Uniformly Continuous — Definition, Formula & Examples

Uniformly continuous means that for any desired closeness in outputs, you can find a single tolerance in inputs that works everywhere on the domain at once — not just at each individual point.

A function $f: D \to \mathbb{R}$ is uniformly continuous on $D$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $x, y \in D$ , whenever $|x - y| < \delta$ it follows that $|f(x) - f(y)| < \varepsilon$ . Crucially, $\delta$ depends only on $\varepsilon$ , not on the choice of $x$ or $y$ .

Key Formula

\forall\,\varepsilon > 0,\;\exists\,\delta > 0 \;\text{ such that }\; |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon \quad \forall\, x, y \in D

Where:

$\varepsilon$ = Maximum allowed difference between outputs
$\delta$ = Input tolerance that works uniformly across all of D
$D$ = Domain of the function f

How It Works

Ordinary (pointwise) continuity lets you pick a different

\delta

at every point in the domain. Uniform continuity strengthens this: one

\delta

must handle every point simultaneously. Functions that "speed up" without bound — like

f(x) = x^2

\mathbb{R}

— are continuous everywhere but not uniformly continuous, because near-equal inputs far from the origin can produce arbitrarily different outputs. A key theorem (Heine–Cantor) guarantees that any continuous function on a closed, bounded interval

[a, b]

is automatically uniformly continuous.

Worked Example

Problem: Prove that f(x) = 3x + 1 is uniformly continuous on all of ℝ.

Set up: Given ε > 0, we need a single δ > 0 so that |f(x) − f(y)| < ε whenever |x − y| < δ for all x, y ∈ ℝ.

|f(x) - f(y)| = |3x + 1 - (3y + 1)| = 3|x - y|

Choose δ: Set δ = ε/3. This choice depends only on ε, not on x or y.

\delta = \frac{\varepsilon}{3}

Verify: If |x − y| < δ = ε/3, then |f(x) − f(y)| = 3|x − y| < 3 · (ε/3) = ε.

3|x - y| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon

Answer: Since δ = ε/3 works for every pair x, y ∈ ℝ, the function f(x) = 3x + 1 is uniformly continuous on ℝ.

Why It Matters

Uniform continuity is essential in real analysis for proving that Riemann integrals exist on closed intervals and that sequences of functions converge nicely. It also underpins the extension of functions to boundary points and completions of metric spaces.

Common Mistakes

Mistake: Assuming every continuous function is uniformly continuous.

Correction: Continuity only guarantees a δ that can vary with the point. For example, f(x) = 1/x is continuous on (0, 1) but not uniformly continuous there. Uniform continuity is guaranteed on closed bounded intervals by the Heine–Cantor theorem, but not in general.