Most Maximally Monotone Operators Have a Unique Zero and a Super-regular Resolvent

0.21 MB·English

by  Xianfu Wang

Most Maximally Monotone Operators Have
a Unique Zero and a Super-regular Resolvent
3
1
0
2
Xianfu Wang∗

n
a
J

8 January 21, 2013
2

]
C
O
. Abstract
h
t
a
m Maximally monotone operators play important roles in optimization, variational
[ analysis and differential equations. Finding zeros of maximally monotoneoperators

has been a central topic. In a Hilbert space, we show that most resolventsare super-
1
v regular,thatmostmaximallymonotoneoperatorshaveauniquezeroandthattheset
3 of strongly monotone mapping is of first category although each strongly monotone
4
operatorhasauniquezero. TheresultsareestablishedbyapplyingtheBaireCategory
4
6 Theoremtothespaceofnonexpansivemappings.
.
1
0
3 2010 Mathematics Subject Classification: Primary 47H05, 47H10; Secondary 54E52, 47H09,
1 54E50.
:
v
i Keywords: Asymptoticregularity,BaireCategory,fixedpoint,graphicalconvergence,maximally
X
monotoneoperator,nonexpansivemapping,resolvent,reflectedresolvent,super-regularity,zeros
r
a ofmonotoneoperator,weaklycontractivemapping.
1 Introduction
Throughout, X is a real Hilbert space whose inner product is denoted by hx,yi and in-
duced inner product norm by kxk := hx,xi for x,y ∈ X. Recall that a set-valued
operator A: X ⇉ X (i.e., (∀x ∈ X) Ax ⊆ X) with graph grA ismonotone if
p
(1) (∀(x,u) ∈ grA)(∀(y,v) ∈ grA) hx−y,u−vi ≥ 0
∗Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail:
[email protected].
1
where grA := {(x,y) ∈ X × X : y ∈ Ax}, and that A is maximally monotone if it is
impossible to find a proper extension of A that is still monotone. We call A : X ⇉ X
strongly monotone [5, 27] if there exists ε > 0 such that A − εId is monotone in which
Id: X → X: x 7→ x denotesthe identity operator.
Weshall work in the spaceof nonexpansive mappingsdefinedon X, i.e.,
N(X) := {T : X → X : kTx−Tyk ≤ kx−yk,∀ x,y ∈ X};
the spaceof firmlynonexpansive mappings
J(X) := {T : X → X : kTx−Tyk2 ≤ hTx−Ty,x−yi,∀ x,y ∈ X};
and the space ofmaximalmonotone operators
M(X) := {A : X ⇉ X : A ismaximally monotone}
endowed with a metric defined in Section 2. The reason to investigate nonexpansive
mappingsdefinedonX isthattheyaredirectlyrelatedtomaximallymonotoneoperators.
In this note, we study generic properties of N(X), M(X) and J(X) by the Baire Category
Theorem. A recent result due to Reich and Zaslavski implies that most nonexpansive mappings
in N(X) are super-regular so that each of them has a unique fixed point. Utilizing Reich and
Zaslavaski’s technique, we show that (i)Most resolvents in J(X) are super regular, thus asymp-
totically regular; (ii) Most maximallymonotone operators in M(X) have a unique zero; (iii) The
setof strongly monotone operators isonly a firstcategorysetin M(X) eventhough itisdense.
While extensive study has been done on N(X) [5, 14, 15, 7, 9, 19, 21, 23, 24] and on
M(X) [1, 5, 27, 29, 31], generic properties on M(X) and J(X) seem new. They are
particularly interesting for the optimization field. Note that De Blasi and Myjak only
considered genericproperties ofcontinuous boundedmonotone operators onabounded
set in [8].
In the reminder of this section, we introduce some definitions, basic facts and prelim-
inary results. For A ∈ M(X), we define its resolvent and reflected resolvent (or Cayley
transform) by
J := (A+Id)−1, R := 2J −Id.
A A A
It is well-known that J + J = Id, R + R = 0, see, e.g., [27], [4, Proposition 4.1].
A A−1 A A−1
Bothresolventandreflectedresolventplayakeyroleintheproximalpointalgorithmand
Douglus-Rachford algorithm [5,25, 11, 12, 17, 4].
The following well-known characterizations about firmly nonexpansive mappings,
nonexpansive mappingsandmaximally monotone operators are crucial.
Fact 1.1 (See, e.g., [5,15, 14].) Let T: X → X. Thenthe followingare equivalent:
(i) T is firmlynonexpansive.
