Lesson 12.pdf - 12 Properties of Continuous Functions Let f D → R be a function defined on an interval D ⊂ R Definition 1 We say that f is bounded

# Lesson 12.pdf - 12 Properties of Continuous Functions Let f...

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12. Properties of Continuous Functions July 24, 2019 Definition Let f : D R be a function defined on an interval D R . 1. We say that f is bounded if there exists a number M such that | f ( x ) | ≤ M, x D. 2. If f is not bounded, we say that f is unbounded over D , that is. for any M > 0, there exists x 0 D such that | f ( x 0 ) | ≥ M . For the bounded case, we have the following definitions. 3. We say that f ( x 0 ) is the minimum value of f on D if x 0 D such that f ( x 0 ) f ( x ) x D. 4. We say that f ( x 1 ) is the maximum value of f on D if x 1 D such that f ( x 1 ) f ( x ) x D. [Examples] 1. f ( x ) = sin x and g ( x ) = cos x are bounded functions on D := R because | sin x | ≤ 1 , | cos x | ≤ 1 , x R . 2. f ( x ) = 1 x is not bounded on (0 , 1]. 3. Any polynomial of degree n 1 defined on R is not bounded. 105

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4. The theorem below is one of the most important results in analysis. The last two examples above show that the condition of [ a, b ] in this theorem cannot be replaced by ( a, b ) or ( a, + ). Theorem 0.1 (Global boundedness) If f : [ a, b ] R is continuous where -∞ < a < b < + , then f is bounded. That is, there exists M > 0 such that | f ( x ) | ≤ M, x [ a, b ] . Proof: By Property 0.4 in § 11, f is locally bounded. Then by Theorem 0.3 in § 5. f is bounded. Theorem 0.2 If f : [ a, b ] R is continuous where -∞ < a < b < + , then f is has a maximum value and a minimum value. That is, there exist points x 0 , x 1 D such that f ( x 0 ) = max x D { f ( x ) } , f ( x 1 ) = min x D { f ( x ) } . Proof: Note that [ a, b ] is compact because [ a, b ] is finite and closed by Heine-Borel theorem. If we can prove two things: We want to show the image of f f ([ a, b ]) = { f ( x ) | x [ a, b ] } is also compact (see Theorem 0.3 below). The image f ([ a, b ]) has a maximum and a minimum (see Lemma 0.4 below), then our proof is complete: In fact, since a maximum y 0 f ([ a, b ]), there exists at least one x 0 [ a, b ] such that y 0 = f ( x 0 ). Similarly, since a minimum y 1 f ([ a, b ]), there exists at least one x 1 [ a, b ] such that y 1 = f ( x 1 ).
• Fall '08
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• Topology, Metric space, Compact space, image f, interval D

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