Proximal point algorithm, Douglas-Rachford algorithm and alternating projections: a case study

0.56 MB·English

by  Heinz H. Bauschke

Proximal point algorithm, Douglas–Rachford algorithm
and alternating projections: a case study
5 HeinzH.Bauschke,∗ MinhN.Dao,† DominikusNoll‡ and HungM.Phan§
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2
January26,2015

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a
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2
Abstract

]
C
O Many iterative methods for solving optimization or feasibility problems have been
invented, and often convergence of the iterates to some solution is proven. Under
.
h favourable conditions, one might have additional bounds on the distance of the iter-
t
a ate to the solution leading thus to worst case estimates, i.e., how fast the algorithm must
m converge.
[ Exactconvergenceestimatesaretypicallyhardtocomeby. Inthispaper,weconsider
the complementary problem of finding best case estimates, i.e., how slow the algorithm
1
hastoconverge,andwealsostudyexactasymptoticratesofconvergence. Ourinvestigation
v
focusesonconvexfeasibilityintheEuclideanplane,whereonesetistherealaxiswhile
3
0 theotheristheepigraphofaconvexfunction.Thiscasestudyallowsustoobtainvarious
6 convergencerateresults. Wefocusonthepopularmethodofalternatingprojectionsand
6 the Douglas–Rachford algorithm. These methods are connected to the proximal point
0
algorithmwhichisalsodiscussed. OurfindingssuggestthattheDouglas–Rachfordal-
.
1 gorithm outperforms the method of alternating projections in the absence of constraint
0
qualifications. Variousexamplesillustratethetheory.
5
1
:
v 2010MathematicsSubjectClassification: Primary65K05;Secondary65K10,90C25.
i
X
Keywords: alternating projections, convex feasibility problem, convex set, Douglas–Rachford algo-
r
a rithm,projection,proximalmapping,proximalpointalgorithm,proximityoperator.
∗Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail:
[e-mail].
†Department of Mathematics and Informatics, Hanoi National University of Education, 136 Xuan Thuy,
Hanoi, Vietnam, and Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail:
[e-mail].
‡InstitutdeMathe´matiques,Universite´deToulouse,118routedeNarbonne,31062Toulouse,France.E-mail:
[e-mail].
§Department of Mathematical Sciences, University of Massachusetts Lowell, 265 Riverside St., Olney Hall
428,Lowell,MA01854,USA.E-mail:hung [e-mail].
1
1 Introduction
Threealgorithms
LetX beaEuclideanspace,withinnerproduct(cid:104)·,·(cid:105)andinducednorm(cid:107)·(cid:107),andlet f: X →
]−∞,+∞] be convex, lower semicontinuous, and proper. A classical method for finding
a minimizer of f is the proximal point algorithm (PPA). It requires using the proximal point
mapping(orproximityoperator)whichwaspioneeredbyMoreau[13]:
Fact1.1(proximalmapping) For every x ∈ X, there exists a unique point p = P (x) ∈ X
f
such that miny∈X f(y)+ 12(cid:107)x−y(cid:107)2 = f(p)+ 12(cid:107)x− p(cid:107)2. The induced operator Pf : X → X is
firmlynonexpansive1,i.e.,(∀x ∈ X)(∀y ∈ X)(cid:107)P (x)−P (y)(cid:107)2+(cid:107)(Id−P )x−(Id−P )y(cid:107)2 ≤
f f f f
(cid:107)x−y(cid:107)2.
TheproximalpointalgorithmwasproposedbyMartinet[12]andfurtherstudiedbyRock-
afellar[16]. Nowadaysnumerousextensionsexist;however,herewefocusonlyonthemost
basicinstanceofPPA:
Fact1.2(proximalpointalgorithm(PPA)) Let f: X → ]−∞,+∞] be convex, lower semicon-
tinuous, and proper. Suppose that Z, the set of minimizers of f, is nonempty, and let x ∈ X. Then
0
thesequencegeneratedby
(1) (∀n ∈ N) xn+1 = Pf(xn)
convergestoapointin Zanditsatisfies
(2) (∀z ∈ Z)(∀n ∈ N) (cid:107)xn+1−z(cid:107)2+(cid:107)xn−xn+1(cid:107)2 ≤ (cid:107)xn−z(cid:107)2.
