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Section 4.5 Uniform Continuity

The extreme value theorem and the intermediate value theorem show that a continuous function on an interval has strong global consequences. There is another global strengthening of continuity that will be useful later: a single choice of \(\delta\) can control the oscillation of the function at every point of the domain at once. This is the idea of uniform continuity.

Subsection 4.5.1 Definition and First Examples

Let \(S \subseteq \R\) and let \(f \colon S \to \R\text{.}\) We say that \(f\) is uniformly continuous on \(S\) if for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that
\begin{equation*} |f(s)-f(t)| \lt \varepsilon \end{equation*}
whenever \(s,t \in S\) and \(|s-t| \lt \delta\text{.}\)
The crucial difference from ordinary continuity is that \(\delta\) depends only on \(\varepsilon\text{,}\) not on a base point \(c \in S\text{.}\) Thus one estimate works simultaneously for all points of the domain.

Proof.
Fix \(c \in S\) and let \(\varepsilon \gt 0\text{.}\) Choose \(\delta \gt 0\) from uniform continuity. If \(x \in S\) and \(|x-c| \lt \delta\text{,}\) then applying the definition with \(s=x\) and \(t=c\) gives \(|f(x)-f(c)| \lt \varepsilon\text{.}\) Therefore \(f\) is continuous at \(c\text{.}\)

Example 4.5.2.

The function \(f(x)=1/x\) on \((0,1)\) is continuous but not uniformly continuous. To see this, take \(\varepsilon = 1\text{.}\) Suppose there were some \(\delta \gt 0\) such that \(|1/s-1/t| \lt 1\) whenever \(s,t \in (0,1)\) and \(|s-t| \lt \delta\text{.}\) Choose \(n \in \N\) so large that \(1/(n(n+1)) \lt \delta\text{,}\) and let
\begin{equation*} s=\frac1n, \qquad t=\frac{1}{n+1}. \end{equation*}
Then \(s,t \in (0,1)\) and
\begin{equation*} |s-t|=\frac{1}{n(n+1)} \lt \delta, \end{equation*}
but
\begin{equation*} \left|\frac1s-\frac1t\right|=|n-(n+1)|=1, \end{equation*}
contradicting the definition of uniform continuity.

Subsection 4.5.2 Closed Bounded Intervals

The same hypotheses that guarantee the extreme value theorem also force uniform continuity.

Proof.
Let \(f \colon [a,b] \to \R\) be continuous. Suppose that \(f\) is not uniformly continuous. Then there exists \(\varepsilon_0 \gt 0\) such that for every \(n \in \N\) we can find \(x_n,x'_n \in [a,b]\) with
\begin{equation*} |x_n-x'_n| \lt \frac1n \qquad\text{but}\qquad |f(x_n)-f(x'_n)| \ge \varepsilon_0. \end{equation*}
The sequence \((x_n)\) lies in the bounded interval \([a,b]\text{,}\) so by the Bolzano-Weierstrass Theorem it has a convergent subsequence \((x_{n_k})\text{.}\) Let \(x_{n_k} \to c\text{.}\) Since \([a,b]\) contains all its cluster points Proposition 4.4.1, \(c \in [a,b]\text{.}\)
As a subsequence of the null sequence \((x_n - x_n')\text{,}\) \((x'_{n_k}-x_{n_k})\) is also null. Thus, \(x'_{n_k}\) converges to \(c\) as well. Since \(f\) is continuous at \(c\text{,}\) Proposition 4.3.1 gives
\begin{equation*} f(x_{n_k}) \to f(c) \qquad\text{and}\qquad f(x'_{n_k}) \to f(c). \end{equation*}
Therefore, \(f(x_{n_k})-f(x'_{n_k}) \to 0\) is null which contradicts \(|f(x_{n_k})-f(x'_{n_k})| \ge \varepsilon_0\) for every \(k\text{.}\) This contradiction shows that \(f\) is uniformly continuous.

