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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.

For prescribed values $F(x_i)$ at ordered points $x_0<\cdots <x_n$, the connect-the-dots interval map is the unique piecewise linear function obtained by linear interpolation between consecutive data points.

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.

The functional precision set of a subdivision scheme consists of functions exactly reproduced by refining their uniform samples with the scheme's consistent parameter placement. Polynomial precision usually specifies the reproduced polynomial space. If exactness is required for every grid spacing and shift, the binary midpoint-interpolation scheme has precision exactly the affine functions. At one fixed grid, its entire reconstruction range is the space of continuous piecewise linear functions on that grid; this larger range should not be confused with polynomial precision or arbitrary-grid reproduction.

The piecewise linear functions $h_n$, for distinct integer indices, are linearly independent over the integers. Restrict a finite integer relation to $y=1$. At $x=n$, only $h_n$ has a jump of one in its slope; hence the coefficient of that function must be zero. This proves that the underlying group of the two-generated free Abelian lattice-ordered group has infinite rank of an abelian group.

Let $f$ have bounded variation, and let $g,h$ be continuous functions on $[a,b]$. In any tagged sum for the Riemann-Stieltjes integral,Taking the limit proves the displayed error bound. Thus approximating $g$ uniformly to accuracy $\epsilon$ controls this integration error by $\epsilon V_a^b(f)$, even when the integrator is not differentiable. For the piecewise linear function in partition example of total variation, accuracy $0.001$ gives an error at most $0.011$. Total variation supplies exactly the movement budget that an endpoint difference would miss.

Every real function on $[a,b]$ has bounded variation if and only if it is a difference of two nondecreasing functions. For a normalized version, let $v(t)=V_a^t(f)$ and $v(a)=0$. Its accumulated positive and negative variations areBoth are nondecreasing functions, $P(a)=N(a)=0$, andThis is the Jordan decomposition of a function of bounded variation.^[ref] Equivalently one may write $f=f_{+}-f_{-}$ with $f_{+}=(v+f)/2$ and $f_{-}=(v-f)/2$; those functions need not vanish at $a$.

For the piecewise linear function in partition example of total variation, the successive values of $P$ are $0,2,2,6,6$, while those of $N$ are $0,0,3,3,5$. At $t=4$, the total rise is $6$ and the total fall is $5$: their difference gives the net change $1$, and their sum gives the total variation $11$. A declining segment increases $N$ while $P$ stays constant; an ascending segment does the opposite.

For a short proof outline, total variation is additive over adjacent intervals, because any two partitions of an interval have a common partition refinement. Thus $v(t)-v(s)=V_s^t(f)\ge |f(t)-f(s)|$ for $s<t$. Substituting this inequality into the formulas for $P$ and $N$ shows their increments are nonnegative. Conversely, if $f=g-h$ for two nondecreasing functions, the triangle inequality bounds each partition sum by $g(b)-g(a)+h(b)-h(a)$. This proves finite total variation without restricting the number of changes of direction.

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.

A linear spline is a continuous piecewise linear function on a specified spline knot sequence. For uniform integer spline knots, the hat basis in the displayed representation interpolates the data $s(i)=c_i$. The derivative is constant on each open span and can jump at a knot. Coinciding neighboring slopes remove a particular jump, but generic data give only $C^0$ regularity.

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]}

Define a piecewise linear function by joining the five points $(0,0),(1,2),(2,-1),(3,3),(4,1)$. Its net change is $1$, but the increments at these knots are $2,-3,4,-2$. Their absolute values giveThis is its exact total variation, not merely one lower bound. Given any partition of an interval, insert all five knots. The triangle inequality ensures that this partition refinement cannot decrease the sum. On each resulting monotone function segment, the absolute increments telescope, so every refined sum is $11$. The knot partition of an interval itself attains $11$.

