133 Theorem 5.4.2 Let f : D → R and let x0 ∈ D. Then f is l.s.c. at x0 if and only if for every sequence {xk} in D that converges to x0, liminf f (xk) ≥ f (x0). k→∞ Similarly, f is u.s.c. at x0 if and only if for every sequence {xk} in D that converges to x0, limsup f (xk) ≤ f (x0). k→∞ Proof: Suppose f is l.s.c. at x0. Then for any ε > 0, there exists δ > 0 such that (5.4) holds. Since {xk} converges to x0, we have xk ∈ B(x0;δ ) when k is suffciently 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 ) ∩ D with f (x0) − ε¯ ≥ f (xδ ). 1 Applying this for δ k = , we obtain a sequence {xk} in D that converges to x0 with k f (x0) − ε¯ ≥ f (xk) for every k ∈ N. This implies f (x0) > f (x0) − ε¯ ≥ liminf f (xk). k→∞ This is a contradiction. □ Defnition 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 D is a compact set of R and f : D → R is lower semicontinuous. Then f has an absolute minimum on D. That means there exists x0 ∈ D such that f (x) ≥ f (x0) for all x ∈ D. Proof: We frst prove that f is bounded below. Suppose by contradiction that for every k ∈ N, there exists xk ∈ D such that f (xk) < −k. Since D is 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 f (xkℓ ) ≥ f (x0). ℓ→∞
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