Topics In Analysis

23.7.4 Approximation Theorems

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

≡ A. Then  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 +

so = 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→∞

. Then here is a simple observation.

Proof: There exists N such that if both m,n ≥ N,a_mn ≤ ε. Then

lim sup a = lim sup a ≤ ε n→ ∞ mn n→∞,n>N mn

Thus also

( ) ( ) lim msu→p∞ lim nsu→p∞ amn = lim m→s∞u,pm ≥N lim nsu→p∞ amn ≤ ε.■

The argument will be based on the following lemma.

Proof: By monotonicity,

lim 〈vn − vm, un − um 〉 = 0 m,n→ ∞

Suppose then that

fails to converge to . Then there is a subsequence, still denoted with subscript n such that → μ≠. Let ε > 0. Then there exists M such that if n,m > M, then

|〈vn,un 〉− μ| < ε,|〈vn − vm,un − um〉| < ε

Then if m,n > M,

|〈v − v ,u − u 〉| = |〈v ,u 〉+ 〈v ,u 〉− 〈v,u 〉− 〈v ,u 〉| < ε n m n m n n m m n m m n

Hence it is also true that

|〈vn,un〉+ 〈vm,um〉 − 〈vn,um 〉− 〈vm, un〉| ≤ |2μ − (〈vn,um〉+ 〈vm,un〉)| < 3ε

Now take a limit first with respect to n and then with respect to m to obtain

|2μ − (〈v,u〉+ 〈v,u〉)| < 3ε

Since ε is arbitrary, μ =

after all. Hence the claim that → is verified. Next suppose ∈G and consider

〈v− y,u− x〉 = 〈v,u〉− 〈v,x〉− 〈y,u〉+ 〈y,x〉

and since is arbitrary, it follows that v ∈ Au.

Next suppose limsup_n→∞

≤ 0. It is not known that ∈G.

Thus limsup_n→∞≤. Now let ∈G

〈v− y,u− x〉 = 〈v,u〉− 〈v,x〉− 〈y,u〉+ 〈y,x〉

Hence ∈G. Now

lim sup 〈vn − v,un − u〉 ≤ 0 ≤ lim inf 〈vn − v,un − u〉 n→∞ n→ ∞

the second coming from monotonicity and the fact that v ∈ Au. Therefore,

lnim→∞ 〈vn − v,un − u〉 = 0

which shows that lim_n→∞

= . ■

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 =

⁻¹x. Thus you are looking at = = A_λx. 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.

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

p−1 ∗ F (xλ − x)+ λ y = 0

Then let x₀ ∈ D

be given.

〈F (xλ − x) ,xλ − x〉+ λp−1〈y∗,xλ − x0〉+ λp−1〈y∗,x0 − x〉 = 0

Then by monotonicity of A,

∥xλ − x∥p + λp−1〈y∗,xλ − x0〉+ λp−1〈y∗,x0 − x〉 ≤ 0 0

It follows that

∥xλ − x∥p ≤ λp− 1∥y∗0∥∥xλ − x0∥ +λp− 1∥y∗∥∥x0 − x∥

Hence if x is in a bounded set, it follows the resulting x_λ = J_λx 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?

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_λx_n → A_λx weakly? The proof will be based on a pair of equations. These are

lim 〈F (Jλxn − xn)− F (Jλxm − xm ),Jλxn − xn − (Jλxm − xm )〉 = 0 m,n→ ∞

and

lim 〈Aλ(xn)− A λ(xm),Jλxn − Jλxm〉 = 0 m,n→∞

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_λx_n to A_λx follows.

Note also that for a given x there is only one solution J_λx to 0 ∈ F + λ^p−1A. By monotonicity of F,

0 ≤ 〈F (Jλxn − xn)− F (Jλxm − xm),xm − xn + Jλxn − Jλxm〉

Then from the above,

Now from the boundedness of these operators, the left side of the above inequality converges to 0 as n,m →∞. Thus

lim inf 〈F (J x − x )− F (J x − x ),J x − J x 〉 ≥ 0 (23.7.56) m,n→∞ λ n n λ m m λ n λ m

(23.7.56)

The expression on the left in the above is non positive. Multiplying by −1,Thus, in fact,the expression in 23.7.56 converges to 0. By boundedness considerations and the strong convergence given,

lim 〈F (J x − x )− F (J x − x ),J x − x − (J x − x )〉 = 0 m,n→ ∞ λ n n λ m m λ n n λ m m (23.7.58)

(23.7.58)

From boundedness again, there is a subsequence still denoted with the subscript n such that

Jλxn − xn → a− x, F (Jλxn − xn) → b both weakly.

Since F is maximal monotone, (Theorem 23.7.9) it follows from Lemma 23.7.34 that

∈G and so in fact F = b. Thus this has just shown that F → F. Next consider 23.7.57. We have J_λx_n → a weakly and A_λ = −λ−

(p−1)

F →−λ−

(p−1)

b weakly. Then from Lemma 23.7.34 again, ∈G so −λ−

(p−1)

b ∈ A so b ∈−λ^p−1A