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If distinct points $x,y$ of a dynamical system on a compact metric space satisfy $d(T^{n}x,T^{n}y)\to 0$ in one compatible metric, no compatible metric can make $T$ an isometry. On a compact space, all compatible metrics give the same asymptotic-pair property by uniform continuity, whereas an isometry preserves the strictly positive distance between distinct points. In a full shift, a constant sequence and a sequence differing at just one coordinate converge to each other under forward shifts, proving this obstruction directly in the product topology.

For a function $f:[0,1]\to \mathbb{R}$ and an integer $n\ge 1$, its Bernstein polynomial isThe Bernstein basis supplies nonnegative weights whose sum is one. Equivalently, the weights are the probabilities of a binomial distribution, and the result is an expected value. At a given $x$, this averages the sampled values $f(k/n)$ with weights concentrated near $k/n=x$.^[ref]

The endpoints are exact: $B_{n}f(0)=f(0)$ and $B_{n}f(1)=f(1)$. Interior sampled points need not be interpolated. For example, binomial distribution moments give $B_n(x)=x$ and $B_n(x^2)=x^2+x(1-x)/n$. Thus every affine function is reproduced exactly, whereas $x^2$ is generally changed. Bernstein polynomial degree preservation is a statement about degree of a polynomial, not equality of the entire polynomial.

For a continuous $f$, the uniform continuity of $f$ and the variance $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$ explain uniform convergence. Split the expected value of the error into $|K/n-x|<\delta$, where uniform continuity makes the change small, and its complement, whose probability is at most $1/(4n{\delta }^2)$ by Chebyshev's inequality. The bounded function $f$ bounds the remaining error. This proves convergence without assuming differentiability.

For $f\in C([0,1{]}^d)$, the tensor-product Bernstein polynomial is the expected value of $f(X_1/n,\dots ,X_d/n)$ for independent random variables $X_j\sim \mathrm{Bin}(n,x_j)$. The same uniform continuity and coordinatewise variance estimates give uniform convergence for each fixed finite $d$.

Let $f$ be bounded on $[a,b]$ and continuous on $(a,b]$. For any positive error tolerance, isolate $[a,a+\delta ]$. Its contribution to the difference of upper Darboux sums and lower Darboux sums is at most $2\Vert f{\Vert }_{\infty }\delta$. On $[a+\delta ,b]$, uniform continuity supplies a partition of an interval with arbitrarily small remaining oscillation sum. The Riemann integrability criterion then proves integrability, even when the values oscillate without a limit at the endpoint.

For a continuous map on a compact metric space, the Bowen metric compares the first $n$ points of two forward orbits. Its radius-$ϵ$ ball is $B_n(x,ϵ)=\{y:d_n(x,y)<ϵ\}$. If $s_n(ϵ)$ is the largest cardinality of an $ϵ$-separated set in this metric, then topological entropy is ${\lim }_{ϵ\downarrow 0}{\mathrm{limsup}}_{n\to \infty }{n}^{-1}\log s_n(ϵ)$. The uniform continuity of the identity between two compatible metrics gives the comparison of separated-set counts needed for independence of the original metric.

For every fixed mesh $1/k$, independent increments with the normal distribution imply increments of absolute value greater than one occur infinitely often almost surely, by the second Borel-Cantelli lemma. A countable intersection over $k$ contradicts uniform continuity on $[0,\infty )$.

For a uniformly elliptic operator $Lu=a^{ij}(x)D_{ij}u$ with continuous coefficients and $W⋐U$,Uniform continuity lets one freeze $a^{ij}$ on sufficiently small open balls and use the perturbation of a constant-coefficient elliptic second-derivative estimate. A partition of unity and cutoff functions reduce the global interior estimate to finitely many such balls; derivatives of the cutoffs produce only the displayed lower-order norm.

Apply the midpoint quadrature rule on each of $n$ equal panels of a partition of an interval, with $h=(b-a)/n$:Each quadrature node is a panel midpoint and each quadrature weight is $h$. For $f\in C^2[a,b]$, put $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$. Summing the panel quadrature errors givesfor some $\xi \in (a,b)$. Each panel contributes at most $K{h}^3/24$; there are $n=(b-a)/h$ panels, explaining the composite midpoint rule error of order $O(h^2)$ in big O notation. On a nonuniform partition of an interval with panel widths $h_j$ and largest width $H$, the corresponding bound is $K{\sum }_j{h}_j^3/24\le K(b-a)H^2/24$.

