25.7. MAXIMAL MONOTONE OPERATORS 891

It follows that if y is given, there exists v∗ ∈ T (x) such that

lim infn→∞⟨v∗n,xn− y⟩ ≥ ⟨v∗,x− y⟩

Now let u∗t ∈ S (x+ t (y− x)) for t > 0. Thus

⟨u∗n−u∗t ,xn− x+ t (x− y)⟩ ≥ 0

⟨u∗n,xn− x+ t (x− y)⟩ ≥ ⟨u∗t ,xn− x+ t (x− y)⟩

Then using the above and the convergence in 25.7.54,

lim infn→∞⟨u∗n + v∗n,xn− y⟩ ≥ lim inf

n→∞⟨u∗t + v∗n,xn− y⟩

= ⟨u∗t ,x− y⟩+ ⟨v∗,x− y⟩

Now as before where it was shown that maximal monotone and defined on X impliedpseudomonotone, and the theorem which says that maximal monotone operators are locallybounded on the interior of their domains, it follows that there exists a sequence, still denotedas u∗t which converges to something called u∗. Then as before, the subspace spanned byx,y is finite dimensional and so from upper semicontinuity, for all t small enough,

u∗t ∈ S (x)+B(0,r)

Note that weak convergence is the same as strong on finite dimensional spaces. Since thisis true for all r and S (x) is closed, it follows that u∗ ∈ S (x) . Thus, passing to a limit ast→ 0 one gets u∗ ∈ S (x) ,v∗ ∈ T (x) , and

lim infn→∞⟨u∗n + v∗n,xn− y⟩ ≥ ⟨u∗+ v∗,x− y⟩

This proves the following generalization of Theorem 25.7.25.

Theorem 25.7.27 Let T,S : X →P (X ′) where X is a strictly convex reflexive Banachspace and suppose T is bounded and pseudomonotone while S is maximal monotone. ThenT +S is pseudomonotone.

Also, there is an interesting result which is based on the obvious observation that if Ais maximal monotone, then so is Â(x)≡ A(x0 + x).

Lemma 25.7.28 Let A be maximal monotone. Then for each λ > 0,

x→ λF (x− x0)+Ax

is onto.

Proof: Let Â(x) ≡ A(x0 + x) so as earlier, Â is maximal monotone. Then let y∗ ∈ X ′.Then there exists y such that Â(y)+λF (y) ∋ y∗. Now define x≡ y+ x0. Then

Â(y)+λF (y) ∋ y∗, Â(x− x0)+λF (x− x0) ∋ y∗, A(x)+λF (x− x0) ∋ y∗