Introduction to Mathematical Analysis I - Second Edition

79 Corollary 3.4.4 is sometimes referred to as the Extreme Value Theorem. It follows immediately from Theorem 3.4.2 , and the fact that the interval [ a , b ] is compact (see Example 2.6.4 ) . The following result is a basic property of continuous functions that is used in a variety of situations. Lemma 3.4.5 Let f : D → R be continuous at c ∈ D . Suppose f ( c ) > 0. Then there exists δ > 0 such that f ( x ) > 0 for every x ∈ B ( c ; δ ) ∩ D . Proof: Let ε = f ( c ) > 0. By the continuity of f at c , there exists δ > 0 such that if x ∈ D and | x − c | < δ , then | f ( x ) − f ( c ) | < ε . This implies, in particular, that f ( x ) > f ( c ) − ε = 0 for every x ∈ B ( c ; δ ) ∩ D . The proof is now complete. Remark 3.4.6 An analogous result holds if f ( c ) < 0. Theorem 3.4.7 Let f : [ a , b ] → R be a continuous function. Suppose f ( a ) · f ( b ) < 0 (this means either f ( a ) < 0 < f ( b ) or f ( a ) > 0 > f ( b ) ). Then there exists c ∈ ( a , b ) such that f ( c ) = 0. Proof: We prove only the case f ( a ) < 0 < f ( b ) (the case f ( a ) > 0 > f ( b ) is completely analogous). Define A = { x ∈ [ a , b ] : f ( x ) ≤ 0 } . This set is nonempty since a ∈ A . This set is also bounded since A ⊂ [ a , b ] . Therefore, c = sup A exists and a ≤ c ≤ b . We are going to prove that f ( c ) = 0 by showing that f ( c ) < 0 and f ( c ) > 0 lead to contradictions. Suppose f ( c ) < 0. Then there exists δ > 0 such that f ( x ) < 0 for all x ∈ B ( c ; δ ) ∩ [ a , b ] . Because c < b (since f ( b ) > 0), we can find s ∈ ( c , b ) such that f ( s ) < 0 (indeed s = min { c + δ / 2 , ( c + b ) / 2 } will do). This is a contradiction because s ∈ A and s > c . Suppose f ( c ) > 0. Then there exists δ > 0 such that f ( x ) > 0 for all x ∈ B ( c ; δ ) ∩ [ a , b ] . Since a < c (because f ( a ) < 0), there exists t ∈ ( a , c ) such that f ( x ) > 0 for all x ∈ ( t , c ) (in fact, t = max { c − δ / 2 , ( a + c ) / 2 } will do). On the other hand, since t < c = sup A , there exists t 0 ∈ A with t < t 0 ≤ c . But then t < t 0 and f ( t 0 ) ≤ 0. This is a contradiction. We conclude that f ( c ) = 0. Theorem 3.4.8 — Intermediate Value Theorem. Let f : [ a , b ] → R be a continuous function. Suppose f ( a ) < γ < f ( b ) . Then there exists a number c ∈ ( a , b ) such that f ( c ) = γ . The same conclusion follows if f ( a ) > γ > f ( b ) .