148 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES

Then by Lemma 5.12.3 there exists a unique x≡ g(y) such that

Hε (g(y) ,y)≡minx∈A

Hε (x,y)

and also, whenever y,z ∈ A,

limt→0+

g(y+ t (z− y)) = g(y) .

Thus Hε (g(y) ,y) = minx∈A Hε (x,y) . But also this shows that y→Hε (g(y) ,y) is the mini-mum of functions which are upper semicontinuous and so this function is also upper semi-continuous. Hence there exists y∗ such that

maxy∈B

Hε (g(y) ,y) = Hε (g(y∗) ,y∗) = maxy∈B

minx∈A

Hε (x,y) (5.11)

Thus from concavity in the second argument and what was just defined, for t ∈ (0,1) ,

Hε (g(y∗) ,y∗)≥ Hε (g((1− t)y∗+ ty) ,(1− t)y∗+ ty)

≥ (1− t)Hε (g((1− t)y∗+ ty) ,y∗)+ tHε (g((1− t)y∗+ ty) ,y)

≥ (1− t)Hε (g(y∗) ,y∗)+ tHε (g((1− t)y∗+ ty) ,y) (5.12)

This is because minx Hε (x,y∗)≡ Hε (g(y∗) ,y∗) so

Hε (g((1− t)y∗+ ty) ,y∗)≥ Hε (g(y∗) ,y∗)

Then subtracting the first term on the right, one gets

tHε (g(y∗) ,y∗)≥ tHε (g((1− t)y∗+ ty) ,y)

and cancelling the t,

Hε (g(y∗) ,y∗)≥ Hε (g((1− t)y∗+ ty) ,y)

Now apply Lemma 5.12.3 and let t→ 0+ . This along with lower semicontinuity yields

Hε (g(y∗) ,y∗)≥ lim inft→0+

Hε (g((1− t)y∗+ ty) ,y) = Hε (g(y∗) ,y) (5.13)

Hence for every x,y

Hε (x,y∗)≥ Hε (g(y∗) ,y∗)≥ Hε (g(y∗) ,y)

Thusmin

xHε (x,y∗)≥ Hε (g(y∗) ,y∗)≥max

yHε (g(y∗) ,y)

and so

maxy∈B

minx∈A

Hε (x,y) ≥ minx

Hε (x,y∗)≥ Hε (g(y∗) ,y∗)

≥ maxy

Hε (g(y∗) ,y)≥minx∈A

maxy∈B

Hε (x,y)