For $f(x)=x^2$ on $[0,1]$, summing the sampled polynomial exactly yieldsDoubling $n$ divides the absolute error by exactly four. For a general $C^2$ function, halving $h$ divides the displayed error bound by four, but the actual absolute error need not follow that ratio: cancellation can make it smaller or zero. Continuous functions still give convergence as the largest panel width tends to zero, by uniform continuity; without the bounded second derivative, the second-order rate is not guaranteed. Noise sensitivity of a quadrature rule explains a separate limitation when samples are inexact.

Every characteristic function is uniformly continuous and satisfies $\phi (0)=1$ and $|\phi (t)|\le 1$. Continuity follows from the dominated convergence theorem applied to $e^{itX}$.

More explicitly, if $t_n\to t$, then $e^{i{t}_{n}X}\to e^{itX}$ pointwise and $|e^{i{t}_{n}X}|=1$. Apply the dominated convergence theorem to the real part and imaginary part separately. There is no finite moment hypothesis. For uniform continuity, the stronger estimatefollows by factoring out $e^{itX}$, whose absolute value is one. The integrand tends pointwise to zero and is bounded by two, so the dominated convergence theorem applies again. The bound is independent of $t$, which supplies uniform continuity on the entire real line.^[ref]

If the first absolute moment is finite, $|e^{iu}-1|\le |u|$ additionally gives $|{\phi }_X(t+h)-{\phi }_X(t)|\le |h|\mathbb{E}[|X|]$, a Lipschitz continuity estimate. The earlier argument matters because continuity survives even when this expected value is infinite. For convergence in distribution, the separate Lévy continuity theorem concerns limits of whole sequences of characteristic functions; the argument here establishes continuity of one characteristic function.

A nonnegative function that is a uniformly continuous function on the interval $[0,\infty )$ and has a finite integral tends to zero. Otherwise uniform continuity gives disjoint intervals of a fixed width on which the function exceeds a fixed positive value, contradicting integrability. This supplies a useful decay argument for Lyapunov functions with merely continuous vector fields.

To make the fixed width explicit, suppose some $\eta >0$ occurs as a lower bound $f(t_n)\ge \eta$ along a sequence with $t_n\to \infty$. Choose $\delta >0$ from uniform continuity for output tolerance $\eta /2$. On every interval $(t_n-\delta /2,t_n+\delta /2)$ lying in $[0,\infty )$, the function is greater than $\eta /2$. Extract a subsequence whose centers are more than $\delta$ apart. Each disjoint interval contributes at least $\eta \delta /2$ to the integral, so the first $N$ contribute at least $N\eta \delta /2$. This contradicts finiteness of the integral. The same input width at every center is the essential step; continuity alone would allow progressively narrower peaks.

Along the dyadic partitions of an interval of $[0,1]$, the sums ${\sum }_i|{\Delta }_{i}M{|}^p$ of a continuous local martingale tend to zero in probability for $p>2$. If $0<p<2$ and these sums are eventually bounded almost surely, then $M$ is constant on $[0,1]$: compare them with the sums defining quadratic variation, using the maximum increment, which tends to zero by uniform continuity.

A function $f:D\to \mathbb{R}$ is continuous at $a\in D$ whenFirst choose the required output tolerance $\epsilon$; then choose an input tolerance $\delta$ that works for every allowed function input in that window. Picking a different $\delta$ separately after seeing each $x$ would reverse the quantifiers and lose this guarantee. There is no need to exclude $x=a$, because the output difference is then zero.

At an isolated point of the domain of a function, choose a small enough window that its only allowed function input is $a$. This proves continuity there directly. At a point with arbitrarily close other inputs, the same definition is the limit of a function condition $f(x)\to f(a)$. For metric spaces, replace the two absolute values by their distance functions; for arbitrary topological spaces, the corresponding definition is a continuous map.

