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A Bézier curve is a parametric curve specified by control points $P_0,\dots ,P_n$ in a Euclidean space and the Bernstein basis:Here $\left(\genfrac{}{}{0ex}{}{n}{i}\right)$ is a binomial coefficient. Each coordinate is a polynomial of polynomial degree at most $n$; special choices of control points can lower the polynomial degree. The ordered control points form the control polygon. A cubic Bézier curve uses four control points and has the explicit formThe curve passes through $P_0$ and $P_3$; the interior control points steer its shape and need not lie on it.^[ref]

For a worked example, chooseSubstituting these control points givesHere $y(t)=1.5-6(t-1/2{)}^2\le 1.5$, so neither interior control point, both at height $2$, lies on the curve. De Casteljau's algorithm computes the same point using only successive linear interpolations.

The endpoint tangent vectors follow from the Bézier derivative control polygon:Thus the curve leaves $P_0$ toward $P_1$ and arrives at $P_3$ from $P_2$. These nonzero derivatives determine the endpoint tangent lines; if an endpoint derivative vanishes, that formula alone does not determine its tangent direction. Endpoint handles are useful when arranging a parametric first-derivative join of Bézier curves.

The parameter $t$ is usually not arc length. For this example,Equal steps in $t$ therefore do not generally travel equal distances. Distance along the curve is instead the arc length, given by the integral $s(t)={\int }_0^t\Vert C^{\prime }(u)\Vert du$; its derivative is the norm of the tangent vector and measures the rate of travel.

Two geometric facts make Bézier curves especially convenient for design. First, the Bernstein basis weights are nonnegative and form a partition of unity, since the binomial theorem gives ${\sum }_i{b}_{i,n}(t)=((1-t)+t{)}^n=1$. Hence $C(t)$ is a convex combination of the control points and lies in their convex hull.^[ref] The convex hull is the smallest convex set containing the control points, not their axis-aligned bounding box. Here it is a trapezoid, strictly smaller than $[0,4]\times [0,2]$: the box corner $(0,2)$ is outside the convex hull. The curve need not fill or touch every edge of either enclosure. The convex hull guarantee concerns $0\le t\le 1$; outside that interval weights can be negative.

Second, applying an affine map to every control point gives exactly the same result as applying it to the entire curve. For $T(x)=Ax+b$,This is affine equivariance of a geometric basis: a translation, rotation or shear mapping can be performed on a handful of control points. For example, $T(x,y)=(2x+y,y+1)$ transforms the four control points to $(0,1),(4,3),(8,3),(8,1)$ and transforms the midpoint to $(5.5,2.5)$, also obtained from the same weights $(1,3,3,1)/8$.

These editable handles explain the usefulness of Bézier curves in computer aided geometric design, vector graphics paths and outline fonts: a small collection of points controls an entire smooth stroke, and resizing the design preserves its mathematical shape. Cambridge's computer-graphics notes describe their use in graphic design, typography and PostScript outlines. The industrial context includes Pierre Bézier's work at Renault and Paul de Casteljau's work at Citroën.^[ref]

// Target: analysis.bigb

For every $c>0$, the stochastic process ${\tilde{B}}_t=c^{-1/2}{B}_{ct}$ is standard Brownian motion. Its Brownian increments remain independent and have variance $c^{-1}c(t-s)=t-s$; continuity and the zero starting value are preserved. Equivalently,Equality here is of whole-process distributions, including every finite-dimensional distribution. It is not an equality of the two expressions along a particular realization. With $c=4$, speeding time up fourfold requires dividing displacement by two to recover the standard law. The figure applies exactly these coordinate changes to four finite samples; the shared colors indicate reused samples, while the dashed lines show pointwise standard deviations, not bounds on entire sample paths.

