Continuity is a local condition: it asks that $f(x)$ be close to $f(a)$ whenever $x$ is close to $a$, but the meaning of "close" --- the size of $\delta$ --- is permitted to depend on the point $a$. For many purposes in analysis, this freedom is harmless. But for others --- extending a function from a dense subset, controlling the oscillation of a function over a partition, or preserving the Cauchy property of sequences --- the dependence of $\delta$ on the point is a fatal obstruction. The difficulty is best seen through a concrete computation. [example: The Failure of a Uniform Bound] Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$. This function is [continuous](/page/Continuity) at every point $a \in \mathbb{R}$: given $\varepsilon > 0$, the estimate \begin{align*} |f(x) - f(a)| = |x^2 - a^2| = |x - a| \cdot |x + a| \leq |x - a| \cdot (2|a| + 1) \end{align*} (valid whenever $|x - a| \leq 1$) shows that $\delta = \min\{1, \varepsilon/(2|a| + 1)\}$ works at the point $a$. But this $\delta$ depends on $a$ and shrinks to $0$ as $|a| \to \infty$. No single $\delta$ can work simultaneously at all points. To see this rigorously, define sequences $x_n = n + \frac{1}{2n}$ and $y_n = n$ for $n \in \mathbb{N}$. Then \begin{align*} |x_n - y_n| = \frac{1}{2n} \to 0, \quad \text{yet} \quad |f(x_n) - f(y_n)| = \left|n + \frac{1}{2n}\right|^2 - n^2 = 1 + \frac{1}{4n^2} > 1. \end{align*} For any candidate "uniform" $\delta > 0$ and any $\varepsilon \leq 1$, the pair $(x_n, y_n)$ with $n > 1/(2\delta)$ satisfies $|x_n - y_n| < \delta$ but $|f(x_n) - f(y_n)| > 1 \geq \varepsilon$. The function $f$ is continuous everywhere, yet no uniform modulus of closeness can be specified. [/example] The issue in this example is not a defect of $f$ at any particular point --- continuity holds at every $a \in \mathbb{R}$. The issue is *global*: the rate at which $f$ distorts distances varies without bound across the domain. **Uniform continuity** is the condition that eliminates this phenomenon. It asks for a single $\delta$ that works at all points simultaneously, turning the local guarantee of continuity into a global one. This upgrade from local to global is not merely cosmetic. Uniform continuity is the precise condition under which: - A [continuous function](/page/Continuity%20(Real%20Analysis)) on a closed bounded interval is [Riemann integrable](/page/Integral) (because uniform continuity controls the oscillation over partition subintervals). - A continuous function defined on a [dense subset](/page/Dense%20Subset) extends uniquely to the ambient [complete metric space](/page/Complete%20Metric%20Space) (because uniform continuity ensures that Cauchy sequences in the domain map to Cauchy sequences in the codomain). - The [completion](/page/Complete%20Metric%20Space) construction works: algebraic operations like addition and scalar multiplication extend from a normed space to its completion precisely because they are uniformly continuous. The [Heine-Cantor theorem](/page/Compact%20Space) reveals that this distinction between continuity and uniform continuity is invisible on [compact](/page/Compact%20Space) domains --- there, every continuous function is automatically uniformly continuous. It is on non-compact domains that the two notions diverge, and where the analyst must pay careful attention. ## Definition The definition of uniform continuity differs from ordinary continuity by a single quantifier swap, but the consequences of that swap are far-reaching. [definition: Uniform Continuity] Let $(X, d_X)$ and $(Y, d_Y)$ be [metric spaces](/page/Metric%20Space). A function $f: X \to Y$ is **uniformly continuous** if for every $\varepsilon > 0$ there exists $\delta > 0$ such that \begin{align*} d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon \quad \text{for all } x, y \in X. \end{align*} [/definition] The critical distinction from [pointwise continuity](/page/Continuity%20(Metric%20Spaces)) lies in the quantifier order. Continuity at