904 CHAPTER 25. NONLINEAR OPERATORS

Using 25.7.60, it follows that⟨Fxλ n + x∗

λ n+Bλ nxλ n −

(Fxλ m + x∗

λ m+Bλ mxλ m

),xλ n − xλ m

⟩= 0

Thus⟨Fxλ n + x∗

λ n−(Fxλ m + x∗

λ m

),xλ n − xλ m

⟩+⟨Bλ nxλ n −Bλ mxλ m ,xλ n − xλ m

⟩= 0 (25.7.62)

Now F +A is surely monotone and so

lim supm,n→∞

⟨Bλ nxλ n −Bλ mxλ m ,xλ n − xλ m

⟩≤ 0

By Proposition 25.7.38, b∗ ∈ Bz and

limm,n→∞

⟨Bλ nxλ n −Bλ mxλ m ,xλ n − xλ m

⟩= 0

Then returning to 25.7.62,

lim supm,n→∞

⟨Fxλ n + x∗

λ n−(Fxλ m + x∗

λ m

),xλ n − xλ m

⟩≤ 0

Now from Lemma 25.7.39, F+A is maximal monotone. Hence Proposition 25.7.38 appliesagain and it follows that u∗+w∗ ∈ Fz+Az. Then passing to the limit as n→ ∞ in

x∗ = Fxλ n +Bλ nxλ n + x∗λ n

it follows thatx∗ = u∗+b∗+w∗ = Fz+Az+Bz

and this shows that A+B is maximal monotone because x∗ was arbitrary.You don’t need to assume all that stuff about 0 ∈ A(0) ,0 ∈ B(0) ,0 on interior of D(A)

and so forth.

Theorem 25.7.42 Suppose A,B are maximal monotone and the interior of D(A) has non-empty intersection with D(B). Then A+B is maximal monotone.

Proof: Let x0 be on the interior of D(A) and also in D(B). Let Â(x) = A(x0 + x)− x∗0where x∗0 ∈ A(x0) . Thus 0 ∈ D

(Â)

and 0 ∈ Â(0). Do the same thing for B to get B̂ definedsimilarly. Are these still maximal monotone? Suppose for all [u,u∗] ∈ G

(Â)

⟨y∗−u∗,y−u⟩ ≥ 0

Does it follow that y∗ ∈ Ây? It is given that u∗ ∈ A(x0 +u) . The above implies for all[u,u∗] ∈ G

(Â)

⟨y∗+ x∗0− (u∗+ x∗0) ,(y+ x0)− (u+ x0)⟩ ≥ 0

and since u+ x0 is a generic element of D(A) for u ∈ D(Â), the above implies y∗+ x∗0 ∈

A(y+ x0) and so y ∈ A(y+ x0)− x∗0 ≡ Â(y). Hence the graph is maximal. Similar forB̂. Thus the lemma can be applied to Â, B̂ to conclude that the sum of these is maximal