25.4. SET-VALUED MAPS, PSEUDOMONOTONE OPERATORS 851

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). (25.4.19)

(This weaker condition says that if the lim sup condition holds for the original sequence,then there is a subsequence such that the lim inf condition holds for all v. In particular, forthis subsequence, the lim sup condition continues to hold.)

Proof: If this is not true, there exists xn→ x, also a weakly open set U, containing T xand zn ∈ T xn, but zn /∈ U . Then, taking a further subsequence, we can assume zn → zweakly and z /∈U . Then the strong convergence implies

lim supn→∞

Re⟨zn,xn− x⟩ ≤ 0

By assumption, there is a subsequence still denoted with n such that for any y,

lim infk→∞

Re⟨zn,xn− y⟩ ≥ Re⟨z(y) ,x− y⟩ , some z(y) ∈ T (x)

Then in particular, for this subsequence,

0≥ lim supn→∞

Re⟨zn,xn− x⟩ ≥ lim infn→∞

Re⟨zn,xn− x⟩ ≥ Re⟨z(x) ,x− x⟩= 0

so for this subsequence,limn→∞

Re⟨zn,xn− x⟩= 0

Therefore, if y ∈ X there exists z(y) ∈ T x such that

Re⟨z,x− y⟩= lim infn→∞

Re⟨zn,xn− y⟩ ≥ Re⟨z(y) ,x− y⟩.

Letting w = x− y, this shows, since y ∈ X is arbitrary, that the following inequality holdsfor every w ∈ X . (If you have w ∈ X , then you just choose y = x−w.)

Re⟨z,w⟩ ≥ Re⟨z(x−w) ,w⟩, z(x−w) ∈ T x.

In particular, we may replace w with −w and obtain

Re⟨z,−w⟩ ≥ Re⟨z(x+w) ,−w⟩,

which impliesRe⟨z(x−w) ,w⟩ ≤ Re⟨z,w⟩ ≤ Re⟨z(x+w) ,w⟩.

Therefore, there exists, λ ∈ [0,1] ,

zλ (y)≡ λ z(x−w)+(1−λ )z(x+w) ∈ Ax

such thatRe⟨z,w⟩= Re⟨zλ (y) ,w⟩.