advertisement

SOLUTION OF HW7 MINGFENG ZHAO October 19, 2012 1. (20 Points) Suppose that f 0 exists and f 0 is continuous on a nonempty, open interval (a, b) with f 0 (x) 6= 0 for all x ∈ (a, b). a. (6 Points) Prove that f is 1-1 on (a, b) and takes (a, b) onto some interval (c, d). b. (10 Points) Show that f −1 ∈ C 1 ((c, d)). c. (4 Points) Using the function f (x) = x3 , show that part b is false if the assumption f 0 (x) 6= 0 fails to hold for some x ∈ (a, b). Proof. Since f 0 (x) 6= 0 for all x ∈ (a, b) and f 0 is continuous on (a, b), then without loss of generality, we can assume f 0 (x) > 0 for all x ∈ (a, b), otherwise, we can consider the function −f (x) on (a, b). Since f 0 (x) > 0 for all x ∈ (a, b), then f is a strictly increase function on (a, b). a. Since f is strictly increase on (a, b), then f is 1-1. Define c = inf x∈(a,b) f (x), and d = sup f (x), x∈(a,b) then −∞ ≤ c < d ≤ ∞. By the definitions of c and d, we know that f ((a, b)) ⊂ [c, d] Claim I: f ((a, b)) ⊂ (c, d). If there exists some x0 ∈ (a, b) such that f (x0 ) = c. Since x0 ∈ (a, b), then there exists some y0 ∈ (a, x0 ). Since f is strictly increase, then f (y0 ) < f (x0 ) = c, which contradicts with the definition of c. If there exists some x0 ∈ (a, b) such that f (x0 ) = d. Since x0 ∈ (a, b), then there exists some y0 ∈ (x0 , b). Since f is strictly increase, then d = f (x0 ) < f (y0 ), which contradicts with the definition of d. 1 2 MINGFENG ZHAO In summary, we know that f ((a, b)) ⊂ (c, d). Claim II: f ((a, b)) = (c, d). For any y ∈ (c, d), then c < y < d. By the definitions of c and d, then there exists some x1 , x2 ∈ (a, b) such that c < f (x1 ) < y < f (x2 ) < d Since f is continuous on (a, b), then f is continuous on [x1 , x2 ], which implies that there exists some x0 ∈ [x1 , x2 ] ⊂ (a, b) such that f (x0 ) = y. Therefore, by the Claim I, then we can conclude that f ((a, b)) = (c, d). b. Claim I: f −1 is continuous on (c, d). For any yn , y0 ∈ (c, d) such that yn → y0 as n → ∞. Let xn = f −1 (yn ) and x0 = f −1 (y0 ). We want to show that xn → x0 as n → ∞, otherwise, there exists some 0 > 0 such that there exists a subsequence of xn , without loss of generality, assume xn itself such that |xn − x0 | ≥ 0 , ∀n ≥ 1. In the sequence xn , we can choose a subsequence without loss of generality, assume xn itself such that xn − x0 ≥ 0 , ∀n ≥ 1, or xn − x0 ≤ −0 , ∀n ≥ 1. Without loss of generality, we assume xn − x0 ≥ 0 , ∀n ≥ 1. xn ≥ x0 + 0 , ∀n ≥ 1. Hence we get Since f is continuous and strictly increasing, then yn = f (xn ) ≥ f (x0 + 0 ) > f (x0 ), ∀n ≥ 1. SOLUTION OF HW7 3 Which contradicts with yn → y0 as n → ∞. Hence we must have xn → x0 as n → ∞. Therefore, g is continuous. Claim II: x → x0 if and only if f (x) → x0 . If x → x0 , since f is continuous, then f (x) → f (x0 ). On the other hand, if f (x) → f (x0 ), applying f −1 on both sides, since g is also continuous, then x = f −1 (f (x)) → x0 = f −1 (f (x0 )). Since f 0 (x0 ) 6= 0, then f 0 (x0 ) > 0. By the result of the Problem 1, we know that there exists some δ > 0 such that f (x) 6= f (x0 ) for all 0 < |x − x0 | < δ. Hence, we know that f −1 (f (x)) − f −1 (f (x0 )) f (x) − f (x0 ) = = → x − x0 f (x) − f (x0 ) 1 f (x)−f (x0 ) x−x0 1 f 0 (x0 ) as x → x0 ⇐⇒ as f (x) → f (x0 ) Since f is increasing and f −1 is continuous. Hence, we know that (f −1 )0 (f (x0 )) exists and (f −1 )0 (f (x0 )) = Since (f −1 )0 (x) = 1 . f 0 (x0 ) 1 for all x ∈ (c, d). Since f −1 is continuous on (c, d) and f is continuous f 0 (f −1 (x)) on (a, b), then (f −1 )0 is continuous on (c, d). Hence we know that f −1 ∈ C 1 (c, d). c. If f (x) = x3 on R, then f 0 (x) = 3x2 ≥ 0 for all x ∈ R with f 0 (0) = 0. Also we know that f −1 (x) = √ 3 1 x = x3 , ∀x ∈ R 1 −2 x 2, 3 ∀x 6= 0 Then (f −1 )0 (x) = But (f 0 )−1 (0) does not exist. 4 MINGFENG ZHAO Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009 E-mail address: [email protected]