826 CHAPTER 25. NONLINEAR OPERATORS

lim supn→∞

⟨Axn,xn−0⟩= lim supn→∞

(−||xn||2

)=−1≤ 0.

If A were pseudomonotone, we would need to be able to conclude that for all y ∈ H,

lim infn→∞⟨Axn,xn− y⟩ ≥ ⟨Ax,x− y⟩= 0.

However,lim inf

n→∞⟨Axn,xn−0⟩=−1 < 0 = ⟨A0,0−0⟩.

The following proposition is useful.

Proposition 25.1.6 Suppose A : V →V ′ is pseudomonotone and bounded where V is sepa-rable. Then it must be demicontinuous. This means that if un→ u, then Aun ⇀ Au. In casethat V is reflexive, you don’t need the assumption that V is separable.

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 25.1.7 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 25.1.8 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⟩