Constants in Approximation Algebras

Why Constants Matter

The Real Stone-Weierstrass Theorem applies to a unital subalgebra of \(C(K)\): an algebra containing the constant functions. But some useful approximation algebras are not unital. In that setting, it matters whether constants can at least be approximated uniformly by elements of the algebra. This tutorial gives a criterion for when they can.

Throughout, let \(K\) be a nonempty compact subset of \(\mathbb{R}\), and let \(A\) be a subalgebra of \(C(K)\). Here “subalgebra” does not assume that \(A\) contains the constant function \(\mathbf{1}\), whose value is \(1\) at every point. If an algebra contains \(\mathbf{1}\), it contains every constant function by scalar multiplication. If it does not, its uniform closure might still contain them.

Definition: The algebra \(A\) has no common zero if, for every \(x\in K\), there is some \(f\in A\) such that \(f(x)\ne 0\). Equivalently, there is no point \(x\in K\) at which every function in \(A\) vanishes.

The function \(f\) in this definition may depend on \(x\). The condition does not require one member of \(A\) to be nonzero everywhere. Compactness will let us combine finitely many functions that are nonzero at different points.

A Criterion for Approximating Constants

Theorem (Constants in the Uniform Closure): Let \(K\) be a nonempty compact set and \(A\) a subalgebra of \(C(K)\), not necessarily unital. Then \(\mathbf{1}\in\overline{A}\) if and only if \(A\) has no common zero. Consequently, if either condition holds, every constant function belongs to \(\overline{A}\).

Proof. First suppose \(\mathbf{1}\in\overline{A}\). By the definition of uniform closure, there is an \(h\in A\) with \(\|h-\mathbf{1}\|_\infty<1/2\). For every \(x\in K\),

|h(x)-1|<\frac12 \quad\Longrightarrow\quad h(x)>\frac12.

Thus \(h(x)\ne 0\) at every point, so in particular there is no point where all members of \(A\) vanish.

For the converse, suppose \(A\) has no common zero. For each \(f\in A\), the set

U_f=\{x\in K:f(x)\ne 0\}

is open relative to \(K\), because \(f\) is continuous. The sets \(U_f\), as \(f\) ranges over \(A\), cover \(K\). Compactness gives a finite subcover \(U_{f_1},\ldots,U_{f_r}\), with each \(f_i\in A\). Define

G=\sum_{i=1}^{r}f_i^2.

Each square and the finite sum belong to \(A\), since \(A\) is an algebra. At any \(x\in K\), at least one of the selected functions is nonzero, so \(G(x)>0\). The Extreme-Value Theorem gives numbers \(m\) and \(M\) such that

0<m\leq G(x)\leq M \qquad (x\in K).

In particular \(M>0\). The function \(t\mapsto 1/t\) is continuous on the interval \([m/2,M+1]\). By the Weierstrass Approximation Theorem, for any \(\eta>0\) there is a polynomial \(q\) satisfying

\left|q(t)-\frac1t\right|<\eta \qquad\left(\frac m2\leq t\leq M+1\right).

Set \(p(t)=tq(t)\). This polynomial has zero constant term, so it can be written as a finite sum of positive powers of \(t\). Therefore \(p(G)\in A\): each term is a scalar multiple of a positive power of \(G\), and \(A\) is closed under products and scalar multiples. For every \(x\in K\), the value \(G(x)\) lies in the interval on which \(q\) approximates the reciprocal. Hence

\begin{aligned} |p(G(x))-1| &=G(x)\left|q(G(x))-\frac{1}{G(x)}\right|\\ &<M\eta. \end{aligned}

Given any \(\varepsilon>0\), choose \(\eta=\varepsilon/(M+1)\). Then \(M\eta<\varepsilon\), so \(\|p(G)-\mathbf{1}\|_\infty<\varepsilon\). This proves \(\mathbf{1}\in\overline{A}\).