Anostensiblyquitedifferenttypeofoptimizationproblemis,fortwogivenclosedconvex
nonemptysubsets AandBofX,tofindapointin A∩B (cid:54)= ∅. Letuspresenttwofundamen-
talalgorithmsforsolvingthisconvexfeasibilityproblem. Thefirstmethodwasproposedby
Bregman[8].
Fact1.3(methodofalternatingprojections(MAP)) Let a ∈ Aandset
0
(3) (∀n ∈ N) an+1 = PAPB(an).
Then(an)n∈N convergestoapoint a∞ ∈ C = A∩B. Moreover,
(4) (∀c ∈ C)(∀n ∈ N) (cid:107)an+1−c(cid:107)2+(cid:107)an+1−PBan(cid:107)2+(cid:107)PBan−an(cid:107)2 ≤ (cid:107)an−c(cid:107)2.
ThesecondmethodisthecelebratedDouglas–Rachfordalgorithm. Thenextresultcanbe
deducedbycombining[11]and[4].
1 Notethatif f = ι istheindicatorfunctionofanonemptyclosedconvexsubsetofX,thenP = P ,where
C f C
theP isthenearestpointmappingorprojectorofC;thecorrespondingreflectorisR =2P −Id.
C C C
2
Fact1.4(Douglas–Rachfordalgorithm(DRA)) Set T = Id−P +P R ,letz ∈ X,andset
A B A 0
(5) (∀n ∈ N) an = PAzn and zn+1 = Tzn.
Then2 (zn)n∈N converges to some point in z∞ ∈ FixT = (A∩B)+NA−B(0), and (an)n∈N con-
vergesto PAz∞ ∈ A∩B.
Again, there are numerous refinements and adaptations of MAP and DRA; however, it
is here not our goal to survey the most general results possible3 but rather to focus on the
speedofconvergence. Wewillmakethispreciseinthenextsubsection.
Goalandcontributions
Most rate-of-convergence results for PPA, MAP, and DRA take the following form: If some
additional condition is satisfied, then the convergence of the sequence is at least as good as some
form of “fast” convergence (linear, superlinear, quadratic etc.). This can be interpreted as a worst
case analysis. In the generality considered here4, we are not aware of results that approach
thisproblemfromtheotherside,i.e.,thataddressthequestion: Underwhichconditionsisthe
convergencenobetterthansomeformof“slow”convergence? Thisconcernsthebestcaseanalysis.
Ideally,onewouldlikeanexactasymptoticrateofconvergenceinthesenseof (14)below.
While we do not completely answer these questions, we do set out to tackle them by
providing a case study when X = R2 is the Euclidean plane, the set A = R×{0} is the
realaxis,andtheset B istheepigraphofaproperlowersemicontinuousconvexfunction f.
We will see that in this case MAP and DRA have connections to the PPA applied to f. We
focus in particular on the case not covered by conditions guaranteeing linear convergence
ofMAPorDRA5.WeoriginallyexpectedthebehaviourofMAPandDRAincasesof“bad
geometry” to be similar6. It came to us as surprise that this appears not to be the case. In
fact, the examples we provide below suggest that DRA performs significantly better than
MAP. Concretely, suppose that B is the epigraph of the function f(x) = (1/p)|x|p, where
1 < p < +∞. Since A = R×{0}, we have that A∩B = {(0,0)} and since f(cid:48)(0) = 0, the
“angle” between A and B at the intersection is 0. As expected MAP converges sublinearly
(evenlogarithmically)to0. However,DRAconvergesfasterinallcases: superlinearly(when
1 < p < 2), linearly (when p = 2) or logarithmically (when 2 < p < +∞). This example is
deducedbygeneralresultsweobtainonexactratesofconvergenceforPPA,MAPandDRA.
2HereFixT=(cid:8)x∈X(cid:12)(cid:12)x=Tx(cid:9)isthesetoffixedpointsofT,andNA−B(0)standsforthenormalconeofthe
(cid:8) (cid:12) (cid:9)
setA−B= a−b(cid:12)a∈ A,b∈B at0.
3See,e.g.,[3]forvariousmoregeneralvariantsofPPA,MAP,andDRA.
4Some results are known for MAP when the sets are linear subspaces; however, the slow (sublinear)
convergencecanonlybeobservedininfinite-dimensionalHilbertspace;see[9]andreferencestherein.
5Indeed,themostcommonsufficientconditionforlinearconvergenceineithercaseisri(A)∩ri(B)(cid:54)=∅;see
[5,Theorem3.21]forMAPand[14]or[6,Theorem8.5(i)]forDRA.
