B.2.4 The Extreme Value Theorem and Uniform Continuity
Here is a proof of the extreme value theorem.
Theorem B.2.25Let C ⊆ ℝ^{p}be closed and bounded and let f : C → ℝ becontinuous. Then f achieves its maximum and its minimum on C. This means there existx_{1},x_{2}∈ C such that for all x ∈ C,
has a convergent
subsequence x_{nk} such that lim_{k→∞}x_{nk} = x ∈ C. Then by continuity and Theorem
B.2.24,
M = kl→im∞ f (xnk) = f (x) .
The case of the minimum value is similar. Just replace sup with inf .■
As in the case of a function of one variable, there is a concept of uniform continuity.
Definition B.2.26A function f : D
(f)
→ ℝ^{q}is uniformly continuous if forevery ε > 0 there exists δ > 0 such that wheneverx,yare points of D
(f)
such that
|x − y|
< δ, it follows
|f (x)− f (y)|
< ε.
Theorem B.2.27Let f : K → ℝ^{q}be continuous at every point of K where Kis a closed and bounded set in ℝ^{p}. Then f is uniformly continuous.
Proof: Suppose not. Then there exists ε > 0 and sequences
{xj}
and
{yj}
of points in K
such that
|xj − yj| < 1
j
but
|f (xj) − f (yj)|
≥ ε. Then Theorem 4.7.2 which says K is sequentially compact, there is a
subsequence
{xnk}
of
{xj}
which converges to a point x ∈ K. Then since
|xnk − ynk|
<
1
k
, it
follows that
{ynk}
also converges to x. Therefore,
ε ≤ kli→m∞|f (xnk)− f (ynk)| = |f (x)− f (x)| = 0,
a contradiction. Therefore, f is uniformly continuous as claimed. ■
Later in the book, I will consider the fundamental theorem of algebra. However, here
is a fairly short proof based on the extreme value theorem. You may have to fill
in a few details however. In particular, note that
(ℂ,|⋅|)
is the same as
(ℝ2,|⋅|)
where
|⋅|
is the standard norm on ℝ^{2}. Thus closed and bounded sets are compact in
(ℂ, |⋅|)
.
Proposition B.2.28Let p
(z)
= a_{0} + a_{1}z +
⋅⋅⋅
+ a_{n−1}z^{n−1} + z^{n}be a nonconstantpolynomial where each a_{i}∈ ℂ. Then there is a root to this polynomial.
Proof: Suppose the nonconstant polynomial p
(z)
= a_{0} + a_{1}z +
⋅⋅⋅
+ z^{n}, has no zero in
ℂ. Since lim_{|z|
→∞}
|p (z)|
= ∞, there is a z_{0} with
|p(z0)| = min |p(z)| > 0
z∈ℂ
Why? (The growth condition shows that you can restrict attention to a closed and bounded
set and then apply the extreme value theorem.) Then let q
(z)
=
p(zp+(zz00))
. This is also a
polynomial which has no zeros and the minimum of
|q(z)|
is 1 and occurs at z = 0. Since
q
(0)
= 1, it follows q
(z)
= 1 + a_{k}z^{k} + r
(z)
where r
(z)
consists of higher order terms. Here a_{k}
is the first coefficient which is nonzero. Choose a sequence, z_{n}→ 0, such that a_{k}z_{n}^{k}< 0. For
example, let −a_{k}z_{n}^{k} =
since it involves higher
order terms. This is a contradiction. ■
Here is another very interesting theorem about continuity and compactness. It says that if
you have a continuous function defined on a compact set K then f
(K)
is also compact. If f is
one to one, then its inverse is also continuous.
Theorem B.2.29Let K be a compact set in F^{p}and let f : K → f
(K )
in F^{p}.Then f
(K )
is compact. If f is one to one, then f^{−1}is also continuous.
Proof:As explained above, compactness and sequential compactness are the same in this
setting. Suppose then that
{f (xk)}
_{k=1}^{∞} is a sequence in f
(K)
. Since K is compact, there is
a subsequence, still denoted as
{xk}
_{k=1}^{∞} such that x_{k}→ x ∈ K. Then by continuity,
f
(xk)
→ f
(x)
and so f
(K )
is compact as claimed.
Next suppose f is one to one. If you have f
(xk)
→ f
(x)
, does it follow that x_{k}→ x? If
not, then by compactness, there is a subsequence, still denoted as
{xk}
_{k=1}^{∞} such that
x_{k}→
ˆx
∈ K,x≠
ˆx
. Then by continuity, it also happens that f
(xk)
→ f
(ˆx)
and so f
(x)
= f
(ˆx )
which is a contradiction. Therefore, x_{k}→ x as desired, showing that f^{−1} is continuous.
■