Recall Theorem 8.10.15 about continuity and convergent sequences. It said roughly that a function f is
continuous at x if and only if whenever x_{k}→x, then f
(xk)
→ f
(x)
. Also recall the following Lemma from
Volume 1, whose proof is summarized below for convenience.
Lemma 9.6.1Let ϕ :
[a,b]
→ ℝ be a continuous function and suppose ϕ is 1 − 1 on
(a,b)
. Thenϕ is either strictly increasing or strictly decreasing on
[a,b]
. Furthermore, ϕ^{−1}is continuous.
Proof:First it is shown that ϕ is either strictly increasing or strictly decreasing on
(a,b)
.
If ϕ is not strictly decreasing on
(a,b)
, then there exists x_{1}< y_{1}, x_{1},y_{1}∈
(a,b)
such that
(ϕ (y1)− ϕ (x1))(y1 − x1) > 0.
If for some other pair of points x_{2}< y_{2} with x_{2},y_{2}∈
(a,b)
, the above inequality does not hold, then since
ϕ is 1 − 1,
(ϕ (y2)− ϕ (x2))(y2 − x2) < 0.
Let x_{t}≡ tx_{1} +
(1− t)
x_{2} and y_{t}≡ ty_{1} +
(1− t)
y_{2}. It follows that x_{t}< y_{t} for all t ∈
[0,1]
. Now
define
h (t) ≡ (ϕ (yt)− ϕ (xt))(yt − xt).
Then h
(0)
< 0,h
(1)
> 0 but by assumption, h
(t)
≠0 for any t ∈
(0,1)
, a contradiction.
This property of being either strictly increasing or strictly decreasing on
(a,b)
carries over to
[a,b]
by
the continuity of ϕ.
It only remains to verify ϕ^{−1} is continuous. If not, there exists s_{n}→ s where s_{n} and s are points of
ϕ
([a,b])
but
| |
|ϕ−1(sn)− ϕ− 1(s)|
≥ ε. By sequential compactness of
[a,b]
, there is a subsequence, still
denoted by n, such that
||ϕ−1(sn)− t1||
→ 0. Thus s_{n}→ ϕ
(t1)
, so s = ϕ
(t1)
, and t_{1} = ϕ^{−1}
(s)
, a
contradiction. ■
Corollary 9.6.2Let f :
(a,b)
→ ℝ be one to one and continuous. Then f
(a,b)
is an open interval
(c,d)
and f^{−1} :
(c,d)
→
(a,b)
is continuous.
Proof: Since f is either strictly increasing or strictly decreasing, it follows that f
(a,b)
is an open
interval
(c,d)
. Assume f is decreasing. Now let x ∈
(a,b)
. Why is f^{−1} is continuous at f
(x)
? Since f is
decreasing, if f
(x)
< f
(y)
, then y ≡ f^{−1}
(f (y))
< x ≡ f^{−1}
(f (x))
and so f^{−1} is also decreasing. Let
ε > 0 be given. Let ε > η > 0 and
(x− η,x+ η)
⊆
(a,b)
. Then f
(x)
∈
(f (x +η) ,f (x− η))
.
Let
δ = min(f (x)− f (x + η),f (x− η) − f (x)).
Then if
|f (z)− f (x)|
< δ, it follows
z ≡ f−1(f (z)) ∈ (x− η,x+ η) ⊆ (x − ε,x+ ε)
which implies
|| −1 || || −1 − 1 ||
f (f (z))− x = f (f (z))− f (f (x))< ε.
This proves the theorem in the case where f is strictly decreasing. The case where f is increasing is similar.
■
Theorem 9.6.3Let f :
[a,b]
→ ℝ be continuous and one to one. Suppose f^{′}
(x1)
exists for somex_{1}∈
[a,b]
and f^{′}
(x1)
≠0. Then
( −1)
f
^{′}
(f (x1))
exists and is given by the formula
( − 1)
f
^{′}
(f (x1))
=
--1--
f′(x1)
.
Proof:By Lemma 9.6.1f is either strictly increasing or strictly decreasing and f^{−1} is continuous on
[a,b]
. Therefore there exists η > 0 such that if 0 <
|f (x1)− f (x)|
< η, then
| |
0 < |x1 − x| = |f−1(f (x1))− f−1 (f (x))| < δ
where δ is small enough that for 0 <
|x1 − x |
< δ,
| |
||--x-−-x1---− ---1--||< ε.
|f (x)− f (x1) f ′(x1)|
It follows that if 0 <
|f (x1)− f (x)|
< η,
| −1 −1 | | |
||f--(f-(x))−-f---(f-(x1))-− --1--|| = ||--x-−-x1---− ---1--||< ε
| f (x)− f (x1) f′(x1)| |f (x)− f (x1) f ′(x1)|