a point $a$ reads: \begin{align*} \forall \varepsilon > 0 \;\; \exists \delta > 0 \;\; \forall x \in X: \quad d_X(x, a) < \delta \implies d_Y(f(x), f(a)) < \varepsilon. \end{align*} Continuity on $X$ universally quantifies over $a$, placing $\forall a \in X$ at the front but *before* the $\exists \delta$: \begin{align*} \forall a \in X \;\; \forall \varepsilon > 0 \;\; \exists \delta > 0 \;\; \forall x \in X: \quad d_X(x, a) < \delta \implies d_Y(f(x), f(a)) < \varepsilon. \end{align*} Uniform continuity moves the universal quantifier $\forall x, y$ to the *outside* of the existential $\exists \delta$: \begin{align*} \forall \varepsilon > 0 \;\; \exists \delta > 0 \;\; \forall x, y \in X: \quad d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon. \end{align*} In the first formulation, $\delta$ may depend on both $\varepsilon$ and the point $a$. In the second, $\delta$ depends on $\varepsilon$ alone. This is a strictly stronger condition: every uniformly continuous function is continuous, but the opening example shows that the converse fails. [remark: Negation of Uniform Continuity] A function $f: (X, d_X) \to (Y, d_Y)$ is **not** uniformly continuous if and only if there exist $\varepsilon_0 > 0$ and sequences $(x_n)_{n=1}^\infty$, $(y_n)_{n=1}^\infty$ in $X$ with $d_X(x_n, y_n) \to 0$ but $d_Y(f(x_n), f(y_n)) \geq \varepsilon_0$ for all $n \in \mathbb{N}$. This sequential characterisation of the failure of uniform continuity is often the most efficient way to prove that a given function is not uniformly continuous: produce a pair of sequences that get closer together while their images stay apart. [/remark] ## The Modulus of Continuity To study uniform continuity quantitatively --- not just as a yes-or-no property, but as a matter of *degree* --- we need a tool that measures the worst-case oscillation of $f$ over pairs of points at a given distance. This leads to the modulus of continuity, which encodes the "rate" at which uniform continuity holds. [definition: Modulus of Continuity] Let $f: (X, d_X) \to (Y, d_Y)$ be a function between [metric spaces](/page/Metric%20Space). The **modulus of continuity** of $f$ is the function $\omega_f: [0, \infty) \to [0, \infty]$ defined by \begin{align*} \omega_f(\delta) := \sup \{ d_Y(f(x), f(y)) : x, y \in X, \; d_X(x, y) \leq \delta \}. \end{align*} [/definition] The function $f$ is uniformly continuous if and only if $\omega_f(\delta) \to 0$ as $\delta \to 0^+$. The modulus $\omega_f$ is always non-decreasing, and it is subadditive: $\omega_f(\delta_1 + \delta_2) \leq \omega_f(\delta_1) + \omega_f(\delta_2)$ for all $\delta_1, \delta_2 \geq 0$. Subadditivity follows because any pair $(x, y)$ with $d_X(x, y) \leq \delta_1 + \delta_2$ can be connected through an intermediate point $z$ with $d_X(x, z) \leq \delta_1$ and $d_X(z, y) \leq \delta_2$ (when such a point exists), and the triangle inequality in $Y$ gives $d_Y(f(x), f(y)) \leq d_Y(f(x), f(z)) + d_Y(f(z), f(y)) \leq \omega_f(\delta_1) + \omega_f(\delta_2)$. The modulus of continuity provides a natural stratification of uniformly continuous functions by their rate of decay: - **[Lipschitz](/page/Continuity%20(Metric%20Spaces)) functions:** $\omega_f(\delta) \leq L\delta$ for some constant $L \geq 0$. The modulus decays linearly. - **Holder continuous functions** (exponent $\alpha \in (0, 1]$): $\omega_f(\delta) \leq C\delta^\alpha$ for some $C \geq 0$. The modulus decays as a power law. - **General uniformly continuous functions:** $\omega_f(\delta) \to 0$, but the rate of decay can be arbitrarily slow. [example: A Uniformly Continuous Function With Arbitrarily Slow Modulus] Define $f: [0, \infty) \to \mathbb{R}$ by $f(x) = \sqrt{x}$. This function is uniformly continuous on $[0, \infty)$: for $x, y \geq 0$, \begin{align*} |\sqrt{x} - \sqrt{y}| = \frac{|x - y|}{\sqrt{x} + \sqrt{y}} \leq \frac{|x - y|}{\max\{\sqrt{x}, \sqrt{y}\}}. \end{align*} When $|x - y| < \delta$, the cruder estimate $|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|} \leq \sqrt{\delta}$ shows that $\omega_f(\delta) \leq \sqrt{\delta}$, so $f$ is Holder continuous with exponent $\alpha = 1/2$. However, $f$ is *not* Lipschitz on any interval containing $0$: the ratio $|\sqrt{x} - \sqrt{0}|/|x - 0| = 1/\sqrt{x} \to \infty$ as $x \to 0^+$. The modulus $\omega_f(\delta) = \sqrt{\delta}$ decays strictly slower than linearly, placing $f$ in the gap between Lipschitz and merely uniformly continuous. For even slower decay, the function $g: [0, 1/e] \to \mathbb{R}$ defined by $g(x) = 1/\log(1/x)$ for $x \in (0, 1/e]$ and $g(0) = 0$ is uniformly continuous on $[0, 1/e]$ (by the Heine-Cantor theorem, since $[0, 1/e]$ is compact and $g$ is continuous), but $\omega_g(\delta) \to 0$ only logarithmically. No power law $C\delta^\alpha$ bounds $\omega_g$ near $\delta = 0$ for any $\alpha > 0$. [/example] ## Compactness and the Heine-Cantor Theorem The opening example demonstrated that continuity does not imply uniform continuity on $\mathbb{R}$. A natural question arises: under what conditions on the domain does continuity *automatically* upgrade to uniform continuity? The answer is [compactness](/page/Compact%20Space). On a compact domain, the local-to-global gap that separates the two notions collapses entirely. The underlying mechanism is the [Lebesgue number lemma](/page/Compact%20Space): for any open cover of a compact metric space, there is a uniform lower bound $\lambda > 0$ (the Lebesgue number) such that every subset of diameter less than $\lambda$ is contained in some member of the cover. Applied to the cover by $\delta_a$-balls from the continuity condition, this produces a single $\delta$ that works everywhere. [quotetheorem:954] The Heine-Cantor theorem is sharp in two respects. First, compactness of the domain cannot be replaced by mere closedness or boundedness in a general metric space. Second, the conclusion says nothing about the *rate* of uniform continuity --- it guarantees existence of a modulus $\omega_f$ with $\omega_f(\delta) \to 0$, but that modulus can decay arbitrarily slowly (as in the logarithmic example above). [example: Compactness Is Necessary] The hypothesis that $X$ is compact cannot be dropped. The function $f: (0, 1) \to \mathbb{R}$ defined by $f(x) = \sin(1/x)$ is continuous on the open interval $(0, 1)$, which is bounded but not compact (it is not closed). The function is not uniformly continuous: define $x_n = \frac{1}{2\pi n}$ and $y_n = \frac{1}{2\pi n + \pi/2}$ for $n \in \mathbb{N}$. Then \begin{align*} |x_n - y_n| &= \frac{1}{2\pi n} - \frac{1}{2\pi n + \pi/2} = \frac{\pi/2}{2\pi n(2\pi n + \pi/2)} \leq \frac{1}{4\pi n^2} \to 0, \end{align*} but $|f(x_n) - f(y_n)| = |\sin(2\pi n) - \sin(2\pi n + \pi/2)| = |0 - 1| = 1$ for all $n$. The pairs $(x_n, y_n)$ converge to each other, yet their images remain a fixed distance apart. The underlying issue is that as $x \to 0^+$, the function oscillates with increasing frequency. On any compact subinterval $[a, 1]$ with $a > 0$, the function *is* uniformly continuous by Heine-Cantor --- it is only the non-compactness near $0$ that allows the modulus of continuity to deteriorate without bound. [/example] The Heine-Cantor theorem also clarifies a common source of confusion: on compact domains, the distinction between continuity and uniform continuity vanishes, so the two concepts appear identical. It is only when working with non-compact domains --- unbounded intervals in $\mathbb{R}$, open subsets of $\mathbb{R}^n$, or infinite-dimensional spaces --- that the distinction becomes visible and consequential. The concept of uniform continuity has a natural counterpart in function spaces: **equicontinuity**. A family $\mathcal{F}$ of functions $f: X \to Y$ is equicontinuous if the $\delta$ in the uniform continuity condition can be chosen independently of *which function* in $\mathcal{F}$ is used, not just independently of the point. The [Arzela-Ascoli theorem](/page/Compact%20Space) characterises