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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.

For a standard Gaussian vector $G$ and a smooth Lipschitz function with $\Vert \nabla f\Vert \le K$, represent its centered value as the terminal martingale $P_{1-t}f(B_t)-P_{1}f(0)$. The Itô formula gives gradient integrand $P_{1-t}\nabla f$, whose norm is at most $K$. Its terminal quadratic variation is therefore at most $K^2$. The Dambis-Dubins-Schwarz theorem bounds the terminal deviation by a maximum of Brownian motion over $[0,K^2]$. Symmetry, the Brownian reflection principle, and the Gaussian tail bound give the displayed inequality. The square on the Lipschitz constant is essential by scaling.

Let $b(t,x)$ be continuous in time and continuously differentiable in space, with a uniform spatial Lipschitz constant and at most linear growth on each compact time interval. Its characteristic flow map exists forward and backward over every bounded time interval. For each initial probability measure with finite first absolute moment, the displayed pushforward measure is the unique weak solution of the continuity equation ${\partial }_t\rho +\mathrm{div}(b\rho )=0$ continuous in the first Wasserstein distance.

The Gronwall inequality bounds the flow by $C_T(1+|x|)$ and therefore preserves finite first absolute moments. The chain rule establishes the weak equation. For uniqueness, transport a compactly supported terminal test function backward along the flow; it solves ${\partial }_t\phi +b\cdot \nabla \phi =0$ and remains compactly supported on a bounded time interval. Inserting it into the weak identity identifies every terminal integral with that of the claimed pushforward measure. A zero divergence makes the corresponding scalar transport equation and the continuity equation coincide.

A clearance bound between trajectory samples converts finitely many separation checks into a guarantee over a continuous trajectory, using a justified bound on how quickly clearance can change. Clear endpoints alone are insufficient: a foot or another robot part can approach an obstacle and move away again between them. The primary planning formulation requires constraints along the entire path.^[ref]

Let nominal clearance $c(s)$ have Lipschitz constant $L$ on a checked interval $[s_i,s_{i+1}]$ of length $h$. If both endpoint clearances are at least $c_{\min }$, then every intermediate point is within $h/2$ of an endpoint. ConsequentlyThis follows directly from $c(s)\ge c(s_i)-L|s-s_i|$ using the nearer endpoint. A uniform additional worst-case geometry or execution allowance $e$ gives actual clearance at least $c_{\min }-Lh/2-e$.

A tighter interval certificate retains the two different endpoint values, $a=c(s_i)$ and $b=c(s_{i+1})$. Writing $u=s-s_i$ gives two lower cones,If $L>0$ and $|a-b|\le Lh$, the cones meet inside the interval. Their common height yieldsIf the endpoint difference exceeds $Lh$, the proposed Lipschitz bound and exact endpoint values are inconsistent; simply applying the formula does not repair that inconsistency. With $L=0$, clearance is constant.

For an original example, take $h=0.10\mathrm{s}$, $L=0.20\mathrm{m}{\mathrm{s}}^{-1}$, $a=18\mathrm{mm}$ and $b=26\mathrm{mm}$. The simpler endpoint-minimum bound gives $8\mathrm{mm}$; the two-cone bound gives $12\mathrm{mm}$. Subtracting a $4\mathrm{mm}$ execution allowance leaves a certified $8\mathrm{mm}$ actual gap, sufficient for a requested $5\mathrm{mm}$ margin. These are lower bounds, not assertions that the trajectory reaches their values.

A counterexample explains why the allowance is necessary. On $[0,1]$, $c(s)=1-4s(1-s)$ has endpoint values one but touches zero at the midpoint. Checking only those endpoints misses the contact. Its derivative is bounded in magnitude by four, so a sufficiently fine sample grid with that valid bound can expose the missing guarantee.

The constant must match the geometry and parameter units. For a point moving relative to a fixed closed obstacle set, the distance to that set changes by at most the point’s displacement: the triangle inequality gives $|d_O(x)-d_O(y)|\le \Vert x-y\Vert$. A speed bound therefore supplies a time-based clearance bound. A rotating body requires a bound on every occupied point, such as the angular clearance bound for a rotating body; terrain-height clearance across a discontinuous edge cannot use a finite slope bound across that jump.

