75.1. AN ASSORTMENT OF NONLINEAR OPERATORS 2527

Proof: Since un→ u is strong convergence and since Aun is bounded, it follows

lim supn→∞

⟨Aun,un−u⟩= limn→∞⟨Aun,un−u⟩= 0.

Suppose this is not so that Aun converges weakly to Au. Since A is bounded, there exists asubsequence, still denoted by n such that Aun ⇀ ξ weak ∗. I need to verify ξ = Au. Fromthe above, it follows that for all v ∈V

⟨Au,u− v⟩ ≤ lim infn→∞⟨Aun,un− v⟩

= lim infn→∞⟨Aun,u− v⟩= ⟨ξ ,u− v⟩

Hence ξ = Au.There is another type of operator which is more general than pseudomonotone.

Definition 75.1.6 Let A : V → V ′ be an operator. Then A is called type M if wheneverun ⇀ u and Aun ⇀ ξ , and

lim supn→∞

⟨Aun,un⟩ ≤ ⟨ξ ,u⟩

it follows that Au = ξ .

Proposition 75.1.7 If A is pseudomonotone, then A is type M.

Proof: Suppose A is pseudomonotone and un ⇀ u and Aun ⇀ ξ , and

lim supn→∞

⟨Aun,un⟩ ≤ ⟨ξ ,u⟩

Thenlim sup

n→∞

⟨Aun,un−u⟩= lim supn→∞

⟨Aun,un⟩−⟨ξ ,u⟩ ≤ 0

Hencelim inf

n→∞⟨Aun,un− v⟩ ≥ ⟨Au,u− v⟩

for all v ∈V . Consequently, for all v ∈V,

⟨Au,u− v⟩ ≤ lim infn→∞⟨Aun,un− v⟩

= lim infn→∞

(⟨Aun,u− v⟩+ ⟨Aun,un−u⟩)

= ⟨ξ ,u− v⟩+ lim infn→∞⟨Aun,un−u⟩ ≤ ⟨ξ ,u− v⟩

and so Au = ξ .An interesting result is the the following which states that a monotone linear function

added to a type M is also type M.

Proposition 75.1.8 Suppose A : V → V ′ is type M and suppose L : V → V ′ is monotone,bounded and linear. Then L+A is type M. Let V be separable or reflexive so that the weakconvergences in the following argument are valid.