A Maximality Theorem for the Sum of Maximal
Monotone Operators in Non-Reexive Banach
Spaces
M.D. Voisei
∗
Abstract
The goal of the present article is to prove that the sum
two maximal monotone operators
A, B
maximal monotone provided that their domains
convex and
D(A) − D(B)
A+B
X
in a general Banach space
D(A), D(B)
of
is
are closed
is absorbing.
1. Introduction
The sum problem for maximal monotone operators in non-reexive Banach
spaces under qualications constraints involving their domains is a long celebrated open problem (see Phelps [4], Simons [6]).
A + B of two maximal monotone opA, B remains maximal monotone provided that the Attouch condition
R+ (coD(A) − coD(B)) = X is fullled (see Attouch [1], Penot [3]). Here coS
stands for the convex hull of S ⊂ X and R+ = (0, ∞).
In reexive Banach spaces the sum
erators
In general Banach spaces this result is still correct if the operators involved
are of subdierential type (see Zalinescu [6]).
Historically, Rockafellar was the rst to consider a stronger qualication
constraint of the form,
intD(A) ∩ D(B) 6= ∅
(1)
and the maximality of the sum was proved under one of the assumptions:
reexive or
A, B
X
is
are subdierentials (see [5,7]).
Also, Heisler showed the maximality of the sum of two maximal operators
A, B
D(A) = D(B) = X
D(A), D(B) are closed convex
under the additional hypothesis that
proof ). We relax this condition to
(see [6] for a
together with
the Attouch condition. Our main result is
Theorem 1.1.
Let
X
be a Banach space and
maximal monotone operators. If
D(A), D(B)
A, B : X → 2X
∗
be two
are closed convex and
R+ (D(A) − D(B)) = X,
∗ Department
of Mathematics, The University of Texas Pan American, TX, USA
1
(2)
A+B
then
is maximal monotone.
X be a real Banach space with topological
∗
A : X → 2X is a multi-valued operator with
Let
ator
dual
X ∗.
A monotone oper-
hx1 − x2 , x∗1 − x∗2 i ≥ 0,
for every
x∗k ∈ Axk , k = 1, 2.
Here
h·, ·i
(3)
stands for the dual product on
X × X ∗.
Recall the notations
D(A) = {x ∈ X; x∗ ∈ Ax for some x∗ ∈ X ∗ },
GraphA = {(x, x∗ ) ∈ X × X ∗ ; x∗ ∈ Ax}.
for the domain and graph of
A
with its graph.
A.
For the sake of notation simplicity we identify
A monotone operator is called maximal monotone if it is
maximal in the sense of inclusion.
Let
Z = X × X ∗.
For
z = (x, x∗ ) ∈ Z , M ⊂ Z ,
and
f : Z → R̄
dene the
transpose operations
z T = (x∗ , x) ∈ X ∗ × X,
M T = {z T ; z ∈ M } ⊂ X ∗ × X,
f T : Z T → R̄, f T (z T ) = f (z), z ∈ Z.
Let
w∗
denote the weak star topology in
denoted by
τ.
X ∗.
Notice that the topological dual of
The strong topology in
Z
is
ZT
if
X∗
X
is
is endowed with
Z ∗ = X ∗ × X ∗∗ for the strong topology.
For a pair of linear spaces E, F in duality the convex conjugate of f : E → R̄
the weak star topology while
is given by
f ∗ (x) = sup{x · e − f (e); e ∈ E}, x ∈ F.
(4)
e · f denotes the dual product in E × F .
Z we have two dual systems (Z, Z T ) and (Z, Z ∗ ). Therefore for
∗
f : Z → R̄ we have two convex conjugates f ∗ : Z ∗ → R̄ given by (4) and f/Z
T.
In addition, we consider the transformation f
: Z → R̄ given by
Here
For
∗
T
f = (f/Z
T) ,
(5)
f (x, x∗ ) = sup{hx, a∗ i + ha, x∗ i − f (a, a∗ ); (a, a∗ ) ∈ Z}.
(6)
or in expanded form
By the Biconjugate Theorem every proper convex
function
f
satises
f
Lemma 1.2. Let
linear continuous, and
τ × w∗
lower semicontinuous
= f.
X , Y, U, V be Banach spaces, A : X → Y , B : U → V be
g : Y × U → R ∪ {+∞} be a lower semicontinuous proper
convex function such that
R+ PY D(g) − R(A) = Y,
2
(7)
PY : Y × U → Y , PY (y, u)
Dene h : X × V → R̄ by
where
= y , (y, u) ∈ Y × U .
h(x, v) = inf{g(Ax, u); u ∈ U, Bu = v}, (x, v) ∈ X × V.