Adaptive clearance sampling subdivides intervals that fail the chosen certificate, concentrating computation where the margin or speed bound is least favorable. Failure of a conservative certificate means the interval needs better bounds or further checking; it does not itself prove a collision. For a proposed robot-foot test, validate the motion and geometry error bounds in one coordinate frame and compare the certified interval minima with dense independent measurements. Use the robust swing-foot clearance margin allowances without counting the same uncertainty twice.

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.

Suppose $f$ is a differentiable convex function with coordinate Lipschitz continuity constant $L_i>0$. Put $\varphi (u)=D_f(u,y)=f(u)-f(y)-\nabla f(y{)}^T(u-y)$, the Bregman divergence from the supporting affine function at $y$. The subgradient inequality gives $\varphi \ge 0$, and its coordinate Lipschitz constant is still $L_i$. The coordinate descent lemma with step size $1/L_i$ yieldsSince ${\partial }_i\varphi (x)={\partial }_{i}f(x)-{\partial }_{i}f(y)$, rearrangement proves the bound. Continuous differentiability suffices; a Hessian matrix is unnecessary.

Target: gradient-descent.bigb, under Lipschitz gradient.

The displayed estimate, for all $x$ and real $t$, defines a coordinate Lipschitz constant $L_i$ for the partial derivative in coordinate $i$. It implies the coordinate descent lemmaby the fundamental theorem of calculus and integration of the Lipschitz continuity estimate. Subtracting an affine function leaves these constants unchanged.

The apparently weaker estimate along the singleton negative gradient direction is equivalent when every real step size is allowed: if the initial partial derivative is nonzero, any displacement can be written $-\alpha {\partial }_{i}f(x)e_i$; if it is zero but the final partial derivative is nonzero, reverse the endpoints; if both vanish the estimate is trivial.

Target: gradient-descent.bigb, under Lipschitz gradient.

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 $f$.

This explicit numerical method is a two-stage Runge-Kutta method and a one-step method. For a sufficiently smooth ordinary differential equation, the chain rule gives $y^{\prime \prime }=f_x+f_{y}f$, and Taylor's theorem shows that its local truncation error is $O(h^3)$. Uniform Lipschitz continuity of $f$ in its solution argument gives an increment-function Lipschitz constant at most $L(1+hL/2)$. The Lipschitz one-step error bound then proves second-order numerical convergence on a fixed time interval. This method uses an intermediate prediction from $y_n$; the explicit midpoint linear multistep method instead requires a separate previous value $y_{n-1}$.

For Picard iteration $y_{n+1}(x)=b+{\int }_a^{x}f(s,y_n(s))ds$, $y_0=b$, assume $|f|\le M$ and a Lipschitz constant $L$ on a rectangle containing every iterate. The first difference is at most $M|x-a|$. Integrating the preceding bound multiplied by $L$ proves the displayed estimate by mathematical induction. Its uniform bounds sum to a convergent exponential series for every finite $Lh$, so the Weierstrass M-test gives uniform convergence without requiring $Lh<1$.

For two arbitrary initial functions $y,z$ in a function set preserved by the Picard iteration operator $T$, the same induction givesThus some iterate is a contraction mapping on any fixed finite interval for which every iterate stays in that set. The iterated contraction argument gives a unique fixed point of $T$ on a nonempty complete metric space, without requiring $Lh<1$. For a rectangle with field bound $M$ and vertical radius $k$, $Mh\le k$ supplies the required self-mapping property.

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.

For a Gaussian random vector $X\sim N(m,\Sigma )$ and an $L$-Lipschitz function $f$ in the Euclidean norm, put $v=L^2\Vert \Sigma {\Vert }_{\mathrm{op}}$. If $v>0$, the Gaussian concentration inequality givesand thereforeWrite $X=m+{\Sigma }^{1/2}G$ with a standard Gaussian random vector $G$. The composed function has Lipschitz constant at most $L\Vert \Sigma {\Vert }_{\mathrm{op}}^{1/2}$. The Gaussian logarithmic Sobolev inequality and Herbst argument give the exponential bound, and the Chernoff bound supplies the tails.^[ref]

For example, $f(x)=\Vert x\Vert$ has $L=1$, so its deviations depend on the largest covariance eigenvalue rather than directly on the number of coordinates. The center $\mathbb{E}\Vert X\Vert$ can still grow with dimension. If $v=0$, $f(X)$ is constant almost surely and every positive deviation has probability zero.

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 $\mathbb{R}$.