4.8: Continuity on Compact Sets. Uniform Continuity
This page is a draft and is under active development.
I. Some additional important theorems apply to functions that are continuous on a compact set (see §6).
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.
(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.
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.
(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.
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. \(\square\)
(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
\[(\forall \varepsilon>0)(\exists \delta>0)(\forall p, x \in B | \rho(x, p)<\delta) \quad \rho^{\prime}(f(x), f(p))<\varepsilon.\]
If (4) is true, we say that \(f\) is uniformly continuous on \(B\).
Clearly, this implies (3), but the converse fails.
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
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. \(\square\)
(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!)