Uniform Continuity

Uniform continuity strengthens ordinary continuity by requiring the same $\delta$ to work for all points simultaneously. While continuity is a local property (checked at each point), uniform continuity is a global property. On compact sets, continuity implies uniform continuity — a fundamental result with applications to integration, approximation, and differential equations.

Definition

Definition4.1Uniform continuity

A function $f : D \to \mathbb{R}$ is uniformly continuous on $D$ if for every $\epsilon > 0$ , there exists $\delta > 0$ such that for all $x, y \in D$ ,

$|x - y| < \delta \implies |f(x) - f(y)| < \epsilon.$

RemarkDifference from continuity

For ordinary continuity, $\delta$ may depend on both $\epsilon$ and the point $c$ . For uniform continuity, $\delta$ depends only on $\epsilon$ and works uniformly for all pairs of points.

ExampleLinear functions are uniformly continuous

Let $f(x) = mx + b$ on $\mathbb{R}$ . Given $\epsilon > 0$ , let $\delta = \epsilon / |m|$ (if $m \neq 0$ ; if $m = 0$ , any $\delta$ works). Then for $|x - y| < \delta$ ,

$|f(x) - f(y)| = |m(x - y)| < |m| \delta = \epsilon.$

The same $\delta$ works for all $x, y \in \mathbb{R}$ .

ExampleLipschitz functions are uniformly continuous

A function $f$ is Lipschitz if there exists $L > 0$ such that $|f(x) - f(y)| \leq L|x - y|$ for all $x, y$ . Every Lipschitz function is uniformly continuous (take $\delta = \epsilon/L$ ).

For example, $f(x) = \sin(x)$ is Lipschitz with $L = 1$ (since $|\sin(x) - \sin(y)| \leq |x - y|$ by the Mean Value Theorem), hence uniformly continuous on $\mathbb{R}$ .

Examplef(x) = x² is not uniformly continuous on R

Let $f(x) = x^2$ on $\mathbb{R}$ . Then $f$ is continuous everywhere but not uniformly continuous on $\mathbb{R}$ .

Proof: Take $\epsilon = 1$ . For any $\delta > 0$ , choose $n$ large enough that $1/\delta < n$ . Let $x = n$ and $y = n + \delta/2$ . Then $|x - y| = \delta/2 < \delta$ , but

$|f(x) - f(y)| = |n^2 - (n + \delta/2)^2| = |n \delta + (\delta/2)^2| > n\delta > 1.$

Thus no $\delta$ works for $\epsilon = 1$ .

Examplef(x) = 1/x is not uniformly continuous on (0, 1)

Let $f(x) = 1/x$ on $(0, 1)$ . Then $f$ is continuous on $(0, 1)$ but not uniformly continuous.

Proof: Take $\epsilon = 1$ . For any $\delta > 0$ , choose $n$ such that $1/n < \delta$ . Let $x = 1/n$ and $y = 1/(n+1)$ . Then

$|x - y| = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} < \delta,$

but

$|f(x) - f(y)| = |n - (n+1)| = 1.$

Thus no $\delta$ works.

Uniform continuity on compact sets

Theorem4.1Continuous functions on compact sets are uniformly continuous

If $K$ is compact and $f : K \to \mathbb{R}$ is continuous, then $f$ is uniformly continuous on $K$ .

RemarkProof sketch

Suppose not. Then there exist $\epsilon > 0$ and sequences $(x_n), (y_n)$ in $K$ with $|x_n - y_n| \to 0$ but $|f(x_n) - f(y_n)| \geq \epsilon$ for all $n$ . By compactness, extract convergent subsequences $x_{n_k} \to c$ and $y_{n_k} \to c'$ . Since $|x_{n_k} - y_{n_k}| \to 0$ , we have $c = c'$ . By continuity, $f(x_{n_k}) \to f(c)$ and $f(y_{n_k}) \to f(c)$ , so $|f(x_{n_k}) - f(y_{n_k})| \to 0$ , contradicting $|f(x_{n_k}) - f(y_{n_k})| \geq \epsilon$ .

ExampleUniform continuity on [a, b]

Since $[a, b]$ is compact, every continuous $f : [a, b] \to \mathbb{R}$ is uniformly continuous. For instance, $f(x) = x^2$ is uniformly continuous on $[0, 10]$ (even though it's not uniformly continuous on $\mathbb{R}$ ).

Applications

ExampleUniform continuity and integration

Uniform continuity is essential for proving that continuous functions on $[a, b]$ are Riemann integrable. If $f$ is uniformly continuous, we can choose partitions fine enough that $|f(x) - f(y)| < \epsilon$ for all $x, y$ in each subinterval, ensuring Riemann sums converge.

ExampleExtension of uniformly continuous functions

If $f : (a, b) \to \mathbb{R}$ is uniformly continuous, then $f$ has a unique continuous extension to $[a, b]$ . The limits $\lim_{x \to a^+} f(x)$ and $\lim_{x \to b^-} f(x)$ exist. This fails for non-uniformly continuous functions (e.g., $f(x) = \sin(1/x)$ on $(0, 1)$ ).

Summary

Uniform continuity is a global strengthening of continuity:

$\delta$ depends only on $\epsilon$ , not on the point.
Every continuous function on a compact set is uniformly continuous.
Lipschitz functions are uniformly continuous.
Applications: Riemann integration, function extension, differential equations.

See Heine-Cantor Theorem and Riemann Integration for applications.