(8)
h∗ (x∗ , v ∗ ) = min{g ∗ (y ∗ , B ∗ v ∗ ); y ∗ ∈ Y ∗ , A∗ y ∗ = x∗ },
(9)
Then
∗ ∗
∗
∗
for every (x , v ) ∈ X × V .
B ∗ : V ∗ → U ∗ , A∗ : Y ∗ → X ∗
Here
h∗ , g ∗
denote the convex conjugate of
stand for the adjoints of
B
and
A,
h, g ,
while min
means that the inmum is attained when it is nite (see Penot [2] or Zalinescu
[6] for a proof ).
2. Simple Duality
We denote the duality product between
X
and
X∗
by
p(x, x∗ ) = hx, x∗ i, (x, x∗ ) ∈ X × X ∗ .
For a monotone operator
A : X → 2X
∗
consider the Fitzpatrick function
hA (x, x∗ ) = sup{hx, a∗ i+ha, x∗ i−ha, a∗ i; (a, a∗ ) ∈ A}, (x, x∗ ) ∈ X ×X ∗ ,
or
h A = p
A,
(10)
where
pA (z) = p(z), for z ∈ A; pA (z) = +∞, otherwise.
ϕA = h
A = cl co pA
semicontinuous function majorized by pA .
In the sequel, for f : Z → R ∪ {+∞},
Also, denote by
the greatest convex
(11)
τ × w∗
lower
the following set notations will be
frequently used:
{f ≤ (=, ≥)p} = {z ∈ Z; f (z) ≤ (=, ≥)p(z)}.
Theorem 2.1. Let
A : X → 2X
∗
(12)
be monotone. The following are equiva-
lent:
(a)
(b)
(c)
A is maximal monotone,
A = {hA ≤ p},
A = {hA = p} and hA ≥ p
on
Z.
Proof. Left as an exercise (see Fitzpatrick [2]).
An operator
semicontinuous
A ⊂ X ×X ∗ is called representable if there exists a τ ×w∗ lower
function h : Z → R̄ such that h ≥ p on Z and A = {h = p}.
Theorem 2.2. Let
A ⊂ X × X∗
be monotone. Then
A is representable if f A = {ϕA = p}.
3
(13)
Proof.
{ϕA = p}
For a monotone operator we always have
ϕA ≥ p
on
Z
A ⊂
and
(see e.g. Penot [3]).
The converse implication is clear since
ϕA
has all desired properties. For the
direct implication we have
pA ≥ ϕA ≥ h ≥ p,
(14)
h is convex lower semicontinous and A ⊂ {h = p}.
{ϕA = p} ⊂ {h = p} = A thus producing the equality. since
Theorem 2.3.
and
hA ≥ p
on
A : X → 2X
∗
This relation implies
is maximal monotone i
A
is representable
Z.
Proof. The direct implication is an easy consequence of Theorem 2.1 since
h = hA
has all the desired properties.
h
For the converse implication only
convex and
A ⊂ {h = p}
suces for
A
to be monotone (see Penot [3]). According to the previous theorem it is enough
to prove that
{hA = p} ⊂ {ϕA = p},
since
A ⊂ {ϕA = p} ⊂ {hA = p}
z = (x, x∗ ) ∈ {hA = p}.
Let
(15)
A.
z T ∈ ∂hA (z), where the
T
system (Z, Z ) (see e.g.
is true for any monotone
Then
∂ is taken with respect to the dual
subdierential
Fitzpatrick [2,
Theorem 2.4, p.60]). A dierent argument can be performed as follows. Take
arbitrary
of
v
v = (u, u∗ ) ∈ Z .
The directional derivative of
satises
h0A (z; v) = lim
t↓0
≥ lim
t↓0
i.e.
z T ∈ ∂hA (z).
hA
at
z
in the direction
hA (z + tv) − hA (z)
t
p(z + tv) − p(z)
= hx, u∗ i + hu, x∗ i = z T · v,
t
(16)
This is equivalent to
hA (z) + ϕA (z) = 2p(z),
and implies
ϕA (z) = p(z)
because
hA (z) = p(z).
(17)
The proof is complete.
3. The Proof of Theorem 1.1.
According to Theorem 2.3 it suces to prove that
on
S
is representable and
hS ≥ p
Z.
