Lattice Version of Stone-Weierstrass

From Algebra to Lattice

The Real Stone-Weierstrass Theorem gives uniform density from the algebraic operations of addition and multiplication, together with constants and point separation. There is another route to density that does not require multiplication. It uses the pointwise maximum and minimum of functions. These operations let us combine local approximations while preserving inequalities elsewhere.

Throughout, let \(K\) be a nonempty compact subset of \(\mathbb{R}\), and write \(C(K)\) for the continuous real-valued functions on \(K\), with the supremum norm. For \(f,g\in C(K)\), the functions \(f\vee g\) and \(f\wedge g\) denote their pointwise maximum and minimum. As established in “Lattice Operations on Functions,” both belong to \(C(K)\).

Definition: A vector sublattice \(L\) of \(C(K)\) is a linear subspace of \(C(K)\) that is closed under pointwise maximum and minimum: if \(f,g\in L\), then \(f\vee g\in L\) and \(f\wedge g\in L\). It is unital if it contains the constant function \(\mathbf{1}\).

A vector sublattice need not be an algebra: it need not be closed under pointwise multiplication. Its key extra operations are instead the lattice operations. The lattice version of Stone-Weierstrass says that a unital vector sublattice that separates points is already rich enough to approximate every continuous function uniformly.

Definition: A subset \(L\subseteq C(K)\) separates points if, whenever \(x,y\in K\) and \(x\ne y\), there is an \(h\in L\) such that \(h(x)\ne h(y)\).

Matching Two Values

The first step is to use a separator to prescribe values at two points. The construction uses only linear combinations and constants, so it is available in any unital vector sublattice that separates points.

Lemma (Two-Point Interpolation in a Lattice): Let \(L\) be a unital vector sublattice of \(C(K)\) that separates points. Given \(f\in C(K)\) and \(x,y\in K\), there is a \(g\in L\) such that \(g(x)=f(x)\) and \(g(y)=f(y)\).

Proof. If \(x=y\), the constant function \(g=f(x)\mathbf{1}\) belongs to \(L\) and has the required value. Suppose instead that \(x\ne y\). Point separation gives \(h\in L\) with \(h(x)\ne h(y)\). Define

g=f(x)\mathbf{1} +\frac{f(y)-f(x)}{h(y)-h(x)}\bigl(h-h(x)\mathbf{1}\bigr).

Because \(L\) is a linear subspace containing \(\mathbf{1}\), this function belongs to \(L\). At \(x\), the second term is zero, so \(g(x)=f(x)\). At \(y\),

\begin{aligned} g(y) &=f(x)+\frac{f(y)-f(x)}{h(y)-h(x)}\bigl(h(y)-h(x)\bigr)\\ &=f(x)+f(y)-f(x)\\ &=f(y). \end{aligned}

Thus the required interpolation holds in both cases. \(\square\)

The lemma is a local construction: for each pair of points, it gives a function that matches \(f\) at those points, but it does not promise to stay close to \(f\) everywhere. Lattice operations and compactness will turn these separate matches into global control.

The Lattice Stone-Weierstrass Theorem

Theorem (Lattice Stone-Weierstrass Theorem): Let \(K\) be a nonempty compact subset of \(\mathbb{R}\), and let \(L\) be a unital vector sublattice of \(C(K)\) that separates points. Then \(L\) is dense in \(C(K)\) in the supremum norm.

Proof. Fix \(f\in C(K)\) and \(\varepsilon>0\). We will find \(H\in L\) such that \(|H(t)-f(t)|<\varepsilon\) at every \(t\in K\).

First fix \(x\in K\). For each \(y\in K\), apply the two-point interpolation lemma to choose \(g_{x,y}\in L\) with

g_{x,y}(x)=f(x) \qquad\text{and}\qquad g_{x,y}(y)=f(y).

The function \(g_{x,y}-f\) is continuous and equals zero at \(y\). Therefore the set

U_{x,y}=\{t\in K:g_{x,y}(t)>f(t)-\varepsilon\}

is open relative to \(K\) and contains \(y\). As \(y\) varies, these sets cover \(K\). Compactness gives a finite subcover \(U_{x,y_1},\ldots,U_{x,y_m}\). Define

H_x=g_{x,y_1}\vee\cdots\vee g_{x,y_m}.

