Uniform Continuity

In the definition of continuity the number $\delta$ depends on both the tolerance $ϵ$ and the point $c$. At a point where the graph is steep, $\delta$ must be small; where it is flat, $\delta$ can be large. Uniform continuity requires a single $\delta$ that works simultaneously at every point of the domain. Whether one exists depends on the shape of the domain as much as on the function.

The definition

The two definitions differ only in quantifier order. Continuity reads for each $c$, for each $ϵ$, there is a $\delta$: $\delta$ may be chosen after seeing $c$. Uniform continuity reads for each $ϵ$, there is a $\delta$, good for all $c$: $\delta$ is chosen before any point is named, and $x$ and $c$ enter symmetrically. A uniformly continuous function is continuous; the converse can fail.

Two decisive examples

In all three failures the domain is either unbounded or not closed, and the slope grows without bound toward a missing endpoint or toward infinity. The next theorem shows that neither can happen on a closed bounded interval.

Continuity on a compact interval is uniform

The proof uses boundedness (to apply Bolzano–Weierstrass) and closedness (to keep $c$ in the domain), the two properties the failing examples lacked. This is the uniform-continuity fact that proves the integrability of continuous functions.³

Preserving Cauchy sequences and extending to endpoints

A uniformly continuous function carries Cauchy sequences to Cauchy sequences, even when the limit lies outside the domain; a merely continuous function need not.

Ordinary continuity fails this: on $(0,1)$ the Cauchy sequence $1/n$ is sent by $1/x$ to $n$, which is not Cauchy.

Lipschitz continuity

The most convenient sufficient condition for uniform continuity is a bound on the secant slopes.

Rearranged for $x\ne y$, the definition says

so every secant line through two points of the graph has slope at most $K$ in absolute value. Equivalently, the graph lies inside a cone of slope $\pm K$ anchored at each of its points.

$\sin$ and $\cos$ are Lipschitz with $K=1$, from the bounds $|\sin x-\sin y|\le |x-y|$ and $|\cos x-\cos y|\le |x-y|$. The square root separates the classes.

Worked example ($\sqrt{x}$ on $[0,\infty )$: uniformly continuous, not Lipschitz). Not Lipschitz: the secant from the origin to $(x,\sqrt{x})$ has slope $\sqrt{x}/x=1/\sqrt{x}\to \infty$ as $x\to 0^{+}$, so no constant $K$ bounds all secants. Still uniformly continuous: on $[0,2]$ the function is continuous on a closed bounded interval, hence uniformly continuous by the theorem; on $[1,\infty )$ it is Lipschitz by the previous example. Given $ϵ>0$, take $\delta$ as the smaller of the two deltas and of $1$; any pair with $|x-y|<\delta$ then lies entirely in $[0,2]$ or entirely in $[1,\infty )$, and either case gives $|\sqrt{x}-\sqrt{y}|<ϵ$.

These two examples separate the three conditions cleanly: on $[0,\infty )$, $\sqrt{x}$ is uniformly continuous but not Lipschitz, and on $\mathbb{R}$, $x^2$ is continuous but not uniformly continuous. The implications run one way only.

The definitions of continuity and uniform continuity used only the distances $|x-c|$ and $|f(x)-f(c)|$, so they transfer verbatim once distance is axiomatized, to maps between metric spaces; the extreme value theorem, the uniform-continuity theorem, and the open-preimage characterization all hold there in general form.