Sampling only at $0,4$ gives $1$. Sampling at $0,1,3,4$ gives $2+1+2=5$. Inserting the remaining knot at $2$ reveals the descent and rise that were hidden inside $[1,3]$, giving $11$. These are nested partitions of an interval: their sums increase because each is a partition refinement of the preceding one. Arbitrarily chosen unrelated partitions of an interval need not produce an increasing sequence of sums.

The equioscillation theorem says that $p_n\in P_n$ is the unique best uniform approximation to a real continuous function $u$ on $[-1,1]$ exactly when its error attains alternating values $\pm E_n$ at $n+2$ ordered distinct points. If $E_n=0$, the error vanishes everywhere and the conclusion is immediate.

For the exponential function, write the candidate line as $p_1(x)=ax+b$. Its error $r(x)=e^x-ax-b$ is strictly convex. Equal endpoint errors require $a=(e-e^{-1})/2=\sinh 1$, and the interior minimum is at $c=\log a$, because $r^{\prime }(c)=e^c-a=0$. Equating the endpoint error to the negative interior error givesIndeed $r(-1)=r(1)=E_1$ and $r(c)=-E_1$; convexity puts every other error value between these extremes. The equioscillation theorem proves optimality.

For the reference-point assertions, put $M=\Vert u-q_a{\Vert }_{\infty }$. The alternating-reference error bound gives $|h_a|\le E_n$. In fact this inequality is strict here. To see the equality obstruction explicitly, let $p_n$ be optimal, $d=p_n-q_a$, andSince $d$ has polynomial degree at most $n$, its order-$(n+1)$ divided difference vanishes: ${\sum }_r{\omega }_{r}d(x_r)=0$. If $E_n=|h_a|>0$, change the overall sign to take $h_a>0$. The error bound for $p_n$ implies $(-1{)}^{r}d(x_r)\ge 0$. The signs of the weights are $(-1{)}^{n+1-r}$, so every term in this sum has the same weak sign. Every term must vanish; $d$ then has $n+2$ distinct zeros, forcing $d=0$. This contradicts $M>E_n$. If $h_a=E_n=0$, $u=p_n$ and the same zero-count argument again forces $q_a=p_n$, also a contradiction. ThusAt every old reference point the absolute error is only $|h_a|$, so a maximizer $\eta$ cannot be one of those points. The printed same-sign choice of $k$ additionally presumes $h_a\ne 0$; when $h_a=0$, none of the reference errors has the nonzero sign of the maximal error.

The printed replacement rule needs an endpoint qualification. For an interior $\eta \in (x_0,x_{n+1})$ and nonzero $h_a$, replacing the adjacent endpoint of its containing reference interval with the same error sign preserves spatial order and alternating signs. The single-point exchange in polynomial minimax approximation then gives the requested strict increase. On an irregular grid, the globally nearest same-sign reference point need not be that adjacent endpoint, so adjacency must be specified. More generally, for any correctly ordered replacement set $z_0<\cdots <z_{n+1}$ on which the old error alternates in sign, let $w_r={\prod }_{s\ne r}(z_r-z_s{)}^{-1}$. The vanished divided difference of $q_a$ and the new alternation equations showThis is a strict weighted average because all the weights are positive in absolute value and $M>|h_a|$.

For an exterior $\eta$ whose error sign is opposite that at the nearest endpoint, merely replacing the nearest same-sign point, as printed, need not preserve alternation. A concrete counterexample uses $n=2$, $q_a=0$, $h_a=1$, and reference pointsLet $u$ be the piecewise linear function through $(-1,-2),(0,1),(1/5,-1),(2/5,1),(3/5,-1),(1,-1)$. Then $M=2$ at $\eta =-1$, and $1<E_2\le 3/2<2$: the upper bound is achieved by the trial constant $-1/2$, while equality at the lower bound is excluded by the preceding zero-count argument. The printed rule selects $k=1$ and hence $z=(0,-1,2/5,3/5)$. Neverthelesssatisfy every printed new alternation equation, with $\boxed{|h_b|=61/64<1=|h_a|}$. Direct substitution at the four points checks this failure; reordering those points would destroy their assigned alternating signs.

