Introduction to Mathematical Analysis I - Second Edition

156 Solutions and Hints for Selected Exercises Since f is lower semicontinuous, by Theorem 3.7.3 , it has an absolute minimum on [ − a , a ] at some point ¯ x ∈ [ − a , a ] . Obviously, f ( x ) ≥ f ( ¯ x ) for all x ∈ [ − a , a ] . In particular, f ( 0 ) ≥ f ( ¯ x ) . If | x | > a , then f ( x ) ≥ f ( 0 ) ≥ f ( ¯ x ) . Therefore, f has an absolute minimum at ¯ x . Observe that in this solution, we can use any number γ in the range of f instead of f ( 0 ) . Since any continuous function is also lower semicontinuous, the result from this problem is applicable for continuous functions. For example, we can use this theorem to prove that any polynomial with even degree has an absolute minimum on R . Since R is a not a compact set, we cannot use the extreme value theorem directly. SECTION 4.1 Exercise 4.1.10 . Use the identity lim n → ∞ f ( a + 1 n ) f ( a ) ! n = lim n → ∞ exp ( n [ ln ( f ( a + 1 n )) − ln ( f ( a )]) . Exercise 4.1.11 . (a) Using the differentiability of sin x and Theorem 4.1.3 , we conclude the function is differentiable at any a 6 = 0. So, we only need to show the differentiability of the function at a = 0. By the definition of the derivative, consider the limit lim x → a f ( x ) − f ( a ) x − a = lim x → 0 x 2 sin ( 1 / x )+ cx x = lim x → 0 [ x sin ( 1 / x )+ c ] . For any x 6 = 0, we have | x sin ( 1 / x ) | = | x | | sin ( 1 / x ) | ≤ | x | , which implies −| x | ≤ x sin ( 1 / x ) ≤ | x | . Since lim x → 0 ( −| x | ) = lim x → 0 | x | = 0, applying the squeeze theorem yields lim x → 0 x sin ( 1 / x ) = 0 . It now follows that f 0 ( 0 ) = lim x → a f ( x ) − f ( a ) x − a = lim x → 0 [ x sin ( 1 / x )+ c ] = c . Using Theorem 4.1.3 and the fact that cos x is the derivative of sin x , the derivative of f can be written explicitly as f 0 ( x ) =   2 x sin 1 x − cos ( 1 / x )+ c , if x 6 = 0; c , if x = 0 .