
# 4.8: Continuity on Compact Sets. Uniform Continuity


I. Some additional important theorems apply to functions that are continuous on a compact set (see §6).

Theorem $$\PageIndex{1}$$

If a function $$f : A \rightarrow\left(T, \rho^{\prime}\right), A \subseteq(S, \rho),$$ is relatively continuous on a compact set $$B \subseteq A,$$ then $$f[B]$$ is a compact set in $$\left(T, \rho^{\prime}\right) .$$ Briefly,

$\text{the continuous image of a compact set is compact.}$

Proof

To show that $$f[B]$$ is compact, we take any sequence $$\left\{y_{m}\right\} \subseteq f[B]$$ and prove that it clusters at some $$q \in f[B]$$.

As $$y_{m} \in f[B], y_{m}=f\left(x_{m}\right)$$ for some $$x_{m}$$ in $$B .$$ We pick such an $$x_{m} \in B$$ for each $$y_{m},$$ thus obtaining a sequence $$\left\{x_{m}\right\} \subseteq B$$ with

$f\left(x_{m}\right)=y_{m}, \quad m=1,2, \ldots$

Now by the assumed compactness of $$B,$$ the sequence $$\left\{x_{m}\right\}$$ must cluster at some $$p \in B .$$ Thus it has a subsequence $$x_{m_{k}} \rightarrow p .$$ As $$p \in B,$$ the function $$f$$ is relatively continuous at $$p$$ over $$B$$ (by assumption). Hence by the sequential criterion $$(§ 2), x_{m_{k}} \rightarrow p$$ implies $$f\left(x_{m_{k}}\right) \rightarrow f(p) ;$$ i.e.,

$y_{m_{k}} \rightarrow f(p) \in f[B].$

Thus $$q=f(p)$$ is the desired cluster point of $$\left\{y_{m}\right\} . \square$$

This theorem can be used to prove the compactness of various sets.

Example $$\PageIndex{1}$$

(1) A closed line segment $$L[\overline{a}, \overline{b}]$$ in $$E^{n}\left(^{*} \text { and in other normed spaces }\right)$$ is compact, for, by definition,

$L[\overline{a}, \overline{b}]=\{\overline{a}+t \vec{u} | 0 \leq t \leq 1\}, \text{ where } \vec{u}=\overline{b}-\overline{a}.$

Thus $$L[\overline{a}, \overline{b}]$$ is the image of the compact interval $$[0,1] \subseteq E^{1}$$ under the $$\operatorname{map} f : E^{1} \rightarrow E^{n},$$ given by $$f(t)=\overline{a}+t \vec{u},$$ which is continuous by Theorem 3 of §3. (Why?)

(2) The closed solid ellipsoid in $$E^{3},$$

$\left\{(x, y, z) | \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \leq 1\right\},$

is compact, being the image of a compact globe under a suitable continuous map. The details are left to the reader as an exercise.

lemma $$\PageIndex{1}$$

Every nonvoid compact set $$F \subseteq E^{1}$$ has a maximum and a minimum.

Proof

By Theorems 2 and 3 of §6, $$F$$ is closed and bounded. Thus $$F$$ has an infimum and a supremum in $$E^{1}$$ (by the completeness axiom), say, $$p=\inf F$$ and $$q=\sup F .$$ It remains to show that $$p, q \in F .$$

Assume the opposite, say, $$q \notin F .$$ Then by properties of suprema, each globe $$G_{q}(\delta)=(q-\delta, q+\delta)$$ contains some $$x \in B$$ (specifically, $$q-\delta<x<q )$$ other than $$q(\text { for } q \notin B, \text { while } x \in B) .$$ Thus

$(\forall \delta>0) \quad F \cap G_{\neg q}(\delta) \neq \emptyset;$

i.e., $$F$$ clusters at $$q$$ and hence must contain $$q$$ (being closed). However, since $$q \notin F,$$ this is the desired contradiction, and the lemma is proved. $$\square$$

The next theorem has many important applications in analysis.

Theorem $$\PageIndex{2}$$

(Weierstrass).

(i) If a function $$f : A \rightarrow\left(T, \rho^{\prime}\right)$$ is relatively continuous on a compact set $$B \subseteq A,$$ then $$f$$ is bounded on $$B ;$$ i.e., $$f[B]$$ is bounded.

(ii) If, in addition, $$B \neq \emptyset$$ and $$f$$ is real $$\left(f : A \rightarrow E^{1}\right),$$ then $$f[B]$$ has a maximum and a minimum; i.e., f attains a largest and a least value at some points of $$B$$.

