As an interesting example of this theorem, here is another result in Barbu [?]. It is
interesting because it is not assumed B is bounded.
Theorem 23.7.15Let B : X → X^{′}be monotone hemicontinuous. Then B ismaximal monotone. If B is coercive, then B is also onto. Here X is a strictly convexreflexive Banach space.
Proof:Suppose B is not maximal monotone. Then there exists
∗
(x0,x0)
∈ X × X^{′}
such that for all x,
〈Bx − x∗0,x− x0〉 ≥ 0
and yet x_{0}^{∗}≠Bx_{0}. This is going to be a contradiction. Let u ∈ X and consider
x_{t}≡ tx_{0} +
(1− t)
u,t ∈
(0,1)
. Then consider
∗
〈Bxt − x 0,xt − x0〉
However, x_{t}− x_{0} = tx_{0} +
(1 − t)
u − x_{0} =
(1− t)
(u− x0)
and so, for each
t ∈
(0,1)
,
0 ≤ 〈Bxt − x∗0,xt − x0〉 = (1− t)〈Bxt − x∗0,u− x0〉
Divide by
(1− t)
and then let t ↑ 1. This yields the following by hemicontinuity.
〈Bx0 − x∗0,u − x0〉 ≥ 0
which holds for all u. Hence Bx_{0} = x_{0}^{∗} after all. Thus B is indeed maximal
monotone.
Next suppose B is coercive. Let F be the duality map (or the duality map
for arbitrary p > 1). Then from Theorem 23.7.13 there exists a solution x_{λ}
to
∗ ′
λF xλ + Bx λ = x0 ∈ X (23.7.51)
(23.7.51)
Then the x_{λ} are bounded because, doing both sides to x_{λ},
λ ∥xλ∥2 + 〈Bxλ,xλ〉 = 〈x∗0,xλ〉
and so
〈Bxλ,xλ〉 ∗
--∥xλ∥---≤ ∥x0∥
Thus the coercivity of B implies that the x_{λ} are bounded. There exists a subsequence
such that
Since y is arbitrary, this shows that x_{0}^{∗} = Bx and so B is onto as claimed.
■
Again, note that it really didn’t matter about the particular duality map used,
although the usual one was featured in the argument.
There are some more things which can be said about maximal monotone
operators. To include some of these, here is a very interesting lemma found in
[?].
Lemma 23.7.16Let X be a Banach space and suppose that
xn → 0, ∥x∗n∥ → ∞
Then denoting by D_{r}the closed disk centered at 0 with radius r. It follows that for everyD_{r}, there exists y_{0}∈ D_{r}and a subsequence with index n_{k}such that
〈 ∗ 〉
xnk,xnk − y0 → − ∞
Proof:Suppose this is not true. Then there exists D_{r} which has the property that
for all u ∈ D_{r},
∗
〈x n,xn − u〉 ≥ Cu
for all n. Now let
Ek ≡ {y ∈ Dr : 〈x∗n,xn − y〉 ≥ − k for all n}
Then this is a closed set, being the intersection of closed sets. Also, by assumption, the
union of these E_{k} equals D_{r} which is a complete metric space. Hence one of these E_{k}
must have nonempty interior by the Bair category theorem, say for k_{0}. Say B
(y,ε)
⊆ D_{r}.
Then for all
∥u − y∥
< ε,
〈x∗n,xn − u〉 ≥ − k0 for all n
Of course −y ∈ D_{r} also, and so there is C such that
〈x∗,xn + y〉 ≥ C for all n
n
Then
〈x∗,2x + y − u〉 ≥ C − k for all n
n n 0
whenever
∥y− u∥
< ε. Now recall that x_{n}→ 0. Consider only u such that
∥y− u∥
< ε∕2.
Therefore, for all n large enough, the expression 2x_{n} + y −u for such u contains a small
ball centered at the origin, say D_{δ}. (The set of all y −u for u closer to y than ε∕2 is the
ball B
(0,ε∕2)
and then the 2x_{n} does not move it by much provided n is large enough.)
