Compositions and convex combinations of asymptotically regular firmly nonexpansive mappings are also asymptotically regular

RESEARCH Open Access Compositions and convex combinations of asymptotically regular firmly nonexpansive mappings are also asymptotically regular Heinz H Bauschke 1* , Victoria Martín-Márquez 2 , Sarah M Moffat 1 and Xianfu Wang 1 * Correspondence: heinz. [email protected] 1 Mathematics, University of British Columbia, Kelowna, BC V1V 1V7, Canada Full list of author information is available at the end of the article Abstract Because of Minty’s classical correspondence between firmly nonexpansive mappings and maximally monotone operators, the notion of a firmly nonexpansive mapping has proven to be of basic importance in fixed point theory, monotone operator theory, and convex optimization. In this note, we show that if finitely many firmly nonexpansive mappings defined on a real Hilbert space are given and each of these mappings is asymptotically regular, which is equivalent to saying that they have or “almost have”fixed points, then the same is true for their composition. This significantly generalizes the result by Bauschke from 2003 for the case of projectors (nearest point mappings). The proof resides in a Hilbert product space and it relies upon the Brezis-Haraux range approximation result. By working in a suitably scaled Hilbert product space, we also establish the asymptotic regularity of convex combinations. 2010 Mathematics Subject Classification: Primary 47H05, 47H09; Secondary 47H10, 90C25. Keywords: asymptotic regularity, firmly nonexpansive mapping, Hilbert space, maximally monotone operator, nonexpansive mapping, resolvent, strongly nonexpansive mapping 1 Introduction and standing assumptions Throughout this article, Xis a real Hilbert s p ace with inner p roduct ·,·  (1) and induced norm || ⋅||. We assume that m∈ { 2, 3, 4, ... } and I:= { 1, 2, ...,m }. (2) Recall that an operator T: X ®Xis firmly nonexpansive (see, e.g., [1-3] for further information) if (∀xÎX)(∀yÎX)||Tx -Ty|| 2 ≤〈x-y, Tx -Ty〉andthataset-valued operator A: X ⇉Xis maximally monotone if it is monotone,i.e.,forall(x, x*)and(y, y*) in the graph of A, we have 〈x - y, x* - y*〉≥0 and if the graph of Acannot be properly enlarged without destroying monotonicity (We shall write dom A={xÎX|Ax≠ Ø} for the domain of A,ranA=A(X)=∪ xÎX Ax for the range of A,andgrAfor the graph of A.) These notions are equivalent (see [4,5]) in the sense that if Ais maximally monotone, then its resolvent J A : = (Id + A) -1 is firmly nonexpansive, and if Tis firmly Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 © 2012 Bauschke et al; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. nonexpansive, then T -1 - Id is maximally monotone. (Here and elsewhere, Id denotes the identity operator on X.) The Minty parametrization (see [4] and also [[1], Remark 23.22(ii)]) states that if Ais maximally monotone, then g rA={ ( JAx,x−JAx ) |x∈X}.