The same square-root normalization explains how a simple random walk approaches Brownian motion. Let $S_k={\sum }_{j=1}^k{\xi }_j$, where the independent steps take values $+1$ and $-1$ with equal probability. Then $R_n(t)=S_{\lfloor nt\rfloor }/\sqrt{n}$ has variance $\lfloor nt\rfloor /n$. Its linear interpolation isand ${\sup }_{0\le t\le 1}|Y_n(t)-R_n(t)|\le 1/\sqrt{n}$. Donsker's theorem says $Y_n\Rightarrow B$ in $C[0,1]$ with the uniform norm: this is convergence in distribution of random paths. The interpolation bound vanishes, but the theorem does not assert pointwise convergence of an arbitrarily fixed walk to an independently chosen Brownian motion. The usefulness is that many-step discrete motion has a universal continuous limit.^[ref]

A coarse-grid correction improves a fine-grid approximation by solving a smaller equation for the part of its approximation error representable on a coarse mesh. Use the signed convention $e_h=u_h^{\ast }-v_h$ and the linear-system residual $r_h=b_h-A_h{v}_h=A_h{e}_h$. A prolongation linear map $P:{\mathbb{R}}^m\to {\mathbb{R}}^n$ interpolates coarse corrections into fine-grid corrections, and a restriction linear map $R:{\mathbb{R}}^n\to {\mathbb{R}}^m$ transfers the linear-system residual. Solve $A_H{z}_H=R{r}_h$ exactly or approximately and update $v_h\leftarrow v_h+P{z}_H$. The plus sign follows from defining the approximation error as exact minus approximate. It is the correction equation's right-hand side $R{r}_h$, rather than merely restricted solution values, that determines the update.

For the multigrid method model $A_h=h^{-2}\mathrm{tridiag}(-1,2,-1)$, take $n=2m+1$, so $H=2h$ and the coarse nodes coincide with every second fine node. Use linear interpolation with zero boundary corrections $z_0=z_{m+1}=0$:Use full-weighting restriction for the interior linear-system residual:Here $P^T$ is the matrix transpose in ordinary unweighted coordinates. With these specific transfers, the Galerkin coarse-grid operator isThus it equals the rediscretized Poisson equation operator on this constant-coefficient one-dimensional mesh. The equality is special to the stated model and transfers. In particular, dropping $h^{-2}$ from the fine matrix while using $H^{-2}$ on the coarse mesh changes the correction unless the right-hand sides and transfers are rescaled consistently.

Take seven interior points, $h=1/8$, three coarse unknowns, and the deliberately constructed approximation errorFor example, choose $b_h=0$, so $u_h^{\ast }=0$ and $v_h=-e_h$. Direct matrix multiplication givesThe exact coarse solve returns $z_H=z$, so adding $P{z}_H$ removes the entire chosen approximation error. A broad triangular error can have a sharply localized linear-system residual: the latter measures discrete curvature, not error height. This is why the two should not be confused. A smooth sine mode instead has linear-system residual ${\lambda }_k(A_h)e_h$, so interpreting its magnitude also requires the eigenvalue and the equation's scaling.

The algebra explains the exact cancellation beyond this example. Assume $P$ has linearly independent columns and choose an invertible Galerkin coarse-grid operator $A_H=R{A}_{h}P$. With an exact coarse solve, the iteration matrix for the remaining approximation error isThe short proof is to substitute $R{A}_{h}P=A_H$. Thus every vector in the column space of $P$ is eliminated, and the corrected linear-system residual restricts to zero. These are exact projection identities, not a statement that every visually smooth vector belongs to that column space. If $A_h$ is a symmetric matrix and a positive-definite matrix, and $R=c{P}^T$ with $c>0$, then $A_H$ is also a positive-definite matrix. Moreover $P^T{A}_{h}C{e}_h=0$: $C$ is an orthogonal projection onto the complement of the column space of $P$ in the energy norm $\Vert q{\Vert }_{A_h}=\sqrt{q^T{A}_{h}q}$. Equivalently, the added correction is the best approximation to $e_h$ from that column space in this energy norm. Without these symmetry and transfer assumptions, the projection need not be orthogonal.^[ref]

