Firmly nonexpansive mappings and maximally monotone operators: correspondence and duality

0.24 MB·English

by  Heinz H. Bauschke

Firmly nonexpansive mappings
and
1 maximally monotone operators:
1
0
correspondence and duality
2

n
a
J

4
2 Heinz H. Bauschke,∗ Sarah M. Moffat,† and Xianfu Wang‡

]
A
January 24, 2011
F
.
h
t
a
m
[
Abstract

1
v
8 Thenotionofafirmlynonexpansivemappingiscentralinfixedpointtheorybecause
8
ofattractiveconvergencepropertiesforiteratesandthecorrespondencewithmaximal
6
4 monotone operators due to Minty. In this paper, we systematically analyze the rela-
1. tionship between properties of firmly nonexpansive mappings and associated maxi-
0 malmonotoneoperators. Dualandself-dualpropertiesarealsoidentified. Theresults
1
areillustratedthroughseveralexamples.
1
:
v
i
X 2010 Mathematics Subject Classification: Primary 47H05, 47H09; Secondary 26B25, 52A41,
r 90C25.
a
Keywords: Banach contraction, convex function, firmly nonexpansive mapping, fixed point,
Hilbert space, Legendre function, maximal monotone operator, nonexpansive mapping, para-
monotone,proximalmap,rectangular,resolvent,subdifferentialoperator.
Mathematics,IrvingK.BarberSchool,UBCOkanagan,Kelowna,BritishColumbiaV1V1V7,Canada.
∗
E-mail: [email protected].
†Mathematics,IrvingK.BarberSchool,UBCOkanagan,Kelowna,BritishColumbiaV1V1V7,Canada.
E-mail: [email protected].
‡Mathematics,IrvingK.BarberSchool,UBCOkanagan,Kelowna,BritishColumbiaV1V1V7,Canada.
E-mail: [email protected].
1
1 Introduction
Throughout thispaper,
(1) X isa real Hilbertspace with innerproduct , and inducednorm .
h· ·i k·k
Recall thata mapping
(2) T: X X
→
is firmlynonexpansive if
(3) ( x X)( y X) Tx Ty 2 + (Id T)x (Id T)y 2 x y 2,
∀ ∈ ∀ ∈ k − k k − − − k ≤ k − k
where Id: X X: x x denotes the identity operator. It is clear that if T is firmly
→ 7→
nonexpansive, then itisnonexpansive, i.e.,Lipschitz continuous with constant 1,
(4) ( x X)( y X) Tx Ty x y ;
∀ ∈ ∀ ∈ k − k ≤ k − k
the converse, however, is false (consider Id). When T is Lipschitz continuous with a
−
constant in [0,1[, then we shall referto T asa Banach contraction.
Returningtofirmlynonexpansive mappings,letusprovide ausefullistofwell-known
characterizations.
Fact 1.1 (See, e.g., [5,15, 16].) Let T: X X. Thenthe followingare equivalent:
→
(i) T is firmlynonexpansive.
(ii) Id T is firmlynonexpansive.
−
(iii) 2T Idis nonexpansive.
−
(iv) ( x X)( y X) Tx Ty 2 x y,Tx Ty .
∀ ∈ ∀ ∈ k − k ≤ h − − i
(v) ( x X)( y X). 0 Tx Ty,(Id T)x (Id T)y .
∀ ∈ ∀ ∈ ≤ h − − − − i
Variousproblemsin the natural sciencesandengineeringcanbeconverted intoa fixed
point problem, where the set ofdesired solutions isthe setof fixedpoints
(5) FixT := x X x = Tx .
∈
(cid:8) (cid:12) (cid:9)
(cid:12)
2
If T isfirmly nonexpansive and FixT = ∅, then the sequenceof iterates
6
(6) (Tnx)n N
∈
iswellknowntoconverge weaklytoafixedpoint[8]—thisisnottrueformappingsthat
are merely nonexpansive; consider, e.g., Id.
−
Firmly nonexpansive mappings are also important because of their correspondence
with maximally monotone operators. Recall that a set-valued operator A: X ⇉ X (i.e.,
( x X) Ax X) with graph grA is monotone if
∀ ∈ ⊆
(7) ( (x,u) grA)( (y,v) grA) x y,u v 0,
∀ ∈ ∀ ∈ h − − i ≥
and that A ismaximallymonotone ifitisimpossible tofind aproper extension of A that
is still monotone. We write domA := x X Ax = ∅ and ranA := A(X) = Ax
x X
∈ 6 ∈
for the domain and range of A, respectively. Monotone operators play a crucial role in
(cid:8) (cid:12) (cid:9) S
modernnonlinearanalysisandoptimization; s(cid:12)ee,e.g.,thebooks[5],[6],[7],[9],[22],[23],
[24], [26], [27], and[28].
