8.10.5 The Extreme Value Theorem And Uniform Continuity

Definition 8.10.16A function f having values in ℝ^{p}is said to be bounded if the set of values off is a bounded set.

Lemma 8.10.17Let C ⊆ ℝ^{p}be closed and bounded and let f : C → ℝ^{s}be continuous. Then f isbounded.

Proof:Suppose not. Then since f is not bounded, there exists x_{n} such that

∏s
f (xn) ∕∈ (− n,n ) ≡ Rn.
i=1

By Corollary 8.10.13, C is sequentially compact, and so there exists a subsequence

{xnk}

which converges
to x ∈ C. Now f

(x)

∈ R_{m} for large enough m. Hence, by continuity of f, it follows f

(xn)

∈ R_{m} for all n
large enough, contradicting the construction. ■

Here is a proof of the extreme value theorem.

Theorem 8.10.18Let 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 such that for allx ∈ C,

f (x1) ≤ f (x) ≤ f (x2 ).

Proof:Let M = sup

{f (x) : x ∈ C}

. Then by Lemma 8.10.17, M is a finite number. Is f

(x2)

= M for
some x_{2}? If not, you could consider the function

1
g (x ) ≡ M-− f-(x)

and g would be a continuous and unbounded function defined on C, contrary to Lemma 8.10.17. Therefore,
there exists x_{2}∈ C such that f

(x2)

= M. A similar argument applies to show the existence of x_{1}∈ C such
that

f (x ) = inf{f (x ) : x ∈ C }. ■
1

As in the case of a function of one variable, there is a concept of uniform continuity.

Definition 8.10.19A function f : D

(f)

→ ℝ^{q}is uniformly continuous if for every ε > 0 thereexists δ > 0 such that wheneverx,yare points of D

(f)

such that

|x− y |

< δ, it follows

|f (x)− f (y)|

< ε.

Theorem 8.10.20Let f : K → ℝ^{q}be continuous at every point of K where K is a closed andbounded 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 by Corollary 8.10.13 on Page 460 which says K is sequentially compact,
there is a subsequence

{xnk}

of

{xj}

which converges to a point x ∈ K. Then since

|xnk − ynk|

<

1k

, 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. ■