5.11. SADDLE POINTS∗ 133

Note that in a metric space, the above definitions of 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 of the saddle point theorem

shorter. It seems fairly interesting also.

Lemma 5.11.3 Suppose H : A×B→R is strictly convex in the first argument and con-cave in the second argument where A,B are compact convex nonempty subsets of Banachspaces E,F respectively and x→H (x,y) is lower semicontinuous while y→H (x,y) is up-per semicontinuous. Let H (g(y) ,y) ≡ minx∈A H (x,y). Then g(y) is uniquely defined andalso 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”, continuity along a line. For allx ∈ A, it follows from the definition 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) (5.6)≤ H (x,y+ t (z− y)) (5.7)

Now let tn → 0. Does g(y+ tn (z− y))→ g(y)? Suppose not. By compactness, each ofg(y+ tn (z− y)) is in a compact set and so there is a further subsequence, still denoted bytn such that g(y+ tn (z− y))→ x̂ ∈ A. Then passing to a limit in 5.7, one obtains, using theupper semicontinuity in one and lower semicontinuity in the other the following inequality.

H (x̂,y) ≤ lim infn→∞

(1− tn)H (g(y+ tn (z− y)) ,y)

+ lim infn→∞

tnH (g(y+ tn (z− y)) ,z)

≤ lim infn→∞

((1− tn)H (g(y+ tn (z− y)) ,y)+tnH (g(y+ tn (z− y)) ,z)

)≤ lim sup

n→∞

H (x,y+ tn (z− y))≤ H (x,y)

This shows that x̂ = g(y) because this holds for every x. Since tn → 0 was arbitrary, thisshows that in fact limt→0+ g(y+ t (z− y)) = g(y) ■

Now with this preparation, here is the min-max theorem.