6ThisexpectationwasfoundedinthesimilarbehaviourofMAPandDRAfortwosubspaces;see[2].
3
Organization
The paper is organized as follows. In Section 2, we provide various auxiliary results on the
convergence of real sequences. These will make the subsequent analysis of PPA, MAP, and
DRAmorestructured. Section3focusesonthePPA.Afterreviewingresultsonfinite,super-
linear, andlinearconvergence, weexhibitacasewheretheasymptoticrateisonlylogarith-
mic. We then turn to MAP in Section 4 and provide results on the asymptotic convergence.
We also draw the connection between MAP and PPA and point out that a result of Gu¨ler is
sharp. InSection5,wedealwithDRA,drawagainaconnectiontoPPAandpresentasymp-
toticconvergence. Thenotationweemployisfairlystandardandfollows,e.g.,[15]and[3].
2 Auxiliary results
InthissectionwecollectvariousresultsthatfacilitatethesubsequentanalysisofPPA,MAP
andDRA.Webeginwiththefollowingusefulresultwhichappearstobepartofthefolklore7.
Fact2.1(generalizedStolz–Cesa`rotheorem) Let (an)n∈N and (bn)n∈N besequencesinRsuch
that (bn)n∈N isunboundedandeitherstrictlymonotoneincreasingorstrictlymonotonedecreasing.
Then
a −a a a a −a
(6) lim n+1 n ≤ lim n ≤ lim n ≤ lim n+1 n,
n→∞ bn+1−bn n→∞ bn n→∞ bn n→∞ bn+1−bn
wherethelimitsmayliein[−∞,+∞].
Setting(bn)n∈N = (n)n∈N inFact2.1,weobtainthefollowing:
Corollary2.2 Thefollowinginequalitiesholdforanarbitrarysequence(xn)n∈N inR:
x x
(7) nl→im∞(xn+1−xn) ≤ nl→im∞ nn ≤ nl→im∞ nn ≤ nl→im∞(xn+1−xn).
Fortheremainderofthissection,weassumethat
(8) g: R++ → R++ isincreasingand H isanantiderivativeof−1/g.
Example2.3(xq) Let g(x) = xq on R++, where 1 ≤ q < ∞. If q > 1, then −1/g(x) =
−x−q and we can choose H(x) = x1−q/(q−1) which has the inverse H−1(x) = 1/((q−
1)x)1/(q−1). If q = 1, then we can choose H(x) = −ln(x) which has the inverse H−1(x) =
exp(−x).
Proposition2.4 Let(βn)n∈N and(δn)n∈N besequencesinR++,andsupposethat
(9) (∀n ∈ N) βn+1 = βn−δng(βn).
Thenthefollowinghold:
7Sincewewereabletolocateonlyanonlinereference,weincludeaproofinAppendixA.
4
β −β g(β )
(i) (∀n ∈ N) δn ≤ H(βn+1)−H(βn) ≤ δn+1βn+n1−βn+n+12 = δng(βn+n1).
H(β ) H(β ) g(β )
(ii) lim δ ≤ lim n ≤ lim n ≤ lim δ n .
n→∞ n n→∞ n n→∞ n n→∞ ng(βn+1)
(iii) If(δn)n∈N isconvergent,sayδn → δ∞,and gg(β(βn+n)1) → 1,then H(nβn) → δ∞.
Proof. Foreveryn ∈ N,wehave
(10a) δn = βng−(ββn)+1 ≤ (cid:90) βn gd(xx) = H(βn+1)−H(βn)
n βn+1
β −β β −β g(β )
(10b) ≤ gn(βn+n1+)1 = δn+1βn+n1−βn+n+12 = δng(βn+n1).
(cid:4)
Hence(i)holds. Combiningwith(7),weobtain(ii). Finally,(iii)followsfrom(ii).
Corollary2.5 Let(xn)n∈N and(δn)n∈N besequencesinR++ suchthat
(11) (∀n ∈ N) xn = xn+1+δng(xn+1).
Thenthefollowinghold:
g(x )
(i) (∀n ∈ N) δn g(xn+)1 ≤ H(xn+1)−H(xn) ≤ δn.
n
g(x ) H(x ) H(x )
(ii) lim δ n+1 ≤ lim n ≤ lim n ≤ lim δ .
n→∞ n g(xn) n→∞ n n→∞ n n→∞ n
(iii) If(δn)n∈N isconvergent,sayδn → δ∞,and gg(x(xn+n)1) → 1,then H(nxn) → δ∞.