(3) In optimization, one main problem is to find zeros of maximally monotone operators–these zeros may correspond to critical points or solutions to optimization problems. In terms of resolvents, the corresponding problem is that of finding fixed points. For background material in fixed point theory and monotone operator theory, we refer the reader to [1-3,6-16]. The aim of this note is to provide approximate fixed point results for compositions and convex combinations of finitely many firmly nonexpansive operators. The first main result (Theorem 4.6) substantially extends a result by Bauschke [17] on the compositions of projectors to the composition of firmly nonexpansive mappings. The second main result (Theorem 5.5) extends a result by Bauschke, Moffat and Wang [18] on the convex combination of firmly nonexpansive operators from Euclidean to Hilbert space. The remainder of this section provides the standing assumptions used throughout the article. Even though the main results are formulated in the given Hilbert space X, it will turn out that the key space to work in is the product space Xm:= {x= ( xi ) i∈I| ( ∀i∈I ) xi∈X} . (4) This product space contains an embedding of the original space Xvia the diagonal subspace := {x= ( x ) i∈I|x∈X} . (5) We also assume that we are given mfirmly nonexpansive operators T 1 ,..., T m ;equivalently, mresolvents of maximally monotone operators A 1 ,..., A m : (∀i∈I)Ti=JA i =(Id+Ai)−1is firmly nonexpansive . (6) We now define various pertinent operators acting on X m . We start with the Cartesian product operators T:Xm→Xm: ( xi ) i∈I→ ( Tixi ) i∈ I (7) and A:Xm⇒Xm: ( xi ) i∈I→ ( Aixi ) i∈I . (8) Denoting the identity on X m by Id, we observe that J A= ( Id +A ) −1=T1×···×Tm=T . (9) Of central importance will be the cyclic right-shift operator R :Xm→Xm: ( x1,x2...,xm ) → ( xm,x1,...,xm−1 ) (10) and for convenience we set M=Id −R . (11) Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 2 of 11 We also fix strictly positive convex coefficients (or weights) (l i ) iÎI , i.e., (∀i∈I)λi∈]0, 1[ and  i∈ I λi=1 . (12) Let us make X m into the Hilbert product space X:=Xm,withx,y= i∈I xi,yi.(13) The orthogonal complement of Δwith respect to this standard inner product is known (see, e.g., [[1], Proposition 25.4(i)]) to be ⊥=  x=(xi)i∈I| i∈I xi=0 . (14) Finally, given a nonempty closed convex subset Cof X,theprojector (nearest point mapping) onto Cis denoted by P C . It is well known to be firmly nonexpansive. 2 Properties of the operator M In this section, we collect several useful properties of the operator M, including its Moore-Penrose inverse (see [19] and e.g., [[1], Section 3.2] for further information.). To that end, the following result–which is probably part of the folklore–will turn out to be useful. Proposition 2.1 Let Y be a real Hilbert space and let B be a continuous linear operator from X to Y with adjoint B*and such that ran B is closed. Then the Moore-Penrose inverse of B satisfies B†=Pr a nB∗◦B−1◦Pr a nB . (15) Proof Take yÎY. Define the corresponding set of least squares solutions (see, e.g., [[1], Proposition 3.25]) by C:=B -1 (P ran B y). Since ran Bis closed, so is ran B* (see, e. g., [[1], Corollary 15.34]); hence, a U:= ( Ker B ) ⊥=ran B∗=ranB ∗ . Thus, C=B † y+ ker B=B † y+U ┴ . Therefore, since ran B † =ranB* (see, e.g., [[1], Proposition 3.28(v)]), P U (C)=P U B † y=B † y, as claimed. Before we present