In fact, these two conditions are equivalent. This is shown in [?]. We give a proof of this
here. First it is necessary to prove a minmax theorem. The proof given follows Brezis [?]
which is where I found it. Here is the minmax theorem. A function f is convex
if
f (λx+ (1− λ) y) ≤ λf (x)+ (1− λ)f (y)
It is concave if the inequality is turned around. It can be shown that in finite dimensions,
convex functions are automatically continuous, similar for concave functions. Recall the
following definition of upper and lower semicontinuous functions defined on a metric
space and having values in
[− ∞, ∞ ]
.
Definition 23.7.3A function is upper semicontinuous if whenever x_{n}→ x, itfollows that f
(x)
≥ limsup_{n→∞}f
(xn)
and it is lower semicontinuous if f
(x)
≤
liminf _{n→∞}f
(xn)
.
Lemma 23.7.4If ℱ is a set of functions which are upper semicontinuous, theng
(x)
≡ inf
{f (x) : f ∈ ℱ }
is also upper semicontinuous. Similarly, if ℱ is a setof functions which are lower semicontinuous, then if g
(x)
≡ sup
{f (x) : f ∈ ℱ}
itfollows that g is lower semicontinuous.
Proof:Let f ∈ℱ where these functions are upper semicontinuous. Then if x_{n}→ x,
and g
(x)
≡ inf
{f (x ) : f ∈ ℱ }
,
f (x ) ≥ lim sup f (xn) ≥ lim sup g(xn)
n→∞ n→∞
Since this is true for each f ∈ℱ, then it follows that you can take the infimum and
obtain g
(x)
≥ limsup_{n→∞}g
(xn)
. Similarly, lower semicontinuity is preserved on taking
sup. ■
Note that in a metric space, the above definitions up upper and lower semicontinuity
in terms of sequences are equivalent to the definitions that
f (x) ≥ lrim→0sup{f (y) : y ∈ B (x,r)}
f (x) ≤ lim inf{f (y) : y ∈ B (x,r)}
r→0
respectively.
Here is a technical lemma which will make the proof shorter. It seems fairly
interesting also.
Lemma 23.7.5Suppose H : A × B → ℝ is strictly convex in the first argument andconcave in the second argument where A,B are compact convex nonempty subsets ofBanach spaces E,F respectively and x → H
(x,y)
is lower semicontinuous whiley → H
(x,y)
is upper semicontinuous. Let
H (g(y),y) ≡ mxi∈nA H (x,y)
Then g
(y)
is uniquely defined and also for t ∈
[0,1]
,
lit→m0 g(y+ t(z − y)) = g (y).
Proof: First suppose both z,w yield the definition of g
(y)
. Then
( )
H z-+-w,y < 1 H (z,y)+ 1 H (w, y)
2 2 2
which contradicts the definition of g
(y)
. As to the existence of g
(y)
this is nothing more
than the theorem that a lower semicontinuous function defined on a compact set achieves
its minimum.
Now consider the last claim about “hemicontinuity”. For all x ∈ 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))
because this holds for every x. Since t_{n}→ 0 was arbitrary, this
shows that in fact
lim g(y+ t(z − y)) = g(y) ■
t→0+
Now with this preparation, here is the min-max theorem. A norm is called strictly convex
if whenever x≠y,
∥x+y∥
∥-2-∥
<
∥x∥
-2-
+
∥y∥
-2-
.
Theorem 23.7.6Let E,F be Banach spaces with E having a strictly convex norm. Alsosuppose that A ⊆ E,B ⊆ F are compact and convex sets and that H : A×B → ℝ is suchthat
x → H (x,y) is convex
y → H (x,y) is concave
Thus H is continuous in each variable in the case of finite dimensional spaces. Hereassume that x → H
(x,y)
is lower semicontinuous and y → H
(x,y)
is uppersemicontinuous. Then
minmax H (x,y) = max min H (x,y)
x∈A y∈B y∈B x∈A
This condition is equivalent to the existence of
(x0,y0)
∈ A × B such that
H (x0,y) ≤ H (x0,y0) ≤ H (x,y0) for all x,y (23.7.42)
(23.7.42)
Proof:One part of the main equality is obvious.
max H (x,y) ≥ H (x,y) ≥ minH (x,y)
y∈B x∈A
and so for each x,
max H (x,y) ≥ maxmin H (x,y)
y∈B y∈Bx∈A
and so
minmax H (x,y) ≥ max min H (x,y) (23.7.43)
x∈A y∈B y∈B x∈A
(23.7.43)
Next consider the other direction.
Define H_{ε}
(x,y)
≡ H
(x,y)
+ ε
∥x∥
^{2} where ε > 0. Then H_{ε} is strictly convex in the
first variable. This results from the observation that