subspaces in Euclideanspace, however, De Pierro proved that the sequence (xn)n∈

� generated by(1.15) converges to PLx; see [13, Theorem 4.3.1]. It would be interesting toknow whether the techniques utilized in the present paper can be modifiedto make them applicable to analyze (1.15).

The purpose of this paper is to extend the known positive results on De Pierro’sconjecture on affine subspaces from Euclidean to general Hilbert space, and to illus-trate the subtleties in infinite-dimensional settings by examples.

Note that, since there are two notions of convergence in Hilbert space (weakand strong), it is not immediately obvious how De Pierro’s conjecture should evenbe formulated. In fact, De Pierro (see [13, first page]) asserts that probably mostof the results in [13] remain valid in Hilbert space with weak convergence. As ourmain results show, the following two strikingly different scenarios occur in the affinesetting:

• If C1, . . . , CN are translates of regular subspaces, a notion we review inSection 5 below, then each set Fλ is nonempty and for every x ∈ X thesequence (Qn

λx)n∈� converges strongly to some point xλ. Moreover, the


resulting curve (xλ)λ∈]0,1] converges strongly to PLx, the least squares so-lution nearest to x (see Theorem 6.4). In other words,

De Pierro’s conjecture is true for translates of regular subspaces.

This result also provides additional support of another conjectureby De Pierro (see Remark 1.7(iv)), since in this setting L and each Fλ

are nonempty.• If C1, . . . , CN are translates of subspaces that are not regular, then it can

happen that L and all sets Fλ are empty (see Example 4.1).

The second irregular case has not been observed previously. This is probably dueto the fact that all arrangements of affine subspaces in Euclidean space are auto-matically regular (see Remark 5.3(iii)).

With an eye towards the general convex case, we have aimed to keep our proofs asgeometrical as possible; for instance, in contrast to [10], the Moore-Penrose inverseis not utilized here.

The paper is organized as follows. In Section 2, we fix the notation for the rest ofthis paper and develop several auxiliary results. The first main result, which con-cerns the asymptotic behaviour of the sequence (Qn

λx)n∈� , is presented in Section 3.

Section 4 contains a systematic construction of two closed affine subspaces in `2(N)that exhibit “bad” behaviour with respect to iterating Qλ. Regularity is the focus ofSection 5. We provide a novel parallel characterization and show that it guaranteeslinear convergence of (Qn

λx)n∈� . Section 6 establishes De Pierro’s conjecture for

translates of regular subspaces. Some illustrative examples in the Euclidean planeconclude the paper.

2. Standing assumptions and auxiliary results

From now on, we assume that the sets

(2.1) C1, . . . , CN are closed affine subspaces.

For each i ∈ {1, . . . , N}, we set

(2.2)

ci := PCi0 ∈ Ci ∩ L⊥

i , so that Ci = ci + Li, and Li is a closed linear subspace.

Further, let

(2.3) L :=⋂N

i=1 Li.

For the reader’s convenience, we present and prove in this section various usefulauxiliary results that are scattered throughout the literature; see, e.g., [3], [11], [14],[15], and [18]. We used implicitly the following result in the formulation of (2.2).

Proposition 2.1. Suppose that S is a nonempty closed convex set in X and thatc ∈ X. Let C := c + S and fix x ∈ X. Then:

(i) PCx = c + PS(x− c).(ii) If S is a closed linear subspace, then C∩S⊥ = {PC0}; consequently, PCx =

PC0 + PSx.


Proof. (i): Note that c + PS(x− c) ∈ C and that, for every s ∈ S,

(2.4) ‖x− (c +PS(x− c))‖ = ‖(x− c)−PS(x− c) ≤ ‖(x− c)− s‖ = ‖x− (c + s)‖;therefore, PCx = c+PS(x−c). (ii): Pick y ∈ C∩S⊥. Then y ∈ C so that C = y+S.Now sup〈C−y, 0−y〉 = sup〈(y+S)−y,−y〉 = sup〈S,−y〉 = 0, since y ∈ S⊥. Using(1.5), we obtain y = PC0. Conversely, PC0 ∈ C and sup〈C − PC0, 0 − PC0〉 ≤ 0.Since C = PC0+S and S is a linear subspace, it follows that PC0 ∈ S⊥. Altogether,C∩S⊥ = {PC0}, as claimed. This identity and (i) imply PCx = PC0+PS(x−PC0) =PC0 + PSx− PSPC0 = PC0 + PSx. �

Corollary 2.2. Let λ ∈ ]0, 1], i ∈ {1, . . . , N}, and x ∈ X. Then:

(i) PCix = ci + PLi

x;(ii) Rλ,Ci

x = λci + Rλ,Lix.

Proof. (i): Combine Proposition 2.1(ii) with the standing assumption that ci = PCi0

(see (2.2)). (ii): Using (i), we have Rλ,Cix = (1− λ)x + λPCi

x = (1− λ)x + λ(ci +PLi

x) = Rλ,Lix + λci. �

Our standing assumptions (2.1)–(2.3) make the following refinement of Fact 1.5possible.

Theorem 2.3. Let x ∈ X. Then:

(i) x ∈ L ⇔ x ∈ Fix∑N

i=11N

PCi⇔ (Id−∑N

i=11N

PLi)x =

∑Ni=1

1N

ci.(ii) L is a closed affine subspace of X.(iii) If L 6= Ø, then the parallel space of L is L, i.e., L+L ⊂ L and L−L ⊂ L.

Proof. (i): The first equivalence follows at once from Fact 1.5. Using Corol-

lary 2.2(i), we observe that x =∑N

i=11N

PCix ⇔ x =

∑Ni=1

1N

(ci + PLix) ⇔

(Id−∑Ni=1

1N

PLi)x =

∑Ni=1

1N

ci.(ii): This is a direct consequence of (i).

(iii): Pick x ∈ L and l ∈ L. Using (i), we have (Id−∑Ni=1

1N

PLi)(x + l) =

(Id−∑Ni=1

1N

PLi)(x)+(Id−∑N

i=11N

PLi)(l) =

∑Ni=1

1N

ci+(l−∑Ni=1

1N

l)=∑N

i=11N

ci.Thus, again by (i), we conclude x + l ∈ L. Hence, L + L ⊂ L. Now, take any

y ∈ L. Utilizing (i) once more, we have (Id−∑N

i=11N

PLi)(x) =

∑Ni=1

1N

ci =

(Id−∑N

i=11N

PLi)(y). Subtraction of the last term from the first yields x − y =∑N

i=11N

PLi(x − y). Using Fact 1.2(i)&(ii), we obtain x − y ∈ Fix

∑Ni=1

1N

PLi=⋂N

i=1 Fix PLi=

⋂Ni=1 Li = L. Consequently, L − L ⊂ L and the proof is com-

plete. �

The remaining results of this section extend results from [5] and [18], where λ = 1.