various useful properties of M, let us recall the notion of a rectangular (which is also known as star or 3* monotone, see [20]) operator. A monotone operator B: X⇉Xisrectangular if (∀(x,y∗)∈dom B×ran B)sup ( z,z∗ ) ∈gr Bx−z,z∗−y<+ ∞ . Theorem 2.2 Define b L :⊥→X:y→ m−1  i =1 m−i mRi−1y . (16) Then the following hold. (i) Mis continuous, linear, and maximally monotone with dom M=X. (ii) Mis rectangular. (iii) ker M= ker M*=Δ. (iv) ran M= ran M*=Δ ┴ is closed. (v) ran L=Δ ┴ . (vi) M◦L=Id| ⊥ . Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 3 of 11 (vii) M−1:X⇒X:y→  Ly +,ify∈⊥; ∅,otherwise . (viii) M†=P⊥◦L◦P⊥=L◦P ⊥ . (ix) M†= m  k =1 m−(2k−1) 2mRk− 1 . Proof.(i):Clearly,domM=Xand (∀xÎX)||Rx|| = ||x||. Thus, Ris nonexpansive and therefore M=Id -Ris maximally monotone (see, e.g., [[1], Example 20.27]). (ii): See [[1], Example 24.14] and [[17], Step 3 in the proof of Theorem 3.1] for two different proofs of the rectangularity of M. (iii): The definitions of Mand Rand the fact that R* is the cyclic left shift operator readily imply that ker M= ker M*=Δ. (iv), (vi), and (vii): Let y=(y 1 , ..., y m )ÎX. Assume first that yÎran M. Then there exists x=(x 1 ,...,x m ) such that y 1 =x 1 -x m ,y 2 =x 2 -x 1 ,...,andy m =x m -x m-1 .It follows that ∑ iÎI y i = 0, i.e, yÎΔ ┴ by [[1], Proposition 25.4(i)]. Thus, ran M ⊆ ⊥ . (17) Conversely, assume now that yÎΔ ┴ . Now set x:= Ly = m− 1  i =1 m−i mRi−1y . (18) It will be notationally convenient to wrap indices around, i.e., y m+1 =y 1 ,y 0 =y m and likewise. We then get (∀i∈I)xi=m−1 m yi+m−2 m yi−1+···+1 m yi+2 . (19) Therefore,  i ∈ I xi=m−1 m i ∈ I yi+m−2 m i ∈ I yi+···+1 m i ∈ I yi=m−1 2 i ∈ I yi=0 . (20) Thus xÎΔ ┴ and ran L ⊆ ⊥.(21) Furthermore, (∀i∈I)xi−xi−1=m−1 m yi−1 m yi−1−1 m yi−2−···− 1 m yi+ 1 (22a) =yi−1 m  j ∈I yj=yi . (22b) Hence Mx =x-Rx =yand thus yÎran M. Moreover, in view of (iii), M−1 y =x+kerM=x+ . (23) We thus have shown ⊥⊆ran M . (24) Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 4 of 11 Combining (17) and (24), we obtain ran M=Δ ┴ . We thus have verified (vi), and (vii). Since ran Mis closed, so is ran M* (by, e.g., [[1], Corollary 15.34]). Thus (iv) holds. (viii)&(v): We have seen in Proposition 2.1 that M†=Pr a nM∗◦M−1◦Pr a nM . (25) Now let zÎX.Then,by(iv),y:= Pr a nMZ=P⊥Z∈ ⊥ .By(vii),M -1 y=Ly +Δ.So M † z=P ran M∗ M −1 P ran M z=P ran M* M −1 y=P  ⊥ (Ly +)=P  ⊥ Ly =Ly =(L◦P  ⊥ ) z because ran L⊆Δ ┴ by (21). Hence (viii) holds. Furthermore, by (iv) and e.g., [[1], Proposition 3.28(v)],ran L=ranL◦P  ⊥=ranM†=ranM∗= ⊥ and so (v) holds. (ix): Note that P  ⊥=Id −P  and that P Δ =m -1 ∑ jÎI R j . Hence P⊥=Id −1 m j ∈I Rj . (26) Thus, by (viii) and (16), M†=L◦P⊥=1 m m−1  i=1 (m−i)Ri−1◦⎛ ⎝ Id −1 m j∈I Rj ⎞ ⎠ (27) =1 m m−1  i=1 (m−i)Ri−1−1 m2 m−1  i=1 (m−i) j ∈I Ri+j−1 . (28) Re-arranging this expression in terms of powers of Rand simplifying leads to M†=(Id −R)†= m  k =1 m−(2k−1) 2mRk−1 . (29) Remark 2.3 Suppose that ˜ L: ⊥→ X satisfies M◦˜ L=Id| ⊥ Then M−1:X⇒X:y→   Ly+,ify∈⊥; ∅,otherwise . (30) One may show that M†=P  ⊥◦˜ L◦P ⊥ and that P  ⊥◦ ˜ L= L (see (16)). Concrete choices for ˜ L and Lare ⊥→X: ( y1,y2,...,ym ) → ( y1,y1+y2,...,y1+y2+y3+...