The input tolerance may depend on $a$. A single choice valid everywhere on the domain of a function is the stronger requirement of uniform continuity. The continuous function $x\mapsto x^2$ on $\mathbb{R}$ shows the difference: for any proposed $\delta >0$, the points $a=1/\delta$ and $b=a+\delta /2$ have $|b-a|<\delta$, yet $b^2-a^2=1+{\delta }^2/4>1$. Thus the fixed output tolerance $\epsilon =1$ admits no global input tolerance on the real line.

Partition $(0,1)$ into $n$ bins of width $\delta =1/n$, and let the discrete random variable $Q_n$ record the bin containing $X$. This is a quantization of $X$. Its probability mass function isand its discrete information entropy is $H_e(Q_n)=-{\sum }_j{p}_j\ln p_j$. Let ${\bar{f}}_j=p_j/\delta$ be the average probability density within bin $j$. Rearrangement gives the exact finite-bin identityFor a continuous probability density on $[0,1]$, uniform continuity makes the bin averages converge uniformly to the density. Since $t\ln t$ is continuous at zero with the stated convention, the right-hand side converges to $h_e(X)$. Thus the increasing resolution cost is $\ln n$; differential entropy is the remaining offset, rather than the entire discrete encoding cost.

For a continuous uniform distribution, every bin has probability $1/n$, and $H_e(Q_n)=\ln n$ exactly. For the continuous uniform distribution's squared transformation above, the exact bin probabilities areThe diagram computes these masses directly, including the bin next to the integrable singularity. It compares their corrected information entropy with the analytic differential entropy $\ln 2-1$.

For $I(u)={\int }_a^{b}f(x,u,u^{\prime })dx$ on $W^{1,\infty }$, continuity of $f,f_u,f_v$ on $[a,b]\times {\mathbb{R}}^2$ gives a bounded Fréchet derivative in the displayed form. Fixing $u$ and restricting $\Vert w{\Vert }_{W^{1,\infty }}$ to a bounded neighborhood confines all intermediate pairs $(u+tw,u^{\prime }+t{w}^{\prime })$ to a compact range up to null sets. Uniform continuity of the partial derivatives bounds the integrated first-order remainder by $(b-a)\omega (\Vert w\Vert )(\Vert w{\Vert }_{\infty }+\Vert w^{\prime }{\Vert }_{\infty })$, where $\omega (s)\to 0$. This is little-oh of the Sobolev norm. The result needs no continuous derivative representative for $u$; essential boundedness suffices.

On the real line, let $T(x)=\max (1-|x|/2,0)$. The Fourier transform of a triangular function gives $\hat{T}(\omega )=2{\sin }^2\omega /{\omega }^2$, with value two at zero. The Gaussian heat kernel has transform $e^{-t{\omega }^2}$. Hence $u=q_t\ast T$ solves the heat equation with initial profile $T$, and the inverse Fourier transform gives the displayed integral. Differentiating for positive time is justified by Gaussian decay. The uniform continuity of $T$ and the vanishing Gaussian mass outside any fixed neighborhood of zero prove uniform recovery of $T$ as $t\downarrow 0$. Convolution preserves total integral two, keeps the solution nonnegative, and smooths the corners for every positive time. Unlike the finite-interval problem in heat flow from a triangular initial temperature, this evolution has no endpoints through which heat can escape.

For a compact metric space with a continuous map $T$, suppose $y$ is a minimal point and $x,y$ are proximal. Every neighborhood $U$ of $y$ admits arbitrarily large $n$ with $T^{n}x,T^{n}y\in U$. First shrink $U$ to an open neighborhood, then choose an open $V$ with $y\in V$ and $\overline{V}\subseteq U$. Minimality and compactness give a finite cover of the orbit closure of $y$ by $T^{-j}V$, $0\le j\le J$. Uniform continuity of these finitely many iterates turns a sufficiently close proximal encounter into a common visit to $U$ after at most $J$ more steps. If the proximal encounters occur only at bounded times, a zero distance at some time makes the two future orbits coincide, and minimality then gives arbitrarily late common visits. Thus injectivity is not needed for this general lemma.