Subsection 4.5.3 Cauchy Sequences and Extension to Endpoints

Uniform continuity is closely tied to the material from the chapter on sequences. Ordinary continuity carries convergent sequences to convergent sequences (Proposition 4.3.1). Uniform continuity does something stronger: it preserves the Cauchy property itself.

Proof.
Let \(\varepsilon \gt 0\text{.}\) Choose \(\delta \gt 0\) from the definition of uniform continuity. Since \((x_n)\) is Cauchy, there exists \(N \in \N\) such that \(|x_n-x_m| \lt \delta\) whenever \(m,n \ge N\text{.}\) Therefore, for \(m,n \ge N\text{,}\)
\begin{equation*} |f(x_n)-f(x_m)| \lt \varepsilon. \end{equation*}
Hence \((f(x_n))\) is Cauchy.

Proof.
First suppose that \(f\) extends to a continuous function \(\widetilde{f} \colon [a,b] \to \R\text{.}\) By Theorem 4.5.3, \(\widetilde{f}\) is uniformly continuous on \([a,b]\text{.}\) Its restriction to \((a,b)\) is therefore uniformly continuous as well, so \(f\) is uniformly continuous on \((a,b)\text{.}\)
Conversely, suppose that \(f\) is uniformly continuous on \((a,b)\text{.}\) Let \((x_n)\) be any sequence in \((a,b)\) with \(x_n \to a\text{.}\) By Proposition 2.4.3, \((x_n)\) is Cauchy, so Proposition 4.5.4 shows that \((f(x_n))\) is Cauchy. By Theorem 2.4.5, \((f(x_n))\) converges.
This limit is independent of the sequence approaching \(a\text{.}\) To see this, let \((x_n)\) and \((x'_n)\) be sequences in \((a,b)\) with \(x_n \to a\) and \(x'_n \to a\text{,}\) and define a new sequence \((z_n)\) by
\begin{equation*} z_{2n-1}=x_n, \qquad z_{2n}=x'_n. \end{equation*}
Then \(z_n \to a\text{,}\) so the argument above shows that \((f(z_n))\) converges. Since \((f(x_n))\) and \((f(x'_n))\) are subsequences of the convergent sequence \((f(z_n))\text{,}\) Proposition 2.1.10 implies that they converge to the same limit. Denote this common limit by \(L_a\text{.}\) In the same way, there is a uniquely determined real number \(L_b\) such that \(f(x_n) \to L_b\) whenever \(x_n \to b\) with \(x_n \in (a,b)\text{.}\)
Define \(\widetilde{f} \colon [a,b] \to \R\) by
\begin{equation*} \widetilde{f}(x)= \begin{cases} L_a & \text{if } x=a,\\ f(x) & \text{if } a \lt x \lt b,\\ L_b & \text{if } x=b. \end{cases} \end{equation*}
On \((a,b)\text{,}\) the function \(\widetilde{f}\) agrees with \(f\text{,}\) so it is continuous there by Proposition 4.5.1.
It remains to prove continuity at the endpoints. We do this at \(a\text{;}\) the proof at \(b\) is identical. Let \((u_n)\) be a sequence in \([a,b]\) with \(u_n \to a\text{.}\) If all but finitely many terms satisfy \(u_n=a\text{,}\) then clearly \(\widetilde{f}(u_n) \to L_a=\widetilde{f}(a)\text{.}\) Otherwise, the subsequence consisting of those terms with \(u_n \gt a\) still converges to \(a\text{.}\) By the definition of \(L_a\text{,}\) the image of that subsequence under \(\widetilde{f}\) converges to \(L_a\text{,}\) while every term with \(u_n=a\) is mapped exactly to \(L_a\text{.}\) Hence the whole sequence \((\widetilde{f}(u_n))\) converges to \(L_a\text{.}\) By the sequential characterization of continuity, \(\widetilde{f}\) is continuous at \(a\text{.}\)

