860 CHAPTER 25. NONLINEAR OPERATORS

u to get the second condition along with the first. Note that liminf gets bigger when yougo to a subsequence so if ∗ holds for the first subsequence, then it holds even better for thesecond.

Hence you would have, for this subsequence depending on v

lim infn→∞⟨wn,un−u⟩ ≥ 0≥ lim sup

n→∞

⟨wn,un−u⟩

and so limn→∞ ⟨wn,un−u⟩ = 0 for this subsequence still denoted with n. Hence for thissubsequence,

lim supn→∞

⟨zn +wn,un−u⟩= lim supn→∞

⟨zn,un−u⟩ ≤ 0

Then using that A is bounded pseudomonotone, limn→∞ ⟨zn,un−u⟩= 0 also. It follows forany v,

lim infn→∞⟨zn,un− v⟩ ≥ ⟨z(v) ,u− v⟩

Then from this it is routine to establish the modified pseudomonotone limit condition forthe sum A+B. For the subsequence just described, still denoted with n,

lim supn→∞

⟨zn +wn,un−u⟩ ≤ 0

and ∗. In fact, you would have

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

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

Then you would get

lim infn→∞⟨zn +wn,un− v⟩ = lim inf

n→∞(⟨zn,un− v⟩+ ⟨wn,un− v⟩)

≥ lim infn→∞

(⟨zn,un− v⟩)+ lim infn→∞

(⟨wn,un− v⟩)

≥ ⟨z(v) ,u− v⟩+ ⟨w(v) ,u− v⟩

and z(v)+w(v) ∈ (A+B)(u).Returning to 25.5.28, the other case to consider is that

lim supn→∞

⟨zn,un−u⟩ ≤ 0

Then in this case, the assumption that A is pseudomonotone implies that for any v,

lim infn→∞⟨zn,un− v⟩ ≥ ⟨z(v) ,u− v⟩ , z(v) ∈ A(u) (***)

No subsequence here. In particular,

0≥ lim infn→∞⟨zn,un−u⟩= ⟨z(u) ,u−u⟩= 0≥ lim sup

n→∞

⟨zn,un−u⟩