902 CHAPTER 25. NONLINEAR OPERATORS

which shows that limn→∞

⟨Aλ nxn,xn

⟩= ⟨x∗,x⟩. Then it follows from this that

limn,m→∞

⟨Aλ nxn−Aλ mxm,xn− xm

⟩= 0

For the rest of this, the usual duality map for p = 2 will be used. It may be that one couldchange this, but I don’t have a need to do it right now so from now on, F will be the usualthing.

25.7.5 Sum Of Maximal Monotone OperatorsTo begin with, here is a nice lemma.

Lemma 25.7.39 Let 0 ∈ D(A) and let A be maximal monotone and let B : X → X ′ bemonotone hemicontinuous, bounded, and coercive. Then B+A is also maximal monotone.Also B+A is onto.

Proof: By Theorem 25.7.9, there exists x ∈ D(A) such that for all [u,u∗] ∈ G (A) ,

⟨Bx+Fx− y∗+u∗,u− x⟩ ≥ 0

Hence for all [u,u∗] ,⟨u∗− (y∗− (Bx+Fx)) ,u− x⟩ ≥ 0

It follows thaty∗− (Bx+Fx) ∈ Ax

and so y∗ ∈ Bx+Ax+Fx showing that B+A is maximal monotone because it added to Fis onto. As to the last claim, just don’t add in F in the argument. Thus for all [u,u∗] ,

⟨Bx− y∗+u∗,u− x⟩ ≥ 0

Then the rest is as before. You find that y∗−Bx ∈ Ax.

Corollary 25.7.40 Suppose instead of 0 ∈ D(A) , it is known that x0 ∈ D(A) and

lim∥x∥→∞

⟨B(x0 + x) ,x⟩∥x∥

= ∞

Then if B is monotone and hemicontinuous and A is maximal monotone, then B+A is onto.

Proof: Let Â(x)≡ A(x0 + x) so in fact 0 ∈ D(Â). Then letting B̂ be defined similarly,

it follows from the above lemma that if y∗ ∈ X ′, there exists x such that

y∗ ∈ Âx+ B̂x≡ A(x0 + x)+B(x0 + x)

Lemma 25.7.41 Let 0 be on the interior of D(A) and also in D(B). Also let 0 ∈ B(0) and0 ∈ A(0). Then if A,B are maximal monotone, so is A+B.