Uniform Continuity

analysis

Definition

Uniform Continuity

A function $f:D\to \mathbb{R}$ is uniformly continuous on $D$ if every requested output tolerance can be met by one input tolerance that works everywhere in the domain. Formally, for every $\epsilon >0$, there exists $\delta >0$ such that, for all $x_1,x_2\in D$,

$$|x_1-x_2|<\delta ⟹|f(x_1)-f(x_2)|<\epsilon .$$

The essential point is that $\delta$ may depend on $\epsilon$, but it may not depend on where $x_1$ and $x_2$ lie in $D$. One horizontal scale must control the whole domain.

Difference from pointwise continuity

Ordinary continuity is local. At each point $a\in D$, and for each $\epsilon >0$, one may choose a radius $\delta$ that depends on both $\epsilon$ and $a$.

Uniform continuity removes that dependence on $a$. After $\epsilon$ is fixed, the same $\delta$ must work for every location in the domain. This prevents the function from becoming more and more sensitive as one moves through the domain.

Symbolically, pointwise continuity allows

$$\delta =\delta (\epsilon ,a),$$

while uniform continuity requires

$$\delta =\delta (\epsilon ).$$

Why compact domains help

The Heine-Cantor theorem states that every continuous real-valued function on a compact domain is uniformly continuous. Compactness prevents the domain from hiding infinitely many places where the required $\delta$ becomes smaller and smaller without limit.

This is why a function can be continuous but fail to be uniformly continuous on an open or unbounded domain. For example, $f(x)=1/x$ is continuous on $(0,1]$, but it is not uniformly continuous there because the function becomes arbitrarily steep near $0$.

Relation to Lipschitz continuity

Every Lipschitz continuous function is uniformly continuous. Lipschitz continuity gives the explicit bound

$$|f(x_1)-f(x_2)|\le L|x_1-x_2|,$$

so one can choose $\delta =\epsilon /L$ when $L>0$.

The converse is false. Uniform continuity controls the same input tolerance across the domain, but it does not require a fixed linear bound on all secant slopes. The function $f(x)=\sqrt{x}$ on $[0,1]$ is uniformly continuous, but not Lipschitz continuous.