This finite maximum belongs to \(L\), since \(L\) is closed under pairwise maxima. For every \(t\in K\), at least one of the selected sets contains \(t\). For its corresponding index \(j\), \(g_{x,y_j}(t)>f(t)-\varepsilon\), and hence

H_x(t)>f(t)-\varepsilon.

At the fixed point \(x\), every function in the maximum has value \(f(x)\). Thus \(H_x(x)=f(x)\).

Now consider the set

V_x=\{t\in K:H_x(t)<f(t)+\varepsilon\}.

It is open relative to \(K\), by continuity, and contains \(x\), because \(H_x(x)=f(x)\). The sets \(V_x\), as \(x\) ranges over \(K\), cover \(K\). Choose a finite subcover \(V_{x_1},\ldots,V_{x_r}\), and put

H=H_{x_1}\wedge\cdots\wedge H_{x_r}.

This finite minimum belongs to \(L\). Every \(H_{x_i}\) is strictly greater than \(f-\varepsilon\) at every point, so their finite minimum satisfies \(H(t)>f(t)-\varepsilon\) for each \(t\in K\). Also, for each \(t\in K\), some \(V_{x_i}\) contains \(t\); then \(H_{x_i}(t)<f(t)+\varepsilon\), so \(H(t)\leq H_{x_i}(t)<f(t)+\varepsilon\). Consequently,

|H(t)-f(t)|<\varepsilon \qquad(t\in K).

The function \(|H-f|\) is continuous on compact \(K\), so by the Extreme-Value Theorem it attains its maximum. Since its value is strictly less than \(\varepsilon\) at every point, that maximum is strictly less than \(\varepsilon\). Therefore \(\|H-f\|_\infty<\varepsilon\). Since \(f\) and \(\varepsilon>0\) were arbitrary, \(L\) is dense in \(C(K)\). \(\square\)

Match pairs of points.
Use point separation and constants to build a function agreeing with the target at a chosen pair.

Take a finite maximum.
For a fixed point, compactness selects finitely many local lower bounds; their maximum stays above the target minus the error everywhere.

Take a finite minimum.
Compactness selects finitely many local upper bounds; their minimum stays below the target plus the error everywhere.

Worked Examples

Worked Example: Piecewise-Linear Functions Are Dense

Let \(L\) be the set of continuous piecewise-linear functions on \([0,1]\), where each function is linear on the intervals of some finite partition. This is a vector sublattice. Sums and scalar multiples are piecewise linear after taking a common refinement of their finite partitions. For maxima and minima, take a common refinement first; on each resulting interval, the difference of the two functions is linear. It either vanishes throughout that interval or has at most one zero there. Splitting at the finitely many such zeros makes the maximum and minimum linear on each piece.

The constants belong to \(L\), and the identity function \(x\mapsto x\) belongs to \(L\) and separates any two distinct points of \([0,1]\). The lattice Stone-Weierstrass theorem therefore shows that \(L\) is dense in \(C[0,1]\). For a direct illustration, consider approximating \(f(x)=x^2\). Divide \([0,1]\) into \(n\) equal intervals, and let \(P_n\) be the linear interpolant of \(x^2\) at the partition points. On an interval \([a,b]\) of this partition, its secant line is

$$ P_n(x)=(a+b)x-ab. $$

Indeed, at \(x=a\) this is \((a+b)a-ab=a^2\), and at \(x=b\) it is \((a+b)b-ab=b^2\). For every \(x\in[a,b]\),

$$ P_n(x)-x^2=(x-a)(b-x). $$

Expanding the right-hand side gives \((a+b)x-ab-x^2\), as required. Since \(b-a=1/n\), the product is nonnegative and is at most \((b-a)^2/4\): writing \(u=x-a\) and \(v=b-x\), we have \(u,v\geq0\), \(u+v=b-a\), and \(uv\leq(u+v)^2/4\). Thus

\|P_n-x^2\|_\infty\leq\frac{1}{4n^2}.

