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Take the continuous function $f(x)=|x-1/2|$ on $[0,1]$. It is Lipschitz continuous with Lipschitz constant one, but is not differentiable at $x=1/2$. For $n=2$, the sample values are $1/2,0,1/2$, soThis is already a useful warning: the Bernstein polynomial is built from grid values but is not generally their polynomial interpolation.

Every finite approximant is differentiable, and symmetry makes its derivative zero at the center. The target instead has left derivative $-1$ and right derivative $1$. Smooth curves can therefore approach a corner in supremum norm while their center derivatives have no target value to approach. The Lipschitz error bound for Bernstein approximation quantifies the improvement in values.

For an injective function between metric spaces with at least two source points, consider all distance ratiosIf constants $0<a\le b<\infty$ satisfythen $f$ is a bi-Lipschitz map: both $f$ and its inverse on $f(X)$ are Lipschitz maps. The bound $b/a$ compares the most expanded pair with the most contracted pair. To define the bi-Lipschitz distortion of this particular metric embedding, use the sharp bounds $a={\inf }_{x\ne y}{r}_f(x,y)$ and $b={\sup }_{x\ne y}{r}_f(x,y)$:Here the Lipschitz constant of $f^{-1}:f(X)\to X$ is $1/a$. If no positive finite bounds exist, the bi-Lipschitz distortion is infinite. For a finite source, every injective function has finite bi-Lipschitz distortion, but its size can depend on the source. Matoušek gives the equivalent definition using one scale factor and a multiplicative error bound. ^[ref]

Changing units is harmless: replacing $d_Y$ by $\lambda d_Y$, with $\lambda >0$, multiplies both $a$ and $b$ by $\lambda$ and leaves $b/a$ unchanged. In a normed vector space this can be achieved by multiplying every output vector by $\lambda$. Thus bi-Lipschitz distortion one means that all distances change by the same factor; an isometric embedding additionally requires that factor to be one. A stretch factor with a specified normalization is a related quantity; when comparing sources, check which overall scale has been fixed.

Take the four corners $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$ of a square, with Euclidean distance in both source and target. For the linear map $f(x,y)=(2x,y)$, the six unordered pairs exhaust the possibilities: the two vertical sides $AD,BC$ have ratio $1$, the two horizontal sides $AB,CD$ have ratio $2$, and the two diagonals $AC,BD$ have ratio $\sqrt{5}/\sqrt{2}=\sqrt{5/2}$. Consequently $a=1$, $b=2$, and $\mathrm{dist}(f)=2$. In contrast, $g(x,y)=(3x,3y)$, three times the identity function, gives ratio $3$ for every pair and has bi-Lipschitz distortion $3/3=1$.

The ratio chart makes the reason visible: a common multiplier moves every ratio together, while unequal coordinate scaling spreads the ratios apart. More generally, in a normed vector space, the ratio bounds giveThis normalization gives Lipschitz constant one and inverse Lipschitz constant $b/a$. It is only a change of scale, so it cannot improve the bi-Lipschitz distortion.

The best possible bi-Lipschitz distortion of a source into a specified target is a separate quantity:The infimum runs over all injective functions, and need not be attained. The notation $c_p(X)$ means the analogous infimum over metric embeddings into an Lp space. A calculation for one chosen $f$ gives an upper bound for $c_Y(X)$; proving optimality also requires a lower bound for every possible $f$.

The bounded Lipschitz norm combines the Lipschitz constant and supremum norm of a bounded real-valued Lipschitz function on a metric space. Both summands satisfy the triangle inequality; the supremum term makes the sum definite even on constant functions. Its unit ball requires the sum of the two bounds to be at most one, a stronger requirement than bounding each separately by one.

Equip the symmetric group $S_n$ with its uniform probability measure and unnormalized Hamming distance. If $f$ has Lipschitz constant $K$, it obeys the displayed two-sided bound, and each one-sided tail has the same exponential without the factor two. Reveal a uniform permutation one image at a time. Two possible next images have completion sets paired by swapping those two values, changing exactly two coordinates. Thus the corresponding conditional means differ by at most $2K$. Each increment of the Doob exposure martingale has conditional range length at most $2K$. Applying the conditional Hoeffding lemma gives $\mathbb{E}{e}^{\lambda (f-\mathbb{E}f)}\le e^{n{\lambda }^2{K}^2/2}$; optimizing the exponential Markov bound proves the bound. A mere absolute-increment bound of $2K$ gives a weaker constant, so using the range length matters.

This initial value problem uses a square root and absolute value to define a continuous function on all of ${\mathbb{R}}^2$. For every finite waiting time $c\ge 0$, a forward solution of a differential equation on $t\ge 0$ isThe constant solution of a differential equation $y_{\infty }(t)=0$ is another choice. For example, at $t=3$, waits $c=0,1,2,3$ give values $9,4,1,0$. These are dimensionless mathematical trajectories; no measured physical waiting time is being supplied.

Verification uses the original ordinary differential equation, including the join. Before departure, both $y_c^{\prime }$ and $2\sqrt{|y_c|}$ vanish. For $t>c$, their common value is $2(t-c)$. For $c>0$, the difference quotient at the join isThus the derivative exists at $c$, equals the rate law there, and is continuous across the join. For $c=0$ the same computation gives the right derivative at the starting endpoint. For $c>0$, the left and right second derivatives at departure are $0$ and $2$: these forward solutions of a differential equation are continuously differentiable functions, without being twice differentiable at departure. The local two-sided statement of the Peano existence theorem is separate from this explicitly forward family.

