25.7. MAXIMAL MONOTONE OPERATORS 869

Proof: Suppose that for all [x,y] ∈ G (A) ,

⟨y− z,x− t⟩ ≥ 0

Does it follow that z ∈ At? By assumption, z+λF (t) = λFx̂+ ξ̂ , ξ̂ ∈ Ax̂. Then replacingy with ξ̂ and x with x̂, ⟨

ξ̂ −(

λFx̂+ ξ̂ −λFt), x̂− t

⟩≥ 0

and soλ ⟨Ft−Fx̂, t− x̂⟩ ≤ 0

which implies from Theorem 25.2.5 that t = x̂ and so the graph of A is indeed maximalmonotone.

z+λF (t) = λFx̂+ ξ̂ ⇒ z = ξ̂ ∈ Ax̂ = At

Note that this would have worked with no change if the duality map had been forarbitrary p > 1.

25.7.1 The minmax TheoremIn fact, these two conditions are equivalent. This is shown in [13]. We give a proof of thishere. First it is necessary to prove a minmax theorem. The proof given follows Brezis [24]which is where I found it. Here is the minmax theorem. A function f is convex if

f (λx+(1−λ )y)≤ λ f (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 thefollowing definition of upper and lower semicontinuous functions defined on a metric spaceand having values in [−∞,∞].

Definition 25.7.3 A function is upper semicontinuous if whenever xn → x, it follows thatf (x)≥ limsupn→∞ f (xn) and it is lower semicontinuous if f (x)≤ liminfn→∞ f (xn) .

Lemma 25.7.4 If F is a set of functions which are upper semicontinuous, then g(x) ≡inf{ f (x) : f ∈F} is also upper semicontinuous. Similarly, if F is a set of functions whichare lower semicontinuous, then if g(x) ≡ sup{ f (x) : f ∈F} it follows that g is lowersemicontinuous.

Proof: Let f ∈F where these functions are upper semicontinuous. Then if xn → x,and g(x)≡ inf{ f (x) : f ∈F} ,

f (x)≥ lim supn→∞

f (xn)≥ lim supn→∞

g(xn)

Since this is true for each f ∈F , then it follows that you can take the infimum and obtaing(x)≥ limsupn→∞ g(xn) . Similarly, lower semicontinuity is preserved on taking sup.