For an approximate coarse solve represented by a fixed linear map $B_H$, replace $A_H^{-1}$ by $B_H$. ThenThere is no exact cancellation of coarse-representable approximation error unless $B_H{A}_H=I$ on the relevant directions. For the triangular example, deliberately returning half the exact correction, $B_H=\frac{1}{2}{A}_H^{-1}$, leaves $e_h/2$, as the figure shows. This controlled mathematical comparison illustrates an incomplete coarse solve; it is not a measured performance result or a prescription for a geometric multigrid V-cycle. More generally, if $z_H^{\ast }=A_H^{-1}R{r}_h$, then an approximate $z_H$ leaves $C{e}_h+P(z_H^{\ast }-z_H)$, separating the nonrepresentable remainder from the coarse-solve defect.

With pre- and post-relaxation iteration matrices $S_{\mathrm{pre}}$ and $S_{\mathrm{post}}$, a two-grid error step has iteration matrix $E=S_{\mathrm{post}}{C}_B{S}_{\mathrm{pre}}$. A geometric multigrid V-cycle supplies $B_H$ by one recursive coarse cycle from zero, with a direct solve only at the bottom. Effective smoothing, a coarse column space that approximates the surviving approximation error, and a sufficiently accurate coarse correction must work together. Coarsening alone supplies no convergence guarantee, and correcting an arbitrary fine-grid vector exactly would require a coarse column space large enough to contain it.

On an equal-panel partition of an interval, with $h=(b-a)/n$, sum the trapezoidal rule over adjacent panels:There are $n+1$ distinct quadrature nodes. Each interior sample belongs to two panels and therefore has quadrature weight $h$, while each endpoint has quadrature weight $h/2$. Equivalently, the composite trapezoidal rule integrates the piecewise linear function obtained by linear interpolation exactly.^[ref]

For $f\in C^2[a,b]$ and $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$, summing panel quadrature errors givesfor some $\xi \in (a,b)$. Thus panel quadrature errors of order $O(h^3)$ accumulate to order $O(h^2)$, using big O notation.^[ref] On a nonuniform partition of an interval with widths $h_j$ and largest width $H$, the bound is $K{\sum }_j{h}_j^3/12\le K(b-a)H^2/12$.

For the polynomial $f(x)=x^2$ on $[0,1]$, direct summation yieldsBoth quadrature errors shrink by exactly four when $n$ doubles, with opposite signs. This exact relation belongs to this polynomial example; the trapezoidal rule is not always twice as inaccurate as the midpoint quadrature rule.

The left plot uses exact polynomial sums; the right adds a common sample bias $\delta =0.002$ to the composite trapezoidal rule. That mathematical model converges to $1/3+\delta$, showing why smaller panels alone do not ensure accuracy for noisy samples. It illustrates noise sensitivity of a quadrature rule, rather than measured data.

De Casteljau's algorithm evaluates a Bézier curve by repeated linear interpolation between neighboring control points. Fix $t\in [0,1]$, start with $P_i^{(0)}(t)=P_i$, and computeEach new row has one fewer point. The final point is $P_0^{(n)}(t)=C(t)$. Use the same $t$ at every stage: it describes the fraction of each line segment traversed from its first endpoint toward its second. A cubic Bézier curve needs three levels containing $3$, $2$, and $1$ linear interpolations, respectively.^[ref]

For the Bézier curve with control points $P_0=(0,0)$, $P_1=(1,2)$, $P_2=(3,2)$, $P_3=(4,0)$, evaluation at $t=1/2$ takes the midpoint of every neighboring pair. The first level givesThe second level givesand the third level givesThe intermediate control polygons shrink toward the evaluated point. Each step is a convex combination, giving a geometric explanation of the convex hull property; every step also commutes with an affine map.