Proof. Indeed,set(∀n ∈ N)ε = δ g(xn+1) andrewritetheupdate
n n g(xn)
g(x )
(12) xn+1 = xn−δn g(xn+)1 g(xn) = xn−εng(xn).
n
(cid:4)
NowapplyProposition2.4.
Definition2.6(typesofconvergence) Let(αn)n∈N beasequenceinR++ suchthatαn → 0,and
supposethereexist1 ≤ q < +∞suchthat
α
(13) n+q1 → c ∈ R+.
α
n
Thentheconvergenceof(αn)n∈N to0is:
(i) withorderqifq > 1andc > 0;
5
(ii) superlinearifq = 1andc = 0;
(iii) linearifq = 1and0 < c < 1;
(iv) sublinearifq = 1andc = 1;
(v) logarithmicifitissublinearand|αn+1−αn+2|/|αn−αn+1| → 1.
If(βn)n∈N isalsoasequenceinR++,itisconvenienttodefine
α
(14) αn ∼ βn ⇔ nl→im∞ βnn ∈ R++.
The following example exhibits a case where we obtain a simple exact asymptotic rate of
convergence.
Example2.7 Let (xn)n∈N and (δn)n∈N be sequences in R++, and let 1 < q < ∞. Suppose
that
x
(15) δn → δ∞ ∈ R++, x n → 1, and (∀n ∈ N) xn = xn+1+δnxnq+1.
n+1
Then x → 0logarithmically,
n
x 1 (cid:16)1(cid:17)1/(q−1)
(16) n → and x ∼ .
(cid:16)1(cid:17)1/(q−1) ((q−1)δ∞)1/(q−1) n n
n
Proof. Suppose that g(x) = xq and note that g(xn+1)/g(xn) = (xn+1/xn)q → 1q = 1. This
implies that x → 0 logarithmically. Finally, (16) follows from Example 2.3, Corollary 2.5,
n
(cid:4)
and(14).
We conclude this section with some one-sided versions which are useful for obtaining
informationabouthowfastorslowasequencemustconverge.
Corollary2.8 Let(βn)n∈N and(ρn)n∈N besequencesinR++,andsupposethat
(17) (∀n ∈ N) βn+1 ≤ βn−ρng(βn) and ρ = lim ρn ∈ R++.
n→∞
Then
(18) (cid:0)∀ε ∈ ]0,ρ[(cid:1)(∃m ∈ N)(∀n ≥ m) β ≤ H−1(cid:0)n(ρ−ε)(cid:1).
n
Proof. Observethat
β −β
(19) (∀n ∈ N) βn+1 = βn−δng(βn), where δn = ng(β n)+1 ≥ ρn.
n
Hence,byProposition2.4,ρ ≤ limn→∞H(βn)/n. Letε ∈]0,ρ[. Thenthereexistsm ∈ Nsuch
that(∀n ≥ m)ρ−ε ≤ H(β )/n ⇔ H−1(n(ρ−ε)) ≥ β . (cid:4)
n n
6
Example2.9 Let (βn)n∈N and (ρn)n∈N be sequences in R++, let 1 ≤ q < ∞, and suppose
that
(20) (∀n ∈ N) βn+1 ≤ βn−ρnβqn and ρ = lim ρn ∈ R++.
n→∞
Let0 < ε < ρ. Thenthereexistsm ∈ Nsuchthatthefollowinghold:
(i) Ifq > 1,then(∀n ≥ m) β ≤ 1 = O(cid:0)1/n1/(q−1)(cid:1).
n (cid:0) (cid:1)1/(q−1)
(q−1)n(ρ−ε)
(ii) Ifq = 1,then(∀n ≥ m) β ≤ γn,whereγ = exp(ε−ρ) ∈ ]0,1[.
n
Consequently,theconvergenceof(βn)n∈Nto0isatleastsublinearifq > 1andatleastlinear
ifq = 1.
(cid:4)
Proof. CombineExample2.3withCorollary2.8.
Remark2.10 Example2.9(i)canalsobededucedfrom[7,Lemma4.1];seealso[1].
Corollary2.11 Let(βn)n∈N and(ρn)n∈N besequencesinR++,andsupposethat
g(β )
(21) (∀n ∈ N) βn ≥ βn+1 ≥ βn−ρng(βn) and ρ = nl→im∞ρng(βn+n1) ∈ R+.
