133 Theorem 5.4.2 Let f : D→Rand let x0 ∈D. Then f is l.s.c. at x0 if and only if for every sequence {xk}inDthat converges to x0, liminf k→∞ f (xk) ≥f (x0). Similarly, f is u.s.c. at x0 if and only if for every sequence {xk}inDthat converges to x0, limsup k→∞ f (xk) ≤f (x0). Proof: Suppose f is l.s.c. at x0. Then for anyε >0, there exists δ >0 such that (5.4) holds. Since {xk}converges to x0, we have xk ∈B(x0;δ) when k is sufficiently large. Thus, f (x0)−ε <f (xk) for such k. It follows that f (x0)−ε ≤liminfk→∞ f (xk). Since ε is arbitrary, it follows that f (x0) ≤ liminfk→∞ f (xk). We now prove the converse. Suppose liminfk→∞ f (xk) ≥f (x0) and assume, by way of contradiction, that f is not l.s.c. at x0. Then there exists ¯ ε >0 such that for every δ >0, there exists xδ ∈B(x0;δ)∩Dwith f (x0)−¯ ε ≥f (xδ). Applying this for δk = 1 k, we obtain a sequence {xk}inDthat converges to x0 with f (x0)−¯ ε ≥f (xk) for every k ∈N. This implies f (x0) >f (x0)−¯ ε ≥liminf k→∞ f (xk). This is a contradiction. □ Definition 5.4.2 Let f : D→R. We say that f is lower semicontinuous on D(or lower semicontinuous if no confusion occurs) if it is lower semicontinuous at every point of D. Theorem 5.4.3 Suppose Dis a compact set of Rand f : D→Ris lower semicontinuous. Then f has an absolute minimum onD. That means there exists x0 ∈Dsuch that f (x) ≥f (x0) for all x ∈D. Proof: We first prove that f is bounded below. Suppose by contradiction that for every k ∈N, there exists xk ∈Dsuch that f (xk) <−k. Since Dis compact, there exists a subsequence {xkℓ} of {xk} that converges to x0 ∈D. Since f is l.s.c., by Theorem 5.4.2 liminf ℓ→∞ f (xkℓ) ≥f (x0).
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