This proves the first part of 2.) To obtain the second part, let δ_{1} be as described
above and let δ_{0}> 0 be such that for
|x− y|
< δ_{0},
|g (x) − g(y)| < |g(x)|∕2
and so by the triangle inequality,
− |g(x)|∕2 ≤ |g(y)|− |g(x)| ≤ |g(x)|∕2
which implies
|g(y)|
≥
|g(x)|
∕2, and
|g(y)|
< 3
|g(x)|
∕2.
Then if
|x− y|
< min
(δ0,δ1)
,
| |
||f (x)− f-(y)||
|g(x) g(y)|
=
| |
||f-(x-)g(y)−-f (y)g(x)||
| g(x)g(y) |
≤
|f (x)g((y)−-f-(y))-g(x)|
|g(x)|2
2
=
2-|f-(x)g-(y)-−-f (y)g(x)|
|g (x)|2
≤
2
-----2
|g (x)|
[|f (x)g (y) − f (y)g(y)+ f (y)g(y)− f (y) g(x)|]
≤
--2---
|g (x)|2
[|g(y)||f (x )− f (y)|+ |f (y)||g (y)− g (x)|]
≤
2
-----2
|g (x)|
[3 ]
2 |g (x)||f (x)− f (y)|+ (1 + |f (x )|)|g (y) − g (x)|
≤
--2--2
|g (x)|
(1+ 2|f (x)|+ 2|g(x)|)
[|f (x)− f (y)|+ |g(y)− g(x)|]
≡ M
[|f (x)− f (y)|+ |g(y)− g(x)|]
where M is defined by
---2--
M ≡ |g(x)|2 (1+ 2|f (x)|+2 |g (x)|)
Now let δ_{2} be such that if
|x− y|
< δ_{2}, then
|f (x)− f (y)| < εM −1
2
and let δ_{3} be such that if
|x− y|
< δ_{3}, then
ε
|g(y)− g(x)| < 2M − 1.
Then if 0 < δ ≤ min
(δ ,δ ,δ,δ )
0 1 2 3
, and
|x− y|
< δ, everything holds and
| |
||f-(x)− f-(y)||≤ M [|f (x)− f (y)|+ |g(y)− g(x)|]
|g (x ) g (y)|
[ ]
< M εM −1 + εM −1 = ε.
2 2
This completes the proof of the second part of 2.)
Note that in these proofs no effort is made to find some sort of “best” δ.
The problem is one which has a yes or a no answer. Either is it or it is not
continuous.
Now consider 3.). If f is continuous at x, f
(x)
∈ D
(g)
⊆ ℝ^{p}, and g is continuous at
f
(x )
,then g ∘f is continuous at x. Let ε > 0 be given. Then there exists η > 0 such that
if
|y− f (x)|
< η and y ∈ D
(g)
, it follows that
|g(y)− g(f (x))|
< ε. From
continuity of f at x, there exists δ > 0 such that if
|x− z|
< δ and z ∈ D
(f)
, then
|f (z)− f (x)|
< η. Then if
|x− z|
< δ and z ∈ D
(g∘ f)
⊆ D
(f)
, all the above hold and
so
|g (f (z))− g (f (x))| < ε.
This proves part 3.)
To verify part 4.), let ε > 0 be given and let δ = ε. Then if
|x− y|
< δ, the triangle
inequality implies
|f (x)− f (y)|
=
||x|− |y||
≤
|x− y|
< δ = ε.
This proves part 4.) and completes the proof of the theorem.
Next here is a proof of the intermediate value theorem.
Theorem 2.2.2Suppose f :
[a,b]
→ ℝ is continuous and suppose f
(a)
< c <f
(b)
. Then there exists x ∈
(a,b)
such that f
(x )
= c.
Proof:Let d =
a+b
2
and consider the intervals
[a,d]
and
[d,b]
. If f
(d)
≥ c, then on
[a,d]
, the function is ≤ c at one end point and ≥ c at the other. On the other hand, if
f
(d)
≤ c, then on
[d,b]
f ≥ 0 at one end point and ≤ 0 at the other. Pick the interval on
which f has values which are at least as large as c and values no larger than c. Now
consider that interval, divide it in half as was done for the original interval and
argue that on one of these smaller intervals, the function has values at least
as large as c and values no larger than c. Continue in this way. Next apply
the nested interval lemma to get x in all these intervals. In the n^{th} interval,
let x_{n},y_{n} be elements of this interval such that f
(xn)
≤ c,f
(yn)
≥ c. Now
|xn − x|
≤
(b− a)
2^{−n} and
|yn − x|
≤
(b− a)
2^{−n} and so x_{n}→ x and y_{n}→ x.
Therefore,
f (x)− c = nl→im∞ (f (xn)− c) ≤ 0
while
f (x)− c = lim (f (yn)− c) ≥ 0.
n→ ∞
Consequently f
(x)
= c and this proves the theorem.
Lemma 2.2.3Let ϕ :
[a,b]
→ ℝ be a continuous function and suppose ϕ is 1 − 1
on
(a,b)
. Then ϕ is either strictly increasing or strictly decreasing on
[a,b]
.
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}. Then x_{t}< y_{t} for all t ∈
[0,1]
because
tx1 ≤ ty1 and (1 − t)x2 ≤ (1 − t)y2
with strict inequality holding for at least one of these inequalities since not both t and
(1 − t)
can equal zero. Now define
h (t) ≡ (ϕ (yt)− ϕ (xt))(yt − xt).
Since h is continuous and h
(0)
< 0, while h
(1)
> 0, there exists t ∈
(0,1)
such that
h
(t)
= 0. Therefore, both x_{t} and y_{t} are points of
(a,b)
and ϕ
(yt)
− ϕ
(xt)
= 0
contradicting the assumption that ϕ is one to one. It follows ϕ is either strictly increasing
or strictly decreasing on
(a,b)
.
This property of being either strictly increasing or strictly decreasing on
(a,b)
carries
over to
[a,b]
by the continuity of ϕ. Suppose ϕ is strictly increasing on
(a,b)
,
a similar argument holding for ϕ strictly decreasing on
(a,b)
. If x > a, then
pick y ∈
(a,x)
and from the above, ϕ
(y)
< ϕ
(x)
. Now by continuity of ϕ at
a,
ϕ(a) = xl→ima+ ϕ(z) ≤ ϕ (y) < ϕ(x).
Therefore, ϕ
(a)
< ϕ
(x)
whenever x ∈
(a,b)
. Similarly ϕ
(b)
> ϕ
(x)
for all x ∈
(a,b)
.
This proves the lemma.
Corollary 2.2.4Let f :
(a,b)
→ ℝ be one to one and continuous. Then f
(a,b)
isan 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