Proof

Indeed, by Theorem $$1, f[B]$$ is compact, so it is bounded, as claimed in $$(i)$$.

If further $$B \neq \emptyset$$ and $$f$$ is real, then $$f[B]$$ is a nonvoid compact set in $$E^{1},$$ so by Lemma $$1,$$ it has a maximum and a minimum in $$E^{1} .$$ Thus all is proved. $$\square$$

Note 1. This and the other theorems of this section hold, in particular, if $$B$$ is a closed interval in $$E^{n}$$ or a closed globe in $$E^{n}\left(^{*} \text { or } C^{n}\right)$$ (because these sets are compact - see the examples in §6). This may fail, however, if $$B$$ is not compact, e.g., if $$B=(\overline{a}, \overline{b}) .$$ For a counterexample, see Problem 11 in Chapter 3, §13.

Theorem $$\PageIndex{3}$$

If a function $$f : A \rightarrow\left(T, \rho^{\prime}\right), A \subseteq(S, \rho)$$, is relatively continuous on a compact set $$B \subseteq A$$ and is one to one on $$B$$ (i.e., when restricted to $$B$$), then its inverse, $$f^{-1}$$, is continuout on $$f[B]$$.

Proof

To show that $$f^{-1}$$ is continuous at each point $$q \in f[B]$$, we apply the sequential criterion (Theorem 1 in §2). Thus we fix a sequence $$\left\{y_{m}\right\} \subseteq f[B], y_{m} \rightarrow q \in f[B]$$, and prove that $$f^{-1}\left(y_{m}\right) \rightarrow f^{-1}(q)$$.

Let $$f^{-1}\left(y_{m}\right)=x_{m}$$ and $$f^{-1}(q)=p$$ so that

$y_{m}=f\left(x_{m}\right), q=f(p), \text { and } x_{m}, p \in B.$

We have to show that $$x_{m} \rightarrow p$$, i.e., that

$(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad \rho\left(x_{m}, p\right)<\varepsilon.$

Seeking a contradiction, suppose this fails, i.e., its negation holds. Then (see Chapter 1, §§1–3) there is an $$\epsilon > 0$$ such that

$(\forall k)\left(\exists m_{k}>k\right) \quad \rho\left(x_{m_{k}}, p\right) \geq \varepsilon,$

where we write “$$m_{k}$$” for “$$m$$” to stress that the $$m_{k}$$ may be different for different $$k$$. Thus by (1), we fix some $$m_{k}$$ for each $$k$$ so that (1) holds, choosing step by step,

$m_{k+1}>m_{k}, \quad k=1,2, \ldots$

Then the $$x_{m_{k}}$$ form a subsequence of $$\{x_{m}\}$$, and the corresponding $$y_{m_{k}}=f(x_{m_{k}})$$ form a subsequence of $$\left\{y_{m}\right\}$$. Henceforth, for brevity, let $$\left\{x_{m}\right\}$$ and $$\left\{y_{m}\right\}$$ themselves denote these two subsequences. Then as before, $$x_{m} \in B, y_{m}=f\left(x_{m}\right) \in f[B]$$, and $$y_{m} \rightarrow q, q=f(p)$$. Also,by(1),

$(\forall m) \quad \rho\left(x_{m}, p\right) \geq \varepsilon\left(x_{m} \text { stands for } x_{m_{k}}\right).$

Now as $$\left\{x_{m}\right\} \subseteq B$$ and $$B$$ is compact, $$\left\{x_{m}\right\}$$ has a (sub)subsequence

$x_{m_{i}} \rightarrow p^{\prime} \text { for some } p^{\prime} \in B.$

As $$f$$ is relatively continuous on $$B$$, this implies

$f\left(x_{m_{i}}\right)=y_{m_{i}}\rightarrow f\left(p^{\prime}\right)$

However, the subsequence $$\left\{y_{m_{i}}\right\}$$ must have the same limit as $$\left\{y_{m}\right\}$$, i.e., $$f(p)$$. Thus $$f\left(p^{\prime}\right)=f(p)$$ whence $$p=p^{\prime}$$ (for $$f$$ is one to one on $$B$$), so $$x_{m_{i}} \rightarrow p^{\prime}=p$$.