Therefore,
〈x∗,v〉 ≥ C − k
n 0
for all
∥v∥
≤ δ. This contradicts the assumption that
∗
∥x n∥
→∞. ■
Corollary 23.7.17Let X be a Banach space and suppose that
x → x, ∥x∗∥ → ∞
n n
Then denoting by D_{r}the closed disk centered at x with radius r. It follows that for everyD_{r}, there exists y_{0}∈ D_{r}and a subsequence with index n_{k}such that
〈 ∗ 〉
xnk,xnk − y0 → − ∞
Proof:It follows that x_{n}−x → 0. Therefore, from Lemma 23.7.16, for every r > 0,
there exists ŷ_{0}∈B
(0,r)
and a subsequence x_{nk} such that
〈x∗ ,(xn − x)− ˆy0〉 → − ∞
nk k
Thus
〈 ∗ 〉
xnk,xnk − (x+ yˆ0) → − ∞
Just let y_{0} = x + ŷ_{0}. Then y_{0}∈ D_{r} and satisfies the desired conditions. ■
Definition 23.7.18A set valued mapping A : D
(A)
→P
(X )
is locally bounded atx ∈D
(A )
if whenever x_{n}→ x, x_{n}∈ D
(A)
it follows that
lim sup {∥x∗n∥ : x∗n ∈ Axn} < ∞.
n→ ∞
Lemma 23.7.19A set valued operator A is locally bounded at x ∈D
(A )
if andonly if there exists r > 0 such that A is bounded on B
(x,r)
∩ D
(A)
.
Proof:Say the limit condition holds. Then if no such r exists, it follows that A
is unbounded on every B
(x,r)
∩ D
(A)
. Hence, you can let r_{n}→ 0 and pick
x_{n}∈ B
(x,rn)
∩D
(A)
with x_{n}^{∗}∈ Ax_{n} such that
∥x∗∥
n
> n, violating the limit condition.
Hence some r exists such that A is bounded on B
(x,r)
∩ D
(A)
. Conversely,
suppose A is bounded on B
(x,r)
∩ D
(A)
by M. Then if x_{n}→ x, it follows that
for all n large enough, x_{n}∈ B
(x,r)
and so if x_{n}^{∗}∈ Ax_{n},
∥x ∗∥
n
≤ M. Hence
limsup_{n→∞}
{∥x∗∥ : x∗ ∈ Axn}
n n
≤ M < ∞ which verifies the limit condition.
■
With this definition, here is a very interesting result.
Theorem 23.7.20Let A : D
(A )
→ X^{′}be monotone. Then if x is an interiorpoint of D
(A)
, it follows that A is locally bounded at x.
Proof:You could use Corollary 23.7.17. If x is an interior point of D
(A )
, and A is
not locally bounded, then there exists x_{n}→ x and x_{n}^{∗}∈ Ax_{n} such that
∗
∥xn∥
→∞.
Then by Corollary 23.7.17, there exists y_{0} close to x, in D
(A)
and a subsequence x_{nk}
such that
〈x∗ ,x − y〉 → − ∞
nk nk 0
Letting y_{0}^{∗}∈ Ay_{0},
〈x∗ − y∗,x − y 〉 ≥ 0
nk 0 nk 0
and so
〈 〉
x∗nk,xnk − y0 ≥ 〈y∗0,xnk − y0〉
and the right side is bounded below because it converges to
〈y∗,x − y〉
0 0
and this is a
contradiction. ■
Does the same proof work if x is a limit point of D
(A)
? No. Suppose x is a limit
point of D
(A)
. If A is not locally bounded, then there exists x_{n}→ x,x_{n}∈ D
(A)
and
x_{n}^{∗}∈ Ax_{n} and
∥x∗∥
n
→∞. Then there is y_{0} close to x such that
〈x∗ ,x − y 〉
nk nk 0
→−∞
but now everything crashes in flames because it is not known that y_{0}∈ D
(A)
.
It follows from the above theorem that if A is defined on all of X and is maximal
monotone, then it is locally bounded everywhere. Now here is a very interesting result
which is like the one which involves monotone and hemicontinuous conditions. It is in
[?].