2
(ii) 2T−Idis nonexpansive.
(iii) (∀x ∈ X)(∀y ∈ X) kTx−Tyk2 ≤ hx−y,Tx−Tyi.
(iv) (∀x ∈ X)(∀y ∈ X) 0 ≤ hTx−Ty,(Id−T)x−(Id−T)yi.
Fact 1.2(Eckstein & Bertsekas, Minty) [18, 31, 13] Let A : X ⇉ X be monotone. Then A is
maximallymonotone if and only if J isfirmlynon-expansive and has afulldomain.
A
For T : X → X, let FixT denote its fixed point set FixT := {x ∈ X : Tx = x}. Facts 1.1,
1.2 allowusto summarize the relationship among N(X),J(X),M(X).
Proposition 1.3 (i) N(X) = {R : A ∈ M(X)},
A
−1
T+Id
M(X) = −Id : T ∈ N(X) .
2
( )
(cid:18) (cid:19)
(ii) J(X) = {J : A ∈ M(X)},
A
M(X) = T−1−Id : T ∈ J(X) .
n o
(iii) N(X) = {2T −Id : T ∈ J(X)},
T+Id
J(X) = : T ∈ N(X) .
2
(cid:26) (cid:27)
(iv) Let A ∈ M(X). ThenFixR = FixJ = A−1(0).
A A
ManynicepropertiesandapplicationsaboutN(X),J(X),M(X) canbefoundin[1,5,
6,23,24]andtheycontinuetoflourish. Wereferreadersto[6]forasystematicrelationship
among these three spaces. Let us now turn to the graphical convergence of set-valued
maximal monotone operstors.
Definition 1.4 [1, page360]Givena sequenceof maximallymonotone operators
{A : n ∈ N},A.
n
g
The sequence {A : n ∈ N} issaidto begraphicallyconvergentto A, writtenas A −→ A, if
n n
for every (x,y) ∈ grA there exists (x ,y ) ∈ grA such that x → x,y → y
n n n n n
strongly in X×X.
In termsof setconvergencegrA ⊂ liminfgrA .
n
3
Proposition 1.5 Thefollowing areequivalent
(i) A sequence of maximally monotone operators (A )∞ in M(X) converges graphically to
k k=1
A;
(ii) (J )∞ convergespointwise to J on X;
Ak k=1 A
(iii) (R )∞ convergespointwise to R on X.
Ak k=1 A
Proof. (i)⇔ (ii) follows from [1, Proposition 3.60, pages 361-362]. (ii)⇔ (iii) is obvious
since R = 2J −Id. (cid:4)
A A
AsetSinacompletemetricspaceY iscalledresidualifthereisasequenceofdenseand
∞ ∞
open sets O ⊂ Y such that O ⊂ S; in this case we call O a dense G set. A
n n=1 n n=1 n δ
classical theorem ofBaire is
T T
Fact 1.6(Baire Category Theorem) [26,page158]LetY beacompletemetricspaceand{O }
n
∞
a countable collectionof denseopensubsets ofY. Then O is denseinY.
n=1 n
T
The technique of Baire Category has been instrumental in studying fixed point of nonex-
pansive mappings; see, e.g.,[7, 8, 9,19, 20, 21, 23, 24].
Thepaperisorganizedasfollows. InSection 2wegivethemainresult. InSection 3we
introduce a class of weakly contractive mappings which contains contractive mappings,
and show thatalthough it isdense,itis onlya setof first category.
Notation. Foraset-valuedmapping A : X ⇉ X,wewritedomA := x ∈ X Ax 6= ∅
andranA := A(X) = Ax forthedomainandrangeof A, respectively. B (x) denotes
x∈X (cid:8) r (cid:12) (cid:9)
N
the closed ball ofradiusr centered at x. standsfor the set ofnatural number(cid:12)s.
S
2 Main results
In this section, using Reich and Zaslavski’s technique on super-regular mappings we es-
tablishagenericpropertyofsuper-regularmappingsincompletesubspacesof(N(X),ρ).
Thisallowsustoshowthatmostresolventsaresuper-regular; mostmaximallymonotone
operators have asuper-regular reflected resolvent anda unique zero.
WestartwiththreecompletemetricspaceswhichsetupthestagefortheBaireCategory
Theorem.
