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 $[0,+\infty).$

See also Example (a) in §6 and Problem 1 below.

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!)