Subsection 4.5.4 Lipschitz Continuity and Gluing Intervals

A useful sufficient condition for uniform continuity is a linear bound on how fast the function can change.
Let \(S \subseteq \R\text{.}\) A function \(f \colon S \to \R\) is Lipschitz continuous on \(S\) if there exists \(K \gt 0\) such that
\begin{equation*} |f(x)-f(x')| \le K|x-x'| \end{equation*}
for all \(x,x' \in S\text{.}\)

Proof.
Suppose \(|f(x)-f(x')| \le K|x-x'|\) for all \(x,x' \in S\text{.}\) Given \(\varepsilon \gt 0\text{,}\) choose \(\delta = \varepsilon/K\text{.}\) Then whenever \(x,x' \in S\) satisfy \(|x-x'| \lt \delta\text{,}\) we have
\begin{equation*} |f(x)-f(x')| \le K|x-x'| \lt K\delta = \varepsilon. \end{equation*}
Thus \(f\) is uniformly continuous.

Example 4.5.7.

The function \(f(x)=\sqrt{x}\) is Lipschitz continuous on \([1,\infty)\text{.}\) Indeed, for \(x,x' \ge 1\text{,}\)
\begin{equation*} |x-x'| = |\sqrt{x}-\sqrt{x'}|(\sqrt{x}+\sqrt{x'}) \ge 2|\sqrt{x}-\sqrt{x'}|, \end{equation*}
so
\begin{equation*} |\sqrt{x}-\sqrt{x'}| \le \frac12 |x-x'|. \end{equation*}
Therefore \(\sqrt{x}\) is uniformly continuous on \([1,\infty)\text{.}\)

Proof.
Let \(\varepsilon \gt 0\text{.}\) Since \(f\) is uniformly continuous on each interval, there exist \(\delta_1,\delta_2 \gt 0\) such that
\begin{equation*} |f(x)-f(x')| \lt \frac{\varepsilon}{2} \end{equation*}
whenever \(x,x' \in I_k\) and \(|x-x'| \lt \delta_k\) (\(k=1,2\)). Let \(\delta=\min\{\delta_1,\delta_2\}\text{.}\)
Take \(x,x' \in I_1 \cup I_2\) with \(|x-x'| \lt \delta\text{.}\) If both points lie in the same interval, then \(|f(x)-f(x')| \lt \varepsilon/2 \lt \varepsilon\text{.}\) So assume that \(x \in I_1 \setminus I_2\) and \(x' \in I_2 \setminus I_1\text{.}\) Choose \(c \in I_1 \cap I_2\text{.}\)
The point \(c\) must lie between \(x\) and \(x'\text{.}\) For if, say, \(x\) lay between \(c\) and \(x'\text{,}\) then since \(c,x' \in I_2\) and \(I_2\) is an interval, we would also have \(x \in I_2\text{,}\) a contradiction. Thus
\begin{equation*} |x-c| \le |x-x'| \qquad\text{and}\qquad |x'-c| \le |x-x'|, \end{equation*}
so both are strictly less than \(\delta\text{.}\)
Since \(x,c \in I_1\) and \(c,x' \in I_2\text{,}\) we obtain
\begin{equation*} |f(x)-f(c)| \lt \frac{\varepsilon}{2} \qquad\text{and}\qquad |f(c)-f(x')| \lt \frac{\varepsilon}{2}. \end{equation*}
The triangle inequality then gives
\begin{equation*} |f(x)-f(x')| \le |f(x)-f(c)| + |f(c)-f(x')| \lt \varepsilon. \end{equation*}
Therefore \(f\) is uniformly continuous on \(I_1 \cup I_2\text{.}\)

Example 4.5.9.

The function \(f(x)=\sqrt{x}\) is uniformly continuous on \([0,\infty)\text{.}\) By Theorem 4.5.3, the function is uniformly continuous on \([0,1]\text{.}\) By Example 4.5.7, it is uniformly continuous on \([1,\infty)\text{.}\) Since these intervals overlap, Proposition 4.5.8 implies that \(\sqrt{x}\) is uniformly continuous on their union \([0,\infty)\text{.}\)
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