Proposition 2.4. Let λ ∈ ]0, 1] and i ∈ {1, . . . , N}. Then:

(i) PLi(L⊥) ⊂ L⊥;

(ii) Rλ,Li(L⊥) =

((1− λ)I + λPLi

)(L⊥) ⊂ L⊥;

(iii) PLi(Id−PL) = PLi∩L⊥ .

Proof. (i): If l ∈ L and l ∈ L⊥, then 0 = 〈l, l〉 = 〈PLil, l〉 = 〈l, PLi

l〉. HencePLi

(L⊥) ⊂ L⊥. (ii): This is a consequence of (i). (iii): (See also [18, Lemma 6].)


For brevity, set T := PLi(Id−PL). If x ∈ Li ∩ L⊥, then Tx = x and hence

ranT ⊃ Li ∩L⊥. Conversely, ranT ⊂ ranPLi= Li and also, using (i), ranT ⊂ L⊥.

Altogether,

(2.5) ranT = Li ∩ L⊥.

This implies

(2.6) T 2 = T.

Now (2.5) also yields T = PLi(Id−PL) = (Id−PL)PLi

(Id−PL), which in turn showsthat

(2.7) T = T ∗.

In view of (2.5)–(2.7), T = PLi(Id−PL) is the orthogonal projector onto ranT =

Li ∩ L⊥. �

Corollary 2.5. Let λ ∈ ]0, 1]. Then:

(i) (Id−PL)Rλ,Li(Id−PL) = Rλ,Li

(Id−PL) = (1− λ)PL⊥ + λPLi∩L⊥;

(ii) Rλ,LN· · ·Rλ,L1(Id−PL) =

(Rλ,LN

(Id−PL))· · ·

(Rλ,L1(Id−PL)

);

(iii) Rλ,LN· · ·Rλ,L1(Id−PL) =

((1 − λ)PL⊥ + λPLN∩L⊥

)· · ·

((1 − λ)PL⊥ +

λPL1∩L⊥

).

Proof. (i): The second identity follows from the definition (see (1.11)) and Propo-sition 2.4(iii): indeed, Rλ,Li

(Id−PL) =((1 − λ) Id+λPLi

)PL⊥ = (1 − λ)PL⊥ +

λPLiPL⊥ = (1 − λ)PL⊥ + λPLi∩L⊥ . Hence ranRλ,Li

(Id−PL) ⊂ L⊥ and the firstidentity is now clear. (ii)&(iii): Using (i) repeatedly, we see that

Rλ,LN· · ·Rλ,L1(Id−PL)(2.8)

= Rλ,LN· · ·Rλ,L3

(Rλ,L2(Id−PL)

)(Rλ,L1(Id−PL)

)

...

=(Rλ,LN

(Id−PL))· · ·

(Rλ,L1(Id−PL)

)

=((1− λ)PL⊥ + λPLN∩L⊥

)· · ·

((1− λ)PL⊥ + λPL1∩L⊥

). �

Proposition 2.6. Let λ ∈ ]0, 1] and i ∈ {1, . . . , N}. Then, for every k ∈ {1, 2, . . .},

(2.9) (Rλ,LN· · ·Rλ,L1)

k = PL ⊕ (Rλ,LN· · ·Rλ,L1PL⊥)k.

Proof. Proposition 2.4(ii) implies that Rλ,Li(L⊥) ⊂ L⊥. We now prove (2.9) by

induction on k. For k = 1, we have Rλ,LN· · ·Rλ,L1 = Rλ,LN

· · ·Rλ,L1(PL ⊕ PL⊥) =PL ⊕ (Rλ,LN

· · ·Rλ,L1PL⊥). If the identity holds true for k ∈ {1, 2, . . .}, then thesame is true for k + 1 because

(Rλ,LN· · ·Rλ,L1)

k+1 = (Rλ,LN· · ·Rλ,L1)

(PL ⊕ (Rλ,LN

· · ·Rλ,L1PL⊥)k)

(2.10)

= PL ⊕((Rλ,LN

· · ·Rλ,L1)(Rλ,LN· · ·Rλ,L1PL⊥)k

)

= PL ⊕ (Rλ,LN· · ·Rλ,L1PL⊥)k+1. �


3. Dichotomy

The results in this section extend their counterparts in [5, Section 5.7], whereλ = 1. For every i ∈ {1, . . . , N}, recall (see (1.11)) that Rλ,Li

= (1 − λ) Id+λPLi.

It will be convenient to abbreviate, for every λ ∈ ]0, 1],

(3.1) Rλ := Rλ,LN· · ·Rλ,L1 ,

and to define

(3.2) Tλ : L⊥1 × · · · × L⊥

N → L⊥ : y = (y1, . . . , yN ) 7→ λ∑N

i=1 Rλ,LN· · ·Rλ,Li+1

yi.

Let us verify that ranTλ ⊂ L⊥. If y = (y1, . . . , yN ) ∈ L⊥1 × · · · × L⊥

N , then each

yi belongs to L⊥; thus, by Proposition 2.4(ii), RLN· · ·RLi+1yi ∈ L⊥ and hence

Tλy ∈ L⊥.We let (see (2.2))

(3.3) c := (c1, . . . , cN ) ∈ L⊥1 × · · · × L⊥

N .

The behaviour of the iterates of Rλ is the topic of the next result.

Proposition 3.1. Let λ ∈ ]0, 1] and x ∈ X. Then:

(i) Rλ is linear and strongly nonexpansive, with Fix Rλ = L;(ii) Rn

λx→ PLx;(iii) (RλPL⊥)nx→ 0.

Proof. Recall that

(3.4) Rλ = Rλ,LN· · ·Rλ,L1 =

((1− λ) Id+λPLN

)· · ·

((1− λ) Id+λPL1

).

It is thus clear that Rλ is linear. Moreover, by Fact 1.2(i)–(iii), the operator Rλ isstrongly nonexpansive with

(3.5) Fix Rλ =⋂N

i=1 Fix Rλ,Li=

⋂Ni=1

(Fix(Id) ∩ Fix(PLi

))

=⋂N

i=1 Li = L.

Hence (i) is verified. (ii): The strong convergence of (Rnλx)n∈

� to some point in Lis guaranteed by Fact 1.2(v) (see also [2, Corollary 2.4]); it remains to determinethis limit. To this end, observe that, for every i ∈ {1, . . . , N},(3.6)

Rλ,Li= R∗

λ,Liand (∀y ∈ X) 〈y,Rλ,Li

y〉 = (1− λ)‖y‖2 + λ‖PLiy‖2 ≥ 0.

