25.1. SOME NONLINEAR SINGLE VALUED OPERATORS 825

Proof of the claim: Since A is monotone,

⟨Aun−Au,un−u⟩ ≥ 0

so⟨Aun,un−u⟩ ≥ ⟨Au,un−u⟩.

Therefore,

0 = lim infn→∞⟨Au,un−u⟩ ≤ lim inf

n→∞⟨Aun,un−u⟩ ≤ lim sup

n→∞

⟨Aun,un−u⟩ ≤ 0.

Now using that A is monotone again, then letting t > 0,

⟨Aun−A(u+ t (v−u)) ,un−u+ t (u− v)⟩ ≥ 0

and so⟨Aun,un−u+ t (u− v)⟩ ≥ ⟨A(u+ t (v−u)) ,un−u+ t (u− v)⟩.

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 25.1.5 Let H be any Hilbert space (complete inner product space, more on theselater) 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⟩=