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
|g-(x-)|2
[|f (x )g(y)− f (y)g (y )+ f (y)g(y)− f (y)g(x)|]
≤
--2--2
|g (x )|
[|g (y)||f (x)− f (y)|+ |f (y)||g(y)− g (x)|]
≤
--2---
|g (x )|2
[ ]
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 ≡ ---2-2 (1 + 2|f (x)|+2 |g (x)|)
|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
|g(y)− g(x)| < εM − 1.
2
Then if 0 < δ ≤ min
(δ0,δ1,δ2,δ3)
, 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 it is or it is not
continuous.
Now begin on (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. It follows from continuity of f at x that 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).
Part (4) says: If f =
(f1,⋅⋅⋅,fq)
: D
(f)
→ ℝ^{q}, then f is continuous if and only if each f_{k} is a continuous
real valued function. Then
( q )1∕2
|f (x)− f (y)| ≤ |f (x )− f (y)| ≡ ∑ |f (x )− f(y)|2
k k i=1 i i
∑q
≤ |fi(x)− fi(y)|. (8.9)
i=1
(8.9)
Suppose first that f is continuous at x. Then there exists δ > 0 such that if
|x − y|
< δ, then
|f (x)− f (y)|
< ε. The first part of the above inequality then shows that for each k = 1,
⋅⋅⋅
,q,
|fk(x)− fk(y)|
< ε. This shows the only if part. Now suppose each function f_{k} is continuous. Then if
ε > 0 is given, there exists δ_{k}> 0 such that whenever
|x − y|
< δ_{k}
|fk(x)− fk(y)| < ε∕q.
Now let 0 < δ ≤ min
(δ1,⋅⋅⋅,δq)
. For
|x− y|
< δ, the above inequality holds for all k and so the last part
of (8.9) implies