Introduction to Mathematical Analysis I - Second Edition

96 3.7 LOWER SEMICONTINUITY AND UPPER SEMICONTINUITY Since ε is arbitrary, we obtain liminf x → ¯ x f ( x ) ≥ f ( ¯ x ) . We now prove the converse. Suppose liminf x → ¯ x f ( x ) = sup δ > 0 h ( δ ) ≥ f ( ¯ x ) and let ε > 0. Since sup δ > 0 h ( δ ) > f ( ¯ x ) − ε , there exists δ > 0 such that h ( δ ) > f ( ¯ x ) − ε . This implies f ( x ) > f ( ¯ x ) − ε for all x ∈ B 0 ( ¯ x ; δ ) ∩ D . Since this is also true for x = ¯ x , the function f is lower semicontinuous at ¯ x . The proof for the upper semicontinuous case is similar. Theorem 3.7.2 Let f : D → R and let ¯ x ∈ D . Then f is l.s.c. at ¯ x if and only if for every sequence { x k } in D that converges to ¯ x , liminf k → ∞ f ( x k ) ≥ f ( ¯ x ) . Similarly, f is u.s.c. at ¯ x if and only if for every sequence { x k } in D that converges to ¯ x , limsup k → ∞ f ( x k ) ≤ f ( ¯ x ) . Proof: Suppose f is l.s.c. at ¯ x . Then for any ε > 0, there exists δ > 0 such that ( 3.12 ) holds. Since { x k } converges to ¯ x , we have x k ∈ B ( ¯ x ; δ ) when k is sufficiently large. Thus, f ( ¯ x ) − ε < f ( x k ) for such k . It follows that f ( ¯ x ) − ε ≤ liminf k → ∞ f ( x k ) . Since ε is arbitrary, it follows that f ( ¯ x ) ≤ liminf k → ∞ f ( x k ) . We now prove the converse. Suppose liminf k → ∞ f ( x k ) ≥ f ( ¯ x ) and assume, by way of contra- diction, that f is not l.s.c. at ¯ x . Then there exists ¯ ε > 0 such that for every δ > 0, there exists x δ ∈ B ( ¯ x ; δ ) ∩ D with f ( ¯ x ) − ¯ ε ≥ f ( x δ ) . Applying this for δ k = 1 k , we obtain a sequence { x k } in D that converges to ¯ x with f ( ¯ x ) − ¯ ε ≥ f ( x k ) for every k . This implies f ( ¯ x ) − ¯ ε ≥ liminf k → ∞ f ( x k ) . This is a contradiction.