SOLUTION OF HW7 October 19, 2012 1. (20 Points) Suppose that f

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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.
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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
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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.
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MINGFENG ZHAO
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: [email protected]
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