The same construction splits the Bézier curve exactly, with no approximation. The outer edges of the triangular array give the new control points: for a cubic, $(P_0,Q_0,R_0,S)$ specify the left piece and $(S,R_1,Q_2,P_3)$ specify the right piece.^[ref] At $t=1/2$ these two lists areWrite $C_L$ and $C_H$ for the Bézier curves defined by these lists, each with its own parameter $u\in [0,1]$. ThenTheir images reproduce the original curve, while their derivatives account for the shorter parameter intervals. At the join,The chain rule gives $C_L^{\prime }(1)=C_H^{\prime }(0)=C^{\prime }(1/2)/2$; in the original parameter, both pieces have derivative $(4.5,0)$. This is the interval scaling in the parametric first-derivative join of Bézier curves.

Repeated midpoint splitting is useful for drawing vector graphics paths: each piece has a smaller control polygon and its own convex hull enclosure. A renderer can refine pieces until they are sufficiently close to line segments at the chosen display scale. Evaluating $C(t)$ and splitting it are thus two results of the same short algorithm.

For a proof outline, expanding the recursion givesPascal's identity proves this by induction; at $r=n$, $i=0$ it is precisely the Bernstein basis formula for the Bézier curve. The geometric construction remains useful even when expanding the polynomial by hand would be cumbersome.

// Target: analysis.bigb

For the constant-coefficient model $x^{\prime }(t)=-kx(t-\tau )$ with $k>0$ and $\tau \ge 0$, the zero equilibrium point is asymptotically stable if and only if $k\tau <\pi /2$. For positive $\tau$, perturbations are measured by the supremum norm of the initial history of delayed feedback; in the stable regime, every continuous history produces a solution tending to zero. At equality, nondecaying sinusoids prevent asymptotic stability. For $k\tau >\pi /2$, growing oscillatory solutions exist. These statements concern this ideal linear time-invariant system, rather than an arbitrary controlled plant. The corresponding linear equation and its threshold are developed in Cain and Reynolds, section 6.4, pages 174–177.

Substitute the exponential function $x(t)=e^{st}$, allowing a complex number $s$. The characteristic equation isFor positive feedback delay this is not a characteristic polynomial: it has infinitely many roots. A root's real part determines growth or decay of its exponential mode, and its imaginary part determines the angular frequency. To find the first boundary, set $s=i\omega$ with $\omega >0$ and use Euler's formula:The smallest positive crossing is $\omega =k$ and $\omega \tau =\pi /2$. Thus at ${\tau }_c=\pi /(2k)$ the complex conjugate roots are $s=\pm ik$. In fact, with the matching initial history of delayed feedback,solves the delay differential equation exactly: a quarter-cycle feedback delay turns the past displacement into precisely the current derivative. Its cycle lasts $2\pi /k=4{\tau }_c$. The amplitude is arbitrary; this linear model does not select a finite-amplitude oscillation.

The roots cross toward growth as feedback delay increases. Differentiating the characteristic equation at fixed $k$ givesThis calculation identifies the direction of the first destabilization. The full threshold also requires ruling out earlier unstable roots: if $s=\alpha +i\beta$ with $\alpha \ge 0$ and $\beta >0$, the characteristic equation implies $\cos (\beta \tau )\le 0$ and $0<\beta \le k$, hence $k\tau \ge \beta \tau \ge \pi /2$. A real nonnegative root is impossible because $s+k{e}^{-s\tau }>0$ there. This supplies the geometric reason the threshold is a quarter-cycle rather than an arbitrary fitting constant.

The tradeoff is concrete. Without feedback delay, increasing $k$ shortens the exponential decay time constant $1/k$. With known $\tau >0$, the same proportional control must satisfy $k<\pi /(2\tau )$. For $\tau =0.2\mathrm{s}$, the threshold is $k\approx 7.854{\mathrm{s}}^{-1}$: $k=5{\mathrm{s}}^{-1}$ is stable and $k=8{\mathrm{s}}^{-1}$ is unstable. A design close to the boundary leaves little room for additional feedback delay; reducing $k$ or reducing measurement latency increases that room. The zero-delay time constant is only a comparison scale, not the decay time of the delayed response. Frank discusses the gain–delay compromise in Control Theory Tutorial, section 13.4 and figure 13.3.