On N(X) we define ametric, for T ,T ∈ N(X)
1 2
∞
1 kT −T k
(2) ρ(T ,T ) := ∑ 1 2 n
1 2 2n1+kT −T k
n=1 1 2 n
4
where kT −T k := sup kT x−T xk. Themetric ρ definesa topology ofpointwise
1 2 n kxk≤n 1 2
convergence on X anduniform convergence on bounded subsets of X.
Proposition 2.1 (N(X),ρ) is acompletemetricspace.
Proof. It is easy to see that ρ is a metric (cf. [16, pages 10-11]). We show that N(X) is
complete. Assumethat(T )∞ isaCauchysequencein(N(X),ρ). Thenforeveryn ∈ N,
k k=1
(T )∞ is a uniform Cauchy sequence on B (0). In particular, (T (x))∞ is Cauchy in
k k=1 n k k=1
X for each x ∈ B (0), so T (x) converges to Tx ∈ X. Moreover, for every n ∈ N,
n k
kT − Tk → 0 as k → ∞. Since each T is nonexpansive, T is nonexpansive, i.e., T ∈
k n k
N(X). It remains to show ρ(T ,T) → 0 as k → ∞. Let ε > 0. Choose M ∈ N large such
k
that
∞
1 ε
∑ <
.
2n 2
n=M+1
For this M, choose a large N ∈ N such that kT −Tk < ε when k > N. Then for k > N
k M 2
we have
M 1 kT −Tk ∞ 1 kT −Tk
(3) ρ(T ,T) = ∑ k n + ∑ k n
k 2n1+kT −Tk 2n1+kT −Tk
n=1 k n n=M+1 k n
M 1 ε/2 ∞ 1
(4) ≤ ∑ + ∑ < ε/2+ε/2 = ε.
2n1+ε/2 2n
n=1 n=M+1
Hence (N(X),ρ) is complete. (cid:4)
Remark2.2 In [21], Reich and Zaslavaski define a unform space (N(X),U) where the
uniformity U isdefinedbythe base
E(n,ε) = {(T,S) ∈ N(X)×N(X) : kT −Sk < ε}
n
for n ∈ N,ε > 0. The topology induced by this uniformity and the metric ρ are exactly
the same.
On M(X) letusdefinea metric
∞
1 kR −R k
(5) ρ˜(A,B) := ρ(R ,R ) = ∑ A B n
A B 2n 1+kR −R k
n=1 A B n
for A,B ∈ M(X).
Proposition 2.3 (i) The space of monotone operators (M(X),ρ˜) is a complete metric space,
and itis isometricto (N(X),ρ).
(ii) When X = RN, the topology on (M(X),ρ˜) is precisely the topology of graphical conver-
gence.
5
Proof. (i)By Facts 1.1, 1.2,underthe mapping A 7→ R
A
(M(X),ρ˜) and (N(X),ρ) are isometric.
Since (N(X),ρ) iscomplete byProposition 2.1, weconclude that (M(X),ρ˜) iscomplete.
(ii)WhenX = RN,onN(X) pointwiseconvergenceanduniformconvergenceoncom-
pact subsets are the same. By Proposition 1.5, we obtain that the topology on (M(X),ρ˜)
is exactlythe topology ofgraphical convergence. (cid:4)
On J(X) let usdefinea metric
∞
1 k2T −2T k
(6) ρˆ(T ,T ) := ρ(2T −Id,2T −Id) = ∑ 1 2 n
1 2 1 2 2n1+k2T −2T k
n=1 1 2 n
for T ,T ∈ J(X).
1 2
Proposition 2.4 Thespaceofresolvents(J(X),ρˆ)isacompletemetricspace,anditisisometric
to (N(X),ρ).
Proof. ByFact 1.1,underthe mapping T 7→ 2T −Id
(7) (J(X),ρˆ) and (N(X),ρ) are isometric.
Since (N(X),ρ) iscomplete byProposition 2.1, the result holds. (cid:4)
Next we study the denseness of contraction mappings and strongly monotone opera-
tors, which are required in laterproofs.
Definition 2.5 Themap T ∈ N(X) iscalledacontractionwith modulus 1 > l ≥ 0if
kTx−Tyk ≤ lkx−yk ∀ x,y ∈ X.
Lemma2.6 (i) (densenessofcontractionmappings)In(N(X),ρ) thesetofcontractions
is dense, i.e., for very ε > 0 and T ∈ N(X) there exists a contraction T ∈ N(X) such
1
that ρ(T,T ) < ε.
1
(ii) (denseness of contractive firmly nonexpansive mappings) In (J(X),ρˆ) the set of
contraction is dense, i.e., for very ε > 0 and T ∈ J(X) there exists a contraction T ∈
1
J(X) suchthat ρˆ(T,T ) < ε.
