Topics In Analysis

23.7.3 Surjectivity Theorems

As an interesting example of this theorem, here is another result in Barbu [?]. It is interesting because it is not assumed B is bounded.

Proof: Suppose B is not maximal monotone. Then there exists

∈ X × X^′ such that for all x,

〈Bx − x∗0,x− x0〉 ≥ 0

and yet x₀^∗≠Bx₀. This is going to be a contradiction. Let u ∈ X and consider x_t ≡ tx₀ +

u,t ∈. Then consider

∗ 〈Bxt − x 0,xt − x0〉

However, x_t − x₀ = tx₀ +

u − x₀ = and so, for each t ∈,

0 ≤ 〈Bxt − x∗0,xt − x0〉 = (1− t)〈Bxt − x∗0,u− x0〉

Divide by

and then let t ↑ 1. This yields the following by hemicontinuity.

〈Bx0 − x∗0,u − x0〉 ≥ 0

which holds for all u. Hence Bx₀ = x₀^∗ 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

xλ → x weakly.

Then from the equation 23.7.51

= λ and so,

Bx λ → x∗strongly. 0

Since B is monotone and hemicontinuous, it satisfies the pseudomonotone condition, Theorem 23.1.4. The above strong convergence implies

lim 〈Bx λ,xλ − x〉 = 0 λ→0

Hence for all y,

lim inf〈Bx λ,xλ − y〉 = lim inf 〈Bx λ,x− y〉 = 〈x∗0,x − y〉 ≥ 〈Bx,x − y〉 λ→0 λ→0

Since y is arbitrary, this shows that x₀^∗ = 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 [?].

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 k₀. Say B

⊆ D_r. Then for all < 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

< ε. Now recall that x_n → 0. Consider only u such that < 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 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

≤ δ. This contradicts the assumption that →∞. ■

Proof: It follows that x_n −x → 0. Therefore, from Lemma 23.7.16, for every r > 0, there exists ŷ₀ ∈B

and a subsequence x_nk such that

〈x∗ ,(xn − x)− ˆy0〉 → − ∞ nk k

Thus

〈 ∗ 〉 xnk,xnk − (x+ yˆ0) → − ∞

Just let y₀ = x + ŷ₀. Then y₀ ∈ D_r and satisfies the desired conditions. ■

Proof: Say the limit condition holds. Then if no such r exists, it follows that A is unbounded on every B

∩ D. Hence, you can let r_n → 0 and pick x_n ∈ B ∩D with x_n^∗∈ Ax_n such that > n, violating the limit condition. Hence some r exists such that A is bounded on B

(x,r)

∩ D. Conversely, suppose A is bounded on B

(x,r)

∩ D by M. Then if x_n → x, it follows that for all n large enough, x_n ∈ B and so if x_n^∗ ∈ Ax_n, ≤ M. Hence limsup_n→∞ ≤ M < ∞ which verifies the limit condition. ■

With this definition, here is a very interesting result.

Proof: You could use Corollary 23.7.17. If x is an interior point of D

, and A is not locally bounded, then there exists x_n → x and x_n^∗∈ Ax_n such that →∞. Then by Corollary 23.7.17, there exists y₀ close to x, in D and a subsequence x_nk such that

〈x∗ ,x − y〉 → − ∞ nk nk 0

Letting y₀^∗∈ Ay₀,

〈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

and this is a contradiction. ■

Does the same proof work if x is a limit point of D

? No. Suppose x is a limit point of D. If A is not locally bounded, then there exists x_n → x,x_n ∈ D and x_n^∗∈ Ax_n and →∞. Then there is y₀ close to x such that →−∞ but now everything crashes in flames because it is not known that y₀ ∈ D.

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 [?].

Proof: Let  be a monotone extension of A. Let

be such that ŵ ∈Â. Now also by assumption, A is not just convex but also closed.

If

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

. 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.

By Proposition 23.7.23 an example of a quasi-bounded operator is a maximal monotone operator G for which 0 ∈ int

.

Then there is a useful result which gives examples of quasi-bounded operators [?].

Proof: From local boundedness, Theorem 23.7.20, there exists δ,C > 0 such that

sup{∥x∗∥ : x∗ ∈ A (x) for ∥x∥ ≤ δ} < C

Now suppose that

,≤ M. Then letting ≤ δ,y^∗∈ Ay,

0 ≤ 〈x∗ − y∗,x − y〉 = 〈x∗,x〉 − 〈x∗,y〉− 〈y∗,x〉+ 〈y∗,y〉

and so for

≤ δ,

∗ ∗ ∗ ∗ 〈x ,y〉 ≤ 〈x ,x〉− 〈y ,x〉+ 〈y ,y〉 ≤ M + M C + Cδ

Hence,

≤ M + MC + Cδ ≡ K_M. ■

This is actually quite a restrictive requirement and leaves out a lot which would be interesting. 

Then here is an interesting result [?].

Proof: Consider the first condition. Say x_i^∗∈ Ax. Let u^∗∈ Au. For λ ∈

,

and so, since is arbitrary, it follows that λx₁^∗ + x₂^∗∈ Ax. Thus Ax is convex. Is it closed? Say x_n^∗∈ Ax and x_n^∗→ x^∗. Is it the case that x^∗∈ D? Let ∈G 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

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 ∈G.

〈x∗ − u∗,x − u〉 = nli→m∞ 〈x∗n − u ∗,xn − u〉 ≥ 0

and since

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

∗ lim sk→up∞ 〈x k,xk − x〉 ≤ 0,

Is it the case that there exists x^∗

∈ Ax such that

lim inf 〈x∗k,xk − y〉 ≥ 〈x∗ (y),x − y〉? k→∞