In view of (3.4) and (3.6), Rλ is the composition of finitely many operators thatare self-adjoint, positive (semidefinite), and nonexpansive. Since Fix Rλ = L, [7,Theorem 2.5] now implies that

(3.7) limn→+∞

Rnλx = PLx,

which completes the proof of (ii). (An alternative proof can be based upon Fejermonotonicity and [6, Fact 2.2].) (iii): Analogous to the derivation of (i), we see thatRλPL⊥ is linear and strongly nonexpansive, with Fix RλPL⊥ = {0}. The conclusionis now a consequence of Fact 1.2(v). �


The following result provides a useful decomposition of the iterates of Qλ.

Theorem 3.2. Let λ ∈ ]0, 1] and n ∈ {1, 2, . . . , }. Then:

(3.8) Qnλx = Rn

λx +∑n−1

k=0 RkλTλc = PLx + (RλPL⊥)nx +

∑n−1k=0(RλPL⊥)kTλc.

Proof. We prove the first identity by induction on n. Recall that Rλ,Ci= λci+Rλ,Li

,for every i ∈ {1, . . . , N} (Corollary 2.2(ii)). This readily implies

(3.9) Qλx = Rλ,LNRλ,LN−1

· · ·Rλ,L1x + Tλc = Rλx + Tλc.

Hence the first identity in (3.8) is true for n = 1. Now assume the first identityholds for n ∈ {1, 2, . . . , }. Then, using (3.9) (with x replaced by Qn

λx), we obtain

Qn+1λ x = Qλ(Qn

λx) = Rλ(Qnλx) + Tλc

= Rλ

(Rn

λx +∑n−1

k=0 RkλTλc

)+ Tλc = Rn+1

λ x +∑n−1

k=0 Rk+1λ Tλc + Tλc

= Rn+1λ x +

∑nk=0 Rk

λTλc.

(3.10)

Thus the formula holds true for n + 1 and the first identity is verified. The secondidentity now follows from Proposition 2.6, the fact that Tc ∈ L⊥, and Proposi-tion 2.4(ii). �

Corollary 3.3. Let λ ∈ ]0, 1], n ∈ N, and x ∈ X. Then:

(i) Fλ = (Id−Rλ)−1(Tλc).(ii) Fλ = y + L, for every y ∈ Fλ. In other words, if Fλ 6= Ø, then its parallel

space is L.(iii) Qn

λx = Rnλ(x− y) + y, for every y ∈ Fλ.

Proof. Recall that Fλ = Fix Qλ (see (1.7)). (i): Using Theorem 3.2, we obtain theequivalences x ∈ Fλ ⇔ x−Qλx = 0⇔ x−(Rλx+Tλc) = 0⇔ (Id−Rλ)x = Tλc⇔x ∈ (Id−Rλ)−1(Tλc). We now turn to the remaining two items. Proposition 3.1(i)yields

(3.11) Fix Rλ = L.

Since (ii)&(iii) are trivially true if Fλ = Ø, let us assume that Fλ 6= Ø and pick anarbitrary y ∈ Fλ. If l ∈ L, then (by (i) and (3.11)) (y + l)−Rλ(y + l) = (y−Rλy)+(l−Rλl) = Tλc and so y+ l ∈ Fλ by (i). We conclude that y +L ⊂ Fλ. Conversely,pick x ∈ Fλ and set h := x−y. Then, by (i), Tλc = x−Rλx = (y+h)−Rλ(y+h) =(y−Rλy)+(h−Rλh) = Tλc+(h−Rλh). Hence h ∈ Fix Rλ = L (see (3.11)), whichimplies Fλ ⊂ y + L. Altogether, we have established (ii). Since y ∈ Fλ, (i) impliesthat y−Rλy = Tλc. We prove the desired identity by induction on n. Clearly, thestatement is true for n = 0 (and also for n = 1, using Theorem 3.2). Assume theidentity holds for some n ∈ N. Utilizing Theorem 3.2, we then obtain

Qn+1λ x = Qλ(Qn

λx) = RλQnλx + Tλc

= Rλ(Rnλ(x− y) + y) + (y −Rλy) = Rn+1

λ (x− y) + y. �

We are now ready for our first main result.

Theorem 3.4 (dichotomy). Let λ ∈ ]0, 1] and x ∈ X. Then exactly one of thefollowing two alternatives holds:


(i) Fλ = Ø and limn→+∞

‖Qnλx‖ = +∞.

(ii) Fλ 6= Ø, limn→+∞

Qnλx = PFλ

x = PLx + PFλ0, and

∑+∞k=0(RλPL⊥)kTλc =

PFλ0 ∈ L⊥.

Proof. If Fλ = Ø, then (i) is precisely Corollary 1.3(iv). Henceforth, we assumethat Fλ is nonempty. Now pick any y ∈ Fλ. By Proposition 3.1(ii),

(3.12) limn→+∞

Rnλ(x− y) = PL(x− y).

Corollary 3.3(iii), (3.12), Proposition 2.1(i), Corollary 3.3(ii), and Proposition 2.1(ii)imply that(3.13)lim

n→+∞Qn

λx = y + limn→+∞

Rnλ(x− y) = y +PL(x− y) = Py+Lx = PFλ

x = PFλ0+PLx.

Now we take the limit in (3.8) of Theorem 3.2 (recall that limn→+∞(RλPL⊥)nx = 0by Proposition 3.1(iii)) and deduce that

(3.14) limn→+∞

Qnλx = PLx +

∑+∞k=0(RλPL⊥)kTλc.

Altogether, (3.13) and (3.14) show that

(3.15) PFλ0 =

∑+∞k=0(RλPL⊥)kTλc.

Finally, since Fλ = y + L (see Corollary 3.3(ii)), Proposition 2.1(ii) yields PFλ0 ∈

L⊥. �

If λ = 1 and each affine subspace is linear, then Theorem 3.4(ii) reduces to theclassical cyclic projections result by von Neumann [19] (for N = 2) and by Halperin[17] (for N ≥ 2). See also [15, Chapter 9] for further information.

In the following section, we provide a striking example where all fixed point setsFλ, as well as L, are empty. This shows that alternative (i) of Theorem 3.4 doesoccur and it also provides additional support for another conjecture of De Pierro’s(see Remark 1.7(iv)). We shall subsequently show that only alternative (ii) can beobserved in the presence of regularity.

4. An irregular example

Example 4.1 (constructing two irregular affine subspaces). Recall thatN = {0, 1, 2, . . .} and consider the Hilbert space of real square-summable sequences,

(4.1) X = `2(N) = {x = (ξn)n∈� ∈ R

�

:∑

n∈� |ξn|2 < +∞}.

For each k ∈ N, the kth standard vector u(k) is defined by u(k)n = 1, if n = k;

u(k)n = 0, otherwise. Then {u(k) : k ∈ N} is the standard orthonormal basis of X.

Consider a decreasing sequence of angles(4.2)

(γn)n∈{1,2,...} in]0, 1

2π[, and let γ∞ := inf

n∈{1,2,...}γn = lim

n∈{1,2,...}γn ∈

[0, 1

2π[.