1
Proof. (i)Let T ∈ N(X) and1 > ε > 0. Choose an integer M sufficiently large such that
∞
1 ε
(8) ∑ ≤ .
2n 2
n=M+1
6
Choose
ε 1
< < <
0 λ
2(1+kTk ) 2
M
and define
T := (1−λ)T.
1
Then T isacontraction with modulus 1/2 < 1−λ < 1. As
1
(9) kT −Tk = sup k(1−λ)Tx−Txk
1 M
kxk≤M
ε
(10) = λ sup kTxk = λkTk < .
M
2
kxk≤M
ε
Using kT −Tk ≤ kT −Tk < when n ≤ M and (8), we have
1 n 1 M
2
M 1 kT −Tk ∞ 1 kT −Tk
(11) ρ(T ,T) = ∑ 1 n + ∑ 1 n
1 2n1+kT −Tk 2n1+kT −Tk
n=1 1 n n=M+1 1 n
M 1 ε ∞ 1
(12) ≤ ∑ + ∑
2n2 2n
n=1 n=M+1
ε ε
(13) < + = ε
2 2
so ρ(T,T ) < ε.
1
(ii) The proof is similar as in (i) by replacing ρ by ρˆ and by observing that T = (1−
1
λ)T ∈ J(X) if T ∈ J(X) and0 ≤ λ ≤ 1. (cid:4)
Tostudy monotone operators, weneed:
Fact 2.7 [6, Corollary 4.7] Let A : X ⇉ X be maximally monotone. Then the following are
equivalent:
(i) Both A and A−1 are strongly monotone;
(ii) Thereexists ε > 0such thatboth (1+ε)J and (1+ε)J are firmlynonexpansive;
A A−1
(iii) R isaBanach contraction.
A
Lemma2.8(densenessof strongly monotone mappings) In (M(X),ρ˜) the set of mono-
tone operators A such that both A and A−1 are strongly monotone is dense, i.e., for very ε > 0
and A ∈ M(X) thereexistsa B ∈ M(X) such thatboth B and B−1 are strongly monotone, and
ρ˜(A,B) < ε. Consequently, thesetof strongly monotone operators isdense in M(X).
7
Proof. Underthe mapping A 7→ R
A
(M(X),ρ˜) and (N(X),ρ) are isometric.
Let A ∈ M(X) and ε > 0. ByLemma 2.6(i), for R there exists a contraction T such that
A 1
ρ(R ,T ) < ε. Proposition 1.3(i) says that there exists B ∈ M(X) such that T = R . By
A 1 1 B
Fact 2.7 both B,B−1 are strongly monotone. The proof is complete by using ρ˜(A,B) =
ρ(R ,R ). (cid:4)
A B
To prove our main results, we require super-regular mappings introduced by Reich and
Zaslavaski [21].
Definition 2.9(Reich-Zaslavski) Amapping T : X → X iscalledsuper-regularifthereexists
a unique x ∈ X suchthatfor each s > 0, whenn → ∞,
T
Tnx → x uniformlyon B (0).
T s
Our nexttwo results collect some elementaryproperties of super-regular mappings.
Proposition 2.10 Assume that T : X → X is super-regular and continuous. Then FixT is a
singleton.
Proof. Let x ∈ X. Usingthe continuity and super-regularity of T, we have
x = lim Tnx = lim T(Tn−1x) = Tx
T T
n→∞ n→∞
soxT ∈ FixT. Letx ∈ FixT. Bythesuper-regularityofTandTnx = x, x = limn→∞Tnx =
x . HenceFixT = {x }. (cid:4)
T T
Proposition 2.11 (i) If T ∈ N(X) isacontraction, then T is super-regular.
(ii) If A ∈ M(X) has both A and A−1 being strongly monotone, then R and J are super-
A A
regular.
Proof. (i) Let s > 0. Let T be a contraction with modulus 0 ≤ l < 1. By the Banach Con-
traction Principle [16, pages 300-302], T has a unique fixed point x , and with arbitrary
T
x ∈ X the error estimate is
ln
kTnx−x k ≤ kx−Txk.
T
1−l
For every x ∈ B (0),
s
ln ln
kTnx−x k ≤ (kxk+kTx−T0k+kT0k) ≤ (s+ls+kT0k).
T
1−l 1−l
8
Therefore,
ln
kTn −x k ≤ (s+ls+kT0k) → 0 when n → ∞.
