134 5.4 Lower Semicontinuity and Upper Semicontinuity This is a contraction because liminfℓ→∞ f (xkℓ) =−∞. This shows f is bounded below. Define 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 Dsuch 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 ℓ→∞ f (ukℓ) =liminf ℓ→∞ f (ukℓ) ≥f (x0) ≥γ. 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 Dis a compact subset of Rand f : D→Ris upper semicontinuous. Then f has an absolute maximum onD. That is, there exists x0 ∈Dsuch 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 everya∈Rdefine 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 inDfor 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 k→∞ f (xk) ≤a. Thus, x0 ∈La(f ). It follows that La(f ) is closed. We now prove the converse. Fix anyx0 ∈Dandε >0. Then the set G={x ∈D: f (x) >f (x0)−ε}=D\Lf (x0)−ε(f ) is open in Dandx0 ∈G. Thus, there exists δ >0 such that B(x0;δ)∩D⊂G.
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