Primeexamplesofmaximallymonotoneoperatorsarecontinuouslinearmonotoneop-
erators and subdifferential operators (in the sense of convex analysis) of functions that
are convex, lower semicontinuous, and proper. Itwill be convenientto set
(8) Γ := f : X ] ∞,+∞] f isconvex, lowersemicontinuous, andproper .
0
→ −
(cid:8) (cid:12) (cid:9)
(cid:12)
See [19], [21], [24], and [5]for background material on convex analysis.
Nowlet A: X ⇉ X be maximallymonotone anddenote the associated resolvent by
(9) J := (Id+A) 1.
A −
When A = ∂f for some f Γ , i.e., A is a subdifferential operator, then we also write
0
∈
J = Prox and, following Moreau [18], we refer to this mapping as the proximal map-
∂f f
ping. In his seminal paper [17], Minty observed that J is in fact a firmly nonexpansive
A
operatorfrom X to X andthat,conversely,everyfirmlynonexpansiveoperatorarisesthis
way:
Fact 1.2(Minty) (See, e.g., [17] or [13].) Let T: X X be firmly nonexpansive, and let
→
A: X ⇉ X be maximallymonotone. Thenthe followinghold.
(i) B = T 1 Idis maximallymonotone (and J = T).
− B
−
3
(ii) J isfirmlynonexpansive (and A = J 1 Id).
A A− −
Therefore, the mapping
(10) T T 1 Id
−
7→ −
from the set offirmly nonexpansive mappingsto the setofmaximal monotone operators
is abijection, with inverse
(11) A J = (Id+A) 1.
A −
7→
Eachclassofobjectshasitsowndualityoperation: ifT: X X isfirmlynonexpansive,
→
then so isthe associated dualmapping(see Fact 1.1)
(12) Id T;
−
note that the dual of the dual is indeed Id (Id T) = T. Analogously, if A: X ⇉ X is
− −
maximally monotone, then soisthe inverse operator
(13) A 1,
−
and we clearly have (A 1) 1 = A. Traversing between the two classes and dualizing is
− −
commutative asthe resolventidentity
(14) Id = J + J
A A 1
−
shows. Wealso havethe Mintyparametrization
(15) grA = J x,(Id J )x x X ,
A A
− ∈
which provides the bijection x ((cid:8)J(cid:0)Ax,x JAx) fro(cid:1)m(cid:12)X onto(cid:9)grA, with inverse (x,u)
7→ − (cid:12) 7→
x+u.
The goal of this paper is threefold. First, we will provide a comprehensive catalogue of corre-
sponding properties of firmly nonexpansive mappings and maximally monotone operators. Sec-
ond, we shall examine duality of these properties, and also identify those properties that are self-
dual, i.e., a property holds for T if and only if the same property holds for Id T, and a corre-
−
spondingpropertyholdsfor Aifandonlyifitholdsfor A 1 inthecontextofmaximallymonotone
−
operators. Third,werevisitsomeofthesepropertiesinmorespecializedsettingsandpresentappli-
cations of our results tooperators occurringinsplittingmethods,includingreflectedresolvents.
The paper is organized as follows. In Section 2, we systematically study properties
offirmly nonexpansivemappingsandcorresponding properties ofmaximallymonotone
operators. Section 3 revisits these properties from a duality point of view. In the final
Section 4, we also touch upon corresponding properties of nonexpansive mappings and
provide a fewpossible applications tomappingsarising in algorithms.
Thenotationweemployisstandardandasin[5],[19],[21],or[24]. Thesebooksprovide
also background material for notions notexplicitly reviewed here.
4
2 Correspondence of Properties
Themainresultinthissectionisthefollowingresult,whichprovidesacomprehensivelist
ofcorresponding propertiesoffirmly nonexpansivemappingsandmaximallymonotone
operators.
Theorem 2.1 Let T: X X be firmly nonexpansive, let A: X ⇉ X be maximally monotone,
→
and supposethat T = J or equivalentlythat A = T 1 Id. Thenthe followinghold:
A −
−
(i) ranT = domA.
