2526 CHAPTER 75. SOME NONLINEAR OPERATORS

Taking the liminf on both sides and using the claim and t > 0,

t lim infn→∞⟨Aun,u− v⟩ ≥ t⟨A(u+ t (v−u)) ,(u− v)⟩.

Next divide by t and use the Hemicontinuity of A to conclude that

lim infn→∞⟨Aun,u− v⟩ ≥ ⟨Au,u− v⟩.

From the claim,

lim infn→∞⟨Aun,u− v⟩= lim inf

n→∞(⟨Aun,un− v⟩+ ⟨Aun,u−un⟩)

= lim infn→∞⟨Aun,un− v⟩ ≥ ⟨Au,u− v⟩.

Monotonicity is very important in the above proof. The next example shows that evenif the operator is linear and bounded, it is not necessarily pseudomonotone.

Example 75.1.4 Let H be any Hilbert space and let A : H→ H ′ be given by

⟨Ax,y⟩ ≡ (−x,y)H .

Then A fails to be pseudomonotone.

Proof: Let {xn}∞

n=1 be an orthonormal set of vectors in H. Then Parsevall’s inequalityimplies

||x||2 ≥∞

∑n=1|(xn,x)|2

and so for any x ∈ H, limn→∞ (xn,x) = 0. Thus xn ⇀ 0≡ x. Also

lim supn→∞

⟨Axn,xn− x⟩=

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⟩.

Now the following proposition is useful.

Proposition 75.1.5 Suppose A : V →V ′ is pseudomonotone and bounded where V is sep-arable. Then it must be demicontinuous. This means that if un→ u, then Aun ⇀ Au.