The irregular case occurs when γ∞ = 0. Now define two closed affine subspaces by

(4.3) C1 :=(1, 1, 1

2 , 13 , . . .

)+ span

{u(1), u(3), u(5), . . .

}


and

(4.4) C2 := span{

cos(γn)u(2n−1) + sin(γn)u(2n) : n ∈ {1, 2, . . .}}.

Let λ ∈ ]0, 1] and x = (ξn)n∈� ∈ X. Then:

(i) PC1x =(1, ξ1,

12 , ξ3,

14 , ξ5,

16 , ξ7, . . .

).

(ii) PC2x =∑

n∈{1,2,...}

(ξ2n−1 cos(γn) + ξ2n sin(γn)

)(cos(γn)u(2n−1) +

sin(γn)u(2n)).

(iii) Fλ 6= Ø ⇔∑

n∈{1,2,...}(cot(γn)/n)2 < +∞; if this is the case, then Fλ is a

singleton with the unique element

(4.5) 1−λ2−λ

u(0) +∑

n∈{1,2,...}12n

(cot(γn)u(2n−1) + u(2n)

).

(iv) L 6= Ø ⇔ ∑n∈{1,2,...}(cot(γn)/n)2 < +∞; if this is the case, then L is a

singleton with the unique element

(4.6) 12u(0) +

∑n∈{1,2,...}

12n

(cot(γn)u(2n−1) + u(2n)

).

(v) The parallel spaces of C1 and C2 are

(4.7) L1 := span{u(1), u(3), u(5), . . .

}

and

(4.8) L2 := C2 = span{

cos(γn)u(2n−1) + sin(γn)u(2n) : n ∈ {1, 2, . . .}},

respectively.(vi) L1 + L2 is closed ⇔ γ∞ > 0.(vii) If γ∞ > 0 or (γn)n∈{1,2,...} converges to 0 “slow enough,” then Fλ and L

are both nonempty. The latter case happens, for instance, when γn :=arccot( 4

√n), for every n ∈ {1, 2, . . .}.

(viii) If (γn)n∈{1,2,...} converges to 0 “fast enough,” then Fλ and L are both empty.

This occurs when γn := arccot(√

n), for every n ∈ {1, 2, . . .}.Proof. (i): Using Proposition 2.1(i), we see that

PC1x =(1, 1, 1

2 , 13 , 1

4 , . . .)

+ Pspan{u(1),u(3),u(5),...}

(ξ0 − 1, ξ1 − 1, ξ2 − 1

2 , ξ3 − 13 , ξ4 − 1

4 , . . .)

=(1, 1, 1

2 , 13 , 1

4 , . . .)

+(0, ξ1 − 1, 0, ξ3 − 1

3 , 0, . . .)

=(1, ξ1,

12 , ξ3,

14 , . . .

).

(4.9)

(ii): The given orthonormal basis for the linear subspace C2,{

cos(γn)u(2n−1) +

sin(γn)u(2n) : n ∈ {1, 2, . . .}}

is orthonormal, so the formula is clear by Fourierexpansion.

It will be instructive and convenient to make a geometric observation beforewe turn to the remaining items. Fix n ∈ {1, 2, . . .}, identify the two-dimensional

subspace Xn := span{u(2n−1), u(2n)} of X with the Euclidean plane R2, and consider

the restrictions to Xn of C1 and C2: C1∩Xn and C2∩Xn. Then the first and secondrestrictions correspond to the lines

(4.10)(

12n−1 , 1

2n

)+ Ru(2n−1),


and

(4.11) R(cos(γn)u(2n−1) + sin(γn)u(2n)

),

respectively. Since γn ∈]0, 1

2π[, the two lines are not parallel; in fact, the unique

point in the intersection of the two lines is

(4.12) 12n

(cot(γn)u(2n−1) + u(2n)

).

Suppose that z = (ζn)n∈� ∈ R

�

.(iii): We assume that z is an algebraic fixed point of Qλ, i.e.,

(4.13) z = Rλ,C2Rλ,C1z,

but postpone the discussion whether z is actually a member of X momentarily. Inview of (i) and (ii), the projections are separable in the sense that (4.13) decouplesinto an equation for ζ0 and into a family of systems of two linear equations in twounknowns ζ2n−1, ζ2n (corresponding to the subspace Xn), for every n ∈ {1, 2, . . .}.Using (i) and (ii), the equation determining ζ0 is

(4.14) ζ0 = (1− λ)2ζ0 + (1− λ)λ; hence ζ0 = 1−λ2−λ

.

It is possible (and tedious) to determine ζ2n−1, ζ2n in the same algebraic fashion(by utilizing (i) and (ii)); however, the above geometric observation allows for thefollowing, much shorter, argument. Because of its separable nature, solving equa-tion (4.13) restricted to Xn amounts to finding fixed points (in Xn) of the com-position of the corresponding relaxed projections onto the lines given by (4.10)and (4.11). Since these two lines meet, Fact 1.2 implies that any fixed point mustcorrespond to a point in the intersection. In view of (4.12), we deduce

(4.15) (∀n ∈ {1, 2, . . . , })(ζ2n−1, ζ2n

)= 1

2n

(cot(γn), 1

).

Now let the coordinates of z be given by (4.14) and (4.15). Since(

1n+1

)n∈

� ∈ X,we have the equivalence

(4.16) z ∈ X ⇔ ∑n∈{1,2,...}

(cot(γn)

n

)2< +∞.

This completes the proof of (iii).The verification of (iv) is analogous, the only difference lying in the equation

determining ζ0, which stems from the characterization z ∈ L if and only if 2z =PC1z + PC2z (see Fact 1.5).

Item (v) is clear.We now turn to (vi). Since L = L1 ∩ L2 = {0}, we have L⊥ = X and Fact 5.2

yields the equivalence

(4.17) L1 + L2 is closed if and only if ‖PL2PL1‖ < 1.

Let z = (ζn)n∈� ∈ X such that ‖z‖ = 1. It is easy to check that

(4.18) ‖PL2PL1z‖2 =∑

n∈{1,2,...} cos2(γn)ζ22n−1.

If γ∞ > 0, then (4.18) implies that ‖PL2PL1‖ ≤ cos(γ∞) < 1 and thus L1 + L2 is

closed. Henceforth, we assume that γ∞ = 0. If z = u(2n−1), then (4.18) shows that‖PL2PL1u

(2n−1)‖ = cos(γn). Supremizing over n ∈ {1, 2, . . .} implies ‖PL2PL1‖ ≥ 1and therefore L1 + L2 is not closed.