This contradicts (2), however, and thus the proof is complete. $$\square$$

Example $$\PageIndex{2}$$

(3) For a fixed $$n \in N,$$ define $$f :[0,+\infty) \rightarrow E^{1}$$ by

$f(x)=x^{n}.$

Then $$f$$ is one to one (strictly increasing) and continuous (being a monomial; see §3). Thus by Theorem 3, $$f^{−1}$$ (the nth root function) is relatively continuous on each interval

$f=[a^{n}, b^{n}].$

hence on $$[0,+\infty).$$

II. Uniform Continuity. If $$f$$ is relatively continuous on $$B$$, then by definition,

$(\forall \varepsilon>0)(\forall p \in B)(\exists \delta>0)\left(\forall x \in B \cap G_{p}(\delta)\right) \quad \rho^{\prime}(f(x), f(p))<\varepsilon.$

Here, in general, $$\delta$$ depends on both $$\epsilon$$ and $$p$$ (see Problem 4 in §1); that is, given $$\epsilon > 0$$, some values of $$\delta$$ may fit a given p but fail (3) for other points.

It may occur, however, that one and the same $$\delta$$ (depending on $$\epsilon$$ only) satisfies (3) for all $$p \in B$$ simultaneously, so that we have the stronger formula

$(\forall \varepsilon>0)(\exists \delta>0)(\forall p, x \in B | \rho(x, p)<\delta) \quad \rho^{\prime}(f(x), f(p))<\varepsilon.$

Definition

If (4) is true, we say that $$f$$ is uniformly continuous on $$B$$.

Clearly, this implies (3), but the converse fails.

Theorem $$\PageIndex{4}$$

If a function $$f : A \rightarrow\left(T, \rho^{\prime}\right), A \subseteq(S, \rho)$$, is relatively continuous on a compact set $$B \subset A$$, then $$f$$ is also uniformly continuous on $$B$$.

Proof

(by contradiction). Suppose $$f$$ is relatively continuous on $$B$$, but (4) fails. Then there is an $$\epsilon > 0$$ such that

$(\forall \delta>0)(\exists p, x \in B) \quad \rho(x, p)<\delta, \text { and } \text { yet } \rho^{\prime}(f(x), f(p)) \geq \varepsilon;$

here $$p$$ and $$x$$ on $$\delta$$. We fix such an $$\epsilon$$ and let
$\delta=1, \frac{1}{2}, \ldots, \frac{1}{m}, \dots$
Then for each $$\delta$$ (i.e., each $$m$$), we get two points $$x_{m}, p_{m} \in B$$ with
$\rho\left(x_{m}, p_{m}\right)<\frac{1}{m}$
and
$\rho^{\prime}\left(f\left(x_{m}\right), f\left(p_{m}\right)\right) \geq \varepsilon, \quad m=1,2, \ldots$
Thus we obtain two sequences, $$\left\{x_{m}\right\}$$ and $$\left\{p_{m}\right\}$$, in $$B$$. As $$B$$ is compact, $$\left\{x_{m}\right\}$$ has a subsequence $$x_{m_{k}} \rightarrow q(q \in B)$$. For simplicity, let it be $$\left\{x_{m}\right\}$$ itself; thus
$x_{m} \rightarrow q, \quad q \in B.$

Hence by (5), it easily follows that also $$p_{m} \rightarrow q$$ (because $$\rho\left(x_{m}, p_{m}\right) \rightarrow 0$$. By the assumed relative continuity of $$f$$ on $$B$$, it follows that

$f\left(x_{m}\right) \rightarrow f(q) \text { and } f\left(p_{m}\right) \rightarrow f(q) \text { in }\left(T, \rho^{\prime}\right).$

This, in turn, implies that $$\rho^{\prime}\left(f\left(x_{m}\right), f\left(p_{m}\right)\right) \rightarrow 0$$, which is impossible, in view of (6). This contradiction completes the proof. $$\square$$

Example $$\PageIndex{1}$$

(a) A function $$f : A \rightarrow \left( T , \rho ^ { \prime } \right) , A \subseteq ( S , \rho )$$, ic called a contraction map (on $$A$$) iff

$\rho ( x , y ) \geq \rho ^ { \prime } ( f ( x ) , f ( y ) ) \text { for all } x , y \in A.$

Any such map is uniformly continuous on A. In fact, given $$\varepsilon > 0$$, we simply take $$\delta = \varepsilon$$. Then $$\forall x , p \in A$$

$\rho ( x , p ) < \delta \text { implies } \rho ^ { \prime } ( f ( x ) , f ( p ) ) \leq \rho ( x , p ) < \delta = \varepsilon,$

as required in (3).

(b) As a special case, consider the absolute value map (norm map) given by

$f ( \overline { x } ) = | \overline { x } | \text { on } E ^ { n } \left( ^ { * } \text { or another normed space } \right).$

It is uniformly continuous on$$E^{n}$$ because

$| | \overline { x } | - | \overline { p } | | \leq | \overline { x } - \overline { p } | , \text { i.e., } \rho ^ { \prime } ( f ( \overline { x } ) , f ( \overline { p } ) ) \leq \rho ( \overline { x } , \overline { p } ),$

which shows that $$f$$ is a contraction map, so Example (a) applies.

