Uniformly continuous function
High school
A function $f : \reals \to \reals$ is uniformly continuous if for every $\varepsilon > 0$, there exists $\delta > 0$ such that
\[\abs{f(x) - f(y)} < \varepsilon \quad \text{for all } \abs{x - y} < \delta.\]The choice of $\delta$ depends only on $\varepsilon$, so the same $\delta$ works at every point.
Undergraduate (metric spaces)
Consider a function $f : X \to Y$ between metric spaces. The function $f$ is uniformly continuous if for every $\varepsilon > 0$, there exists $\delta > 0$ such that
\[d(f(x), f(y)) < \varepsilon \quad \text{for all } x, y \in X \text{ with } d(x, y) < \delta.\]The choice of $\delta$ depends only on $\varepsilon$, so the same $\delta$ works at every point.
Examples
Examples
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$x \mapsto \abs{x}$ on $[-1, 1]$
High school
$x \mapsto \abs{x}$ on $[-1, 1]$ is Hölder continuous (with $\alpha = 1$) and uniformly continuous, because it is Lipschitz continuous.
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$x \mapsto x^2$ on $[0, 1]$
High school
$x \mapsto x^2$ on $[0, 1]$ is uniformly continuous because every continuous function on a compact set is uniformly continuous.
- $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$ Undergraduate
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The Cantor function on $[0, 1]$
Undergraduate
The Cantor function is uniformly continuous because it is continuous on the compact interval $[0, 1]$.
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$x \mapsto \sqrt{x}$ on $[0, 1]$
Undergraduate
$x \mapsto \sqrt{x}$ is Hölder continuous with $\alpha = \frac{1}{2}$, since $\abs{\sqrt{x} - \sqrt{y}} \leq \abs{x - y}^{1/2}$, and so it is uniformly continuous.
Non-examples
Things that may be mistaken for examples.
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$f(x) = x^2$ on $\reals$
High school
$f(x) = x^2$ is not uniformly continuous on $\reals$, because its graph gets steeper and steeper, so no single closeness of inputs works everywhere.
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$x \mapsto \sin(1/x)$ on $(0, 1]$
Undergraduate
$x \mapsto \sin(1/x)$ on $(0, 1]$ is not uniformly continuous, because points that are arbitrarily close together have values that differ by $2$. For each natural number $n$, let \[x_n = \frac{1}{\frac{\pi}{2} + 2 \pi n} \quad \text{and} \quad y_n = \frac{1}{\frac{3\pi}{2} + 2 \pi n}.\] Then $\sin(1/x_n) = 1$ and $\sin(1/y_n) = -1$, so $\abs{\sin(1/x_n) - \sin(1/y_n)} = 2$ for every $n$, but $\abs{x_n - y_n} \to 0$ as $n \to \infty$. Therefore, for $\varepsilon = 1$, no $\delta > 0$ satisfies the definition of uniform continuity: some pair $x_n, y_n$ is closer than $\delta$, but the values of the pair differ by more than $\varepsilon$.
Properties
- Every uniformly continuous function is continuous. On a compact domain, every continuous function is uniformly continuous (the Heine–Cantor theorem).
- A uniformly continuous function maps Cauchy sequences to Cauchy sequences.
Distinctions involving this term
- Types of continuity: There are many ways to describe the continuity of a function. On a compact interval, each is a special case of the next, from continuously differentiable to continuous.