relatively compact subsets of $C(X, Y)$ (with $X$ compact) as precisely the bounded and equicontinuous families. This is the function-space analogue of Heine-Cantor: compactness in the function space corresponds to a uniform version of the regularity condition. ## The Regularity Hierarchy Uniform continuity occupies a specific position in a chain of increasingly strong regularity conditions on functions between metric spaces. Understanding this hierarchy reveals what uniform continuity *buys* and what it *costs* relative to its neighbours. The inclusions \begin{align*} \text{Lipschitz} \;\subsetneq\; \text{Holder (exponent } \alpha \in (0,1)) \;\subsetneq\; \text{uniformly continuous} \;\subsetneq\; \text{continuous} \end{align*} are all strict. The opening example ($f(x) = x^2$ on $\mathbb{R}$) separates continuous from uniformly continuous. The modulus-of-continuity section provided $f(x) = \sqrt{x}$ on $[0, \infty)$ as a function that is uniformly continuous (and Holder with $\alpha = 1/2$) but not Lipschitz. To complete the picture, we need an example separating uniformly continuous from Holder. [example: Uniformly Continuous But Not Holder] Let $g: [0, 1/e] \to \mathbb{R}$ be defined by $g(0) = 0$ and \begin{align*} g(x) = \frac{1}{\log(1/x)} \quad \text{for } x \in (0, 1/e]. \end{align*} The function $g$ is continuous on $[0, 1/e]$ (since $g(x) \to 0$ as $x \to 0^+$) and therefore uniformly continuous by the Heine-Cantor theorem. However, $g$ is not Holder continuous for any exponent $\alpha > 0$ at $x = 0$: the ratio \begin{align*} \frac{|g(x) - g(0)|}{|x - 0|^\alpha} = \frac{1}{x^\alpha \log(1/x)} \end{align*} diverges as $x \to 0^+$ for every $\alpha > 0$. To verify this, note that $\log(1/x) = o(x^{-\alpha})$ for any $\alpha > 0$ (logarithms grow slower than any positive power), so $1/(x^\alpha \log(1/x)) \to \infty$. This shows that the modulus of continuity of $g$ decays slower than $C\delta^\alpha$ for every $\alpha > 0$. Uniform continuity accommodates arbitrarily slow decay rates; Holder continuity imposes a power-law lower bound on the rate. [/example] [explanation: Where Uniform Continuity Sits in the Hierarchy] The regularity hierarchy for functions between metric spaces has additional levels above Lipschitz. On $\mathbb{R}$, the full chain (from strongest to weakest) is: \begin{align*} C^1 \text{ with bounded derivative} \;\subsetneq\; \text{Lipschitz} \;\subsetneq\; \text{Holder} \;\subsetneq\; \text{uniformly continuous} \;\subsetneq\; \text{continuous.} \end{align*} The [Mean Value Theorem](/theorems/186) provides the bridge between differentiability and Lipschitz continuity: if $f: [a, b] \to \mathbb{R}$ is differentiable with $|f'(x)| \leq L$ for all $x \in (a, b)$, then $|f(x) - f(y)| \leq L|x - y|$ for all $x, y \in [a, b]$. Conversely, [Rademacher's theorem](/page/Rademacher's%20Theorem) asserts that a Lipschitz function on $\mathbb{R}^n$ is differentiable $\mathcal{L}^n$-almost everywhere. Each level in the hierarchy carries a different quantitative commitment. Lipschitz continuity says that the modulus of continuity $\omega_f(\delta)$ is controlled by $L\delta$ --- the function distorts distances by at most a constant factor. Holder continuity with exponent $\alpha$ allows polynomial distortion $C\delta^\alpha$. Uniform continuity demands only that $\omega_f(\delta) \to 0$, imposing no rate. Continuity does not even demand a uniform modulus. [/explanation] ## Preservation of Cauchy Sequences One of the most consequential properties of uniformly continuous functions --- and the property that explains why uniform continuity is the "right" condition for extension theorems --- is that uniformly continuous functions preserve [Cauchy sequences](/page/Cauchy%20Sequence). Merely continuous functions do not. [quotetheorem:1052] The proof is a direct application of the definition: given $\varepsilon > 0$, uniform continuity provides $\delta > 0$ such that $d_X(x, y) < \delta$ implies $d_Y(f(x), f(y)) < \varepsilon$. Since $(x_n)$ is Cauchy, there exists $N \in \mathbb{N}$ with $d_X(x_m, x_n) < \delta$ for all $m, n \geq N$. Then $d_Y(f(x_m), f(x_n)) < \varepsilon$ for all $m, n \geq N$, so $(f(x_n))$ is Cauchy. The result fails for merely continuous functions. The following example shows that a continuous function can map a Cauchy sequence to a sequence that is not Cauchy. [example: Continuity Does Not Preserve Cauchy Sequences] Consider the function $f: (0, 1) \to \mathbb{R}$ defined by $f(x) = 1/x$. This function is continuous on $(0, 1)$. The sequence $x_n = 1/n$ for $n \geq 2$ is Cauchy in $(0, 1)$ (since $|1/m - 1/n| \to 0$ as $m, n \to \infty$). However, $f(x_n) = n$, and the sequence $(n)_{n=2}^\infty$ is unbounded, hence not Cauchy. The function $f$ fails to be uniformly continuous on $(0, 1)$: define $x_n = 1/n$ and $y_n = 1/(n+1)$. Then $|x_n - y_n| = 1/(n(n+1)) \to 0$, but $|f(x_n) - f(y_n)| = |n - (n+1)| = 1$ for all $n$, verifying failure of uniform continuity directly. [/example] This preservation property is the engine behind extension theorems. When we want to extend a function $f: D \to Y$ from a [dense subset](/page/Dense%20Subset) $D$ of a metric space $X$ to all of $X$, the natural strategy is: for each $x \in X$, pick a sequence $(d_n) \subset D$ converging to $x$ and define $\bar{f}(x) = \lim_{n \to \infty} f(d_n)$. For this to work, we need two things: (1) the sequence $(f(d_n))$ must converge, which requires it to be Cauchy (and $Y$ to be complete), and (2) the limit must be independent of the choice of approximating sequence. Uniform continuity guarantees both. ## Extension from Dense Subsets A recurring construction in analysis is the need to extend a function from a "nice" subset to a larger space. The Fourier transform, for instance, is initially defined on $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ and then extended to all of $L^2(\mathbb{R}^n)$. The [completion](/page/Complete%20Metric%20Space) of a metric space adds limit points of Cauchy sequences. Algebraic operations on a normed space (addition, scalar multiplication) extend to the completion. In each case, the mechanism that makes the extension possible is uniform continuity. [quotetheorem:964] The theorem has three essential hypotheses, and each is necessary. **Uniform continuity of $f$ cannot be weakened to continuity.** The function $h: \mathbb{Q} \to \mathbb{R}$ defined by \begin{align*} h(q) = \begin{cases} 0 & \text{if } q < \sqrt{2}, \\ 1 & \text{if } q > \sqrt{2}, \end{cases} \end{align*} is continuous on $\mathbb{Q}$ (since $\sqrt{2} \notin \mathbb{Q}$, the function is locally constant near every rational). But $h$ has no continuous extension to $\mathbb{R}$: any extension would need to be both $0$ and $1$ "at" $\sqrt{2}$. The jump is hidden in $\mathbb{Q}$ but revealed upon passing to the completion. Uniform continuity forbids such hidden jumps. **Completeness of $Y$ is essential.** The identity function $\operatorname{id}: \mathbb{Q} \to \mathbb{Q}$ is an isometry (hence uniformly continuous), and $\mathbb{Q}$ is dense in $\mathbb{R}$. But there is no continuous function $\bar{f}: \mathbb{R} \to \mathbb{Q}$ extending $\operatorname{id}$, because the [Intermediate Value Theorem](/theorems/180) would force $\bar{f}$ to take irrational values. The codomain $\mathbb{Q}$ lacks the limits that the extension needs to hit. **Density of $D$ is used for both existence and uniqueness.