25.5. SUM OF PSEUDOMONOTONE OPERATORS 855

25.5 Sum Of Pseudomonotone OperatorsOne of the nice properties of pseudomonotone maps is that when you add two of them, youget another one. I will give a proof in the case that the two pseudomonotone maps are bothbounded. It is probably true in general, but as just noted, it is less trouble to verify if youdon’t have to worry about as many conditions. I will also assume the spaces are all real soit will not be necessary to constantly write the real part. Actually, we do a slightly moregeneral version which says that a bounded pseudomonotone added to a modified boundedpseudomonotone is a modified bounded pseudomonotone. First is the theorem about thesum of two bounded set valued pseudomonotone operators.

Theorem 25.5.1 Say A,B are set valued bounded pseudomonotone operators. Then theirsum is also a set valued bounded pseudomonotone operator. Also, if un→ u weakly, zn→ zweakly, zn ∈ A(un) , and wn→ w weakly with wn ∈ B(un) , then if

lim supn→∞

⟨zn +wn,un−u⟩ ≤ 0,

it follows that

lim infn→∞⟨zn +wn,un− v⟩ ≥ ⟨z(v)+w(v) ,u− v⟩ , z(v) ∈ A(u) ,w(v) ∈ B(u) ,

and z ∈ A(u) ,w ∈ B(u).

Proof: Say zn ∈ A(un) ,wn ∈ B(un) ,un→ u weakly and

lim supn→∞

⟨zn +wn,un−u⟩ ≤ 0.

Claim: Both of limsupn→∞ ⟨zn,un−u⟩ , limsupn→∞ ⟨wn,un−u⟩ are no larger than 0.Proof of the claim: Suppose limsupn→∞ ⟨wn,un−u⟩ = δ > 0. Then take a subse-

quence such that the limsup equals lim . The limsup only gets smaller when you go to asubsequence. Thus, continuing to denote the subsequence with n we still have

lim supn→∞

⟨zn +wn,un−u⟩ ≤ 0

But from the fact that we just took a subsequence for which the limsup = lim,

lim supn→∞

⟨zn +wn,un−u⟩ = lim supn→∞

⟨zn,un−u⟩+ limn→∞⟨wn,un−u⟩

= lim supn→∞

⟨zn,un−u⟩+δ ≤ 0

and so limsupn→∞ ⟨zn,un−u⟩=−δ < 0. Therefore by the limit condition,

lim infn→∞⟨zn,un−u⟩ ≥ ⟨z(u) ,u−u⟩= 0

and so0 >−δ ≥ lim sup

n→∞

⟨zn,un−u⟩ ≥ lim infn→∞⟨zn,un−u⟩ ≥ 0