The corrected exterior rule inserts $\eta$ as the new extreme reference point and drops the farthest endpoint when its sign is opposite the nearest endpoint. If its sign agrees with the nearest endpoint, replace that endpoint. Interior exchanges replace the adjacent reference point with the same sign. These rules preserve an ordered alternating set, so the strict-increase formula applies in every case.

The resulting Remez algorithm solves the $n+2$ linear alternation equations for the polynomial coefficients and reference height, locates a global maximum of the error, and performs a valid exchange unless $M=|h_a|$. The interval $|h_a|\le E_n\le M$ is an explicit optimality certificate; a small gap $M-|h_a|$ gives a stopping test. Strict increase of bounded reference heights alone does not prove convergence, so that stronger conclusion needs the additional analysis of the chosen exchange algorithm.

Write the spline interpolation operator as $P_{\mathbf{x}}=T{A}_{\mathbf{x}}^{-1}R$, where $R$ samples at the interpolation sites and $T$ synthesizes the normalized B-spline expansion. Sampling has operator norm at most one, and nonnegative B-splines with sum at most one give $\Vert T\Vert \le 1$. HenceHere the matrix norm is the maximum absolute row sum. Existence of the inverse matrix requires the usual ordered distinct interpolation sites and the Schoenberg–Whitney theorem support conditions, with endpoint values interpreted by their one-sided limits. If the displayed diagonal condition is read without the implicit ordering and distinctness, it is insufficient: all three sites equal to $1/2$ in the quadratic example give positive diagonal entries but three identical rows. The estimate applies whenever the interpolating linear map in the question is defined uniquely.

For the quadratic Bernstein basis, $N_1=(1-x{)}^2$, $N_2=2x(1-x)$ and $N_3=x^2$. Sampling and inverting giveThe three cardinal Lagrange interpolation polynomials areThe Lebesgue constant of interpolation gives the exact operator norm: the upper bound follows from $|Pg(x)|\le \Vert g{\Vert }_{\infty }{\sum }_i|{\ell }_i(x)|$. To attain it at a maximizing point, prescribe at the three sites the signs of the corresponding cardinal polynomials and extend those values by a continuous piecewise linear function of norm one.

On $0\le x\le 1/2$, the first two cardinal polynomials are nonnegative and the last is nonpositive. Since their sum is one, their absolute sum is $1-2{\ell }_2(x)=1+2x-4x^2$. Its maximum is $5/4$ at $x=1/4$. Reflection gives the same maximum at $3/4$ on the other half. In particular the data $(1,1,-1)$ produce $1+2x-4x^2$, attaining $5/4$. ThusThe coefficient estimate loses cancellation among the B-splines, which explains why it is not sharp for the resulting function norm.

Fix $\epsilon >0$ and put $\eta =\epsilon /2$. For each pair $s,t\in T$, choose $a_{s,t}\in A$ whose errors at these two points are smaller than $\eta$. By continuity, there is a neighborhood $U_{s,t}$ of $t$ on which $a_{s,t}>f-\eta$. For a fixed $s$, these neighborhoods cover the compact space $T$. Choose a finite subcover $U_{s,t_1},\dots ,U_{s,t_m}$ and formThen $b_s>f-\eta$ everywhere, while $b_s(s)<f(s)+\eta$, because every chosen function has error smaller than $\eta$ at $s$. Since $b_s$ is continuous, there is a neighborhood $W_s$ of $s$ on which $b_s<f+\eta$.

A second use of compactness supplies $W_{s_1},\dots ,W_{s_q}$ covering $T$. SetEvery $b_{s_i}$ is above $f-\eta$, so their minimum is also above it. At each point at least one $b_{s_i}$ is below $f+\eta$, and hence so is their minimum. This proves $\Vert a-f{\Vert }_{\infty }\le \eta <\epsilon$. The two finite operations are the essential mechanism in lattice approximation from two-point approximation; no closure under addition or multiplication has been assumed.

