866 CHAPTER 25. NONLINEAR OPERATORS

Then if α ≥ 0,x∗0 (αy) = α f 0 (x,y) = f 0 (x,αy) .

If α < 0,x∗0 (αy)≡ α f 0 (x,y) =

lim infµ→0+ h→0

(−α) f (x+h)− (−α) f (x+h+µy)µ

=

(−α) lim infµ→0+ h→0

f (x+h−µy)− f (x+h)µ

≤

(−α) f 0 (x,−y) = f 0 (x,αy) .

Therefore, x∗0 (αy)≤ f 0 (x,αy) for all α. By the Hahn Banach theorem there is an extensionof x∗0 to all of V, x∗ which satisfies,

x∗ (y)≤ f 0 (x,y)

for all y. It remains to verify x∗ is continuous. This follows easily from

|x∗ (y)|= max(x∗ (−y) ,x∗ (y))≤

max(

f 0 (x,y) , f 0 (x,−y))≤ Lipx ( f ) ||y|| ,

which verifies 25.6.33 and proves the lemma.This lemma has verified the first condition needed in the definition of pseudomonotone.

The next lemma verifies that these generalized subgradients satisfy the second of the con-ditions needed in the definition. In fact somewhat more than is needed in the definition isshown.

Lemma 25.6.3 Let U be weakly open in V ′ and suppose ∂ f (x) ⊆ U. Then ∂ f (z) ⊆ Uwhenever z is close enough to x.

Proof: Suppose to the contrary there exists zn→ x but z∗n ∈ ∂ f (zn)\U. From the firstlemma, we may assert that ||z∗n|| ≤ 2Lip( f ) for all n large enough. Therefore, there is asubsequence, still denoted by n such that z∗n converges weakly to z∗ /∈U.

Claim: f 0 (x,y)≥ limsupn→∞ f 0 (xn,y) .Proof of the claim: There exists δ > 0 such that if µ, ||h||< δ , then

ε + f 0 (x,y)≥ f (x+h+µy)− f (x+h)µ

.

Thus, for ||h||< δ ,

ε + f 0 (x,y)≥ f (xn +(x− xn)+h+µy)− f (xn +(x− xn)+h)µ

.