928 CHAPTER 25. NONLINEAR OPERATORS

Then⟨L∗u,u⟩= lim

ε→0⟨L∗u,uε⟩= lim

ε→0⟨Luε ,u⟩= lim

ε→0

1ε⟨−F (uε −u) ,u⟩

= limε→0

1ε⟨F (uε −u) ,uε −u⟩− lim

ε→0

1ε⟨F (uε −u) ,uε⟩

Both of these last terms are nonnegative, the first obviously and the second from the abovewhere it was shown that ⟨F (uε −u) ,uε⟩ ≤ 0.

In the hypotheses of Theorem 25.8.8, one could have simply said that L is closed, linear,densely defined and maximal monotone. One can also show that if L is maximal monotone,then it must be densely defined. This is done in [23].

One can go further in obtaining a perturbation theorem like the above. Let linear Lbe densely defined with L closed and L,L∗ monotone. In short, L is densely defined andmaximal monotone, L : X→ X ′. Let A be a set valued L pseudomonotone operator which iscoercive and bounded. Also let B : D(B)→P (X) be maximal monotone. It is of interestto consider whether L+A+B is onto X ′. In considering this, I will add further assumptionsas needed. First note that ⟨Lx,x⟩= ⟨Lx−L0,x−0⟩ ≥ 0.

Definition 25.8.10 Define limsupm,n→∞ am,n ≡ limk→∞ sup{am,n : min(m,n)≥ k}

Then

lim supm,n→∞

am,n ≥ lim supm→∞

(lim sup

n→∞

am,n

).

To see this, suppose a > limsupm,n→∞ am,n. Then there exist k such that whenever m,n > k,

am,n < a

It follows that for m≥ k,lim sup

n→∞

am,n ≤ a

Hence

lim supm→∞

(lim sup

n→∞

am,n

)≤ a

Since a > limsupm,n→∞ am,n is arbitrary, it follows that

lim supm→∞

(lim sup

n→∞

am,n

)≤ lim sup

m,n→∞

am,n.

Then the following lemma is useful. I found this result in a paper by Gasinski, Migorski andOchal [54]. They begin with the following interesting lemma or something like it whichis similar to some of the ideas used in the section on approximation of maximal monotoneoperators.

Lemma 25.8.11 Suppose A is a set valued operator, A : X→P (X) and u∗n ∈Aun. Supposealso that un→ u weakly and u∗n→ u∗ weakly. Suppose also that

lim supm,n→∞

⟨u∗n−u∗m,un−um⟩ ≤ 0