For the corollary take $A$ to consist of all continuous piecewise linear functions on $[0,1]$. On a common finite partition, the maximum or minimum of two affine pieces can change which piece it selects only at their crossing. Adding those finitely many crossings to the partition shows that $A$ is closed under both operations. For two distinct points, the affine polynomial through the two prescribed values belongs to $A$ and matches $f$ exactly there; for equal points use a constant function. The proved criterion therefore gives density of continuous piecewise linear functions in $C[0,1]$ in the supremum norm. Concretely, interpolation on a partition of maximum gap $h$ also gives error at most $\omega (f,h)$ by uniform continuity.

The Weierstrass approximation theorem asserts that every real continuous function on a compact interval is a uniform limit of polynomials. An affine change of variable reduces the interval to $[0,1]$. There are two complementary routes: construct positive approximation operators, or construct an algebra dense enough to approximate absolute values and hence pointwise maxima and minima.

For the first route, the Korovkin theorem reduces convergence of positive linear operators on continuous functions $L_n$ to the three tests $1,t,t^2$. Here is its proof. Given $f\in C[0,1]$ and $\epsilon >0$, uniform continuity supplies a $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ for $|t-x|<\delta$. For all $t,x$ one then hasApply positivity to the two real inequalities obtained from this bound. At $x$ this givesThe expression in brackets is nonnegative and tends uniformly to zero by the three test assumptions. Also $L_{n}1\to 1$ uniformly. Taking the upper limit of the supremum norm and then letting $\epsilon \downarrow 0$ proves convergence for every $f$. This is the quadratic barrier estimate for positive approximation operators.

The Bernstein polynomial supplies the constructive corollaryThe binomial theorem makes its nonnegative weights sum to one. Differentiating the binomial theorem, or using the first two binomial distribution moments, givesThe Korovkin theorem proves $B_{n}f\to f$ uniformly, and each $B_{n}f$ is a polynomial, proving the Weierstrass approximation theorem. The same criterion gives the Schoenberg spline operator convergence in Solution 3. For periodic functions its counterpart uses $1,\cos t,\sin t$: the nonnegative barrier $1-\cos (t-x)$ is bounded away from zero outside any fixed small arc. This periodic Korovkin test set gives convergence of the normalized Fejér sums and density of trigonometric polynomials.

The Lebesgue convergence criterion for approximation operators addresses stability. Suppose bounded linear operators $L_n:C[a,b]\to C[a,b]$ converge to the identity on every polynomial. Then they converge on every continuous function if and only if ${\sup }_n\Vert L_n\Vert <\infty$. Sufficiency is the extension of approximation operators from a dense subspace: for a polynomial $p$ with $\Vert f-p\Vert <\eta$ and a common bound $M$,First let $n\to \infty$, then $\eta \downarrow 0$. Necessity follows from the Uniform boundedness principle, since a convergent $L_{n}f$ is bounded for every $f$. Applied to approximation projectors, the Lebesgue inequality for approximation projectors isIndeed $P_{n}v=v$ makes $f-P_{n}f=(f-v)-P_n(f-v)$; take the infimum. Thus density of the ranges and a uniform bound on the Lebesgue constants of interpolation imply convergence. Density alone does not control a sequence of interpolation projectors.

There is also a kernel version of this stability criterion, useful for trigonometric polynomial approximation. If integrable periodic kernels $K_n$ have uniformly bounded $L^1$ norms, their integrals tend to one, and ${\int }_{|t|\ge \delta }|K_n(t)|dt\to 0$ for every $\delta >0$, then convolution with $K_n$ converges uniformly to every continuous periodic function. Subtract $f(x)\int K_n$ and split the error into a small arc, bounded by $\omega (f,\delta ){\sup }_n\Vert K_n{\Vert }_1$, and its complement, bounded by $2\Vert f{\Vert }_{\infty }{\int }_{|t|\ge \delta }|K_n|$. The remaining mass error tends to zero. This approximate identity argument shows directly why nonnegative unit-mass Fejér kernels work even though ordinary Fourier partial sums need not converge uniformly for every continuous target.

