25.4. SET-VALUED MAPS, PSEUDOMONOTONE OPERATORS 853

then there exists a subsequence, still denoted as {uk} such that if u∗k ∈ Tuk then for allv ∈V , there exists u∗ (v) ∈ Tu such that

lim infk→∞

Reu∗k (uk− v)≥ Reu∗ (v)(u− v). (25.4.21)

In this limit condition, there is a subsequence which works for all v. However, thepreservation of lower semicontinuity happens under even less.

Definition 25.4.12 Say T : V →P (V ′) is generalized bounded pseudomonotone if thefollowing conditions hold.

Tu is closed, nonempty, convex. (25.4.22)

T is bounded meaning it takes bounded sets to bounded sets.If uk ⇀ u and if

lim supk→∞

Reu∗k (uk−u)≤ 0,

then if v is given there exists a subsequence, still denoted as {uk} possibly depending on vsuch that if u∗k ∈ Tuk then, there exists u∗ (v) ∈ Tu such that

lim infk→∞

Reu∗k (uk− v)≥ Reu∗ (v)(u− v). (25.4.23)

This is more general because in this situation, the subsequence depends on the choice of v.

In case T is single valued, this condition is equivalent to type M.

Proposition 25.4.13 A single valued bounded operator T : V →V ′,V reflexive is general-ized bounded pseudomonotone then it is bounded and type M.

Proof: Suppose that un→ u weakly and Tun→ ξ weakly and

lim supn→∞

⟨Tun,un⟩ ≤ ⟨ξ ,u⟩ .

Thenlim sup

n→∞

⟨Tun,un−u⟩ ≤ 0

and so there is a subsequence depending on v and a further one depending on u such that

lim infn→∞⟨Tun,un−u⟩ ≥ ⟨Tu,u−u⟩= 0

lim infn→∞⟨Tun,un− v⟩ ≥ ⟨Tu,u− v⟩

The first in the above shows with the limsup condition that limn→∞ ⟨Tun,un−u⟩ = 0.Therefore, the second condition implies

⟨ξ ,u− v⟩ ≥ ⟨Tu,u− v⟩

since v is arbitrary, it follows that ξ = Tu. Thus T is type M.If T is generalized bounded pseudomonotone, then it is upper semicontinuous from the

strong to the weak topology.