5.11. SADDLE POINTS∗ 135

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 5.11.3 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) (5.12)

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)

Thus, letting C ≡max{∥x∥ : x ∈ A}

εC2 +maxy∈B

minx∈A

H (x,y)≥minx∈A

maxy∈B

H (x,y)

Since ε is arbitrary, it follows that

maxy∈B

minx∈A

H (x,y)≥minx∈A

maxy∈B

H (x,y)

This proves the first part because it was shown above in 5.9 that

minx∈A

maxy∈B

H (x,y)≥maxy∈B

minx∈A

H (x,y)

Now consider 5.8 about the existence of a “saddle point” given the equality of minmaxand maxmin. Let

α = maxy∈B

minx∈A

H (x,y) = minx∈A

maxy∈B

H (x,y)

Then fromy→min

x∈AH (x,y) and x→max

y∈BH (x,y)

being upper semicontinuous and lower semicontinuous respectively, there exist y0 and x0such that

α = minx∈A

H (x,y0) =

.

maxy∈B

minimum of u.s.cminx∈A

H (x,y) = minx∈A

maximum of l.s.c.maxy∈B

H (x,y) = maxy∈B

H (x0,y)