906 CHAPTER 25. NONLINEAR OPERATORS

Proof: The existence of a solution to the inclusion 25.7.64 comes from the above dis-cussion. The last claim follows from almost a repeat of the last part of the proof of theabove theorem. Since {Bλ xλ} is given to be bounded for λ ∈ (0,δ ) , there is a sequence,λ n→ 0 such that

xλ n → z weakly

x∗λ→ w∗ weakly

Fxλ → u∗ weakly

Bλ nxλ n → b∗ weakly

Using 25.7.64, 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 (25.7.65)

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 m xλ m ,xλ n − xλ m

⟩= 0

Then returning to 25.7.65,

lim supm,n→∞

⟨Fxλ n + x∗

λ n−(Fxλ m + x∗

λ m

),xλ n − xλ m

⟩≤ 0

Now from Corollary 25.7.40, F +A is maximal monotone (In fact, F +A is onto). HenceProposition 25.7.38 applies again and it follows that u∗+w∗ ∈ Fz+Az. Then passing tothe limit as n→ ∞ in

x∗ = Fxλ n +Bλ nxλ n + x∗λ n

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

25.7.6 Convex Functions, An Example

As before, X will be a Banach space in what follows. Sometimes it will be a reflexiveBanach space and in this case, it will be assumed that the norm is strictly convex.