The rate law tends to zero as $y\to 0$, but its secant slopes from zero are unbounded:No finite local Lipschitz constant exists at zero. At $\epsilon =10^{-2},10^{-4},10^{-6}$ the quotient is $20,200,2000$, even though the rates are shrinking to zero. Continuity controls the rate itself; Lipschitz continuity would control its change relative to the state displacement.

There is also a numerical lesson. The Forward Euler method starting exactly at zero gives $Y_{k+1}=Y_k+2\Delta t\sqrt{|Y_k|}=0$ at every step. It reproduces the constant solution of a differential equation for every positive step size, while the moving trajectories remain equally valid. A computation producing one trajectory cannot establish uniqueness. Likewise, the separable differential equation calculation that divides by $\sqrt{|y|}$ only describes the moving pieces: checking the original equation recovers the zero pieces and their joins.

On a separable metric space, a sequence uniformly bounded in bounded Lipschitz norm has a pointwise convergent subsequence. Enumerate a countable dense subset and use the Bolzano-Weierstrass theorem and diagonal subsequence argument to obtain convergence there. Uniform control of the Lipschitz constant then makes the subsequence Cauchy at every point. The pointwise closure of the bounded Lipschitz unit ball preserves the original norm bound after rescaling.

For a real continuous function $f$ on $[0,1]$, let $G_n$ be its secant slope on each length-$2^{-n}$ dyadic cell. Regard $([0,1],\mathcal{B},dt)$ as a probability space and use the filtration of half-open dyadic cells with the endpoint $\{1\}$ as a separate null cell. The mean of the two child slopes equals the parent slope, so $(G_n)$ is a martingale. Its integral gives the linear interpolation of $f$ on that grid. If $f$ is Lipschitz continuous, these slopes are bounded by its Lipschitz constant; the Lp martingale convergence theorem then supplies a bounded integral density for $f$.

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

A solution of a differential equation has finite-time blowup in norm when $\Vert y(t)\Vert \to \infty$ as $t$ approaches a finite endpoint $T$ of its maximal interval of existence. This is stronger than merely approaching an excluded boundary of an open state domain while staying bounded. The finite-dimensional continuation criterion distinguishes these obstructions; for a locally Lipschitz continuous ordinary differential equation on $I\times {\mathbb{R}}^d$, a finite maximal endpoint inside $I$ necessarily entails finite-time blowup.

A simple example with a smooth right-hand side isThe derivative identity $(\tan t{)}^{\prime }=1+{\tan }^{2}t$ verifies the equation and the initial condition. Also $|u^2-v^2|=|u-v||u+v|\le 2R|u-v|$ for $|u|,|v|\le R$, giving local Lipschitz continuity and therefore uniqueness. Yet the positive blowup time is $T=\pi /2\approx 1.57079633$. Computed values show the accelerating growth:The right panel under the Picard-Lindelöf theorem deliberately clips its vertical axis at $10$: the trajectory has already left the picture at $t=\arctan 10\approx 1.47112767$, before reaching $T$. A plot ceiling is a display limit, not a finite value at blowup.

The tangent function also quantifies the last part of the trajectory. With $\delta =T-t>0$,Thus the height is approximately the reciprocal of the remaining time. At remaining times $\delta =0.1,0.01,0.001$, the heights are approximately $9.96664442,99.99666664,999.99966667$. These follow from the sine and cosine limits at zero and explain the steep rise without relying on the appearance of a clipped function graph.

The distinction is useful when interpreting a solution of a differential equation: uniqueness certifies a trajectory while it exists, while growth estimates decide how far it can be followed. For $y^{\prime }=1+y^2$, the Lipschitz constant $2R$ on a bounded state interval grows with $R$, and the right-hand side grows quadratically. There is no global Lipschitz constant or linear-growth bound of the kind used by the finite-dimensional continuation criterion. Smoothness at every finite state is entirely compatible with finite-time blowup.

On a Polish space $(X,\rho )$, probability measures with finite first moments satisfyThe supremum runs over functions with Lipschitz constant at most one. Fixing $f(x_0)=0$ ensures integrability from the moment assumption.

A Lipschitz bound is any valid inequality of the displayed form with a finite constant $L$. It need not use the least Lipschitz constant. For $f(x)=3x-2$, both $L=3$ and $L=5$ are valid, but $L=2$ fails. The tighter bound gives a sharper error estimate: input error at most $0.01$ guarantees output error at most $0.03$ using $L=3$, compared with the valid but weaker $0.05$ using $L=5$. The bound constrains errors; it does not state that every pair achieves that amplification.

The Lipschitz constant is the least nonnegative number that works in the defining Lipschitz bound, equivalently the displayed supremum of distance ratios. A constant function has constant zero; for a domain with at most one point we also set it to zero. An infinite supremum means there is no finite Lipschitz bound. The least constant need not be attained by one pair of inputs.

For $f(x)=3x-2$ on the real line, every distinct pair has ratio $3$, so the Lipschitz constant is $3$. For $f(x)=x^2$ on $[-R,R]$, with $R>0$,Pairs $x=R$, $y=R-\epsilon$ have ratio $2R-\epsilon$, approaching $2R$. Thus the Lipschitz constant is $2R$, although no distinct pair realizes that exact value. Reporting the domain is essential: the same function has no finite global Lipschitz constant on $\mathbb{R}$.

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.