(vii)&(viii): In view of (iii)&(iv), this depends precisely on whether or not theseries

(4.19)∑

n∈{1,2,...}cot2(γn)

n2

converges. Clearly, this series converges if γ∞ > 0. It remains to verify the proposedconcrete assignments for the sequences of angles. If γn = arccot( 4

√n), for every n ∈

{1, 2, . . .}, then (4.19) turns into the convergent series∑

n∈{1,2,...} n−1.5. Similarly,

if γn = arccot(√

n), for every n ∈ {1, 2, . . .}, then (4.19) becomes the harmonicseries. �

Remark 4.2. Some comments on Example 4.1 are in order.

(i) If X is a Euclidean space, then Fλ and L are always nonempty. In essence,this holds because subspaces are automatically closed and the alternativeseen in item (viii) can never occur.

(ii) If γ∞ = 0 (the irregular case), then L1+L2 is not closed (see (vi)); however,no further conclusion can be drawn on whether Fλ and L are both nonempty(see (vii)&(viii)).

(iii) If γ∞ > 0 (the regular case), then L1+L2 is closed (see (vi)) and De Pierro’sconjecture is true in this setting (combine Remark 5.3(i) with Theorem 6.4).

5. Regularity

We now recall the notion of regular subspaces, which plays an important role inthe study of projection methods (see [4, Section 5]).

Definition 5.1 (regular subspaces). We say that the subspaces L1, . . . , LN areregular, if

(5.1) d(xn, L1 ∩ · · · ∩ LN )→ 0,

whenever (xn)n∈� is a sequence in X such that maxi∈{1,...N} d(xn, Li)→ 0.

Consider the product Hilbert space X := XN , with inner product 〈x,y〉 :=∑Ni=1

1N〈xi, yi〉 and induced norm ‖x‖2 :=

∑Ni=1

1N‖xi‖2 for x = (x1, . . . , xN ) and

y = (y1, . . . , yN ) in X. Now define the product set L and the diagonal ∆ by

(5.2) L := L1 × L2 × · · · × LN and ∆ := {(x, x, · · · , x) ∈ X : x ∈ X},respectively. Then regularity of L1, . . . , LN is characterized as follows.

Fact 5.2. The following are equivalent.

(i) L1, . . . , LN are regular.(ii) L1 ∩ L⊥, . . . , LN ∩ L⊥ are regular.(iii) L⊥

1 + · · ·+ L⊥N is closed.

(iv) ‖PLN· · ·PL1PL⊥‖ = ‖PLN∩L⊥ · · ·PL1∩L⊥‖ < 1.

(v) L,∆ are regular in X.

Proof. The equivalences (i)⇔(iii)⇔(v) follow from [4, Lemma 5.18 andTheorem 5.19]. Next, [5, Proposition 3.7.3] shows that (i)⇔(ii). For the equivalenceof (iii) and (iv), combine Corollary 2.5(iii) (for λ = 1) with [5, Theorem 3.7.4]. �


Remark 5.3. Some comments regarding Fact 5.2 are in order.

(i) If N = 2, then (by [4, Proposition 5.16] or by [15, Theorem 9.35]) regularityof L1, L2 is also equivalent to

(5.3) L1 + L2 is closed.

(If N ≥ 3, then the closedness of L1 + · · · + LN is in general independentof the closedness of L⊥

1 + · · ·+ L⊥N ; see [4, Remark 5.20].)

(ii) Let us define the angle in [0, 12π] of the (ordered) subspaces L1, . . . , LN by

its cosine as follows:

(5.4)cos γ(L1, · · · , LN ) := ‖PLN

· · ·PL1PL⊥‖ = ‖PLN∩L⊥ · · ·PL1∩L⊥‖ ∈ [0, 1].

Note that L1, . . . , LN are regular if and only if γ(L1, . . . , LN ) > 0. More-over, the angle coincides with the classical Friedrichs angle when N = 2.We refer the reader to [5, Section 3.7] and [15, Chapter 9] for further infor-mation.

(iii) Each of the following conditions implies (see [5, Proposition 3.7.7]) that thesubspaces L1, . . . , LN are regular:• at least one Li ∩ L⊥ is finite-dimensional;• all Li, except possibly one, are finite-codimensional;• X is finite-dimensional;• each Li is a hyperplane.

The next result can be viewed as the parallel counterpart of Fact 5.2(iv).

Theorem 5.4. The subspaces L1, . . . , LN are regular if and only if

(5.5) ‖PL1∩L⊥ + · · ·+ PLN∩L⊥‖ = ‖PL1PL⊥ + · · ·+ PLNPL⊥‖ < N.

Proof. Suppose first that L1, . . . , LN are regular. In view of Fact 5.2, the subspacesL,∆ are regular in X and hence

(5.6) ‖P∆PLP(∆∩L)⊥‖ < 1.

Pick an arbitrary x ∈ X and set x = (x, . . . , x) ∈ X. It is straightforward to verifythat(5.7)

P∆PLP(∆∩L)⊥x = y, where y = (y, . . . , y) and y = 1N

(PL1 + · · ·+ PLN)PL⊥x.

Using (5.6), we deduce that

(5.8) ‖y‖2 = ‖y‖2 ≤ ‖P∆PLP(∆∩L)⊥‖2‖x‖2 = ‖P∆PLP(∆∩L)⊥‖2‖x‖2.Altogether, (5.6)—(5.8) imply

(5.9) 1N‖(PL1 + · · ·PLN

)PL⊥‖ < 1.

Therefore, (5.5) now follows from (5.9) and Proposition 2.4(iii).Now suppose that L1, . . . , LN are not regular. Fact 5.2 shows that L1 ∩ L⊥, . . . ,

LN ∩ L⊥ are not regular either. Since⋂N

i=1 Li ∩ L⊥ = {0}, there exists a sequence(xn)n∈

� in X such that

(5.10) ‖xn‖ = d(xn, {0}) ≡ 1,


while each d(xn, Li ∩ L⊥)→ 0. Hence

xn − PL1∩L⊥xn → 0,

...

xn − PLN∩L⊥xn → 0.

(5.11)

Adding yields

(5.12) Nxn − (PL1∩L⊥ + · · ·+ PLN∩L⊥)xn → 0.

The nonexpansivity of each PLi∩L⊥ , (5.10), the triangle inequality, and (5.12) imply

0 ≤ N − ‖(PL1∩L⊥ + · · ·+ PLN∩L⊥)xn‖= ‖Nxn‖ − ‖(PL1∩L⊥ + · · ·+ PLN∩L⊥)xn‖≤ ‖Nxn − (PL1∩L⊥ + · · ·+ PLN∩L⊥)xn‖ → 0.

