Next is the theorem about the graph being maximal being equivalent to the operator
being maximal monotone. It is a very convenient result to have. The proof is a modified
version of one in Barbu [?]. It is based on the following lemma also in Barbu. This is a
little like the Browder lemma but is based on the min max theorem above. It is also a
very interesting argument.
Lemma 23.7.7Let E be a finite dimensional Banach space and let K be a convex andcompact subset of E. Let G
(A)
be a monotone subset of E ×E^{′}such that D
(A )
⊆ K andB is a single valued monotone and continuous operator from E to E^{′}. Then there existsx ∈ K such that
〈Bx + v,u− x〉E′,E ≥ 0 for all [u,v] ∈ G (A ).
If B is coercive
〈Bx, x〉
∥xli∥m→∞ --∥x∥--= ∞,
and 0 ∈ D
(A)
, then one can assume only that K is convex and closed.
Proof:Let T : E → K be the multivalued operator defined by
{ }
Ty ≡ x ∈ K : 〈By + v,u − x〉E ′,E ≥ 0 for all [u,v] ∈ G (A)
Here y ∈ E and it is desired to show that Ty≠∅ for all y ∈ K. For
[u,v]
∈G
(A)
,
let
{ }
Ku,v = x ∈ K : 〈By +v,u − x〉 ′ ≥ 0
E,E
Then K_{u,v} is a closed, hence compact subset of K. The thing to do is to show that
∩_{}
[u,v]
∈G
(A)
K_{u,v}≡ Ty≠∅ whenever y ∈ K. Then one argues that T is set valued, has
convex compact values and is upper semicontinuous. Then one applies the Kakutani fixed
point theorem to get x ∈ Tx.
Since these sets K_{u,v} are compact, it suffices to show that they satisfy the finite
intersection property. Thus for
{[ui,vi]}
_{i=1}^{n} a finite set of elements of G
(A )
, it is
necessary to show that there exists a solution x to the inequalities
〈ui − x,By + vi〉 ≥ 0,i = 1,2,⋅⋅⋅,n
and then it follows from finite intersection property that there exists
x ∈ ∩[u,v]∈G(A)Ku,v
which is what was desired. Let P_{n} be all
⃗
λ
=
(λ1,⋅⋅⋅,λn)
such that each λ_{k}≥ 0 and
∑_{k=1}^{n}λ_{k} = 1. Let H : P_{n}× P_{n}→ ℝ be given by
equal 1 in the i^{th} position and 0 elsewhere and conclude that for all
i = 1,
⋅⋅⋅
,n,
〈 〉
∑n 0
By + vi,ui − λjuj ≥ 0
j=1
so you let x = ∑_{j=1}^{n}λ_{j}^{0}u_{j} and this shows that Ty≠∅ because the sets K_{u,v} have the
finite intersection property.
Thus T : K →P
(K )
and for each y ∈ K,Ty≠∅. In fact this is true for any y but we
are only considering y ∈ K. Now Ty is clearly a closed subset of K. It is also clearly
convex. Is it upper semicontinuous? Let y_{k}→ y and consider Ty + B
(0,r)
. Is
Ty_{k}∈ Ty + B
(0,r)
for all k large enough? If not, then there is a subsequence,
denoted as z_{k}∈ Ty_{k} which is outside this open set Ty + B
(0,r)
. Then taking a
further subsequence, still denoted as z_{k}, it follows that z_{k}→ z
∕∈
Ty + B
(0,r)
.
Now
〈Byk +v,u − zk〉 ≥ 0 all [u,v] ∈ G(A)
Therefore, from continuity of B,
〈By +v,u − z〉 ≥ 0 all [u,v] ∈ G(A )
which means z ∈ Ty contrary to the assumption that T is not upper semicontinuous.
Since T is upper semicontinuous and maps to compact convex sets, it follows from
Theorem 23.4.4 that T has a fixed point x ∈ Tx. Hence there exists a solution x
to
〈Bx + v,u− x〉 ≥ 0 all [u,v] ∈ G(A)
Next suppose that K is only closed and convex but B is coercive and 0 ∈ D
(A )
. Then
let K_{n}≡B
(0,n)
∩K and let A_{n} be the restriction of A to B
(0,n)
. It follows that there
exists x_{n}∈ K_{n} such that for all
[u,v]
∈G
(A )
n
,
〈Bxn + v,u− xn〉 ≥ 0
Then since 0 ∈ D
(A)
, one can pick v_{0}∈ A0 and obtain
〈Bx + v ,− x 〉 ≥ 0, 〈v ,− x 〉 ≥ 〈Bx ,x 〉
n 0 n 0 n n n
from which it follows from coercivity of B that the x_{n} are bounded independent of n. Say
∥xn∥
< C. Then there is a subsequence still denoted as x_{n} such that x_{n}→ x ∈ K, thanks
to the assumption that K is closed and convex. Let
[u,v]
∈G
(A)
. Then for all n large
enough
∥u∥
< n and so
〈Bxn + v,u− xn〉 ≥ 0
Then letting n →∞ and using the continuity of B,
〈Bx + v,u− x〉 ≥ 0
Since
[u,v]
was arbitrary, this proves the lemma. ■
Observation 23.7.8If you have a monotone set valued function, thenits graphcan always be considered a subset of the graph of a maximal monotone graph. IfA is monotone, then let ℱ be G
(B )
such that G
(B)
⊇G
(A )
and B is monotone.Partially order by set inclusion. Then let C be a maximal chain. Let G
( ˆ)
A
= ∪C.If
[xi,yi]
∈ G
( )
ˆA
, then both are in some B ∈ C. Hence
(y1 − y2,x1 − x2)
≥ 0
so monotone and must be maximal monotone because if
〈z − v,x − u〉
≥ 0 for all
[u,v]
∈G
( ˆ)
A
and
[x,z]
∈∕
Â, then you could include this ordered pair and contradictmaximality of the chain C.