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 $f$.

For a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.

A finite element mesh partitions a domain into simple regions called elements, such as intervals, triangles or tetrahedra. In a conforming partition, distinct element interiors are disjoint, their closures cover the domain closure, and any nonempty intersection is a common face, edge or vertex of both elements. Chen's author notes give this compatibility condition for simplicial meshes.^[ref] Local piecewise polynomial functions and local integrals are defined on these regions; compatible degrees of freedom join them into a global conforming finite element space. The assembled stiffness matrix and mass matrix then inherit the local overlap pattern.

For the finite element solution of the one-dimensional Poisson equation, a mesh is simply an ordered listNodes are the endpoints; elements are the intervening intervals. Four equal elements have five nodes but only three unknown nodal values after imposing the zero Dirichlet boundary conditions. A nonuniform partition such as $0,1/8,1/4,1/2,1$ has element lengths $1/8,1/8,1/4,1/2$ and $h=1/2$, so counting elements alone does not measure the largest unresolved scale.

Refining a finite element mesh replaces some or all elements by smaller ones. For continuous piecewise linear functions in one dimension, retaining all old nodes gives nested conforming finite element spaces $V_h\subset V_{h/2}$ when every interval is bisected. The old approximation remains available, and the new interior nodes allow additional changes of slope. With a symmetric coercive bilinear form and exact integration, best approximation from Galerkin orthogonality means that the energy norm error cannot increase under such nested refinement.

For the finite element solution of the one-dimensional Poisson equation, uniform refinement from four to eight elements increases the unknown count from three to seven. The exact formulas give maximum errors $0.0078125$ and $0.001953125$, and energy norm errors approximately $0.0721688$ and $0.0360844$. Refinement near a region of strong curvature can spend unknowns where the linear interpolation error matters most; convergence in a general problem still needs approximation and regularity assumptions such as those in the finite element interpolation estimate.

Consider the Poisson equation on an interval with homogeneous Dirichlet boundary conditions:Two integrations and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a test function $v$ that vanishes at both endpoints. Integration by parts givesThe boundary term is zero. Thus the weak formulation is to find $u\in V=H_0^1(0,1)$ such thatThe zero-trace Sobolev space $H_0^1(0,1)$ consists of square-integrable functions with square-integrable weak derivatives and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous piecewise linear functions are admissible despite corners. The Poincaré inequality makes $a$ a coercive bilinear form on $V$. Strang derives this one-dimensional weak formulation, the piecewise-linear hat functions, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.^[ref]

Take the uniform finite element mesh $x_j=jh$, $j=0,\dots ,n$, with $h=1/n$. The continuous $P_1$ trial space isHere $P_1$ means polynomials of degree at most one on each interval. This is a conforming finite element space in $V$, with $n-1$ interior piecewise-linear hat functions ${\varphi }_i$ as a basis. Their endpoint values enforce the Dirichlet boundary conditions automatically. Set $u_h={\sum }_{i=1}^{n-1}{c}_i{\varphi }_i$ and test the weak formulation against each ${\varphi }_i$:Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared interval, giving $-1/h$. All other derivative products vanish. The stiffness matrix is the tridiagonal matrixThe load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The piecewise-linear hat function article explains how the same entries arise by adding local two-node contributions.

With four mesh intervals, $h=1/4$, there are three unknowns:For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The coefficients are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, linear interpolation gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.