(ii) T is surjectiveifand only if domA = X.
(iii) Id T is surjectiveifand only if A is surjective.
−
(iv) T is injectiveifand only if A is atmostsingle-valued.
(v) T is anisometry, i.e.,
(16) ( x X)( y X) Tx Ty = x y
∀ ∈ ∀ ∈ k − k k − k
if and only ifthere existsz X such that A: x z, inwhichcase T: x x z.
∈ 7→ 7→ −
(vi) T satisfies
(17) ( x X)( y X) Tx = Ty Tx Ty 2 < x y,Tx Ty
∀ ∈ ∀ ∈ 6 ⇒ k − k h − − i
if and only if A isstrictlymonotone, i.e.,
(18) ( (x,u) grA)( (y,v) grA) x = y x y,u v > 0.
∀ ∈ ∀ ∈ 6 ⇒ h − − i
(vii) T is strictlymonotone ifand only if A is atmostsingle-valued.
(viii) T is strictlyfirmlynonexpansive, i.e.,
(19) ( x X)( y X) x = y Tx Ty 2 < x y,Tx Ty
∀ ∈ ∀ ∈ 6 ⇒ k − k h − − i
if and only if A isatmostsingle-valuedand strictlymonotone.
(ix) T is strictlynonexpansive, i.e.,
(20) ( x X)( y X) x = y Tx Ty < x y
∀ ∈ ∀ ∈ 6 ⇒ k − k k − k
if and only if A isdisjointly injective1, i.e.,
(21) ( x X)( y X) x = y Ax Ay = ∅.
∀ ∈ ∀ ∈ 6 ⇒ ∩
1In some of the literature, A is simply referred to as “injective” but this is perhaps a little imprecise:
“injectivity of A” could be interpreted as “[ x,y domA and x = y] Ax = Ay” which is different
{ } ⊆ 6 ⇒ 6
fromwhatwerequirehere.
5
(x) T is injectiveand strictlynonexpansive, i.e.,
(22) ( x X)( y X) x = y 0 < Tx Ty < x y
∀ ∈ ∀ ∈ 6 ⇒ k − k k − k
if and only if A isatmostsingle-valuedand disjointlyinjective.
(xi) Suppose that ε ]0,+∞[. Then (1 + ε)T is firmly nonexpansive if and only if A is
∈
stronglymonotonewithconstantε,i.e., A εIdismonotone,inwhichcaseT isaBanach
−
contraction withconstant (1+ε) 1.
−
(xii) Suppose that γ ]0,+∞[. Then (1+γ)(Id T) is firmly nonexpansive if and only if A
∈ −
is γ-cocoercive, i.e.,
(23) ( (x,u) grA)( (y,v) grA) x y,u v γ u v 2.
∀ ∈ ∀ ∈ h − − i ≥ k − k
(xiii) Suppose that β ]0,1[. Then T is a Banach contraction with constant β if and only if A
∈
satisfies
(24)
1 β2
( (x,u) grA)( (y,v) grA) − x y 2 2 x y,u v + u v 2.
∀ ∈ ∀ ∈ β2 k − k ≤ h − − i k − k
(xiv) Suppose that φ: [0,+∞[ [0,+∞] isincreasingand vanishesonly at0. Then T satisfies
→
(25) ( x X)( y X) Tx Ty,(x Tx) (y Ty) φ Tx Ty
∀ ∈ ∀ ∈ h − − − − i ≥ k − k
if and only if A isuniformly monotone with modulus φ, i.e., (cid:0) (cid:1)
(26) ( (x,u) grA)( (y,v) grA) x y,u v φ x y .
∀ ∈ ∀ ∈ h − − i ≥ k − k
(cid:0) (cid:1)
(xv) T satisfies
(27)
Tx = T Tx+y Ty
( x X)( y X) Tx Ty 2 = x y,Tx Ty −
∀ ∈ ∀ ∈ k − k h − − i ⇒ (Ty = T(cid:0)Ty+x Tx(cid:1)
−
if and only if A isparamonotone [10], i.e., (cid:0) (cid:1)
(28)
( (x,u) grA)( (y,v) grA) x y,u v = 0 (x,v),(y,u) grA.