Then
(22) (∀ε ∈ R++)(∃m ∈ N)(∀n ≥ m) βn ≥ H−1(cid:0)n(ρ+ε)(cid:1).
Proof. Observethat
β −β
(23) (∀n ∈ N) βn+1 = βn−δng(βn), where δn = ng(β n)+1 ≤ ρn.
n
Hence, by Proposition 2.4, limn→∞H(βn)/n ≤ ρ. Let ε ∈ R++. Then there exists m ∈ N
suchthat(∀n ≥ m)ρ+ε ≥ H(β )/n⇔ H−1(n(ρ+ε)) ≤ β . (cid:4)
n n
Example2.12 Let (βn)n∈N and (ρn)n∈N be sequences in R++, let 1 ≤ q < ∞, and suppose
that
q
β
(24) (∀n ∈ N) βn ≥ βn+1 ≥ βn−ρnβqn and ρ = nl→im∞ρnβqn ∈ R+.
n+1
Letε ∈ R++. Thenthereexistsm ∈ Nsuchthatthefollowinghold:
1
(i) Ifq > 1,then(∀n ≥ m) β ≥ .
n (cid:0) (cid:1)1/(q−1)
(q−1)n(ρ+ε)
(ii) Ifq = 1,then(∀n ≥ m) β ≥ γn,whereγ = exp(−ρ−ε) ∈ ]0,1[.
n
Consequently,theconvergenceof(βn)n∈N to0isatbestsublinearifq > 1andatbestlinear
ifq = 1.
(cid:4)
Proof. CombineExample2.3withCorollary2.11.
7
3 Proximal point algorithm (PPA)
Thissectionfocusesontheproximalpointalgorithm. Weassumethat
(25) f: R → ]−∞,+∞] isconvex,lowersemicontinuous,proper,
with
(26) f(0) = 0and f(x) > 0when x (cid:54)= 0.
Given x ∈ R,wewillstudythebasicproximalpointiteration
0
(27) (∀n ∈ N) xn+1 = Pf(xn).
Notethatifx > 0andy < 0,then f(y)+1|x−y|2 > f(0)+1|x−0|2 ≥ f(P x)+1|x−P x|2.
2 2 f 2 f
Hencethebehaviourof f|R−− isirrelevantforthedeterminationofPf|R++ (andananalogous
statementholdsforthedeterminationofPf|R−−)! Forthisreason,werestrictourattentionto
thecasewhen
(28) x0 ∈ R++
isthestartingpointoftheproximalpointalgorithm. Thegeneraltheory(Fact1.2)thenyields
(29) x ≥ x ≥ ··· ≥ x ↓ 0.
0 1 n
Inthissection,itwillbeconvenienttoadditionallyassumethat
(30) f isanevenfunction;
although, as mentioned, the behaviour of f|R−− is actually irrelevant because x0 ∈ R++.
Combining the assumption that 0 is the unique minimizer of f with [15, Theorem 24.1], we
learnthat
(31) 0 ∈ ∂f(0) = [f(cid:48) (0), f(cid:48) (0)]∩R = [−f(cid:48) (0), f(cid:48) (0)]∩R.
− + + +
Westartourexplorationbydiscussingconvergenceinfinitelymanysteps.
Proposition3.1(finiteconvergence) We have x → 0 in finitely many steps, regardless of the
n
startingpoint x ∈ R ,ifandonlyif
0 ++
(32) 0 < f(cid:48) (0),
+
inwhichcase P x = 0⇔ x ≤ f(cid:48) (0).
f n n +
Proof. Let x > 0. Then P x = 0⇔ x ∈ 0+∂f(0)⇔ x ≤ f(cid:48) (0)by(31).
f +
Supposefirstthat f(cid:48) (0) > 0. Then,by(31),0 ∈ int∂f(0)and,using(29),thereexistsn ∈ N
+
suchthatxn ≤ f+(cid:48) (0). Itfollowsthatxn+1 = xn+2 = ··· = 0. (Alternatively,thisfollowsfrom
amuchmoregeneralresultofRockafellar;see[16,Theorem3]andalsoRemark3.4below.)