In this model the Galerkin method solution equals the nodal approximation $I_{h}u$ obtained by linear interpolation on any finite element mesh. Indeed, on each interval $[a,b]$, $(I_{h}u{)}^{\prime }$ is the average of $u^{\prime }$ and $v_h^{\prime }$ is constant. Therefore ${\int }_a^b(u^{\prime }-(I_{h}u{)}^{\prime })v_h^{\prime }dx=0$. Summing proves Galerkin orthogonality for $I_{h}u$, and uniqueness gives $u_h=I_{h}u$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for finite element methods with other operators or integration rules.

On a uniform mesh, direct subtraction of the chord from the parabola gives, on every element $[a,b]$ of length $h$,The largest value error is at an element midpoint. Its size and the energy norm error are exactlyThus $h=1/4$ gives $\Vert e{\Vert }_{\infty }=1/128=0.0078125$ and $\Vert e{\Vert }_a=1/(4\sqrt{12})\approx 0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx 0.0360844$. Halving the mesh spacing quarters the maximum value error but only halves the energy norm error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.

A gel electrophoresis size calibration relates migration distance to known fragment lengths on the same gel. Here is an explicitly chosen model for linear double-stranded DNA, not measured data or a universal law. Let $L$ be the numerical length in base pairs and $d$ the numerical migration distance in millimetres from the wells:The logarithm has a dimensionless argument: $L/100$ is the length divided by 100 base pairs. The chosen DNA ladder fragments $L=100,250,500,1000$ give $d=55.00,50.22,46.61,43.00$, respectively. The gel bands therefore appear in the reverse order of length along the direction of migration.

To invert the calibration, rearrange the logarithm:Our chosen unknown $L=375$ gives $d=48.1116248\dots$, between the 250-base pair and 500-base pair references. Inserting that distance into the inverse recovers 375; retaining only $d=48.11$ gives about $375.12$ base pairs. The extra digits reflect rounding in this model, not the resolving power of a real gel.

For linear interpolation between those two reference gel bands, use the logarithm of length. Here the fraction of the distance from the 500-base pair reference toward the 250-base pair reference is $t={\log }_{10}(500/375)/{\log }_{10}2\approx 0.4150$, giving $L=500(250/500{)}^t=375$. Direct linear interpolation of lengths instead gives approximately $396.24$ base pairs. This difference explains why the choice of calibration coordinates matters. A real calibration must be checked over its usable range; folded RNA or circular DNA does not automatically share the linear double-stranded DNA calibration.

Interpolation constructs a function agreeing exactly with prescribed values at sample points. The resulting function is an interpolant. Choosing a family determines how missing values are filled: polynomial interpolation selects a polynomial, while linear interpolation joins adjacent samples by straight segments. The McShane extension theorem instead permits any Lipschitz function with a specified bound and provides envelopes controlling all allowed choices.

Between two prescribed values, linear interpolation assigns $(1-t)f(a)+tf(b)$ at $(1-t)a+tb$, for $0\le t\le 1$. Joining successive samples produces a piecewise linear function. For a continuous function on a compact interval, interpolants along grids with mesh tending to zero converge in the uniform norm; their error is bounded by the modulus of continuity at the mesh size.

Let $A$ be a nonempty subset of a metric space $X$, and let $f:A\to \mathbb{R}$ satisfy a Lipschitz bound $L$. Thenare real-valued Lipschitz functions on all of $X$, agree with $f$ on $A$, and obey the same bound $L$. Every other extension $H$ satisfying that bound lies between them, $G\le H\le F$. Despite the infimum in its formula, $F$ is the largest such extension: every given sample supplies an upper constraint $H(x)\le f(a)+Ld(x,a)$.

Take three real samples $f(0)=0$, $f(1)=1/2$, $f(2)=0$, and allow the bound $L=1$. Their least Lipschitz constant is only $1/2$, so this is deliberately a looser permitted bound. The two extensions areAt $x=1/2$, the allowable interval is $0=G(1/2)\le H(1/2)\le F(1/2)=1/2$. The interpolant obtained by linear interpolation is a piecewise linear function and has value $1/4$ there. Extending it constantly outside $[0,2]$ obeys the same bound on the entire real line. At $x=5/2$, the allowable interval is $[-1/2,1/2]$. Thus the data and the bound control possible values without selecting a unique curve.