Next is an interesting theorem which comes from this lemma. It is an infinite
dimensional version of the above lemma.
Theorem 23.7.9Let X be a reflexive Banach space and let K be a closed convexsubset of X. Let A,B be monotone such that
D
(A)
⊆ K,0 ∈ D
(A)
.
B is single valued, hemicontinuous, bounded and coercive mapping X to X^{′}.
Then there exists x ∈ K such that
〈Bx + v,u− x〉 ′ ≥ 0 for all [u,v] ∈ G (A)
X ,X
Before giving the proof, here is an easy lemma.
Lemma 23.7.10Let E be finite dimensional and let B : E → E^{′}be monotoneand hemicontinuous. Then B is continuous.
Proof:The space can be considered a finite dimensional Hilbert space (ℝ^{n}) and so
weak and strong convergence are exactly the same. First it is desired to show that B is
bounded. Suppose it is not. Then there exists
∥xk∥
_{E} = 1 but
∥Bxn ∥
_{E′}→∞.
Since finite dimensional, there is a subsequence still denoted as x_{k} such that
x_{k}→ x,
∥x ∥
_{E} = 1.
〈Bxk − Bx,xk − x〉 ≥ 0
Hence
〈 〉
Bxk-−-Bx-,xk − x ≥ 0
∥Bxk∥E′
Then taking another subsequence, written with index k, it can be assumed
that
∗ ′ ∗
Bxk ∕∥Bxk ∥ → y ∈ E ,∥y ∥E′ = 1
Hence,
〈y∗,xk − x 〉 ≥ 0
for all x ∈ E, but this requires that y^{∗} = 0, a contradiction. Thus B is monotone,
hemicontinuous, and bounded. It follows from Theorem 23.1.4 which says that monotone
and hemicontinuous operators are pseudomonotone and Proposition 23.1.6 which
says that bounded pseudomonotone operators are demicontinuous that B is
demicontinuous, hence continuous because, as just noted above, weak and strong
convergence are the same for finite dimensional spaces. In case B is bounded, then this
follows from Proposition 23.1.6 above. It is pseudomonotone and bounded hence
demicontinuous and weak and strong convergence is the same in finite dimensions.
■
be an increasing sequence of finite dimensional
subspaces. Let Â be maximal monotone on ∪_{n}X_{n} and extending A. By this is meant that
the graph of Â contains the graph of A restricted to ∪_{n}X_{n},Â is monotone and there is no
other larger graph with these properties. See the above observation. Let j_{n} : X_{n}→ X be
the inclusion map and j_{n}^{∗} : X^{′}→ X_{n}^{′} be the dual map. Then j_{n}^{∗}Âj_{n}≡ A_{n} and
j_{n}^{∗}Bj_{n}≡ B_{n} have monotone graphs from X_{n} to P
(X ′)
n
with B_{n} being continuous and
single valued. This follows from the hemicontinuity and the above lemma which states
that on finite dimensional spaces, hemicontinuity and monotonicity imply continuity.
Then
[u,v] ∈ G (A )
n
means
∗ˆ ∗ ˆ
u ∈ D (A )∩Xn and v ∈ jnAjn(u) = jnA (u) since u ∈ Xn
Then from Lemma 23.7.7, there exists x_{n}∈ X_{n} such that
〈Bnxn + vn,un − xn〉X′,X ≥ 0 all [un,vn] ∈ G (An )
That is, there exists x_{n}∈ K ∩ X_{n} such that for all u ∈ D
( )
Aˆ
∩ X_{n},
[u,v]
∈G
( )
ˆA
〈Bxn + v,u− xn〉X′,X ≥ 0 (23.7.49)
(23.7.49)
Then
〈v,u − xn〉 ≥ 〈Bxn,xn − u〉 (23.7.50)
(23.7.50)
From the assumption that 0 ∈ D
( ˆ)
A
, one can let u = 0 and then pick v_{0}∈Â0. Then the
above reduces to
〈v0,− xn〉 ≥ 〈Bxn,xn〉
By coercivity of B, these x_{n} are all bounded and so by the Eberlien Smulian theorem,
there is a subsequence