Now assume that there exists n ∈ N such that P x = 0 and x > 0. By the above,
f n n
x ≤ f(cid:48) (0)andthus f(cid:48) (0) > 0. (cid:4)
n + +
8
Anextremecaseoccurswhen f(cid:48) (0) = +∞inProposition3.1:
+
Example3.2(ι andtheprojector) Suppose that f = ι . Then P = P and (∀n ≥ 1)
{0} {0} f {0}
x = 0.
n
Example3.3(|x|1 andthethresholder) Suppose that f = |·| in which case ∂f(0) = [−1,1]
and f+(0) = 1. Proposition 3.1 guarantees finite convergence of the PPA. Indeed, either a
directargumentor[3,Example14.5]yields
(cid:40)
x− x , if|x| > 1;
(33) P : x (cid:55)→ |x|
f
0, otherwise,
Consequently, x = 0ifandonlyifn ≥ (cid:100)x (cid:101).
n 0
Remark3.4 In [16, Theorem 3], Rockafellar provided a very general sufficient condition for
finiteconvergenceofthePPA(whichworksactuallyforfindingzerosofamaximallymono-
toneoperatordefinedonaHilbertspace). Inourpresentsetting,hisconditionis
(34) 0 ∈ int∂f(0).
ByProposition3.1,thisisalsoaconditionthatisnecessaryforfiniteconvergence.
Thus, we assume from now on that f(cid:48) (0) = 0, or equivalently (since f is even and by
+
(31)),that
(35) f(cid:48)(0) = 0.
inwhichcasefiniteconvergencefailsandthus
(36) x > x > ··· > x ↓ 0.
0 1 n
We now have the following sufficient condition for linear convergence. The proof is a
refinementoftheideasofRockafellarin[16].
Proposition3.5(sufficientconditionforlinearconvergence) Supposethat
f(x)
(37) λ = lim ∈ ]0,+∞].
x2
x↓0
Thenthefollowinghold:
(i) Ifλ < +∞,thenthereexistsα ∈ (cid:2) 1 , 1(cid:3)suchthat
0 2λ λ
α
(38) (∀ε > 0)(∃m ∈ N)(∀n ≥ m) |xn+1| ≤ (cid:113) 0 |xn|.
1+α2(1+2λ−2ε)
0
9
(ii) Ifλ = +∞,then
α
(39) (∀α > 0)(∀ε > 0)(∃m ∈ N)(∀n ≥ m) |xn+1| ≤ (cid:112) |xn|.
1+α2(1+2λ−ε)
Proof. By [16, Remark 4 and Proposition 7], there exists α ∈ (cid:2) 1 , 1(cid:3) such that (∂f)−1 is
0 2λ λ
Lipschitz continuous at 0 with every modulus α > α . Let α > α . Then there exists τ > 0
0 0
suchthat
(40) (∀|x| < τ)(∀z ∈ (∂f)−1(x)) |z| ≤ α|x|.
Since x → 0 by [16, Theorem 2] (or (36)), there exists m ∈ N such that (∀n ≥ m) |x −
n n
xn+1| ≤ τ. Letn ≥ m. Noticingthat xn ∈ (Id+∂f)(xn+1),wehave
(41) xn+1 ∈ (∂f)−1(xn−xn+1).
Itfollowsby(40)that
(42) |xn+1| ≤ α|xn−xn+1|.
Since xn−xn+1 ∈ ∂f(xn+1),wehave
(43) (cid:104)xn−xn+1,xn+1(cid:105) = (cid:104)xn−xn+1,xn+1−0(cid:105) ≥ f(xn+1)− f(0) = f(xn+1).
Now for every ε > 0, employing (37) and increasing m if necessary, we can and do assume
that
(cid:16) ε(cid:17)
(44) (∀n ≥ m) (cid:104)xn−xn+1,xn+1(cid:105) ≥ λ− |xn+1|2.
2
Letn ≥ m. Combining(42)and(44),weobtain
(45a) |xn|2 = |xn+1|2+|xn−xn+1|2+2(cid:104)xn−xn+1,xn+1(cid:105)
1
(45b) ≥ |xn+1|2+ α2|xn+1|2+(2λ−ε)|xn+1|2
(cid:16)1+α2(1+2λ−ε)(cid:17)
(45c) = α2 |xn+1|2.
Thisgives
α
(46) |xn+1| ≤ (cid:112) |xn|
1+α2(1+2λ−ε)
andhence(39)holds. Nowassumethatλ < +∞sothatα > 0. Sinceα (cid:55)→ √ α =
0
1+α2(1+2λ−ε)
(cid:113)
(cid:113) 1 +(11+2λ−ε) is strictly increasing on R+, we note that the choice α = α0/ 1−εα20 > α0
α2 (cid:4)
yields(38).
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