The construction is useful when data are known only on a subset: it extends them without increasing an allowed Lipschitz bound. A short justification explains why the envelopes work. Fix $a_0\in A$. The triangle inequality and the bound on $f$ implyso the infimum is finite below; the term $a=a_0$ makes it finite above. At a sample $x=a_0$, all upper terms are at least $f(a_0)$, while that term gives equality. Finally, $d(x,a)\le d(y,a)+d(x,y)$ gives $F(x)\le F(y)+Ld(x,y)$ after taking an infimum; exchanging $x,y$ proves the bound. The lower envelope follows by applying the same argument to $-f$.

E. J. McShane's paper Extension of range of functions appeared in the Bulletin of the American Mathematical Society, volume 40, in 1934, pp. 837–842. ^{[ref] [ref]}

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

Work on $([0,1],\mathcal{B},dt)$ as a probability space, with Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the filtration generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the dyadic slope martingale byThe endpoint $t=1$ may be assigned any value, since it is a null set for Lebesgue measure. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb{E}[G_{n+1}\mid {\mathcal{F}}_n]=G_n$ almost everywhere, so this is a martingale. The Lipschitz condition gives $|G_n|\le K$ everywhere except possibly at the freely chosen endpoint, where we take zero.

Apply the Lp martingale convergence theorem with $p=2$. Its limit $g$ has $|g|\le K$ almost everywhere and $G_n\to g$ in L1 norm as well, by Cauchy-Schwarz inequality. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded measurable function on the entire interval.

Set $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the linear interpolation of $f$ on the dyadic grid. The Lipschitz condition gives $\Vert f_n-f{\Vert }_{\infty }\le K{2}^{-n}\to 0$. AlsoThe two uniform limits coincide, giving the absolutely continuous function representationThe chosen bound holds for every $t$ after the null-set modification; the integral identity holds for every $x$ simultaneously.

Use the same dyadic slope martingale $G_n$ and dyadic filtration as in (c). Each $G_n$ is integrable, since it takes finitely many finite values. On each dyadic cell, the absolute slope is $2^n$ times the absolute endpoint increment. Consequently the hypothesis in the PDF is exactlyThus $(G_n)$ is uniformly integrable. It is also bounded in L1 norm: choose a finite ${\lambda }_0>0$ at which the supremum of the tails is finite, and use $\Vert G_n{\Vert }_1\le {\lambda }_0+{\sup }_j{V}_j(f,{\lambda }_0)$. The uniformly integrable martingale convergence theorem supplies $g\in L^1([0,1])$ with $G_n\to g$ in L1 norm.

The functions $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$ are again the dyadic linear interpolations of $f$. Since $f$ is continuous on a compact interval, it is uniformly continuous, and $\Vert f_n-f{\Vert }_{\infty }\le {\omega }_f(2^{-n})\to 0$, where ${\omega }_f$ is its modulus of continuity. On the other hand, the integral of $G_n-g$ is uniformly bounded in absolute value by $\Vert G_n-g{\Vert }_1\to 0$. Hence the dyadic slope-tail criterion for absolute continuity givesNo boundedness of $g$ is asserted here; the tail condition permits integrable densities that are unbounded.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.

Integrate the ordinary differential equation across one step, using the fundamental theorem of calculus:The midpoint quadrature rule approximates the integral by $hf(x_n+h/2,y(x_n+h/2))$. Approximate the unknown midpoint state by the midpoint value of linear interpolation between the endpoint states:Combining these approximations gives the implicit midpoint ruleIt is an implicit time-stepping method because its new state occurs inside the evaluation of $f$. An equivalent derivation uses a straight-line approximation on the step and imposes the ordinary differential equation at its midpoint, a one-stage collocation method.