∀ ∈ ∀ ∈ h − − i ⇒ { } ⊆
(xvi) (Bartz etal.) T is cyclically firmlynonexpansive, i.e.,
n
∑
(29) x Tx ,Tx Tx 0,
i i i i+1
h − − i ≥
i=1
for every setof points x ,...,x X, where n 2,3,... and x = x , if and only
1 n n+1 1
{ } ⊆ ∈ { }
if A isasubdifferentialoperator,i.e., thereexists f Γ suchthat A = ∂f.
0
∈
6
(xvii) T satisfies
(30) ( x X)(y X) inf Tx Tz,(y Ty) (z Tz) > ∞
∀ ∈ ∈ z Xh − − − − i −
∈
if and only if A is rectangular [23, Definition 31.5](thisis alsoknown as3 monotone),
∗
i.e.,
(31) ( x domA)( v ranA) inf x z,v w > ∞.
∀ ∈ ∀ ∈ (z,w) grAh − − i −
∈
(xviii) T is linearifand only if A is alinear relation, i.e.,grA isalinear subspaceof X X.
×
(xix) T is affineifand only if A is anaffine relation, i.e.,grA isan affinesubspaceof X X.
×
(xx) (Zarantonello) ranT = FixT =: C if and only if A is a normal cone operator, i.e., A =
∂ι ; equivalently, T isaprojection(nearestpoint)mapping P .
C C
(xxi) T is sequentiallyweakly continuous ifand only if grA is sequentiallyweakly closed.
Proof. Let x,y,u,v be in X.
(i): Clear.
(ii): Thisfollows from (i).
(iii): Clearfrom the Minty parametrization (15).
(iv): Assume first that T is injective and that u,v Ax. Then x + u,x + v
{ } ⊆ { } ⊆
(Id+A)x andhence x = T(x+u) = T(x+v). Since T isinjective,itfollowsthat x+u =
x+v and hence that u = v. Thus, A is at most single-valued. Conversely, let us assume
that Tu = Tv = x. Then u,v (Id+A)x = x + Ax and hence u x,v x Ax.
{ } ⊆ { − − } ⊆
Since Aisatmostsingle-valued,wehaveu x = v xandsou = v. Thus, T isinjective.
− −
(v): Assume first that T is an isometry. Then Tx Ty 2 = x y 2 Tx Ty 2 +
k − k k − k ≥ k − k
(Id T)x (Id T)y 2. Itfollowsthatthereexists z X such that T: w w z. Thus,
k − − − k ∈ 7→ −
T 1: w w+z. On the other hand, T 1 = Id+A: w w + Aw. Hence A: w z,
− −
7→ 7→ 7→
as claimed. Conversely, let us assume that there exits z X such that A: w z. Then
∈ 7→
Id+A: w w+z and hence T = J = (Id+A) 1: w w z. Thus, T isanisometry.
A −
7→ 7→ −
(vi): Assume first that T satisfies (17), that (x,u),(y,v) grA, and that x = y. Set
{ } ⊆ 6
p = x +u and q = y+v. Then (x,u) = (Tp,p Tp) and (y,v) = (Tq,q Tq). Since
− −
x = y, itfollows that Tp = Tq andtherefore that Tp Tq 2 < p q,Tp Tq because
T6satisfies (17). Hence 06 < (p Tp) (q Tqk),Tp− Tkq =h u− v,x −y .iThus, A
h − − − − i h − − i
is strictly monotone. Conversely, let us assume that A is strictly monotone and that x =
Tu = Tv = y. Then (x,u x),(y,v y) grA. Sincex = yand Aisstrictlymonotone,
6 { − − } ⊆ 6
7
we have x y,(u x) (v y) > 0, i.e., x y 2 < x y,u v , i.e., Tu Tv 2 <
h − − − − i k − k h − − i k − k
Tv Tu,u v . Thus, T satisfies(17).
h − − i
(vii): In view of (vi) it suffices to show that T is injective if and only if T is strictly
monotone. Assume first that T is injective and that x = y. Then Tx = Ty and hence
0 < Tx Ty 2 x y,Tx Ty . Thus, T is strictly6 monotone. Con6 versely, assume
thatTkiss−trictlykm≤onhoto−neand−thatix = y. Then x y,Tx Ty > 0andhenceTx = Ty.
6 h − − i 6
Thus, T isinjective.
(viii): Observe that T is strictly firmly nonexpansive if and only if T is injective and T
satisfies (17). Thus, the result follows from combining (iv) and (vi).