Consider
ϕA = h
A , ϕB = hB ,
and
g : X × X × X ∗ × X ∗ → R ∪ {+∞}
given
by
g(u, v, u∗ , v ∗ ) = ϕA (u, u∗ ) + ϕB (v, v ∗ ), u, v ∈ X, u∗ , v ∗ ∈ X ∗ .
4
(18)
X = X , Y = X × X , U = Y ∗ = X ∗ × X ∗, V = X ∗,
Ax = (x, x), x ∈ X , B(u∗ , v ∗ ) = A∗ (u∗ , v ∗ ) = u∗ + v ∗ , (u∗ , v ∗ ) ∈ X ∗ × X ∗ , and
Apply Lemma 1.2. for
h(x, x∗ ) = inf{g(Ax, u); u ∈ U, Bu = x∗ } =
= inf{ϕA (x, u∗ ) + ϕB (x, v ∗ ); u∗ + v ∗ = x∗ },
Notice that condition (7) is fullled. Indeed, for every
a ∈ D(A), b ∈ D(B), λ > 0,
x = λa − x1 = λb − x2 . Then
(2), there exist
Denote by
x1 , x2 ∈ X , by condition
x1 − x2 = λ(a − b).
such that
(x1 , x2 ) = λ(a, b) − (x, x),
(19)
R+ D(A) × D(B) − R(A) = X × X . This translates into (7),
A ⊂ {ϕA = p}, B ⊂ {ϕB = p} we know that D(A) × D(B) ⊂
∗
∗∗
∗
∗∗
every (x , x ) ∈ X × X , we get
which shows that
because from
PY D(g).
For
h∗ (x∗ , x∗∗ ) = min{g ∗ (u∗ , v ∗ , A∗∗ x∗∗ ); A∗ (u∗ , v ∗ ) = x∗ } =
= min{ϕ∗A (u∗ , x∗∗ ) + ϕ∗B (v ∗ , x∗∗ ); u∗ + v ∗ = x∗ },
since
∗∗ ∗∗
A x
∗∗
∗∗
∗∗
= (x , x ), x
∈X
∗∗
. When
∗∗
x
= x ∈ X,
(20)
this yields
h (x, x∗ ) = min{hA (x, u∗ ) + hB (x, v ∗ ); u∗ + v ∗ = x∗ }, (x, x∗ ) ∈ Z.
(21)
h is proper convex τ × w∗ lower semicontinuous, h ≥ p on Z , and
S = {h = p}, that is, S is representable.
For a closed convex C ⊂ X we denote by NC = ∂iC the normal cone to C .
Here iC (x) = 0, if x ∈ C ; iC (x) = ∞, otherwise, is the indicator function of C .
Clearly,
We have
A = A + ND(A) , B = B + ND(B) ,
since
A, B
are maximal monotone and
Because
D(A) − D(B)
D(A), D(B)
(22)
are closed convex.
is absorbing in the Banach space
X
we know that
ND(A) + ND(B) = ND(A)∩D(B) .
(23)
S = S + ND(S) .
(24)
Hence
Let
∗
z = (x, x ) ∈ {hS ≤ p},
that is, for every
∗
(s, s ) ∈ S ,
we have
hx − s, x∗ − s∗ i ≥ 0.
From (24) and the fact that, for every
(25)
s ∈ D(S), ND(S) (s)
is a cone, we nd
∗
hx − s, n i ≤ 0,
(26)
s ∈ D(S), n∗ ∈ ND(S) (s).
Therefore x ∈ D(S), since ND(S) is maximal monotone. From
hS , x ∈ D(S) easily implies that z ∈ {hS ≥ p}. We proved
for every
of
{hS ≤ p} ⊂ {hS = p},
and this yields
hS ≥ p
on
Z.
The proof is complete.
5
the denition
(27)
References
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ators, Nonlinear Analysis 5 (1981), no. 2, 143147.
[2] Fitzpatrick, S., Representing monotone operators by convex functions, Workshop on Functional Analysisand Optimization, Canberra, 1988, 5965.
[3] Penot, J.P., The relevance of convex analysis for the study of monotonicity,
Nonlinear Analysis 58 (2004), 855871.
[4] Phelps, R.R., Convex functions, monotone operators, and dierentiability,
second edition, Lecture Notes in Mathematics 1364, Springer-Verlag Berlin,
1993.
[5] Rockafellar, R.T., On the maximality of sums of nonlinear monotone oper-
ators, Trans. Amer. Math. Soc. 149 (1970), 7588.
[6] Simons, S., Minimax and monotonicity, Lecture Notes in Mathematics 1693,
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Publishing Co., Inc., River Edge, NJ, 2002.
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