Lebesgue's constructive approach to the Weierstrass approximation theorem starts instead with the absolute value. On $[-1,1]$ define the polynomial recurrence for the absolute valueInduction gives $0\le p_r(x)\le |x|$, monotonicity, andThe pointwise limit satisfies $p(x{)}^2=x^2$ and is nonnegative, hence equals $|x|$. Dini's theorem makes the convergence uniform. More explicitly, the continuous decreasing errors have, for any positive tolerance, open sublevel sets covering the compact interval; a finite subcover supplies one index valid everywhere. Every continuous piecewise linear function has the formThe $c_j$ are its changes in slope. Scaling and translating the absolute-value approximation shows that $L$ is uniformly approximable by polynomials. Solution 4, or polygonal interpolation and uniform continuity, makes these piecewise linear functions dense in $C[a,b]$. Approximating $f$ first by $L$ and then by a polynomial is another complete proof of the Weierstrass approximation theorem.

The Stone-Weierstrass theorem generalizes from intervals to a compact Hausdorff space $K$: a real subalgebra $A\subseteq C(K,\mathbb{R})$ containing the constants and separating points is uniformly dense. Its proof uses the same absolute-value mechanism. The uniform closure of a function algebra is an algebra. If $a$ belongs to this closure, approximation of $|t|$ by polynomials on a bounded interval containing its range proves that $|a|$ also belongs to the closure. Therefore the closure containsPoint separation and constants allow exact interpolation of any two real values: choose $h\in A$ with $h(s)\ne h(t)$ and solve for $\alpha ,\beta$ in $\alpha h+\beta$. The lattice approximation from two-point approximation proved in Solution 4 now makes the closure all of $C(K,\mathbb{R})$.

The complex version requires a unital, point-separating subalgebra closed under complex conjugation. Its real-valued members separate points using real or imaginary parts, so the real result approximates both parts of every complex target. Conjugation closure is essential: polynomials in $z$ alone on the closed unit disk cannot approximate $\overline{z}$ uniformly, because ${\int }_{|z|=1}p(z)dz=0$ whereas ${\int }_{|z|=1}\overline{z}dz=2\pi i$.

Among the corollaries are density of multivariate polynomials on compact subsets of real Euclidean space, since coordinates separate points, and density of trigonometric polynomials on the circle, since sine and cosine separate points there. On a compact interval, polynomials with rational coefficients are a countable dense family, giving separability of $C[a,b]$. The real and complex approximation results above concern uniform approximation of continuous targets; neither asserts convergence of arbitrary interpolation schemes or arbitrary Fourier series.

A concrete model of the two-generated free Abelian lattice-ordered group is the lattice subgroup of real-valued functions on ${\mathbb{R}}^2$ generated by the coordinate projections $X(x,y)=x$, $Y(x,y)=y$. Its elements are finite joins of finite meets of integer linear forms $ax+by$. Thus they are continuous positively homogeneous piecewise linear functions, with pointwise addition and lattice operations. This is the piecewise integer-linear representation of a free Abelian lattice-ordered group.

For the construction, first form the term algebra on two symbols using addition, negation, zero, join and meet, then identify terms that agree in every Abelian lattice-ordered group. Evaluation under any assignment of the two generators factors through this quotient, proving its universal property. To obtain the concrete function model, use the identity-detection theorem that the real line generates the variety of Abelian lattice-ordered groups. Its proof passes a failed identity to a prime ordered quotient, chooses its finitely many active linear pieces, and uses the rational-inequality alternative underlying Farkas' lemma to find a real witness. This makes evaluation on ${\mathbb{R}}^2$ injective as well as surjective onto the generated function group.

Its underlying group is not ${\mathbb{Z}}^n$ for any finite $n$. The integer-linear hinge functions have infinite independent rank: $h_k(x,y)=(x-ky)\vee 0$ are integer-linearly independent for distinct integers $k$. On $y=1$, only $h_k$ has a slope jump at $x=k$, so each coefficient in a finite relation must vanish. The underlying group therefore has infinite rank of an abelian group despite having two lattice generators.