(c) Other examples of contraction maps are

(1) constant maps (see §1, Example (a)) and

(2) projection maps (see the proof of Theorem 3 in §3).

Verify!

(d) Define $$f : E ^ { 1 } \rightarrow E ^ { 1 }$$ by

$f ( x ) = \sin x$

By elementary trigonometry, $$| \sin x | \leq | x |$$. Thus $$\left( \forall x , p \in E ^ { 1 } \right)$$

\begin{aligned} | f ( x ) - f ( p ) | & = | \sin x - \sin p | \\ & = 2 \left| \sin \frac { 1 } { 2 } ( x - p ) \cdot \cos \frac { 1 } { 2 } ( x + p ) \right| \\ & \leq 2 \left| \sin \frac { 1 } { 2 } ( x - p ) \right| \\ & \leq 2 \cdot \frac { 1 } { 2 } | x - p | = | x - p | \end{aligned},

and $$f$$ is a contraction map again. Hence the sine function is uniformly continuous on $$E^{1}$$; similarly for the cosine function.

(e) Given $$\emptyset \neq A \subseteq ( S , \rho ) ,$$ define $$f : S \rightarrow E ^ { 1 }$$ by

$f ( x ) = \rho ( x , A ) \text { where } \rho ( x , A ) = \inf _ { y \in A } \rho ( x , y )$

It is easy to show that

$( \forall x , p \in S ) \quad \rho ( x , A ) \leq \rho ( x , p ) + \rho ( p , A )$

i.e.,

$f ( x ) \leq \rho ( p , x ) + f ( p ) , \text { or } f ( x ) - f ( p ) \leq \rho ( p , x )$

Similarly, $$f ( p ) - f ( x ) \leq \rho ( p , x ) .$$ Thus

$| f ( x ) - f ( p ) | \leq \rho ( p , x )$

i.e., $$f$$ is uniformly continuous (being a contraction map).

(f) The identity map $$f : ( S , \rho ) \rightarrow ( S , \rho ) ,$$ given by

$f ( x ) = x$

is uniformly continuous on $$S$$ since

$\rho ( f ( x ) , f ( p ) ) = \rho ( x , p ) \text { (a contraction map!) }$

However, even relative continuity could fail if the metric in the domain space $$S$$ were not the same as in $$S$$ when regarded as the range space (e.g., make $$\rho ^ { \prime }$$ discrete!)

(g) Define $$f : E ^ { 1 } \rightarrow E ^ { 1 }$$ by

$f ( x ) = a + b x \quad ( b \neq 0 ).$

Then

$\left( \forall x , p \in E ^ { 1 } \right) \quad | f ( x ) - f ( p ) | = | b | | x - p |;$

i.e.,

$\rho ( f ( x ) , f ( p ) ) = | b | \rho ( x , p ).$

Thus, given $$\varepsilon > 0 ,$$ take $$\delta = \varepsilon / | b | .$$ Then

$\rho ( x , p ) < \delta \Longrightarrow \rho ( f ( x ) , f ( p ) ) = | b | \rho ( x , p ) < | b | \delta = \varepsilon,$

proving uniform continuity.

(h) Let

$f ( x ) = \frac { 1 } { x } \quad \text { on } B = ( 0 , + \infty ).$

Then $$f$$ is continuous on $$B ,$$ but not uniformly so. Indeed, we can prove the negation of $$( 4 ) ,$$ i.e.

$( \exists \varepsilon > 0 ) ( \forall \delta > 0 ) ( \exists x , p \in B ) \quad \rho ( x , p ) < \delta \text { and } \rho ^ { \prime } ( f ( x ) , f ( p ) ) \geq \varepsilon.$

Take $$\varepsilon = 1$$ and any$$\delta > 0 .$$ We look for $$x , p$$ such that

$| x - p | < \delta \text { and } | f ( x ) - f ( p ) | \geq \varepsilon,$

i.e.,

$\left| \frac { 1 } { x } - \frac { 1 } { p } \right| \geq 1,$

This is achieved by taking

$p = \min \left( \delta , \frac { 1 } { 2 } \right) , x = \frac { p } { 2 } . \quad ( \text { Verify! } )$

Thus $$( 4 )$$ fails on $$B = ( 0 , + \infty ) ,$$ yet it holds on $$[ a , + \infty )$$ for any $$a > 0$$ .
(Verify!)