** Without density, the values of $\bar{f}$ at points far from $D$ are undetermined --- there is no approximating sequence to take a limit along. [explanation: How the Extension Theorem Builds Completions] The extension theorem is the key ingredient in the construction of the [completion](/page/Complete%20Metric%20Space) $(\hat{X}, \hat{d})$ of a metric space $(X, d)$. After constructing $\hat{X}$ as the set of equivalence classes of Cauchy sequences in $X$ and embedding $X$ isometrically as a dense subset, every uniformly continuous function $f: X \to Y$ (with $Y$ complete) extends uniquely to $\hat{f}: \hat{X} \to Y$. In particular, algebraic operations like vector addition \begin{align*} +: X \times X &\to X \\ (u, v) &\mapsto u + v \end{align*} are uniformly continuous (since $\|u + v - (u_0 + v_0)\| \leq \|u - u_0\| + \|v - v_0\|$) and therefore extend uniquely to the completion. This is why the completion of a normed vector space is automatically a [Banach space](/page/Banach%20Space), and the completion of an inner product space is a [Hilbert space](/page/Hilbert%20Space): the algebraic structure transfers because the algebraic operations are uniformly continuous. [/explanation] In functional analysis, the extension theorem specialises to a result about bounded linear operators. [quotetheorem:965] The BLT theorem follows from the general extension theorem because a bounded linear operator $T$ with $\|Tx - Ty\| = \|T(x - y)\| \leq \|T\| \cdot \|x - y\|$ is Lipschitz, hence uniformly continuous. The additional content of the BLT theorem is that the extension preserves linearity and the exact operator norm. ## Uniform Continuity and Integrability The connection between uniform continuity and [Riemann integration](/page/Integral) is one of the oldest and most important applications of the concept. The fundamental question of integration theory is: for which functions $f: [a, b] \to \mathbb{R}$ do the Riemann sums converge? Uniform continuity provides a clean sufficient condition by controlling the oscillation of $f$ over subintervals of a partition. Given a partition $\mathcal{P} = \{a = t_0 < t_1 < \cdots < t_N = b\}$ of $[a, b]$, the **oscillation** of $f$ on the $k$-th subinterval is \begin{align*} \omega_k = \sup_{x, y \in [t_{k-1}, t_k]} |f(x) - f(y)|. \end{align*} The difference between the upper and lower Riemann sums is \begin{align*} U(f, \mathcal{P}) - L(f, \mathcal{P}) = \sum_{k=1}^{N} \omega_k \cdot (t_k - t_{k-1}). \end{align*} For $f$ to be Riemann integrable, this difference must be made arbitrarily small by refining the partition. If $f$ is uniformly continuous, then for any $\varepsilon > 0$ there exists $\delta > 0$ such that $|x - y| < \delta$ implies $|f(x) - f(y)| < \varepsilon / (b - a)$. Choosing any partition with mesh $\|\mathcal{P}\| = \max_k (t_k - t_{k-1}) < \delta$ gives $\omega_k < \varepsilon / (b - a)$ for every $k$, and therefore \begin{align*} U(f, \mathcal{P}) - L(f, \mathcal{P}) = \sum_{k=1}^{N} \omega_k \cdot (t_k - t_{k-1}) < \frac{\varepsilon}{b - a} \sum_{k=1}^{N} (t_k - t_{k-1}) = \varepsilon. \end{align*} The uniform bound on the oscillation --- which depends on the mesh of the partition alone, not on the particular subintervals --- is precisely what uniform continuity provides. [quotetheorem:282] The proof combines two results: the [Heine-Cantor theorem](/page/Compact%20Space) upgrades continuity on $[a, b]$ to uniform continuity, and the oscillation estimate above converts uniform continuity into the Riemann integrability condition $U(f, \mathcal{P}) - L(f, \mathcal{P}) < \varepsilon$. Without the Heine-Cantor theorem, we would only know that each subinterval has *some* $\delta$ controlling its oscillation, but these $\delta$ values might vary across subintervals and could be impossible to satisfy simultaneously with a single partition mesh. ## Standard Arguments Using Uniform Continuity The techniques for working with uniform continuity appear repeatedly throughout analysis. This section collects the standard argument patterns. ### The Sequential Negation Argument To prove that a function $f: X \to Y$ is *not* uniformly continuous, use the sequential characterisation from the negation: exhibit $\varepsilon_0 > 0$ and sequences $(x_n)$, $(y_n)$ with $d_X(x_n, y_n) \to 0$ but $d_Y(f(x_n), f(y_n)) \geq \varepsilon_0$. [example: The Exponential Is Not Uniformly Continuous] Consider $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = e^x$. Define $x_n = \log n$ and $y_n = \log(n + 1)$ for $n \in \mathbb{N}$. Then \begin{align*} |x_n - y_n| = |\log n - \log(n+1)| = \log\left(1 + \frac{1}{n}\right) \leq \frac{1}{n} \to 0, \end{align*} where the inequality uses the concavity estimate $\log(1 + t) \leq t$ for $t > 0$. However, \begin{align*} |f(x_n) - f(y_n)| = |n - (n+1)| = 1 \quad \text{for all } n. \end{align*} Taking $\varepsilon_0 = 1$ confirms that $f$ is not uniformly continuous on $\mathbb{R}$. [/example] ### The Restriction-to-Compact-Subsets Technique Many results in analysis are first proved for compact domains (where uniform continuity is automatic) and then extended. The Heine-Cantor theorem guarantees that any continuous function on a compact metric space is uniformly continuous, so compactness arguments can often substitute for direct uniform continuity proofs. [example: Uniform Convergence on Compact Subsets] When proving that a sequence of functions $f_n: U \to \mathbb{R}$ converges uniformly on compact subsets $K \subset U$, one often uses the fact that each $f_n$ is uniformly continuous on $K$ (by Heine-Cantor). This gives a single $\delta$ that works for all points in $K$, enabling uniform estimates over the entire compact set. Concretely, if $f: U \to \mathbb{R}$ is continuous on an open set $U \subset \mathbb{R}^n$ and $K \subset U$ is compact, then for each $\varepsilon > 0$, there is a single $\delta > 0$ (depending on $f$, $K$, and $\varepsilon$, but not on any particular point in $K$) such that $|f(x) - f(y)| < \varepsilon$ whenever $x, y \in K$ with $|x - y| < \delta$. This uniform estimate is what makes mollification arguments work: the mollified function $f * \eta_\varepsilon$ converges uniformly to $f$ on $K$ as $\varepsilon \to 0$ precisely because the modulus of continuity of $f|_K$ tends to zero uniformly. [/example] ### The Extension-by-Cauchy-Sequences Method To extend a uniformly continuous function $f: D \to Y$ from a dense subset $D \subset X$ to the full space $X$ (with $Y$ complete): 1. For each $x \in X \setminus D$, choose a sequence $(d_n) \subset D$ with $d_n \to x$. 2. Verify that $(f(d_n))$ is Cauchy in $Y$ using uniform continuity and the Cauchy property of $(d_n)$. 3. Define $\bar{f}(x) = \lim_{n \to \infty} f(d_n)$, using completeness of $Y$. 4. Verify that the limit is independent of the choice of sequence: if $(d_n')$ is another sequence in $D$ with $d_n' \to x$, then $d_X(d_n, d_n') \to 0$, so uniform continuity gives $d_Y(f(d_n), f(d_n')) \to 0$, and the limits agree. 5. Verify that $\bar{f}$ is uniformly continuous by showing that the same $\delta$ (from the uniform continuity of $f$ on $D$) works for $\bar{f}$ on $X$, passing through the approximating sequences. This method is used in the construction of the [completion](/page/Complete%20Metric%20Space) of a metric space, the extension of [bounded linear operators](/page/Linear%20Operators%20on%20Banach%20Spaces) from dense subspaces, and the definition of the Fourier transform on $L^2$. ### Verifying Uniform Continuity via Lipschitz Estimates The most common way to prove that a function *is* uniformly continuous is to establish a Lipschitz or Holder estimate. If $f: X \to Y$ satisfies $d_Y(f(x), f(y)) \leq C \cdot d_X(x, y)^\alpha$ for some $C > 0$ and $\alpha > 0$, then $f$ is uniformly continuous with $\delta = (\varepsilon / C)^{1/\alpha}$. In settings where a direct Lipschitz estimate is unavailable, one can sometimes establish uniform continuity by showing that $f$ extends continuously to a compactification of $X$, then applying Heine-Cantor. For example, a continuous function $f: (0, 1) \to \mathbb{R}$ that has finite limits as $x \to 0^+$ and $x \to 1^-$ extends to a continuous function on $[0, 1]$ and is therefore uniformly continuous. ## References - Rudin, W., *Principles of Mathematical Analysis*, 3rd edition (1976). - Munkres, J., *Topology*, 2nd edition (2000). - Pugh, C. C., *Real Mathematical Analysis*, 2nd edition (2015). - Aliprantis, C. D. and Burkinshaw, O., *Principles of Real Analysis*, 3rd edition (1998).