(ix): Assume first that T is strictly nonexpansive, that x = y, and to the contrary that
6
u Ax Ay. Then x+u (Id+A)x and y+u (Id+A)y; equivalently, T(x +u) =
∈ ∩ ∈ ∈
x = y = T(y+u). SinceT isstrictlynonexpansive,wehave x y = T(x+u) T(y+
u)6 < (x +u) (y + u) = x y , which is absurd. Tkhus−, Akis dkisjointly in−jective.
k k − k k − k
Conversely, assume that A is disjointly injective, that u = v, and to the contrary that
6
Tu Tv = u v . Since T is firmly nonexpansive, we deduce that u Tu = v Tv.
k − k k − k − −
Assume that x = u Tu = v Tv. Then Tu = u x and Tv = v x; equivalently,
− − − −
u x (Id+A)u and v x (Id+A)v. Thus, x Au Av, which contradicts the
− ∈ − ∈ − ∈ ∩
assumption on disjoint injectivity of A.
(x): Combine (iv) and(ix).
(xi): (See also [5, Proposition 23.11].) Assume first that (1+ ε)T is firmly nonexpan-
sive and that (x,u),(y,v) grA. Then x = T(x + u) and y = T(y + v). Hence,
{ } ⊆
(x+u) (y+v),x y (1+ε) x y 2 x y,u v ε x y 2. Thus, A εId
h − − i ≥ k − k ⇔ h − − i ≥ k − k −
is monotone. Conversely, assume that A εId is monotone and that (x,u),(y,v)
− { } ⊆
grT. Then (u,x u),(v,y v) grA and hence u v,(x u) (y v) ε u
{ − − } ⊆ h − − − − i ≥ k −
v 2;equivalently, x y,u v (1+ε) u v 2. Thus,(1+ε)T isfirmlynonexpansive.
k h − − i ≥ k − k
(xii): Applying (xi) to Id T and A 1, we see that (1+γ)(Id T) is firmly nonexpan-
−
− −
sive ifandonly if A 1 γIdis monotone, which isequivalentto A being γ-cocoercive.
−
−
(xiii): Assume first that T is a Banach contraction with constant β and that
(x,u),(y,v) grA. Set p = x + u and y = y + v. Then (x,u) = (Tp,p Tp),
{ } ⊆ −
(y,v) = (Tq,q Tq), and Tp Tq β p q , i.e.,
− k − k ≤ k − k
(32) x y 2 β2 (x+u) (y+v) 2 = β2 (x y)+(u v) 2
k − k ≤ k − k k − − k
= β2 x y 2+2 x y,u v + u v 2 .
k − k h − − i k − k
(cid:0) (cid:1)
Thus, (24)holds. Theconverse isproved similarly.
(xiv): Theequivalence isimmediatefrom the Minty parametrization (15).
8
(xv): Assume first that T satisfies (27) and that (x,u),(y,v) grA with
{ } ⊆
x y,u v = 0. Set p = x + u and q = y + v. Then (x,u) = (Tp,p Tp),
h − − i −
(y,v) = (Tq,q Tq), and Tp Tq,(p Tp) (q Tq) = 0, i.e., Tp Tq 2 =
− h − − − − i k − k
p q,Tp Tq . By (27), Tp = T(Tp + q Tq) and Tq = T(Tq + p Tp), i.e.,
h − − i − −
x = T(x + v) and y = T(y + u), i.e., x + v x + Ax and y + u y + Ay, i.e.,
∈ ∈
v Ax and u Ay. Thus, A is paramonotone. Conversely, assume that A is para-
∈ ∈
monotone, that Tu Tv 2 = u v,Tu Tv , that x = Tu, and that y = Tv. Then
k − k h − − i
(x,u x),(y,v y) grA and x y,(u x) (v y) = 0. Since A is para-
{ − − } ⊆ h − − − − i
monotone, we deduce that v y Ax and u x Ay, i.e., x y + v (Id+A)x
− ∈ − ∈ − ∈
and y x + u (Id+A)y, i.e., x = T(x y + v) and y = T(y x + u), i.e., Tu =
− ∈ − −
T(Tu+v Tv) and Tv = T(Tv+u Tu). Thus, T satisfies (27).
− −
(xvi): Thisfollows from [2, Theorem 6.6].
(xvii): Theequivalence isimmediatefrom the Minty parametrization (15).
(xviii): Indeed, T = J is linear (A+Id) 1 is a linear relation A+Id is a linear
A −
⇔ ⇔
relation A isa linearrelation (see [11]for more on linearrelations).
⇔
(xix): Thisfollows from (xviii).