+ym ); (31) however, the range of the latter operator is not equal Δ ┴ whenever X≠{0}. Remark 2.4 Denoting the symmetric part of Mby M+=1 2 M+1 2 M∗and defining the quadratic form associated with Mby qM:x→1 2 x, Mx  , we note that [[17], Proposition 2.3] implies that c ran M+= domq∗ M = ⊥ . Fact 2.5 (Brezis-Haraux) (See [20] and also, e.g., [[1], Theorem 24.20].) Suppose A and B are monotone operators on X such that A + B is maximally monotone, dom A⊆ dom B, and B is rectangular. Then int ran(A+B)=int(ran A+ran B)and ran(A+B)=ran A+ranB. Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 5 of 11 Applying the Brezis-Haraux result to our given operators Aand M, we obtain the following. Corollary 2.6 The operator A+Mis maximally monotone and ran ( A+M ) =⊥+ran A Proof. Since each A i is maximally monotone and recalling Theorem 2.2(i), we see that Aand Mare maximally monotone. On the other hand, dom M=X. Thus, by the well known sum theorem for maximally monotone operators (see, e.g., [[1], Corollary 24.4 (i)]), A+Mis maximally monotone. Furthermore, by Theorem 2.2(ii) and (iv), Mis rectangular and ran M=Δ ┴ . The result therefore follows from Fact 2.5. 3 Composition We now use Corollary 2.6 to study the composition. When m= 2, then Theorem 3.1 (v) also follows from [[21], p. 124]. Theorem 3.1 Suppose that ( ∀i∈I ) 0∈ran ( Id −Ti ) .Then the following hold. (i) 0∈ran ( A+M ) . (ii) (∀ε>0)(∃(b, x)ÎX×X)||b|| ≤εand x=T(b+Rx). (iii) (∀ε>0)(∃(c, x)ÎX×X)||c|| ≤εand x=c+T(Rx). (iv) (∀ε>0)(∃xÎX)(∀iÎI)||T i-1 ···T 1 x m -T i T i-1 ···T 1 x m -x i-1 +x i || ≤(2i- 1)ε, where x 0 =x m . (v) (∀ε>0)(∃xÎX)||x-T m T m-1 ···T 1 x|| ≤m 2 ε. Proof. (i): The assumptions and (3) imply that ( ∀i∈I ) 0∈ran Ai. Hence, 0 ∈ ran A . Obviously, 0ÎΔ ┴ .Itfollowsthat0 ∈ ⊥+ran A .Thus,byCorollary2.6, 0∈ran ( A+M ) . (ii): Fix ε>0. In view of (i), there exists xÎXand bÎXsuch that ||b|| ≤εand b ÎAx +Mx. Hence b+Rx Î(Id +A)xand thus x=J A (b +Rx) =T(b +Rx). (iii): Let ε>0. By (ii), there exists (b, x)ÎX×Xsuch that ||b|| ≤εand x=T(b + Rx). Set c=x-T(Rx) =T(b +Rx) -T(Rx). Then, since Tis nonexpansive, ||c|| = ||T(b+Rx)-T(Rx)|| ≤||b|| ≤ε. (iv): Take ε>0. Then, by (iii), there exists xÎXand cÎXsuch that ||c|| ≤εand x=c+T(Rx). Let iÎI. Then x i =c i +T i x i-1 . Since ||c i || ≤||c|| ≤εand T i is nonexpansive, we have  TiTi−1···T1x0−xi  ≤  TiTi−1···T1x0−Tixi−1  +  Tixi−1−xi  (32a) ≤  TiTi−1···T1x0−Tixi−1  +ε . (32b) We thus obtain inductively TiTi−1···T1x0−xi≤iε.(33) Hence, Ti−1···T1x0−xi−1≤ ( i−1 ) ε . (34) Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 6 of 11 The conclusion now follows from adding (33) and (34), and recalling the triangle inequality (v): Let ε>0. In view of (iv), there exists xÎXsuch that ( ∀i∈I ) Ti−1···T1xm−TiTi−1···T1xm−xi−1+xi≤ ( 2i−1 )ε (35) where x 0 =x m .Nowset(∀iÎI)e i =T i-1 ···T 1 x m -T i T i-1 ···T 1 x m -x i-1 +x i .Then (∀iÎI)||e i || ≤(2i-1)ε. Set x=x m . Then m  i =1 ei= m  i =1 Ti−1...T1xm−TiTi−1...T1xm−xi−1+x i (36) =x− TmTm −1... T 1x . (37) This, (35), and the triangle inequality imply that x−TmTm−1···T1x≤ m  i =1 ei≤ m  i =1 (2i−1)ε=m2ε . (38) This completes the proof. Corollary 3.2 Suppose that ( ∀i∈I ) 0∈ran ( Id −Ti ) .Then 0∈ran ( Id −TmTm−1···T1 ) . Proof. This