(5.13)

Therefore, ‖(PL1∩L⊥ + · · · + PLN∩L⊥)xn‖ → N , which (recall again the nonexpan-sivity of each PLi∩L⊥ and (5.10)) yields ‖PL1∩L⊥ + · · ·+ PLN∩L⊥‖ = N . �

Proposition 5.5. Suppose that L1, . . . , LN are regular and let λ ∈ ]0, 1]. Recallthat the angle γ(L1, . . . , LN ) is defined in Remark 5.3(ii). Then:

(i) ‖Rλ(Id−PL)‖ ≤ λN cos(γ(L1, . . . , LN )

)+ 1− λN < 1;

(ii)1

1− ‖RλPL⊥‖ ≤1

λN(1− cos γ(L1, . . . , LN )

) .

Proof. Expanding Corollary 2.5(iii) shows that

Rλ(Id−PL) = Rλ,LN· · ·Rλ,L1(Id−PL)

=((1− λ)PL⊥ + λPLN∩L⊥

)· · ·

((1− λ)PL⊥ + λPL1∩L⊥

)

= λNPLN∩L⊥ · · ·PL1∩L⊥ + (1− λN )S,

(5.14)

for some nonexpansive linear operator S : X → X. Thus, using Fact 5.2 and (5.4),we deduce that

‖Rλ(Id−PL)‖ = ‖λNPLN∩L⊥ · · ·PL1∩L⊥ + (1− λN )S‖≤ λN‖PLN∩L⊥ · · ·PL1∩L⊥‖+ (1− λN )

= λN cos(γ(L1, . . . , LN )

)+ (1− λN )

< λN + (1− λN )

= 1.

(5.15)

Hence (i) is verified, and (ii) follows readily. �

We are now ready for our second main result which states that in the presenceof regularity, all fixed point sets Fλ are nonempty and the convergence guaranteedby Theorem 3.4(ii) is linear. This coincides with [5, Theorem 5.7.8] when λ = 1.

Theorem 5.6 (regularity implies linear convergence). Suppose L1, . . . , LN are reg-ular, let λ ∈ ]0, 1] and x ∈ X. Then Fλ 6= Ø, ‖RλPL⊥‖ < 1, and for everyn ∈ {1, 2, . . .},(5.16) ‖Qn+1

λ x− PFλx‖ ≤ ‖RλPL⊥‖‖Qn

λx− PFλx‖.


In other words, (Qnλx)n∈

� converges linearly to PFλx with a rate no worse than

‖RλPL⊥‖ < 1.

Proof. On the one hand, Theorem 3.2 implies

(5.17) (∀n ∈ {1, 2, . . .}) Qnλx = PLx + (RλPL⊥)nx +

∑n−1k=0(RλPL⊥)kTλc.

On the other hand, Proposition 5.5(i) yields ‖RλPL⊥‖ < 1. Therefore, we takethe limit as n → +∞ in (5.17) (see also Proposition 3.1(iii)) and conclude thatQn

λx → PLx + (Id−RλPL⊥)−1(Tλc) (see, e.g., [8, Theorem 12.1]). In view ofTheorem 3.4, we further deduce Fλ 6= Ø and

(5.18) Qnλx→ PLx + (Id−RλPL⊥)−1(Tλc) = PFλ

x = PLx + PFλ0.

Now fix n ∈ {1, 2, . . .}. By (5.17) and (5.18),

Qn+1λ x− PFλ

x = (RλPL⊥)n+1x +∑n

k=0(RλPL⊥)kTλc− (Id−RλPL⊥)−1(Tλc)

= (RλPL⊥)n+1x−∑+∞k=n+1(RλPL⊥)kTλc

= (RλPL⊥)((RλPL⊥)nx−

∑+∞k=n(RλPL⊥)kTλc

)

= (RλPL⊥)(Qnλx− PFλ

x).

(5.19)

Therefore, ‖Qn+1λ x− PFλ

x‖ = ‖(RλPL⊥)(Qnλx− PFλ

x)‖ ≤ ‖RλPL⊥‖‖Qnλx− PFλ

x‖.�

6. Main result

The following result will be utilized later.

Proposition 6.1. Let {λ, µ} ⊂ ]0, 1]. Then ‖RλPL⊥ − RµPL⊥‖ ≤ ‖Rλ − Rµ‖ ≤N |λ− µ|.

Proof. The first inequality is clear. We establish the second inequality by inductionon N , the number of subspaces. Fix x ∈ X. Then

‖Rλ,L1x−Rµ,L1x‖ =∥∥(

(1− λ) Id+λPL1

)x−

((1− µ) Id+µPL1

)x∥∥

= |µ− λ|‖x− PL1x‖≤ |λ− µ|‖x‖,

(6.1)

and the conclusion holds for N = 1. So let us assume that the desired inequalityis verified for N − 1. Abbreviate Sλ :=

((1− λ) Id+λPLN−1

)· · ·

((1− λ) Id+λPL1

)

and similarly for Sµ. Then

Rλx−Rµx = Rλ,LNSλx−Rµ,LN

Sµx

=(Rλ,LN

Sλx−Rµ,LNSλx

)+

(Rµ,LN

Sλx−Rµ,LNSµx

)

= (µ− λ)(Id−PLN)Sλx + (1− µ)(Sλx− Sµx) + µPLN

(Sλx− Sµx).

(6.2)


Now take the norm, apply the triangle inequality, recall that (Id−PLN)Sλ and PLN

are both nonexpansive, and use the induction hypothesis to conclude that

‖Rλx−Rµx‖ ≤ |λ− µ|‖x‖+ (1− µ)(N − 1)|λ− µ|‖x‖+ µ(N − 1)|λ− µ|‖x‖= N |λ− µ|‖x‖.

(6.3)

Since x ∈ X was chosen arbitrarily, it follows that ‖Rλ −Rµ‖ ≤ N |λ− µ|. �

Suppose L1, . . . , LN are regular and set, for every λ ∈ ]0, 1],

(6.4) fλ := PFλ0 = (Id−RλPL⊥)−1(Tλc) =

∑+∞k=0(RλPL⊥)kTλc ∈ L⊥.

Note that these identities and inclusions are justified by Theorem 3.4(ii) and The-orem 5.6. The next result provides a useful estimate of the distance between twopoints on the curve (fλ)λ∈]0,1].

Proposition 6.2. Suppose that L1, . . . , LN are regular and let {λ, µ} ⊂ ]0, 1]. Then:

(6.5) ‖fλ − fµ‖ ≤|λ− µ|N

(‖fµ‖+

∑Ni=1 ‖ci‖

)

1− ‖RλPL⊥‖ ≤ |λ− µ|N(‖fµ‖+

∑Ni=1 ‖ci‖

)

λN(1− cos γ(L1, . . . , LN )

) .

