892 CHAPTER 25. NONLINEAR OPERATORS

Definition 25.7.29 Let A : D(A)→P (X ′) be maximal monotone. Let A−1 : A(D(A))→P (X ′) be defined as follows.

x ∈ A−1x∗ if and only if x∗ ∈ Ax

Observation 25.7.30 A−1 is also maximal monotone. This is easily seen as follows. [x,y]∈G (A) if and only if [y,x] ∈ G

(A−1

).

Earlier, it was shown that if B is monotone and hemicontinuous and coercive, then itwas onto. It was not necessary to assume that B is bounded. The same thing holds for Amaximal monotone. This will follow from the next result. Recall that a maximal monotoneoperator is locally bounded at every interior point of its domain which was shown above.Also it appears to not be possible to show that a maximal monotone operator is locallybounded at a limit point of D(A). The following result is in [13] although he claims abetter result than what I am proving here in which it is only necessary to verify A−1 islocally bounded at every point of A(D(A)). However, I was unable to follow the argumentand so I am proving another theorem with the same argument he uses. It looks like a typoto me but I often have trouble following hard theorems so I am not sure. Anyway, thefollowing is the best I can do. I think it is still a very interesting result.

Theorem 25.7.31 Suppose A−1 is locally bounded at every point of A(D(A)). Then in factA(D(A)) = X ′ and in fact A(D(A)) = A(D(A)) .

Proof: This is done by showing that A(D(A)) is both open and closed. Since it isnonempty, it must be all of X ′ because X ′ is connected. First it is shown that A(D(A)) isclosed. Suppose yn ∈ Axn and yn→ y. Does it follow that y ∈ A(D(A))? Since y is a limitpoint of A(D(A)) , it follows that A−1 is locally bounded at y. Thus there is a subsequencestill denoted by yn such that yn→ y and for xn ∈A−1yn or in other words, yn ∈Axn, it followsthat xn is bounded. Hence there exists a subsequence, still denoted with the subscript n suchthat xn→ x weakly and yn→ y strongly. Hence if [u,v] ∈ G (A) ,

⟨y− v,x−u⟩= limn→∞⟨yn− v,xn− x⟩ ≥ 0

Since [u,v] is arbitrary and A is maximal monotone, it follows that y∈ Ax or in other words,x ∈ A−1y and y ∈ A(D(A)). Thus A(D(A)) is closed.

Next consider why A(D(A)) is open. Let y0 ∈ A(D(A)) . Then there exists Dr ≡B(y0,r) centered at y0 such that A−1 is bounded on Dr. Since A is maximal montone,for each y ∈ X ′ there is a solution xε to the inclusion

y ∈ εF (xε − x0)+Axε , yε ≡ y− εF (xε − x0) ∈ Axε

Consider only y ∈ B(y0,

r2

).

⟨(y− εF (xε − x0))− y0,xε − x0⟩ ≥ 0

Then using ⟨Fz,z⟩= ∥z∥2 ,

∥y− y0∥∥xε − x0∥ ≥ ⟨y− y0,xε − x0⟩ ≥ ε ∥xε − x0∥2