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An explicit piecewise linear function isThe adjacent formulas agree at $1/3$ and $2/3$, so the function is continuous, and $f(0)=f(1)=0$. Any pair of points at distance $2/3$ can be written $t,t+2/3$ with $0\le t\le 1/3$. For every such $t$,Thus their values are never equal. The horizontal chord of reciprocal-integer length conclusion does not extend to this distance.

Evaluate the generators as coordinate projections on ${\mathbb{R}}^n$. The lattice-ordered group they generate consists of continuous positively homogeneous piecewise linear functions with finitely many integer-linear pieces. Equivalently its elements are finite joins of finite meets of integer linear forms. The universal property follows because the real line detects every identity of Abelian lattice-ordered groups: if an identity fails, pass to a prime ordered quotient and select the finitely many active linear pieces. The resulting rational linear inequalities have a real solution by the alternative underlying Farkas' lemma, giving a real witness to failure. This is a statement about integer coefficients and positive homogeneity, not arbitrary real coefficients or affine constant terms.

A piecewise-linear hat function is a continuous piecewise linear function that selects one node of a finite element mesh: it is one there and zero at every other node. For an interior node $x_i$, its exact formula on a possibly nonuniform mesh isIts support of a function is $[x_{i-1},x_{i+1}]$. It rises on the left interval and falls on the right, producing its hat-shaped graph. The interior hat functions form a basis of the continuous $P_1$ conforming finite element space with homogeneous Dirichlet boundary conditions. Chen's author notes define this nodal basis and illustrate it in one and two dimensions.^[ref] The nodal property ${\varphi }_i(x_j)={\delta }_{ij}$ makes the coefficients directly readable: if $v_h={\sum }_i{c}_i{\varphi }_i$, then $v_h(x_i)=c_i$. Unlike the global Lagrange cardinal polynomials, each of these functions is confined to two neighboring elements.

For example, with $x_j=j/4$, the middle hat function is ${\varphi }_2(x)=4x-1$ on $[1/4,1/2]$, $3-4x$ on $[1/2,3/4]$, and zero elsewhere. Hence ${\varphi }_2(3/8)=1/2$, while ${\varphi }_1(3/8)=1/2$ and ${\varphi }_3(3/8)=0$. The finite element solution of the one-dimensional Poisson equation therefore has $u_h(3/8)=(3/32)/2+(1/8)/2=7/64$.

Local assembly makes this locality computational. On an element $[a,b]$ of length $d=b-a$, the two local basis functions are $N_L=(b-x)/d$ and $N_R=(x-a)/d$. For the Poisson equation with unit forcing, their local stiffness matrix and load vector areAdd these four matrix entries and two loads to the rows and columns belonging to the element's endpoints, then impose the prescribed boundary values. An interior node receives contributions from its two neighboring elements. Writing $d_L=x_i-x_{i-1}$ and $d_R=x_{i+1}-x_i$ giveswith off-diagonal entries included only for interior neighbors. On a uniform mesh these reduce to $2/h$, $-1/h$ and $h$. Nonadjacent hat functions have no derivative overlap of positive length, so all distant entries vanish. This produces a sparse matrix with only three diagonals: the global coupling is built entirely from local two-node interactions.

Starting with $p_0=0$ on $[-1,1]$, this polynomial recurrence gives $0\le p_r(x)\le |x|$ and $p_{r+1}(x)\ge p_r(x)$. Indeed the error satisfies $|x|-p_{r+1}=(|x|-p_r)[1-(|x|+p_r)/2]$. Its pointwise limit solves $p^2=x^2$, so it is $|x|$. Dini's theorem gives uniform convergence. Each continuous piecewise linear function is an affine function plus a finite linear combination of $(x-t{)}_{+}=(x-t+|x-t|)/2$. Scaling and translating the recurrence therefore gives polynomial approximation of every such function, and their density gives the Weierstrass approximation theorem.