2.2.4 The Extreme Value Theorem and Uniform Continuity
Here is a proof of the extreme value theorem.
Theorem 2.2.25Let C ⊆ ℝ^{p}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 such that for allx ∈ C,
has a convergent subsequence x_{nk} such that
lim_{k→∞}x_{nk} = x ∈ C. Then by continuity and Theorem 2.2.24,
M = lk→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 2.2.26A function f : D
(f)
→ ℝ^{q}is uniformly continuous if for every ε > 0 there existsδ > 0 such that wheneverx,yare points of D
(f)
such that
|x− y|
< δ, it follows
|f (x) − f (y )|
< ε.
Theorem 2.2.27Let f : K → ℝ^{q}be continuous at every point of K where K is a compact set inℝ^{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
1
|xj − yj| < j
but
|f (xj)− f (yj)|
≥ ε. Then Theorem 2.2.10 which says K is sequentially compact, there is a
subsequence
{xn }
k
of
{xj}
which converges to a point x ∈ K. Then since
|xn − yn |
k k
<
1
k
, it follows that
{yn }
k
also converges to x. Therefore,
ε ≤ lim |f (xnk)− f (ynk)| = |f (x) − f (x )| = 0,
k→ ∞
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,|⋅|)
. Also, the above theorems apply for ℝ^{p} and so they also apply for ℂ^{p} because it is
the same as ℝ^{2p}.
Proposition 2.2.28Let p
(z)
= a_{0}+a_{1}z +
⋅⋅⋅
+a_{n−1}z^{n−1}+z^{n}be a nonconstant polynomial whereeach 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(pz+(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 2.2.29Let K be a sequentially compact set and let f : K → f
(K )
be continuous. Thenf
(K )
is sequentially 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. ■
Note that this theorem avoids context completely. It says that the continuous image of a sequentially
compact set will be sequentially compact. As shown in Problem 11, in a normed linear space, you can write
either sequentially compact or compact in the above.