Mathematics.

limits and continuity

Uniform Continuity

Real Analysis40 minDifficulty5 out of 10

You should know: epsilon delta

Overview

A function f is continuous at each point of its domain if, near every point a, a suitable δ can be found for each ε (δ may depend on a). Uniform continuity strengthens this: a single δ must work simultaneously for every point in the domain, for a given ε. Every uniformly continuous function is continuous, but the converse fails — f(x) = 1/x on (0,1) is continuous at every point yet not uniformly continuous, because the required δ shrinks without bound as x approaches 0. A key theorem (Heine–Cantor) guarantees that continuity and uniform continuity coincide on compact sets, such as closed bounded intervals [a,b].

Intuition

Ordinary continuity lets you pick a fresh δ tailored to each point a — you're allowed to 'zoom in' more aggressively wherever the function is steep or wild. Uniform continuity denies you that luxury: you must commit to one δ that works everywhere in the domain at once. This is why f(x)=1/x fails uniform continuity on (0,1): near x=0 the function is arbitrarily steep, so no matter how small a δ you pick, points x, y within δ of each other near 0 can still have wildly different f-values — you'd need an ever-smaller δ as you approach 0, but uniform continuity forbids letting δ depend on where you are.

Formal Definition

Definition

A function f: D → ℝ is uniformly continuous on D if:

∀ε>0, ∃δ>0 such that ∀x,y∈D, ∣x−y∣<δ  ⟹  ∣f(x)−f(y)∣<ε\forall \varepsilon > 0,\ \exists \delta > 0 \text{ such that } \forall x, y \in D,\ |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon

δ depends only on ε, not on the location of x or y in D

Definition of uniform continuity
(ordinary continuity): ∀a∈D, ∀ε>0, ∃δ(ε,a)>0, …\text{(ordinary continuity):}\ \forall a \in D,\ \forall \varepsilon>0,\ \exists \delta(\varepsilon, a) > 0,\ \ldots
Contrast: pointwise continuity allows δ to depend on a
f continuous on compact [a,b]  ⟹  f uniformly continuous on [a,b]f \text{ continuous on compact } [a,b] \implies f \text{ uniformly continuous on } [a,b]
Heine–Cantor theorem

Worked Examples

  1. Compute |f(x)-f(y)| directly in terms of |x-y|.

    ∣f(x)−f(y)∣=∣2x+1−2y−1∣=2∣x−y∣|f(x)-f(y)| = |2x+1-2y-1| = 2|x-y|

  2. Given ε, choose δ = ε/2, independent of x and y.

    δ=ε2\delta = \frac{\varepsilon}{2}

  3. Then |x-y|<δ forces |f(x)-f(y)| = 2|x-y| < 2(ε/2) = ε for all x,y ∈ ℝ.

    2⋅ε2=ε2\cdot\frac{\varepsilon}{2} = \varepsilon

Answer: δ = ε/2 works uniformly for all x, y ∈ ℝ, so f is uniformly continuous.

Practice Problems

Difficulty 5/10

Show f(x) = x² is uniformly continuous on [0,3] but find why the same δ argument fails on all of ℝ.

Difficulty 4/10

The Heine–Cantor theorem guarantees that a continuous function is automatically uniformly continuous when its domain is:

Difficulty 7/10

Prove that a uniformly continuous function on a bounded interval maps Cauchy sequences to Cauchy sequences.

Quiz

The key difference between continuity and uniform continuity is:
Which function is continuous on (0,1) but NOT uniformly continuous there?

Summary

  • Uniform continuity requires ∀ε>0 ∃δ>0 such that ∀x,y in the domain, |x-y|<δ ⟹ |f(x)-f(y)|<ε — δ depends only on ε.
  • Every uniformly continuous function is continuous, but the converse fails (e.g. f(x)=1/x on (0,1)).
  • The Heine–Cantor theorem guarantees continuity implies uniform continuity on any compact domain, such as a closed bounded interval.

References

  1. WebsiteWikipedia — Uniform continuity