This explicit estimate confirms uniform convergence for this target and is consistent with the density conclusion for every continuous function.

Worked Example: A Clipped Ramp in a Generated Lattice

On \([0,1]\), let \(L\) be the smallest unital vector sublattice of \(C[0,1]\) containing the identity function \(u(x)=x\). It separates points because \(u(x)\ne u(y)\) whenever \(x\ne y\). The theorem implies that \(L\) is dense in \(C[0,1]\), even though its definition uses no multiplication.

For a concrete use of its operations, define

r=(2u-\mathbf{1})\vee 0.

This belongs to \(L\), since \(L\) is a vector subspace and is closed under maxima. Evaluating pointwise gives

r(x)=\max\{2x-1,0\} = \begin{cases} 0,&0\leq x\leq \frac12,\\ 2x-1,&\frac12\leq x\leq1. \end{cases}

At \(x=\frac12\), both formulas give zero, so the function is continuous there. The lattice operation has clipped the negative part of the affine function \(2x-1\) to zero. More generally, expressions such as \((u-a\mathbf{1})\vee0\) create continuous ramps that are zero up to \(a\) and increase afterward. This illustrates how lattice operations construct useful local shapes while the theorem supplies density.

Worked Example: Constants Alone Do Not Separate

Let \(K=[0,1]\) and let \(L\) consist only of the constant functions. It is a unital vector sublattice: maxima and minima of two constants are constant. But it does not separate points, so the lattice Stone-Weierstrass theorem does not apply. In fact, \(L\) is not dense in \(C[0,1]\).

For the function \(f(x)=x\), the distance to a constant \(c\) is

\|x-c\|_\infty=\max\{|c|,|1-c|\}.

The equality follows because \(x-c\) is linear, so its absolute value on \([0,1]\) has its maximum at an endpoint. The triangle inequality gives \(1=|1-0|\leq |c|+|1-c|\), so at least one of these two endpoint errors is at least \(1/2\). Taking \(c=1/2\) makes both errors equal to \(1/2\). Therefore

\inf_{c\in\mathbb{R}}\|x-c\|_\infty=\frac12.

Thus no sequence of constants can approximate \(x\) uniformly. The example shows why the point-separation hypothesis is not a technical formality: lattice operations can combine functions already present, but cannot distinguish points that all those functions treat identically.

Why the Lattice Operations Matter

The proof uses a precise division of labor. Point separation gives the two-point interpolation lemma. The lemma supplies functions that match the target at selected points. Continuity turns each match into a local inequality, and compactness reduces the resulting cover to finitely many sets. The maximum combines lower bounds; the minimum combines upper bounds. Both operations preserve membership in the lattice.

A common pitfall is to assume that one interpolating function already approximates the target on all of \(K\). The lemma says nothing of the kind: its function can behave poorly away from the two selected points. The proof avoids that problem by using one finite maximum to secure a global lower bound and a separate finite minimum to secure a global upper bound. Neither step can simply be omitted.

The argument also clarifies the role of constants. They allow a separator to be shifted and scaled to match the target at two points. The lattice operations then build the global approximant. Multiplication, which is central to the algebraic version of Stone-Weierstrass, is not used.

Takeaway: A unital vector sublattice of \(C(K)\) that separates points is uniformly dense. Two-point interpolation, followed by finite maxima and minima over compact covers, converts local matches into a global approximation.

Check Your Understanding

Use the definitions and proof to answer the following questions.

Which operations must a vector sublattice be closed under, and which algebraic operation is not required?
Why does the two-point interpolation formula require \(h(y)-h(x)\ne0\)?
For a fixed \(x\), why does the finite maximum \(H_x\) remain greater than \(f-\varepsilon\) everywhere on \(K\)?
Why is the second compactness argument needed after constructing \(H_x\)?
How does the identity function show that the piecewise-linear sublattice on \([0,1]\) separates points?
What fails in the density conclusion for the lattice of constant functions?