Proof. First, observe that the quotients are well-defined by Proposition 5.5(i). Bydefinition, we have fλ = Tλc+RλPL⊥fλ and fµ = Tµc+RµPL⊥fµ. Hence fλ−fµ =(Tλc−Tµc) + (RλPL⊥fλ −RµPL⊥fµ) and thus

(6.6) ‖fλ − fµ‖ ≤ ‖Tλc−Tµc‖ + ‖RλPL⊥fλ −RµPL⊥fµ‖.On the one hand (recall (3.2)),

Tλc−Tµc = (λ− µ)∑N

i=1 Rλ,LN· · ·Rλ,Li+1

ci

+ µ∑N

i=1

((Rλ,LN

· · ·Rλ,Li+1)− (Rµ,LN

· · ·Rµ,Li+1))ci.

(6.7)

Therefore, using the triangle inequality, the fact that each Rλ,Lkis nonexpansive,

and Proposition 6.1, we obtain

‖Tλc−Tµc‖ ≤ |λ− µ|∑Ni=1 ‖ci‖+ µ

∑Ni=1(N − i)|λ− µ|‖ci‖

≤ |λ− µ|N ∑Ni=1 ‖ci‖.

(6.8)

On the other hand, RλPL⊥fλ −RµPL⊥fµ = (RλPL⊥fλ −RλPL⊥fµ) + (RλPL⊥fµ −RµPL⊥fµ). Thus, using Proposition 6.1,

(6.9) ‖RλPL⊥fλ −RµPL⊥fµ‖ ≤ ‖RλPL⊥‖‖fλ − fµ‖+ |λ− µ|N‖fµ‖.Combining (6.6), (6.8), and (6.9) yields altogether

(6.10) ‖fλ − fµ‖ ≤ ‖RλPL⊥‖‖fλ − fµ‖+ |λ− µ|N(‖fµ‖+

∑Ni=1 ‖ci‖

).

This implies the first inequality of (6.5); the second then follows from Proposi-tion 5.5(ii). �

We now follow the points along the curve (fλ)λ∈]0,1] as λ approaches 0 and 1.

Theorem 6.3. Suppose that L1, . . . , LN are regular. Then:

(i) f0 := limλ→0+ fλ =(Id−∑N

i=11N

PLi∩L⊥)−1(∑N

i=11N

ci

);


(ii) f1 = limλ→1− fλ = (Id−PLN· · ·PL1PL⊥)−1(T1c).

Moreover, the corresponding map [0, 1]→ L⊥ : λ 7→ fλ is continuous, and (fλ)λ∈[0,1]

is a connected compact curve in L⊥.

Proof. Recall (see (3.3)) that c = (c1, . . . , cN ), where each ci ∈ L⊥i ⊂ L⊥, and (see

(6.4)) that each fλ belongs to L⊥. Define, for each i ∈ {1, . . . , N},(6.11) Ki := Li ∩ L⊥

and fix λ ∈ ]0, 1]. With the help of Proposition 2.4(ii) and Corollary 2.5, we obtainthe following identities.

(∀y ∈ L⊥) Rλy = RλPL⊥y

= Rλ,KN· · ·Rλ,K1y

=((1− λ) Id+λPKN

)· · ·

((1− λ) Id+λPK1

)y

=(Id+λ(PKN

− Id))· · · (Id+λ(PK1 − Id)

)y

=(Id+λ

∑Ni=1(PKi

− Id) + Aλ

)y,

(6.12)

where Aλ : L⊥ → L⊥ is a bounded linear operator and

(6.13) ‖Aλ‖ ≤ λ2(N2

)+ λ3

(N3

)+ · · · + λN

(NN

)= (1 + λ)N − (1 + Nλ).

Thus

(6.14) (∀y ∈ L⊥) 1λ(y −Rλy) = − 1

λAλy +

∑Ni=1(Id−PKi

)y.

Since 1λ‖Aλ‖ ≤ 1

λ

((1 + λ)N − 1

)−N → 0 as λ→ 0+, we see that

(6.15) (∀y ∈ L⊥) limλ→0+1λ(Id−Rλ)y =

(∑Ni=1(Id−PKi

))y.

On the other hand, the regularity of L1, . . . , LN and Theorem 5.4 imply‖∑N

i=1 PKi‖ < N , hence ‖ 1

N

∑Ni=1 PKi

‖ < 1. Thus Id− 1N

∑Ni=1 PKi

has a bounded

inverse (by [8, Theorem 12.1]), and the same is true for∑N

i=1(Id−PKi). In view of

(6.15) and the continuity of inversion (see, e.g., [8, Corollary 12.3]), this yields thefollowing operator limit identity

(6.16) limλ→0+(Id−Rλ)−1(λ Id) =(∑N

i=1(Id−PKi))−1

on L⊥.

Furthermore, we note that

(6.17) limλ→0+1λTλc = limλ→0+

∑Ni=1 Rλ,LN

· · ·Rλ,Li+1ci =

∑Ni=1 ci,

and (since {fλ,Tλc} ⊂ L⊥ and Rλ = RλPL⊥ on L⊥) that

(6.18) fλ = (Id−Rλ)−1(Tλc),

where (Id−Rλ)−1 is viewed as an operator on L⊥. Combining (6.16), (6.17), and(6.18) now results in

(6.19) limλ→0+ fλ = limλ→0+(Id−Rλ)−1(λ( 1

λTλc)

)

=(∑N

i=1(Id−PKi))−1(∑N

i=1 ci

)

and (i) follows.


(ii): clearly limλ→1− RλPL⊥ = PLN· · ·PL1PL⊥ , and this limit has norm less than

1 by Proposition 5.5(i). It follows that fλ = (Id−RλPL⊥)−1Tλc →(Id−PLN

· · ·PL1PL⊥)−1T1c.Now fix an arbitrary µ ∈ ]0, 1]. Using Proposition 6.2, we see that

(6.20) (∀λ ∈ [ 12µ, 1]) ‖fλ − fµ‖ ≤|λ− µ|N2N

(‖fµ‖+

∑Ni=1 ‖ci‖

)

µN(1− cos γ(L1, . . . , LN )

) .

Therefore, the map λ 7→ fλ is continuous at µ, and hence on ]0, 1]. In view of (i), letus extend this map continuously on [0, 1]. The final sentence of the theorem is nowclear, since the curve (fλ)λ∈[0,1] is the continuous image of the interval [0, 1]. �

We are now in a position to state and prove our main result.

Theorem 6.4 (De Pierro’s conjecture is true for translates of regular subspaces).Suppose that L1, . . . , LN are regular and let x ∈ X. Then the following strong limitsexist.

(6.21) (∀λ ∈ ]0, 1]) xλ := limn→+∞

Qnλx = PFλ

x = PLx + PFλ0 = PLx + fλ.

Moreover, as λ → 0+, (xλ)λ∈]0,1] converges strongly to the least squares solutionnearest to x:

(6.22) PLx = limλ→0+

xλ.

