You saw continuous functions in beginning calculus. It is no different for vector valued functions.
Definition 2.2.12A function f : D
(f)
⊆ ℝ^{p}→ ℝ^{q}is continuous at x ∈ D
(f)
if for each ε > 0 thereexists δ > 0 such that whenever y ∈ D
(f)
and
|y − x| < δ
it follows that
|f (x)− f (y)| < ε.
f is continuous if it is continuous at every point of D
(f)
.
By equivalence of norms, this is equivalent to the same statement with
∥⋅∥
_{∞} in place of
|⋅|
or any other
norm equivalent to
|⋅|
and it will be shown that this includes all of them.
Proposition 2.2.13If f : D
(f)
→ ℝ^{q}as above, then if it is continuous on D
(f)
, it follows thatfor any open set V in ℝ^{q},f^{−1}
(V )
is open in D
(f)
which means that ifx ∈ f^{−1}
(V)
, then there existsδ > 0 such that f
(B(x,δ)∩ D (f))
⊆ V .
Proof:It comes from the definition. f
(x)
∈ V and V is open. Hence, there is ε > 0 such that
B
(f (x),ε)
⊆ V. Then the existence of the δ in the above claim is nothing more than the definition of
continuity. ■
Note that the converse is also true. It is nothing more than specializing to let V be B
(f (x),ε)
.
If the function is defined on ℝ^{p} this just says the inverse image of open sets is open. Similarly the
inverse image of closed sets is closed. Indeed, if C is a closed set, and f continuous, then
−1 −1 ( C) −1 C
f (C) = f U = f (U)
for some open U which shows that f^{−1}
(C)
is closed, being the complement of an open set. Thus we have
the following corollary.
Corollary 2.2.14f : ℝ^{p}→ ℝ^{q}is continuous if and only if f^{−1}
(V)
is open in ℝ^{p}whenever V isopen in ℝ^{q}and f^{−1}
(C)
is closed whenever C is closed in ℝ^{q}.
Note that there is no change in any of the above if you generalize to replace ℝ^{p} and ℝ^{q} with a normed
vector space. If you don’t know what one of these is, don’t worry about it or see the problems at the end of
the chapter.
In particular,
{ } { }
x : dist(x,S ) > 1 is open, x : dist(x,S) ≥ 1 is closed
k k
and so forth. This follows from Proposition 2.2.13.
Now here are some basic properties of continuous functions.
Theorem 2.2.15The following assertions are valid.
The function af + bg is continuous at x when f, g are continuous at x ∈ D
(f)
∩ D
(g)
anda,b ∈ ℝ.
If and f and g are each real valued functions continuous at x, thenfg is continuous at x. If,in addition to this, g
(x )
≠0, then f∕g is continuous at x.
If f is continuous at x, f
(x)
∈ D
(g)
⊆ ℝ^{p}, and g is continuous at f
(x)
, then g∘f is continuousat x.
If f =
(f1,⋅⋅⋅,fq)
: D
(f)
→ ℝ^{q}, then f is continuous if and only if each f_{k}is a continuousreal valued function.
The function f : ℝ^{p}→ ℝ, given by f
(x)
=
|x|
is continuous.
Proof: Begin with (1). Let ε > 0 be given. By assumption, there exist δ_{1}> 0 such that whenever
|x − y|
< δ_{1}, it follows
|f (x)− f (y)|
<
----ε----
2(|a|+ |b|+1)
and there exists δ_{2}> 0 such that whenever
|x − y|
< δ_{2}, it follows that
|g(x)− g (y )|
<
---ε-----
2(|a|+|b|+1)
. Then let 0 < δ ≤ min
(δ1,δ2)
. If
|x − y|
< δ, then
everything happens at once. Therefore, using the triangle inequality
|af (x)+ bf (x)− (ag(y)+ bg (y))|
≤
|a|
|f (x) − f (y)|
+
|b|
|g(x)− g(y)|
<
|a|
( )
-----ε-------
2(|a|+ |b|+ 1)
+
|b|
( )
------ε------
2(|a|+ |b|+ 1)
< ε.
Now begin on (2). There exists δ_{1}> 0 such that if
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
2
M ≡ -----2 (1 + 2|f (x)|+2 |g (x)|)
|g(x)|
Now let δ_{2} be such that if
|x − y|
< δ_{2}, then
ε −1
|f (x)− f (y )| < 2M
and let δ_{3} be such that if
|x− y|
< δ_{3}, then
ε − 1
|g(y)− g(x)| < 2M .
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