134 5.4 Lower Semicontinuity and Upper Semicontinuity This is a contraction because liminfℓ →∞ f (xkℓ )= −∞. This shows f is bounded below. Defne γ = inf{ f (x) : x ∈ D}. Since the set { f (x) : x ∈ D} is nonempty and bounded below, we see that γ ∈ R. Let {uk} be a sequence in D such that { f (uk)} converges to γ. By the compactness of D, the sequence {uk} has a convergent subsequence {ukℓ } that converges to some x0 ∈ D. Then γ = lim f (ukℓ )= liminf f (ukℓ ) ≥ f (x0) ≥ γ. ℓ→∞ ℓ→∞ This implies γ = f (x0) and, hence, f (x) ≥ f (x0) for all x ∈ D. The proof is now complete. □ The following theorem can be proved by a similar way. Theorem 5.4.4 Suppose D is a compact subset of R and f : D → R is upper semicontinuous. Then f has an absolute maximum on D. That is, there exists x0 ∈ D such that f (x) ≤ f (x0) for all x ∈ D. We now proceed to characterize upper and lower semicontinuity of a function f in terms of preimages of certain intervals under f . Given f : D → R, for every a ∈ R defne La( f )= {x ∈ D : f (x) ≤ a} = f − 1((−∞,a]) and Ua( f )= {x ∈ D : f (x) ≥ a} = f − 1([a,∞)). Theorem 5.4.5 Let f : D → R. Then f is lower semicontinuous if and only if La( f ) is closed in D for every a ∈ R. Similarly, f is upper semicontinuous if and only if Ua( f ) is closed in D for every a ∈ R. Proof: Suppose f is lower semicontinuous. Using Corollary 5.1.11, we will prove that for every sequence {xk} in La( f ) that converges to a point x0 ∈ D, we get x0 ∈ La( f ). For every k, since xk ∈ La( f ), f (xk) ≤ a. Since f is lower semicontinuous at x0, f (x0) ≤ liminf f (xk) ≤ a. k→∞ Thus, x0 ∈ La( f ). It follows that La( f ) is closed. We now prove the converse. Fix any x0 ∈ D and ε > 0. Then the set G = {x ∈ D : f (x) > f (x0) − ε} = D\ Lf (x0) −ε( f ) is open in D and x0 ∈ G. Thus, there exists δ > 0 such that B(x0;δ ) ∩ D ⊂ G.
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