T s
1−l
>
Since s 0 wasarbitrary, T is super-regular.
(ii) By Fact 2.7, R is a contraction. Since A is strongly monotone, J is a contraction
A A
[25]. Hence (i)applies. (cid:4)
The proof ideas to Proposition 2.12 and Theorem 2.13 below are due to Reich and Za-
slaski [21, 20]. We adopt them to our complete metric space setting, and to subspaces of
N(X). For two metrics ρ,d on F ⊂ N(X), if ρ(T ,T) ≤ d(T ,T) for all T ,T ∈ F we
1 1 1
write ρ ≤ d.
Proposition 2.12 Assume that F ⊆ N(X), (F,d) is complete and d ≥ ρ. Let T ∈ F be
super-regular and ε,s be positive numbers. Then there exists δ > 0 and n ≥ 2 such that when
0
d(T ,T) < δ and n ≥ n we have
1 0
(14) kTnx−x k < ε forevery x ∈ B (0),
1 T s
i.e., kTn −x k < ε.
1 T s
< <
Proof. We may and do assume that 0 ε 1/2. Let x denote the unique fixed point of
T
T. Choose an integer M > 1+2s+4kx k so that
T
M 1 M
(15) s < , +s+2kx k < .
T
2 2 2
As T issuper-regular, there exists n ≥ 2 such that
0
ε
(16) kTnx−x k < whenever x ∈ B (0) and n ≥ n .
T M 0
8
Put
1 (8n )−1ε
0
δ := .
2M 1+(8n )−1ε
(cid:18) 0 (cid:19)
Wewill show that (14)holdswhen d(T ,T) < δ and n ≥ n .
1 0
Letd(T ,T) < δ. Thenρ(T ,T) < δ. Usingthedefinitionofρandthatt 7→ t isstrictly
1 1 1+t
increasing on [0,+∞), we have
(17) kT −Tk < (8n )−1ε.
1 M 0
Claim1.Whenever x ∈ B (0) and 1 ≤ n ≤ n ,
M/2 0
(18) kTnx−Tnxk < n(8n )−1ε,
1 0
1
(19) kTnxk < +kxk+2kx k < M.
1 2 T
9
Weprove this byinduction. As T ∈ N(X) and Tx = x , for every n ∈ N,
T T
(20) kTnx−Tnxk ≤ kTnx−TTn−1xk+kTTn−1x−Tnxk
1 1 1 1
(21) ≤ kTnx−TTn−1xk+kTn−1x−Tn−1xk,
1 1 1
and
(22) kTnx−x k ≤ kTnx−Tnxk+kTnx−x k
1 T 1 T
(23) ≤ kTnx−Tnxk+kx−x k
1 T
(24) ≤ kTnx−Tnxk+kxk+kx k.
1 T
Now when n = 1,(18)follows from (17); for (19), by(22)and (17)
1
kT xk ≤ kT x−x k+kx k ≤ kT x−Txk+kxk+2kx k < +kxk+2kx k.
1 1 T T 1 T T
2
Assume that (18)-(19)hold for 1 ≤ n < n , i.e.,
0
(25) kTnx−Tnxk < n(8n )−1ε,
1 0
1
(26) kTnxk < +kxk+2kx k < M.
1 2 T
Using(20)for n+1, (25),(17), kTnxk < M and n < n , weobtain
1 0
(27) kTn+1x−Tn+1xk ≤ kTn+1x−TTnxk+kTnx−Tnxk
1 1 1 1
(28) < (8n )−1ε+n(8n )−1ε = (n+1)(8n )−1ε.
0 0 0
Using(22)for n+1, (27),
(29) kTn+1xk ≤ kTn+1x−x k+kx k
1 1 T T
(30) ≤ kTn+1x−Tn+1xk+kxk+2kx k
1 T
1
(31) < (n+1)(8n )−1ε+kxk+2kx k < +kxk+2kx k.
0 T T
2
This establishes(18)-(19).
Claim2.
(32) kTny−x k < ε whenever y ∈ B (0) and n ≥ n .
1 T s 0
This isdone againby induction. When n = n ,as kyk ≤ s < M/2, by(16)and (18)
0
kTn0y−x k ≤ kTn0y−Tn0yk+kTn0y−x k < ε/8+ε/8 < ε.
1 T 1 T
Assume that (32) holdsfor all n ≤ n ≤ k. For i = 1,...,n , (19) and(15)give
0 0
(33) kTiyk < 1/2+kyk+2kx k < 1/2+s+2kx k < M/2;
1 T T
10