25.7. MAXIMAL MONOTONE OPERATORS 907

Definition 25.7.44 Let φ : X → (−∞,∞]. Then φ is convex if whenever t ∈ [0,1] ,x,y ∈ X ,

φ (tx+(1− t)y)≤ tφ (x)+(1− t)φ (y)

The epigraph of φ is defined by

epi(φ)≡ {(x,y) : y≥ φ (x)}

When epi(φ) is closed in X × (−∞,∞], we say that φ is lower semicontinuous, l.s.c. Thefunction is called proper if φ (x)<∞ for some x. The collection of all such x is called D(φ) ,the domain of φ .

This definition of lower semicontinuity is equivalent to the usual definition.

Lemma 25.7.45 The above definition of lower semicontinuity is equivalent to the assertionthat whenever xn → x, it follows that φ (x) ≤ liminfn→∞ φ (xn) . In case that φ is convex,lower semicontinuity is equivalent to weak lower semicontinuity. That is epi(φ) is closedif and only if epi(φ) is weakly closed. In this case, the limit condition: If xx → x weakly,then φ (x)≤ liminfn→∞ φ (xn) is valid.

Proof: Suppose the limit condition holds. Why is epi(φ) closed? Why is X×(−∞,∞]\epi(φ) ≡ epi(φ)C open? Let (x,α) ∈ epi(φ)C . Then α < φ (x) ,α + δ < φ (x) . ConsiderB(x,r)×

(α− δ

2 ,α + δ

2

). If every such open set contains a point of epi(φ) , then there

exists xn→ x,yn < α + δ

2 ,yn ≥ φ (xn) . Hence, from the limit condition,

φ (x)≤ lim infn→∞

φ (xn)≤ lim infn→∞

yn ≤ α +δ

2< α +δ < φ (x)

a contradiction. It follows that there exists r > 0 such that B(x,r)×(

α− δ

2 ,α + δ

2

)∩

epi(φ) = /0. Since epi(φ)C is open, it follows that epi(φ) is closed.Next suppose epi(φ) is closed. Why does the limit condition hold? Suppose xn → x.

Then (xn,φ (xn)) ∈ epi(φ). There is a subsequence such that

α ≡ lim infn→∞

φ (xn) = limk→∞

φ(xnk

)and so

(xnk ,φ

(xnk

))→ (x,α). Since epi(φ) is closed, this means (x,α) ∈ epi(φ). Hence

α ≡ lim infn→∞

φ (xn)≥ φ (x) .

Consider the last claim. In this case, epi(φ) is convex. If it is closed, then it is weaklyclosed thanks to separation theorems: If (x,α) ∈ epi(φ)C , then α < ∞ and so there exists(x∗,β ) ∈ (X×R)′ and l such that for all (t,γ) ∈ epi(φ) ,

x∗ (t)+βγ > l > x∗ (x)+αβ