79 For every n ∈ N, there exists xn ∈ [a,b] such that α − 1/n < f (xn) ≤ α. Since a ≤ xn ≤ b for every n, the sequence is bounded. Applying the Bolzano-Weierstrass theorem (Theorem 2.4.1), there exists a convergent subsequence, say, {xnk }, and x0 such that lim xnk = x0. k→∞ Note that x0 ∈ [a,b] by the Comparison theorem (Theorem 2.1.2). From the continuity of f on [a,b] we have lim f (xnk )= f (x0). k→∞ On the other hand, 1 α − < f (xnk ) ≤ α for every k nk and applying the Squeeze theorem (Theorem 2.1.3) we get that limk →∞ f (xnk )= α. Since f is a continuous function and limk →∞ xnk = x0, we have limk →∞ f (xnk )= f (x0). Therefore f (x0)= α and f (x0) ≥ f (x) for every x ∈ [a,b]. Thus, f has an absolute maximum at x0. Let xM = x0. Similarly, considering − f instead of f , we can show the existence of the absolute minimum. The result follows. □ The following result is a basic property of continuous functions that is used in a variety of situations. Lemma 3.4.3 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 ∈ D such that |x − c| < δ f (c) Proof: Let ε = > 0. By the continuity of f at c, there exists δ > 0 such that if x ∈ D and 2 |x − c| < δ , then | f (x) − f (c)| < ε. This implies, in particular, that f (c) f (x) > f (c) − ε = > 0. 2 Pick this δ , then the result follows. □ Remark 3.4.1 An analogous result holds if f (c) < 0. Theorem 3.4.4 Let f : [a,b] → R be a continuous function. Suppose 0 is strictly between f (a) and f (b) (that is, 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). Defne A = {x ∈ [a,b] : f (x) ≤ 0}.
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