This section continues following Barbu [?]. Always it is assumed that the situation is of a
real reflexive Banach space X having strictly convex norm and its dual X^{′}. As observed
earlier, there exists a solution x_{λ} to the inclusion
0 ∈ F (x λ − x)+ λA (xλ)
To see this, you consider Â
(y)
≡ A
(x + y)
. Then Â is also maximal monotone and so
there exists a solution to
0 ∈ F (ˆx)+ λAˆ(ˆx) = F (ˆx)+ λA (x+ ˆx)
Now let x_{λ} = x +
xˆ
so
xˆ
= x_{λ}− x. Hence
0 ∈ F (xλ − x)+ λAx λ
Here you could have F the duality map for any given p > 1.
The symbol limsup_{n,n→∞}a_{mn} means lim_{N→∞}
( )
supm ≥N,n≥N amn
. Then here is a
simple observation.
Lemma 23.7.33Suppose limsup_{n,n→∞}a_{mn}≤ 0. Then
limsup_{m→∞}
(lim supn→∞ amn )
≤ 0.
Proof:There exists N such that if both m,n ≥ N,a_{mn}≤ ε. Then
This is for F a duality map with p > 1. Thus for the usual duality map, you wouldhave
A (x) = − λ− 1F (J x − x )
λ λ
Recall how this x_{λ}is defined. In general,
0 ∈ F (Jλx− x)+ λp−1Ax λ
Thus, from the definition,
A (x) ∈ A(J x)
λ λ
Formally, and to help remember what is going on, you are looking at a generalization
of
( )
Aλx = --A---x = 1- x− (I + λA )−1x
1 +λA λ
This is in the case where F = I to keep things simpler. You have 0 = x_{λ}−x + λAx_{λ} and so
formally x_{λ} =
(I + λA)
^{−1}x. Thus you are looking at
1λ
(x − xλ)
=
1λ
( )
x− (I + λA )−1x
= A_{λ}x.
In fact, this is exactly what you do when you are in a single Hilbert space.
This is just a generalization to mappings between Banach spaces and their
duals.
Then there are some things which can be said about these operators. It is presented
for the general duality map for p > 1.
Theorem 23.7.36The following hold. Here X is a reflexive Banach space withstrictly convex norm. A : D
(A )
→P
(X ′)
is maximal monotone. Then
J_{λ}and A_{λ}are bounded single valued operators defined on X. Bounded meansthey take bounded sets to bounded sets. Also A_{λ}is a monotone operator.
A_{λ},J_{λ}are demicontinuous. That is, strongly convergent sequences are mappedto weakly convergent sequences.
For every x ∈ D
(A)
,
∥A λ(x)∥
≤
|Ax|
≡ inf
∗ ∗
{∥y ∥ : y ∈ Ax}
. For everyx ∈conv
(D (A))
, it follows that lim_{λ→0}J_{λ}
(x)
= x. The new symbol meansthe closure of the convex hull. It is the closure of the set of all convexcombinations of points of D
(A )
.
Proof: 1.)It is clear that these are single valued operators. What about the
assertion that they are bounded? Let y^{∗}∈ Ax_{λ} such that the inclusion defining x_{λ}
becomes an equality. Thus
Hence if x is in a bounded set, it follows the resulting x_{λ} = J_{λ}x remain in a
bounded set. Now from the definition of A_{λ}, it follows that this is also a bounded
operator.
Why is A_{λ} monotone?
0 ≤ 〈Aλx− A λy,x− y〉 = 〈A λx− Aλy,x − Jλx− (y− Jλy)〉
+ 〈A x− A y,J x − J y〉
〈 λ λ λ λ 〉
= λ−(p−1)F (Jλx− x) − λ −(p−1)F (Jλy − y),Jλx− x − (Jλy − y)
+ 〈A x− A y,J x − J y〉
λ λ λ λ
and both terms are nonnegative, the first because F is monotone so indeed A_{λ} is
monotone.
2.) What of the demicontinuity of A_{λ}? This one is really tricky. Suppose x_{n}→ x.
Does it follow that A_{λ}x_{n}→ A_{λ}x weakly? The proof will be based on a pair of equations.
These are
When these have been established, Lemma 23.7.34 is used to get the desired result for a
subsequence. It will be shown that every sequence has a subsequence which gives the
right sort of weak convergence and from this the desired weak convergence of A_{λ}x_{n} to
A_{λ}x follows.
0 ∈ F (Jλxn − xn) + λp− 1A (Jλxn)
p− 1
0 ∈ F (Jλx− x) +λ A (Jλx)
− λ−(p−1)F (Jλx − x) ≡ Aλ(x) ∈ A (Jλx)
− λ−(p−1)F (J x − x ) ≡ A (x ) ∈ A (J x )
λ n n λ n λ n
Note also that for a given x there is only one solution J_{λ}x to 0 ∈ F