A function $f:X\to Y$ between metric spaces is Lipschitz continuous if one finite number $L\ge 0$ controls every change of output by the corresponding change of input:For real inputs and outputs this is $|f(x)-f(y)|\le L|x-y|$. The same bound must work for every pair in the specified domain of a function. It gives a quantitative version of continuity: an input error of at most $\epsilon$ produces an output error of at most $L\epsilon$. The least possible bound is the Lipschitz constant. Lebl presents the real and metric-space definitions and their connection with uniform continuity. ^[ref]

Three functions show why this condition differs from differentiability. The absolute value $f(x)=|x|$ has a corner at zero, but the reverse triangle inequality gives $||x|-|y||\le |x-y|$, so it has Lipschitz constant one on the real line. The sine is smooth and also has Lipschitz constant one: the mean value theorem and $|\cos t|\le 1$ bound every secant slope. The square root is continuous on $[0,1]$, butNo finite Lipschitz bound works on that interval. At $h=1,1/4,1/16,1/64$, the ratios are $1,2,4,8$. Smaller input changes near zero therefore require larger amplification factors.

For a real function, Lipschitz continuity means that the absolute slopes of all secant lines are bounded. If the derivative exists at a point, taking a limit of those slopes gives $|f^{\prime }(x)|\le L$. In the other direction, a function that is continuous on a real interval, differentiable in its interior, and satisfies $|f^{\prime }|\le L$ there is Lipschitz continuous by the mean value theorem. The interval hypothesis matters: the theorem uses the segment connecting the two inputs.

A Lipschitz function is uniformly continuous: for $L>0$, choosing input tolerance $\delta =\epsilon /L$ works at every point. If $L=0$, the function is constant. The converse fails; the square root on $[0,1]$ is uniformly continuous, and satisfies the weaker Hölder continuity bound $|\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}$, without a finite Lipschitz constant. These distinctions are developed in Lebl's worked examples. ^[ref]

The usefulness is controlled sensitivity. Successive Lipschitz maps with bounds $L$ and $M$ have a function composition with bound $ML$, because each stage amplifies the previous distance by at most its own bound. For an ordinary differential equation, Lipschitz continuity in the unknown state is the condition that makes nearby trial solutions controllably close in the Picard-Lindelöf theorem. For incomplete data, the McShane extension theorem fills in unobserved values while preserving the same bound. This one inequality connects error estimates, uniqueness, and interpolation.

For nonnegative partition-normalized order-$k$ B-splines, let $Vf(t)={\sum }_{i}f({\tau }_i)N_i(t)$ with each sample point inside $[t_i,t_{i+k}]$. If $N_i(t)$ is nonzero, both $t$ and ${\tau }_i$ lie in an interval of length at most $k|\Delta |$. Their function values differ by at most the modulus of continuity at that length. Summing with nonnegative weights of total one proves the displayed bound. With fixed order and vanishing maximum knot gap, uniform continuity gives uniform convergence. The same argument works for bounded piecewise-continuous outputs if full interior knot multiplicities are allowed; their individual continuity should not then be asserted.

If $\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1\to 0$, thenIn finite dimension this follows from multipartite superadditivity of quantum relative entropy, additivity of quantum relative entropy, contraction of trace distance under partial trace, and uniform continuity on the compact state space.

The modulus of continuity of a function $f$ isIt quantifies uniform continuity by recording the largest oscillation of $f$ across pairs of points at distance at most $\delta$.

Let $U\subseteq \mathbb{C}$ be open, let $L$ be a straight line, and let $f:U\to \mathbb{C}$ be continuous on $U$ and holomorphic on $U\setminus L$. Then $f$ is holomorphic on all of $U$. Near a point of $L$, work in a disk whose closure lies in $U$. Split a small filled triangle along $L$ and triangulate the two resulting convex polygons. Translate each triangle slightly into its corresponding open half-plane, keeping it in $U$. Its contour integral is zero by Cauchy's integral theorem. On each segment the integral is the endpoint difference times the integral of $f$ along the unit-interval parametrization; uniform continuity on a containing compact disk therefore makes the contour integral continuous as the vertices move back. Each untranslated piece consequently has zero integral. Internal oriented edges cancel when the pieces are added, so every small triangle has zero integral. Morera's theorem proves the claim locally along the line. The continuity across $L$ is essential; holomorphicity on the two sides alone is insufficient.