908 CHAPTER 25. NONLINEAR OPERATORS

Then B(x∗,β ) ((x,α) ,δ ) is a weakly open set containing (x,α). For δ small enough, it doesnot intersect epi(φ) since if not so, there would exist (tn,γn) ∈ epi(φ)∩B(x∗,β )

((x,α) , 1

n

)and so

x∗ (tn)+βγn→ x∗ (x)+αβ

contrary to the above inequality. Thus epi(φ) is weakly closed. Also, if epi(φ) is weaklyclosed, then it is obviously strongly closed.

What of the limit condition using weak convergence instead of strong convergence? Sayxn→ x weakly. Does it follow that if epi(φ) is weakly closed that φ (x)≤ liminfn→∞ φ (xn)?It is just as above. There is a subsequence such that

α ≡ lim infn→∞

φ (xn) = limk→∞

φ(xnk

)and so

(xnk ,φ

(xnk

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

epi(φ). Henceα ≡ lim inf

n→∞φ (xn)≥ φ (x) .

There is also another convenient characterization of what it means for a function to belower semicontinuous.

Lemma 25.7.46 Let φ : X → (−∞,∞]. Then φ is lower semicontinuous if and only ifφ−1 ((a,∞]) is open for any a ∈ R.

Proof: Suppose first that epi(φ) is closed. Consider x ∈ φ−1 ((a,∞]) . Thus φ (x) > a.

Thus (x,a) ∈ epi(φ)C because a < φ (x) . Since epi(φ) is closed, there exists r,ε > 0 suchthat

B(x,r)× (a− ε,a+ ε)⊆ epi(φ)C

Hence if y ∈ B(x,r) , it follows that φ (y)≥ a+ε since otherwise there would be a point ofepi(φ)C in this open set B(x,r)× (a− ε,a+ ε). Hence B(x,r)⊆ φ

−1 ((a,∞]).Conversely, suppose φ

−1 ((a,∞]) is open for any a and let (x,b) ∈ epi(φ)C. Thenφ (x) > b. Thus there exists B(x,r) such that for y ∈ B(x,r) , it follows that φ (y) > b.That is, y ∈ φ

−1 ((b,∞]). So consider B(x,r)× (−∞,b). If (y,α) ∈ B(x,r)× (−∞,b), thensince φ (y) > b,α < φ (y) and so there is no point of intersection between epi(φ) and thisopen set B(x,r)× (−∞,b).

Of course one can define upper semicontinuous the same way that φ−1 (−∞,a) is open.

Thus a function is continuous if and only if it is both upper and lower semicontinuous.In case X is reflexive, the limit condition implies that epi(φ) is weakly closed. Suppose

(x,α) is a weak limit point of epi(φ) . Then by the Eberlein Smulian theorem, there is asubsequence of points of X ,(xn,αn) which converges weakly to (x,α) . Thus if the limitcondition holds,

φ (x)≤ lim infn→∞

φ (xn)≤ lim infn→∞

αn = α

and so (x,α) ∈ epi(φ). If X is not reflexive, this isn’t all that clear because it is not clearthat a limit point is the limit of a sequence. However, one could consider a limit conditioninvolving nets and get a similar result.