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

$(\forall \varepsilon > 0)\ (\exists \delta > 0)\ (\forall x, x' \in A)\ \bigl[\ |x - x'| < \delta \ \Longrightarrow\ |f(x) - f(x')| < \varepsilon\ \bigr],$

with $\varepsilon$ 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 $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point) is

$\forall \varepsilon\ \forall c \in A\ \exists \delta\ \forall x \in A ,$

and uniform continuity is

$\forall \varepsilon\ \exists \delta\ \forall x' \in A\ \forall x \in A .$

Moving $\exists \delta$ 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 $\varepsilon > 0$, take the $\delta$ supplied by uniform continuity and, at a point $c \in A$, apply the condition with $x' := c$: every $x \in A$ with $|x - c| < \delta$ satisfies $|f(x) - f(c)| < \varepsilon$. 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: $x \mapsto 1/x$ is uniformly continuous on $[1,\infty)$ 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 $x = x'$, at which the implication reads $|f(x) - f(x)| = 0 < \varepsilon$ (Basic properties of the absolute value) and is automatic, and the condition is symmetric in the two points because $|x - x'| = |x' - x|$.

• A single $\delta$, and a picture. For a uniformly continuous $f$ one may slide a rectangle of width $\delta$ and height $2\varepsilon$ 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 $x \mapsto 1/x$ over $(0,1)$ it must be narrowed without limit.

• Neighbourhood form. The condition says $f\bigl(A \cap N_{\delta}(x')\bigr) \subseteq N_{\varepsilon}(f(x'))$ for every $x' \in A$, with one $\delta$ (The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-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. 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 $x \mapsto 1/x$ 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: $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1/(k+2)$ and $1/(k+3)$ 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 $k+1+1/(k+1)$ 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 $d(x,y) = |x-y|$; 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 $d(x,y) = |x-y|$, 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.