Proof. The statements concerning (6.21) are implied by Theorem 5.6,Theorem 3.4(ii), and (6.4). As in Theorem 6.3(i), we let

(6.23) f0 := limλ→0+

fλ ∈ L⊥

so that (Id−∑Ni=1

1N

PLi∩L⊥)f0 =∑N

i=11N

ci. Using Proposition 2.4(iii) and the fact

that f0 ∈ L⊥, we obtain

(6.24)∑N

i=11N

ci = (Id−∑N

i=11N

PLi∩L⊥)f0

= (Id−∑Ni=1

1N

PLiPL⊥)f0 = (Id−∑N

i=11N

PLi)f0.

Theorem 2.3(i) yields f0 ∈ L; thus altogether f0 ∈ L ∩ L⊥. Hence f0 = PL0 and

(6.25) PLx = f0 + PLx

by Theorem 2.3(iii) and Proposition 2.1(ii). Using (6.21), (6.23), and (6.25), we seethat

�(6.26) limλ→0+

xλ = PLx + limλ→0+

fλ = PLx + f0 = PLx.

We conclude with some concrete examples in the Euclidean plane that illustratethe possible nonlinearity of the curve (fλ)λ∈]0,1]. See also [1] for some explicitcomputations of f1 for hyperplanes in Euclidean space.


10-1

1

0

-1

Figure 1. Three nonlinear curves (fλ)λ∈]0,1] for u1, u2, u3.

Example 6.5 (singletons). Suppose that the affine subspaces are all singletons,i.e., Li = {0}, for each i ∈ {1, . . . , N}. Let λ ∈ ]0, 1]. Then Fλ is a singleton and itsunique element is

(6.27) fλ = λ(1− λ)N−1c1 + (1− λ)N−2c2 + · · ·+ (1− λ)cN−1 + cN

1− (1− λ)N.

Consequently, for every x ∈ X, limn→+∞ Qnλx = fλ and limλ→0+ fλ =

∑Ni=1

1N

ci.

Proof. Each Ci is a singleton, hence Li = {0} and thus L = {0}. By Remark 5.3(iii),the subspaces L1, . . . , LN are regular, and Theorem 5.6 now implies that Fλ 6= Ø.In view of Corollary 3.3(ii), the set Fλ is a singleton and its only element is fλ (see(6.4)). The formula for fλ presented in (6.27) is a consequence of expanding andsolving

fλ =((1− λ) Id +λPCN

)· · ·

((1− λ) Id+λPC1

)fλ

=((1− λ) Id +λcN

)· · ·

((1− λ) Id+λc1

)fλ

(6.28)

for fλ. Furthermore, Theorem 2.3(i) results in L = {∑Ni=1

1N

ci}. The remainingstatements now follow from Theorem 6.4. �

Remark 6.6. We now illustrate the nonlinearity in the formula for fλ given by(6.27). In the setting of Example 6.5, let X = R

2, identified with C, and N = 3.Let u1, u2, u3 be all three cube roots of unity. Then L = {0} and Figure 1 showsthree curves (fλ)λ∈]0,1], obtained by assigning u1, u2, u3 to c1, c2, c3 in three differentorderings. Note that these curves meet at the unique least squares solution 0, aspredicted by Example 6.5.


H4

H3

H2

H1

10

1

0

Figure 2. The hyperplanes H1, H2, H3, and H4.

In our last example, we show that the nature of the fixed point curves is highlydependent on the order of the sets.

Example 6.7. Let X = R2 and N = 4. Further, set H1 := {(ξ1, ξ2) ∈ X : ξ2 = 0},

H2 := {(ξ1, ξ2) ∈ X : ξ1 = 0}, H3 := {(ξ1, ξ2) ∈ X : ξ1 = ξ2}, and H4 := {(ξ1, ξ2) ∈X : ξ1 + ξ2 = 2}; see Figure 2. Now take λ ∈ ]0, 1] and let

(6.29) C1 × C2 ×C3 × C4 := H1 ×H2 ×H3 ×H4.

Then L = {0}, L = {( 12 , 1

2 )}, and Fλ = {fλ}, where

(6.30) fλ =

(1

2− λ,

1

2− λ

).

Similarly, the fixed points corresponding to the three remaining cyclic permutationsare:

fλ =

(1

2− λ,1− λ

2− λ

), for C1 × C2 × C3 × C4 := H2 ×H3 ×H4 ×H1;(6.31)

fλ =

(1− λ

2− λ,1− λ

2− λ

), for C1 × C2 × C3 × C4 := H3 ×H4 ×H1 ×H2;(6.32)

fλ =

(1− λ

2− λ,1− λ

2− λ

), for C1 × C2 × C3 × C4 := H4 ×H1 ×H2 ×H3.(6.33)

Notice that the four curves corresponding to (6.30)–(6.33), depicted in Figure 3, areall linear. In contrast, the four fixed point curves given by

(6.34) fλ =

(2(4− 3λ + λ2)

16− 16λ + 5λ2,

2(4− λ)

16− 16λ + 5λ2

),

for C1 × C2 × C3 × C4 := H1 ×H3 ×H2 ×H4;


(6.33)(6.32)(6.31)(6.30)

10

1

0

Figure 3. Four linear fixed point curves (fλ)λ∈]0,1].

(6.35) fλ =

(2(4− 3λ + λ2)

16− 16λ + 5λ2,2(4 − 5λ + λ2)

16− 16λ + 5λ2

),

for C1 × C2 × C3 × C4 := H3 ×H2 ×H4 ×H1;

(6.36) fλ =

(2(4− 3λ)

16− 16λ + 5λ2,2(4 − 5λ + 2λ2)

16− 16λ + 5λ2

),

for C1 × C2 × C3 × C4 := H2 ×H4 ×H1 ×H3;

(6.37) fλ =

(2(4− 7λ + 3λ2)

16− 16λ + 5λ2,2(4 − 5λ + 2λ2)

16− 16λ + 5λ2

),

for C1 × C2 × C3 × C4 := H4 ×H1 ×H3 ×H2.

are all nonlinear ; see Figure 4.


(6.37)(6.36)(6.35)(6.34)

10

1

0

Figure 4. Four nonlinear fixed point curves (fλ)λ∈]0,1].

Acknowledgments. The authors wish to thank an anonymous referee for carefulreading and constructive comments. H. H. Bauschke’s work was partially supportedby the Natural Sciences and Engineering Research Council of Canada and by aNATO Collaborative Linkage grant. M. R. Edwards’ work was partially supportedby the Natural Sciences and Engineering Research Council of Canada.

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Manuscript received June 25, 2004

revised October 16, 2004

Heinz H. Bauschke

Department of Mathematics and Statistics, University of Guelph, Guelph, Ontario N1G 2W1,Canada

E-mail address: [email protected]

Mclean R. Edwards

Department of Mathematics and Statistics, University of Guelph, Guelph, Ontario N1G 2W1,Canada

E-mail address: [email protected]

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