follows from Theorem 3.1(v). Remark 3.3 The converse implication in Corollary 3.2 fails in general: indeed, consider the case when X≠{0}, m=2,andvÎX\ {0}. Now set T 1 X®X: x ↦x+vand set T 2 X®X: x ↦x-v. Then 0/∈ran ( Id −T1 ) ={−v } and 0/∈ran ( Id −T2 ) ={v } however, T 2 T 1 = Id and ran ( Id −T2T1 ) ={0 } . Remark 3.4 Corollary 3.2 is optimal in the sense that even if (∀iÎI)wehave0Î ran(Id - T i ), we cannot deduce that 0 Îran(Id - T m T m-1 ···T 1 ): indeed, suppose that X = R2 and m=2.SetC 1 : = epi exp and C2:= R× { 0 } . Suppose further that T1=PC 1 and T2=PC 2 .Then(∀iÎI)0Îran(Id - T i ); however, 0∈ran ( Id −T2T1 ) \ran ( Id −T2T1 ) . 4 Asymptotic regularity The following notions (taken from Bruck and Reich’s seminal article [22]) will be very useful to obtain stronger results. Definition 4.1 ((strong) nonexpansiveness and asymptotic regularity) Let S: X ® X. Then: (i) Sisnonexpansive if (∀xÎX)(∀yÎX)||Sx - Sy|| ≤||x - y||. (ii) Sisstrongly nonexpansive if S is nonexpansive and whenever (x n ) nÎN and (y n ) nÎN are sequences in X such that (x n -y n ) nÎN is bounded and ||x n -y n ||- ||Sx n - Sy n || ®0, it follows that (x n -y n )-(Sx n -Sy n )®0. (iii) S is asymptotically regular if (∀xÎX)S n x-S n+1 x®0. The following result illustrates that strongly nonexpansive mappings generalize the notion of a firmly nonexpansive mapping.Inaddition,theclassofstrongly Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 7 of 11 nonexpansive mappings is closed under compositions. (In contrast, the composition of two (necessarily firmly nonexpansive) projectors may fail to be firmly nonexpansive.) Fact 4.2 (Bruck and Reich) The following hold. (i) Every firmly nonexpansive mapping is strongly nonexpansive. (ii) The composition of finitely many strongly nonexpansive mappings is also strongly nonexpansive. Proof. (i): See [[22], Proposition 2.1]. (ii): See [[22], Proposition 1.1]. The sequences of iterates and of differences of iterates have striking convergence properties as we shall see now. In passing, we note that Fact 4.3(i) also appears in [[21], Theorem 3.7(b)] even in certain Banach spaces. Fact 4.3 (Bruck and Reich) Let S: X ®X be strongly nonexpansive and let x ÎX. Then the following hold. (i) The sequence (S n x-S n+1 x) nÎN converges strongly to the unique element of least norm in ran ( Id −S ) . (ii) If Fix S =Ø,then ||S n x||®+∞. (iii) If Fix S≠Ø, then (S n x) nÎN converges weakly to a fixed point of S. Proof (i): See [[22], Corollary 1.5]. (ii): See [[22], Corollary 1.4]. (iii): See [[22], Corollary 1.3]. Suppose S: X ®Xis asymptotically regular. Then, for every xÎX,0¬S n x-S n+1 x =(Id-S)S n xÎran(Id - S) and hence 0∈ran ( Id −S ) . The opposite implication fails in general (consider S= - Id), but it is true for strongly nonexpansive mappings. Under the assumption that Sis firmly nonexpansive, the following result also follows from [[23], Corollary 2]. Corollary 4.4 Let S: X ®X be strongly nonexpansive. Then S is asymptotically regular if and only if 0∈ran ( Id −S ) . Proof.