Theorem 23.7.21Let A : X → P
(X ′)
be monotone and satisfies the followingconditions:
If λ_{n}→ λ,λ_{n}∈
[0,1]
and z_{n}∈ A
(u+ λ (v − u))
n
, then if B is anyweakly open set containing 0, z_{n}∈ A
(u)
+ B for all n large enough. (Uppersemicontinuous into weak topology along a line segment)
A
(x )
is closed and convex.
Then one can conclude that A is maximal monotone.
Proof:Let Â be a monotone extension of A. Let
[ˆu,w ˆ]
be such that ŵ∈Â
(ˆu)
. Now
also by assumption, A
(x)
is not just convex but also closed.
If
[ˆu, ˆw]
is not in the graph of A, then by separation theorems, there is u such
that
〈x∗,u〉 < 〈ˆw,u〉 for all x∗ ∈ A(ˆu)
Then for λ > 0, let x_{λ}≡û + λu, x_{λ}^{∗}∈ A
(x )
λ
. Then from monotonicity of
Â,
∗ ∗
0 ≤ 〈xλ −wˆ, xλ − ˆu〉 = λ〈xλ − wˆ,u〉
Thus
〈x∗λ − ˆw,u〉 ≥ 0
By Theorem 23.7.20, the monotonicity of A on X implies A is locally bounded also.
Thus in particular, Ax_{λ} for small λ is contained in a bounded set. Now by that
hemicontinuity assumption, you can get a subsequence λ_{n}→ 0 for which x_{λn}^{∗}
converges weakly to x^{∗}∈ Aû. Therefore, passing to the limit in the above, we
get
〈x∗ − ˆw,u〉 ≥ 0
〈x ∗,u〉 ≥ 〈wˆ, u〉 > 〈x∗,u〉
a contradiction. Thus there is no proper extension and this shows that A is maximal
monotone. ■
Recall the definition of a pseudomonotone operator.
Definition 23.7.22A set valued operator B is quasi-bounded if whenever x ∈ D
(B )
and x^{∗}∈ Bx are such that
|〈x∗,x〉|, ∥x∥ ≤ M,
it follows that
∥x∗∥
≤ K_{M}. Bounded would mean that if
∥x∥
≤ M, then
∥x∗∥
≤ K_{M}.Here you only know this if there is another condition.
By Proposition 23.7.23 an example of a quasi-bounded operator is a maximal
monotone operator G for which 0 ∈int
(D (G ))
.
Then there is a useful result which gives examples of quasi-bounded operators
[?].
Proposition 23.7.23Let A : D
(A)
⊆ X → P
(X ′)
be maximal monotone andsuppose 0 ∈int
(D (A))
. Then A is quasi-bounded.
Proof: From local boundedness, Theorem 23.7.20, there exists δ,C > 0 such
that
x_{2}^{∗}∈ Ax. Thus Ax is
convex. Is it closed? Say x_{n}^{∗}∈ Ax and x_{n}^{∗}→ x^{∗}. Is it the case that x^{∗}∈ D
(A)
? Let
∗
[u,u ]
∈G
(A )
be arbitrary. Then
∗ ∗ ∗ ∗
〈x − u ,x − u〉 = nli→m∞ 〈xn − u ,xn − u〉 ≥ 0
and so Ax is also closed.
Consider the second condition. It is to show that if x_{n}→ x in V a finite dimensional
subspace and if U is a weakly open set containing 0, then eventually Ax_{n}⊆ Ax + U.
Suppose then that this is not the case. Then there exists x_{n}^{∗} outside of Ax + U but in
Ax_{n}. Since A is locally bounded at x, it follows that the
∥x∗∥
n
are bounded. Thus there is
a subsequence, still denoted as x_{n} and x_{n}^{∗} such that x_{n}^{∗}→ x^{∗} weakly and x^{∗}
∕∈
Ax + U.
Now let
[u,u∗]
∈G
(A )
.
〈x∗ − u∗,x − u〉 = nli→m∞ 〈x∗n − u ∗,xn − u〉 ≥ 0
and since
[u,u∗]
is arbitrary, it follows that x^{∗}∈ Ax and so is inside Ax + U. Thus the
second condition holds also.
Consider the third. Say x_{k}→ x weakly and letting x_{k}^{∗}∈ Ax_{k},suppose