870 CHAPTER 25. NONLINEAR OPERATORS

Note that in a metric space, the above definitions up upper and lower semicontinuity interms of sequences are equivalent to the definitions that

f (x) ≥ limr→0

sup{ f (y) : y ∈ B(x,r)}

f (x) ≤ limr→0

inf{ f (y) : y ∈ B(x,r)}

respectively.Here is a technical lemma which will make the proof shorter. It seems fairly interesting

also.

Lemma 25.7.5 Suppose H : A×B→R is strictly convex in the first argument and concavein the second argument where A,B are compact convex nonempty subsets of Banach spacesE,F respectively and x→ H (x,y) is lower semicontinuous while y→ H (x,y) is uppersemicontinuous. Let

H (g(y) ,y)≡minx∈A

H (x,y)

Then g(y) is uniquely defined and also for t ∈ [0,1] ,

limt→0

g(y+ t (z− y)) = g(y) .

Proof: First suppose both z,w yield the definition of g(y) . Then

H(

z+w2

,y)<

12

H (z,y)+12

H (w,y)

which contradicts the definition of g(y). As to the existence of g(y) this is nothing morethan the theorem that a lower semicontinuous function defined on a compact set achievesits minimum.

Now consider the last claim about “hemicontinuity”. For all x ∈ A, it follows from thedefinition of g that

H (g(y+ t (z− y)) ,y+ t (z− y))≤ H (x,y+ t (z− y))

By concavity of H in the second argument,

(1− t)H (g(y+ t (z− y)) ,y)+ tH (g(y+ t (z− y)) ,z) (25.7.39)≤ H (x,y+ t (z− y)) (25.7.40)

Now let tn→ 0. Does g(y+ tn (z− y))→ g(y)? Suppose not. By compactness, the expres-sion g(y+ tn (z− y)) is in a compact set and so there is a further subsequence, still denotedby tn such that

g(y+ tn (z− y))→ x̂ ∈ A

Then passing to a limit in 25.7.40, one obtains, using the upper semicontinuity in one andlower semicontinuity in the other the following inequality.

H (x̂,y)≤ lim infn→∞

(1− tn)H (g(y+ tn (z− y)) ,y)+