Uniform continuity of $f:A\to \mathbb{R}$: one $\delta$ serving every pair of points of $A$

Let $A\subseteq \mathbb{R}$ and let $f:A\to \mathbb{R}$. Then $f$ is uniformly continuous on $A$ when

$(\mathrm{\forall }\epsilon >0)(\mathrm{\exists }\delta >0)(\mathrm{\forall }x,x^{\mathrm{\prime }}\in A)[\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }<\delta ⟹\mathrm{\mid }f(x)-f(x^{\mathrm{\prime }})\mathrm{\mid }<\epsilon ],$

with $\epsilon$ and $\delta$ ranging over the positive reals.

The whole content is in the order of the quantifiers. Written out, continuity on $A$ (Continuity of $f:A\to \mathbb{R}$ at a point of $A$ and on $A$: the $\epsilon$-$\delta$ condition, its agreement with ${\lim }_{x\to c}f(x)=f(c)$ at a limit point, and continuity at an isolated point) is

$\mathrm{\forall }\epsilon \mathrm{\forall }c\in A\mathrm{\exists }\delta \mathrm{\forall }x\in A,$

and uniform continuity is

$\mathrm{\forall }\epsilon \mathrm{\exists }\delta \mathrm{\forall }{x}^{\mathrm{\prime }}\in A\mathrm{\forall }x\in A\mathrm{.}$

Moving $\mathrm{\exists }\delta$ to the left of the point quantifier is the entire difference: for continuity the radius may shrink from point to point, for uniform continuity one radius must serve the whole of $A$ at once. This is the same distinction, for the same reason, that Uniform continuity of a map of metric spaces: one $\delta$ serving every point draws for maps of metric spaces.

Uniform continuity implies continuity. Given $\epsilon >0$, take the $\delta$ supplied by uniform continuity and, at a point $c\in A$, apply the condition with $x^{\mathrm{\prime }}:=c$: every $x\in A$ with $\mathrm{\mid }x-c\mathrm{\mid }<\delta$ satisfies $\mathrm{\mid }f(x)-f(c)\mathrm{\mid }<\epsilon$. So the same $\delta$ witnesses continuity at every point of $A$ simultaneously. The converse fails, and the failure is not marginal: FALSE: every continuous real function is uniformly continuous on its domain refutes it on this page, and the companion page works two witnesses out in full.

Uniform continuity is a property of the pair $(f,A)$, not of $f$ alone. The same formula may be uniformly continuous on one set and not on another: $x\mapsto 1\mathrm{/}x$ is uniformly continuous on $[1,\mathrm{\infty })$ and not on $(0,1)$, and $x\mapsto x^2$ is uniformly continuous on every bounded interval and not on $\mathbb{R}$. Restricting the domain therefore never destroys uniform continuity, since the condition then quantifies over fewer pairs; enlarging it may.

The two points are unordered and may coincide. Nothing above excludes $x=x^{\mathrm{\prime }}$, at which the implication reads $\mathrm{\mid }f(x)-f(x)\mathrm{\mid }=0<\epsilon$ (Basic properties of the absolute value) and is automatic, and the condition is symmetric in the two points because $\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }{x}^{\mathrm{\prime }}-x\mathrm{\mid }$.

• A single $\delta$, and a picture. For a uniformly continuous $f$ one may slide a rectangle of width $\delta$ and height $2\epsilon$ along the graph and never have the graph leave it through the top or bottom. For a merely continuous $f$ the rectangle must be narrowed as one moves, and on $x\mapsto 1\mathrm{/}x$ over $(0,1)$ it must be narrowed without limit.

• Neighbourhood form. The condition says $f(A\cap N_{\delta }(x^{\mathrm{\prime }}))\subseteq N_{\epsilon }(f(x^{\mathrm{\prime }}))$ for every $x^{\mathrm{\prime }}\in A$, with one $\delta$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$). That is continuity on $A$ with the radius independent of the centre.

• On an interval this is the notion the extension theorem needs. Assuming the Axiom of Choice (The Axiom of Choice), a uniformly continuous function on a set $D$ extends to one on the closure of $D$ (A uniformly continuous real function on a subset $D\subseteq \mathbb{R}$ extends uniquely to a uniformly continuous function on the closure of $D$); mere continuity does not suffice, since $x\mapsto 1\mathrm{/}x$ on $(0,1)$ has no continuous extension to $[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).

• The two standard witnesses, for orientation. The converse fails in two independent ways, each worked out on the companion page: $x\mapsto 1\mathrm{/}x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1\mathrm{/}(k+2)$ and $1\mathrm{/}(k+3)$ defeating every $\delta$ ↗ on a bounded domain that is not closed, and $x\mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1\mathrm{/}(k+1)$ defeating every $\delta$ ↗ on a closed domain that is not bounded. Both are named here for orientation only; nothing in this definition rests on them.

• Lipschitz and Hölder conditions are stronger still, and are not redefined here. They are Lipschitz map, $\alpha$-Hölder map for rational $0<\alpha \le 1$, and contraction instantiated at $A\subseteq \mathbb{R}$ with $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$; the dictionary that makes that instantiation legitimate, and that transports the implications of Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent to real functions, is Dictionary: for $A\subseteq \mathbb{R}$ with the metric $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, continuity and uniform continuity of $f:A\to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, immediately below.