||f (x ) f (y )||
||g-(x-) − g-(y)-|| ≤ M [|f (x) − f (y)|+ |g(y)− g (x )|]
[ ]
< 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)
⊆ F^{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)
→ F^{q}, then f is continuous if and only if each
f_{k} is a continuous F valued function. Then
|fk(x)− fk (y)| ≤ |f (x)− f (y)|
( )
∑q 2 1∕2
≡ |fi(x)− fi(y)|
i=1
∑q
≤ |fi(x) − fi(y)|. (5.10.11)
i=1
(5.10.11)
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 5.10.11 implies
|f (x)− f (y)|
≤∑_{i=1}^{q}
|fi(x )− fi(y )|
<∑_{i=1}^{q}
ε
q
= ε.
This proves part 4.)
To verify part 5.), 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 5.) and completes the proof of the theorem.
Here is a multidimensional version of the nested interval lemma.
The following definition is similar to that given earlier. It defines what is meant by a
sequentially compact set in F^{p}.
Definition 5.10.2A set, K ⊆ F^{p}is sequentially compact if and only if whenever
{xn}
_{n=1}^{∞}is a sequence of points in K, there exists a point, x ∈ K and asubsequence,
{xnk}
_{k=1}^{∞}such that x_{nk}→x.
It turns out the sequentially compact sets in F^{p}are exactly those which are closed and
bounded. Only half of this result will be needed in this book and this is proved next.
First note that ℂ can be considered as ℝ^{2}. Therefore, ℂ^{p} may be considered as
ℝ^{2p}.
Theorem 5.10.3Let C ⊆ F^{p}be closed and bounded. Then C is sequentiallycompact.
Proof:Let
{an}
⊆ C. Then let a_{n} =
( n n)
a1,⋅⋅⋅,ap
. It follows the real and imaginary
parts of the terms of the sequence,
{ }
anj
_{n=1}^{∞} are each contained in some sufficiently
large closed bounded interval. By Theorem 2.0.3 on Page 43, there is a subsequence of
the sequence of real parts of
{ }
anj
_{n=1}^{∞} which converges. Also there is a further
subsequence of the imaginary parts of
{ }
anj
_{n=1}^{∞} which converges. Thus there is a
subsequence, n_{k} with the property that a_{j}^{nk} converges to a point, a_{
j}∈ F. Taking further
subsequences, one obtains the existence of a subsequence, still called n_{k} such that for
each r = 1,
⋅⋅⋅
,p, a_{r}^{nk} converges to a point, a_{
r}∈ F as k →∞. Therefore, letting
a ≡
(a1,⋅⋅⋅,ap)
,lim_{k→∞}a^{nk} = a. Since C is closed, it follows a ∈ C. This proves the
theorem.
Here is a proof of the extreme value theorem.
Theorem 5.10.4Let C be closed and bounded and let f : C → ℝ be continuous. Then fachieves its maximum and its minimum on C. This means there exist, x_{1},x_{2}∈ C suchthat for all x ∈ C,
f (x1) ≤ f (x) ≤ f (x2).
Proof:Let M = sup
{f (x) : x ∈ C}
. Recall this means +∞ if f is not bounded
above and it equals the least upper bound of these values of f if f is bounded above.
Then there exists a sequence,
{xn}
such that f
(xn)
→ M. Since C is sequentially
compact, there exists a subsequence, x_{nk}, and a point, x ∈ C such that x_{nk}→x. But
then since f is continuous at x, it follows from Theorem 5.7.10 on Page 306 that
f
(x )
= lim_{k→∞}f
(xn )
k
= M. This proves f achieves its maximum and also shows
its maximum is less than ∞. Let x_{2} = x. The case of a minimum is handled
similarly.
Recall that a function is uniformly continuous if the following definition
holds.
Definition 5.10.5Let f : D
(f)
→ F^{q}. Then f is uniformly continuous if forevery ε > 0 there exists δ > 0 such that whenever
|x − y|
< δ, it follows
|f (x)− f (y)|
< ε.
Theorem 5.10.6Letf :C → F^{q}be continuous where C is a closed and boundedset in F^{p}. Then f is uniformly continuous on C.
Proof:If this is not so, there exists ε > 0 and pairs of points, x_{n} and y_{n}
satisfying
|xn − yn|
< 1∕n but
|f (xn)− f (yn)|
≥ ε. Since C is sequentially
compact, there exists x ∈ C and a subsequence,
{xnk}
satisfying x_{nk}→ x. But
|xnk − ynk|
< 1∕k and so y_{nk}→ x also. Therefore, from Theorem 5.7.10 on Page
306,
ε ≤ lim |f (xnk) − f (ynk )| = |f (x)− f (x)| = 0,
k→∞