25.7. MAXIMAL MONOTONE OPERATORS 903

Proof: Note that, since 0∈ A(0) , if x∗ ∈ Ax, then ⟨x∗,x⟩ ≥ 0. Also note that ∥Bλ (0)∥≤|B(0)|= 0 and so also ⟨Bλ x,x⟩ ≥ 0. It is necessary to show that F+A+B is onto. However,Bλ is monotone hemicontinuous, bounded and coercive. Hence, by Lemma 25.7.39, Bλ +Ais maximal monotone. If x∗ ∈ X ′ is given, there exists a solution to

x∗ ∈ Fxλ +Bλ xλ +Axλ

Do both sides to xλ and let x∗λ∈ Axλ be such that equality holds in the above.

x∗ = Fxλ +Bλ xλ + x∗λ

(25.7.60)

Then

⟨x∗,xλ ⟩= ∥xλ∥2 +≥0⟨

x∗λ,xλ

⟩It follows that

∥xλ∥ ≤ ∥x∗∥ ,⟨x∗

λ,xλ

⟩≤ ⟨x∗,xλ ⟩ ≤ ∥x∗∥∥xλ∥ ≤ ∥x∗∥2 (25.7.61)

Next, 0 is on the interior of D(A) and so from Theorem 25.7.20, there exists ρ > 0 suchthat if y∗ ∈ Ax for ∥x∥ ≤ ρ, then ∥y∗∥< M and in fact, all such x are in D(A). Now let

yλ =1

2∥∥x∗

λ

∥∥F−1 (x∗λ

)so ∥yλ∥< ρ

Thus yλ ∈ D(A) and if y∗λ∈ Ayλ , then

∥∥y∗λ

∥∥< M. Then for such bounded y∗λ,

0≤⟨y∗

λ− x∗

λ,yλ − xλ

⟩=⟨y∗

λ,yλ

⟩−⟨x∗

λ,yλ

⟩−⟨y∗

λ,xλ

⟩+⟨x∗

λ,xλ

⟩Then

12

∥∥x∗λ

∥∥=⟨x∗λ,

12∥∥x∗

λ

∥∥F−1 (x∗λ

)⟩=⟨x∗

λ,yλ

⟩≤⟨y∗

λ,yλ

⟩−⟨y∗

λ,xλ

⟩+⟨x∗

λ,xλ

⟩≤Mρ +M ∥xλ∥+

⟨x∗

λ,xλ

⟩From 25.7.61, ∥∥x∗

λ

∥∥≤ 2(

Mρ +M ∥x∗∥+∥x∗∥2)

Thus from 25.7.61, xλ ,x∗λ ,Fxλ are all bounded. Hence it follows from 25.7.60 that Bλ xλ

is also bounded. Therefore, 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