42.1. BASIC THEORY 1371

and this says x1 ∈ D(A) because Jλ maps into D(A). Also it says

x1 +Ax1 ∋ x1 +λy1

and so y1 ∈ Ax1.This makes the last claim pretty easy. Suppose xk → x where xk ∈ D(A) and that yk ∈

Axk and yk ⇀ y. I need to verify y = Ax and x ∈ D(A) . Let [u,v] ∈ G (A). Then

(y− v,x−u) = limk→∞

(yk− v,xk−u)≥ 0

and so, by the first part, x ∈ D(A) and y ∈ Ax. Why does that limit hold? It is because

|(y− v,x−u)− (yk− v,xk−u)|

≤ |(y− v,x−u)− (yk− v,x−u)|+ |(yk− v,xk− x)|

The second term is no larger than

|yk− v| |xk− x|

which converges to 0 since yk is weakly convergent, hence bounded. The first term con-verges to 0 because of the assumption that yk converges weakly to y. This proves thelemma.

What about the sum of maximal monotone operators? This might not be maximalmonotone but what you can say is the following.

Proposition 42.1.6 Let A be maximal monotone and let B be Lipschitz and monotone. ThenA+B is maximal monotone.

Proof: First suppose B has a Lipschitz constant less than 1. The monotonicity is obvi-ous. I need to show that for any y there exists x ∈ D(A) such that

y ∈ x+Bx+Ax

This hapens if and only ify−Bx ∈ (I +A)x

if and only if x = (I +A)−1 (y−Bx). Let

T x≡ (I +A)−1 (y−Bx)

Then T is clearly a contraction mapping because (I +A)−1 is Lipschits with Lipschitzconstant 1. Therefore, there exists a unique fixed point and this shows A+ B is maxi-mal monotone. Now the same argument applied to A+B shows that A+ 2B is maximalmonotone. Continuing this way A+ nB is maximal monotone. Now for arbitrary B letn be large enough that n−1B has Lipschitz constant less than 1. Then as just explained,A+n

(n−1B

)= A+B is maximal monotone. This proves the proposition.

The following is a useful result for determining conditions under which A+B is max-imal monotone or more particularly whether a given y is in (I +A+B)(H) where A,B areboth maximal monotone.