(xx): Thisfollows from [25, Corollary 2 on page 251].
(xxi): Assume first that T is sequentially weakly continuous. Let (xn,un)n N be a se-
∈
quence in grA that converges weakly to (x,u) X X. Then (xn +un)n N converges
∈ × ∈
weakly to x + u. On the other hand, Id T is sequentially weakly continuous because
−
T is. Altogether, (xn,un)n N = (T(xn +un),(Id T)(xn +un))n N converges weakly to
∈ − ∈
(x,u) = (T(x + u),(Id T)(x + u)). Thus, (x,u) grA and therefore grA is sequen-
− ∈
tially weakly closed. Conversely, let us assume that grA is sequentially weakly closed.
Let (xn)n N be a sequence in X that is weakly convergent to x. Our goal is to show that
Txn⇀Tx∈. SinceTisnonexpansive,thesequence(Txn)n Nisbounded. Afterpassingtoa
∈
subsequenceandrelabelingifnecessary, wecananddoassumethat(Txn)n N converges
∈
weakly to some point y X. Now (Txn,xn Txn)n N lies in grA, and this sequence
∈ − ∈
converges weakly to (y,x y). Since grA is sequentially weakly closed, it follows that
−
(y,x y) grA. Therefore, x y Ay x (Id+A)y y = Tx, which implies the
− ∈ − ∈ ⇔ ∈ ⇔
result. (cid:4)
Example2.2 Concerningitems(xi)and(xiii)inTheorem2.1,itisclearthatif Aisstrongly
monotone, then T is a Banach contraction. (This result is now part of the folklore; for
perhapsthe firstoccurrence see [20, p.879f]andalso[21, Section 12.H]forfurther results
in this direction.) The converse, however, is false: indeed, consider the case X = R2 and
set
0 1
(33) A = − .
1 0
(cid:18) (cid:19)
9
Then ( z X) z,Az = 0so A cannot be strongly monotone. On the other hand,
∀ ∈ h i
1 1 1
(34) T = J = (Id+A) 1 =
A − 2 1 1
(cid:18)− (cid:19)
is linear and Tz 2 = 1 z 2, which implies that T is a Banach contraction with constant
k k 2k k
1/√2.
Corollary 2.3 Let A: X X be continuous, linear, and maximallymonotone. Thenthe follow-
→
ing hold.
(i) If J isaBanach contraction, then A is(disjointly)injective.
A
(ii) If ranA isclosedand A is(disjointly)injective,then J isaBanach contraction.
A
Proof. The result istrivial if X = 0 so weassume that X = 0 .
{ } 6 { }
(i): Assume that J is a Banach contraction, say with constant β [0,1[. If β = 0, then
A
J 0 A = N , which contradicts the single-valuedness of A.∈Thus, 0 < β < 1. By
A ≡ ⇔ 0
{ }
Theorem 2.1(xiii),
1 β2
(35) ( x X)( y X) − x y 2 2 x y,Ax Ay + Ax Ay 2.
∀ ∈ ∀ ∈ β2 k − k ≤ h − − i k − k
If x = y, then the left side of (35) is strictly positive, which implies that Ax = Ay. Thus,
6 6
A is(disjointly) injective.
(ii): Letusassume thatranA isclosedandthat A is(disjointly) injective. ThenkerA =
0 and hence, by e.g. [12, Theorem 8.18], there exists ρ ]0,+∞[ such that ( z X)
{ } ∈ ∀ ∈
Az ρ z . Thus, ( z X) Az 2 ρ2 z 0. Set β = 1/ 1+ρ2. Then ρ2 =
k k ≥ k k ∀ ∈ k k − k k ≥
(1 β2)/β2 and hence
− p
(36)
1 β2
( x X)( y X) − x y 2 Ax Ay 2 2 x y,Ax Ay + Ax Ay 2.
∀ ∈ ∀ ∈ β2 k − k ≤ k − k ≤ h − − i k − k
Again byTheorem 2.1(xiii), J isaBanach contraction with constant β ]0,1[. (cid:4)
A
∈
Initem (ii) ofCorollary 2.3, the assumption that ranA isclosed iscritical:
Example2.4 Suppose that X = ℓ , the space of square-summable sequences, i.e., x =
2
(x ) X ifand only if ∑∞ x 2 < +∞, and set
n ∈ n=1| n|
(37) A: X X: (x ) 1x .
→ n 7→ n n
(cid:0) (cid:1)
10