2528 CHAPTER 75. SOME NONLINEAR OPERATORS

Proof: Suppose un ⇀ u and Aun +Lun ⇀ ξ and also that

lim supn→∞

⟨Aun +Lun,un⟩ ≤ ⟨ξ ,u⟩

Does it follow that ξ = Au+ Lu? Suppose not. There exists a further subsequence, stillcalled n such that Lun ⇀ Lu. This follows because L is linear and bounded. Then frommonotonicity,

⟨Lun,un⟩ ≥ ⟨Lun,u⟩+ ⟨L(u) ,un−u⟩

Hence with this further subsequence, the limsup is no larger and so

lim supn→∞

⟨Aun,un⟩+ limn→∞

(⟨Lun,u⟩+ ⟨L(u) ,un−u⟩)≤ ⟨ξ ,u⟩

and solim sup

n→∞

⟨Aun,un⟩ ≤ ⟨ξ −Lu,u⟩

It follows since A is type M that Au = ξ −Lu, which contradicts the assumption that ξ ̸=Au+Lu.

There is also the following useful generalization of the above proposition.

Corollary 75.1.9 Suppose A : V → V ′ is type M and suppose L : V → V ′ is monotone,bounded and linear. Then for u0 ∈V define M (u)≡ L(u−u0) . Then M+A is type M. LetV be separable or reflexive so that the weak convergences in the following argument arevalid.

Proof: Suppose un ⇀ u and Aun +Mun ⇀ ξ and also that

lim supn→∞

⟨Aun +Mun,un⟩ ≤ ⟨ξ ,u⟩

Does it follow that ξ = Au+Mu? Suppose not. By assumption, un−u0 ⇀ u−u0 and so,since L is bounded, there is a further subsequence, still called n such that

Mun = L(un−u0)⇀ L(u−u0) = Mu.

Since M is monotone,⟨Mun−Mu,un−u⟩ ≥ 0

Thus⟨Mun,un⟩−⟨Mun,u⟩−⟨Mu,un⟩+ ⟨Mu,u⟩ ≥ 0

and so⟨Mun,un⟩ ≥ ⟨Mun,u⟩+ ⟨Mu,un−u⟩

Hence with this further subsequence, the limsup is no larger and so

lim supn→∞

⟨Aun,un⟩+ limn→∞

(⟨Mun,u⟩+ ⟨M (u) ,un−u⟩)≤ ⟨ξ ,u⟩

and solim sup

n→∞

⟨Aun,un⟩ ≤ ⟨ξ −Mu,u⟩