For the usual spline interpolation problem the sites are distinct and ordered; the Schoenberg–Whitney theorem then makes $A_x$ invertible under the stated support positivity. More generally, the following argument applies whenever the interpolation map in the question is well defined and $A_x$ is invertible.

Factor the spline interpolation operator as $P_x=B{A}_x^{-1}R$, whereSampling has operator norm at most one from the supremum norm to ${\ell }_{\infty }$. Nonnegativity and the subpartition of unity for B-splines giveConsequently $\Vert P_x\Vert \le \Vert A_x^{-1}{\Vert }_{{\ell }_{\infty }}$.

For the reverse bound, select a row $r$ of $A_x^{-1}$ with maximal absolute row sum, and choose $y_j=\mathrm{sgn}((A_x^{-1}{)}_{rj})$, assigning any value of modulus at most one when the entry is zero. Then $\Vert y{\Vert }_{{\ell }_{\infty }}=1$ andAt distinct sites there is a continuous piecewise linear function $f$ with $f(x_j)=y_j$ and $\Vert f{\Vert }_{\infty }=1$: interpolate linearly between the ordered sites and extend constantly to the endpoints. The given uniform-norm stability of a B-spline basis therefore impliesThus the B-spline interpolation operator norm satisfiesPositivity $N_i(x_i)>0$ alone, without the usual distinct-site convention, is insufficient to define an interpolation operator. For example, take order-two B-splines with knots $0,1,2,3$ and $x_1=x_2=3/2$. Both diagonal values are $1/2$, but the B-spline collocation matrix has identical rows $(1/2,1/2)$ and is singular. The displayed proof uses the well-defined interpolation map posited in the question; ordered distinct sites supply its standard existence guarantee.

Work in the complete Banach space $C([0,1])$ with the uniform norm. Each $E_m$ is closed. Indeed if $f_j\to f$ uniformly and $x_j$ witnesses membership in $E_m$, pass to a subsequence with $x_j\to x$. For each $y\in [0,1]$, uniform convergence and continuity giveThus $x$ witnesses membership for $f$.

Each $E_m$ has empty interior. Approximate an arbitrary $f$ uniformly by a piecewise linear function $q$ on a uniform mesh, and let $D$ bound the absolute slopes of $q$. On a sufficiently fine uniform refinement of mesh $\delta$, add a triangular oscillation of fixed small amplitude $b$, alternating between $b$ and $-b$ at successive vertices. The resulting function $g$ is still as close to $f$ as desired, and every segment has absolute slope at least $2b/\delta -D$. For any $x$, choose the farther endpoint of a segment containing $x$. Its distance $h$ is between $\delta /2$ and $\delta$, sowhen $\delta$ is small. Therefore $g\notin E_m$.

By the Baire category theorem, the complement of ${\bigcup }_m{E}_m$ is dense and nonempty. If the local limsup in the question were finite at some $x$, its nearby differences would satisfy a finite bound. The remaining differences, with $|h|\ge \delta >0$, are bounded by $2\Vert f{\Vert }_{\infty }{\delta }^{-\alpha }$. Combining the bounds would put $f$ in some $E_m$, a contradiction. Hence the required limsup is infinite at every point, including one-sided approaches at the endpoints.

TakeNowsoThe recurrent pieces of this covering graph are the two-cycle $I_0\leftrightarrow I_2$ and the loop at $I_1$; neither contains two competing closed routes. Alsoso this ordering forces no 3-cycle.

The bound is attained by the piecewise linear functionIndeed,is a 4-cycle with the required spatial order. On $I_0\cup I_2$, the fourth iterate is the identity function. Within $I_1$, the fixed point is $5/9$ andso every other point eventually leaves $I_1$ and enters $I_0\cup I_2$. The map therefore has no horseshoe in any iterate and is not chaotic. The minimum number of distinct 3-cycles is consequently $\boxed{0}$.