You saw continuous functions in beginning calculus. It is no different for vector valued
functions.
Definition B.2.12A function f : D
(f)
⊆ ℝ^{p}→ ℝ^{q}is continuous at x ∈ D
(f)
if foreach ε > 0 there exists 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
|⋅|
.
Proposition B.2.13If f : D
(f)
→ ℝ^{q}as above, then if it is continuous on D
(f)
,it follows that for any open set V in ℝ^{q},f^{−1}
(V)
is open in D
(f)
which means that ifx ∈ f^{−1}
(V)
, then there exists δ > 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 B.2.14f : ℝ^{p}→ ℝ^{q}is continuous if and only if f^{−1}
(V)
is open in ℝ^{p}whenever V is open 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.
In particular,
{ 1} { 1 }
x : dist(x,S ) > k is open, x : dist(x,S) ≥ k is closed
and so forth. This follows from Proposition B.2.13.
Now here are some basic properties of continuous functions.
Theorem B.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)
and a,b ∈ ℝ.
If and f and g are each real valued functions continuous at x, thenfg is continuousat 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 )
, theng ∘ f is continuous at x.
If f =
(f1,⋅⋅⋅,fq)
: D
(f)
→ ℝ^{q}, then f is continuous if and only if each f_{k}is acontinuous real 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 )|
<
2(|a|ε+|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
|y− x|
< δ_{1}, then
|f (x) − f (y)|
< 1. Therefore, for such y,
|f (y)| < 1 +|f (x)|.
It follows that for such y,
|fg(x)− f g(y)| ≤ |f (x)g (x)− g(x)f (y)|+ |g(x)f (y)− f (y )g(y)|
≤
|g(x)|
|f (x )− f (y)|
+
|f (y)|
|g (x )− g(y)|
≤
(1+ |g(x)|+|f (y)|)
[|g(x)− g (y )|+ |f (x)− f (y )|]
.
Now let ε > 0 be given. There exists δ_{2} such that if
||f (x) f (y)||
||----− ----|| ≤ M [|f (x)− f (y)|+ |g (y )− g(x)|]
g(x) g(y)
[ε −1 ε −1]
< M 2M + 2M = ε.
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 =
(f ,⋅⋅⋅,f )
1 q
: D
(f)
→ ℝ^{q}, then f is continuous if and only if each f_{k} is
a continuous real valued function. Then