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)