“⇒": Clear. “⇐": Fact 4.3(i). Corollary 4.5 Set S =T m T m-1 ···T 1 .Then S is asymptotically regular if and only if 0∈ran ( Id −S ) . Proof Since each T i is firmly nonexpansive, it is also strongly nonexpansiveby Fact 4.2 (i). By Fact 4.2(ii), Sis strongly nonexpansive. Now apply Corollary 4.4. Alternatively, 0∈ran ( Id −S ) by Corollary 3.2 and again Corollary 4.4 applies. We are now ready for our first main result. When m=2,thentheconclusionalso follows from [[21], p. 124]. Theorem 4.6 Suppose that each T i is asymptotically regular. Then T m T m-1 ···T 1 is asymptotically regular as well. Proof. Theorem 3.1(v) implies that 0∈ran ( Id −TmTm−1···T1 ) . The conclusion thus follows from Corollary 4.5. As an application of Theorem 4.6, we obtain the main result of [17]. Example 4.7 Let C 1 , ..., C m be nonempty closed convex subsets of X. Then the composition of the corresponding projectors, PC m PC m −1...PC1is asymptotically regular. Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 8 of 11 Proof. For every iÎI, the projector PC i is firmly nonexpansive, hence strongly nonexpansive, and Fix PC i =Ci= ∅ . Suppose that (∀i∈I)Ti=PC i , which is thus asymptotically regular by Corollary 4.4. Now apply Theorem 4.6. 5 Convex combination In this section, we use our fixed weights (l i ) iÎI (see (12)) to turn X m into a Hilbert product space different from Xconsidered in the previous sections. Specifically, we set Y:=Xmwith x,y=  i∈ I λixi,yi  (39) so that ||x|| 2 =∑ iÎI l i ||x i || 2 . We also set Q:Xm→Xm:x→ (¯ x)i∈I,where ¯ x:=  i∈ I λixi . (40) Fact 5.1 (See [[1], Proposition 28.13].) In the Hilbert product space Ywe have P Δ = Q. Corollary 5.2 In the Hilbert product space Ythe operator Qis firmly nonexpansive and strongly nonexpansive. Furthermore,FixQ=Δ≠Ø, 0Îran(Id -Q), and Qis asymptotically regular. Proof By Fact 5.1, the operator Qis equal to the projector P Δ and hence firmly nonexpansive. Now apply Fact 4.2(i) to deduce that Qis strongly nonexpansive. It is clear that Fix Q=Δand that 0Îran(Id -Q). Finally, recall Corollary 4.4 to see that Qis asymptotically regular. Proposition 5.3 In the Hilbert product space Ythe operator Tis firmly nonexpansive. Proof. Since each T i is firmly nonexpansive, we have (∀x=(x i ) iÎI ÎY)(∀y=(y i ) iÎI Î Y)||T i x i -T i y i || 2 ≤〈x i -y i ,T i x i -T i y i 〉⇒||Tx -Ty|| 2 =∑ iÎI l i ||T i x i -T i y i || 2 ≤∑ iÎI l i 〈x i -y i ,T i x i -T i y i 〉=〈x-y, Tx -Ty〉. Theorem 5.4 Suppose that ( ∀i∈I ) 0∈ran ( Id −Ti ) .Then the following hold in the Hilbert product space Y. (i) 0∈ran ( Id −T ) . (ii) Tis asymptotically regular. (iii) Q○Tis asymptotically regular. Proof. (i): This follows because (∀x=(x i ) iÎI )||x-Tx|| 2 =∑ iÎI l i ||x i -T i x i || 2 . (ii): Combine Fact 4.2(i) with Corollary 4.4. (iii): On the one hand, Qis firmly nonexpansive and asymptotically regular by Corollary 5.2. On the other hand, Tis firmly nonexpansive and asymptotically regular by Proposition 5.3 and Theorem 5.4(ii). Altogether, the result follows from Theorem 4.6. We are now ready for our second main result, which concerns convex combinations of firmly nonexpansive mappings. For further results in this direction-namely convex Bauschke et al.Fixed Point Theory and Applications 2012, 2012:53 http://www.fixedpointtheoryandapplications.com/content/2012/1/53 Page 9 of 11