Introduction to Mathematical Analysis I - Second Edition

97 Definition 3.7.2 Let f : D → R . We say that f is lower semicontinuous on D (or lower semicontinu- ous if no confusion occurs) if it is lower semicontinuous at every point of D . Theorem 3.7.3 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 ¯ x ∈ D such that f ( x ) ≥ f ( ¯ x ) for all x ∈ D . Proof: We first prove that f is bounded below. Suppose by contradiction that for every k ∈ N , there exists x k ∈ D such that f ( x k ) < − k . Since D is compact, there exists a subsequence { x k ` } of { x k } that converges to x 0 ∈ D . Since f is l.s.c., by Theorem 3.7.2 liminf ` → ∞ f ( x k ` ) ≥ f ( x 0 ) . This is a contraction because liminf ` → ∞ f ( x k ` ) = − ∞ . 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, γ ∈ R . Let { u k } be a sequence in D such that { f ( u k ) } converges to γ . By the compactness of D , the sequence { u k } has a convergent subsequence { u k ` } that converges to some ¯ x ∈ D . Then γ = lim ` → ∞ f ( u k ` ) = liminf ` → ∞ f ( u k ` ) ≥ f ( ¯ x ) ≥ γ . This implies γ = f ( ¯ x ) and, hence, f ( x ) ≥ f ( ¯ x ) for all x ∈ D . The proof is now complete. The following theorem is proved similarly. Theorem 3.7.4 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 ¯ x ∈ D such that f ( x ) ≤ f ( ¯ x ) for all x ∈ D . For every a ∈ R , define L a ( f ) = { x ∈ D : f ( x ) ≤ a } and U a ( f ) = { x ∈ D : f ( x ) ≥ a } . Theorem 3.7.5 Let f : D → R . Then f is lower semicontinuous if and only if L a ( f ) is closed in D for every a ∈ R . Similarly, f is upper semicontinuous if and only if U a ( f ) is closed in D for every a ∈ R .