Finally, if \(c\in\mathbb{R}\), scalar multiplication of approximants to \(\mathbf{1}\) gives approximants to the constant function \(c\mathbf{1}\). For \(c=0\), the zero function is already in \(A\); for \(c\ne0\), multiply an approximant to \(\mathbf{1}\) by \(c\). Thus every constant belongs to \(\overline{A}\). \(\square\)

The proof has two distinct ingredients. The no-common-zero condition supplies a positive function \(G\in A\) with a positive minimum. Polynomial approximation to \(1/t\) on the range of \(G\) then allows \(Gq(G)\) to behave like \(1\), while still belonging to \(A\). The factor \(G\) is important: it ensures that \(tq(t)\) has zero constant term, so its substitution into \(G\) uses only algebra operations available in a possibly nonunital algebra.

Worked Examples

Worked Example: Approximating One with an Algebra That Is Not Unital

Let \(K=[1,2]\) and \(A=x\mathbb{R}[x]\), the algebra of polynomials of the form \(xp(x)\), where \(p\) is a real polynomial. The function \(x\) belongs to \(A\), and \(x\ne0\) everywhere on \(K\); thus \(A\) has no common zero. Yet \(\mathbf{1}\) does not belong to \(A\). If \(xp(x)=1\) throughout \([1,2]\), the polynomial \(xp(x)-1\) would vanish at every point of that interval and hence be identically zero. Evaluating it at \(x=0\) gives \(-1\), a contradiction.

Even though \(A\) is not unital, constants can be approximated explicitly. For each integer \(N\geq0\), define

P_N(x)=\frac{x}{2}\sum_{k=0}^{N}\left(1-\frac{x}{2}\right)^k.

This is \(x\) times a polynomial, so \(P_N\in A\). The finite geometric-sum identity gives

P_N(x)=1-\left(1-\frac{x}{2}\right)^{N+1}.

Indeed, setting \(u=1-x/2\) gives \(x/2=1-u\), and \((1-u)\sum_{k=0}^N u^k=1-u^{N+1}\). For \(x\in[1,2]\), \(0\leq1-x/2\leq1/2\), so

|P_N(x)-1| =\left|1-\frac{x}{2}\right|^{N+1} \leq\left(\frac12\right)^{N+1}.

The right-hand side tends to zero, proving uniform convergence to \(\mathbf{1}\). This example separates two claims: \(A\) need not contain constants for its closure to contain them.

Worked Example: A Common Zero Prevents Approximation

On \(K=[0,1]\), let

A=\{p\in\mathbb{R}[x]:p(0)=0\}.

This is a subalgebra: sums and scalar multiples still vanish at \(0\), and if \(p(0)=q(0)=0\), then \((pq)(0)=p(0)q(0)=0\). It separates points because it contains the identity function \(x\mapsto x\), which takes different values at distinct points of \([0,1]\). But every member of \(A\) vanishes at \(0\), so \(A\) has a common zero.

For each \(p\in A\),

\|p-\mathbf{1}\|_\infty \geq |p(0)-1| =1.

Thus no sequence in \(A\) can converge uniformly to \(\mathbf{1}\). In fact, the zero function belongs to \(A\) and has distance exactly \(1\) from \(\mathbf{1}\). This example shows why separation alone does not ensure that an algebra’s closure contains constants.

Worked Example: Why Compactness Matters

Let \(K=(0,1]\), which is not compact, and consider \(A=x\mathbb{R}[x]\) as an algebra of bounded continuous functions on \(K\). The function \(x\in A\) is nonzero at every point of \(K\), so \(A\) has no common zero. Nevertheless, \(\mathbf{1}\) cannot be uniformly approximated by members of \(A\).

Every member has the form \(xp(x)\). Since \(p\) is a polynomial, \(xp(x)\to0\) as \(x\to0^+\). It follows that

\lim_{x\to0^+}|xp(x)-1|=1.

The supremum of \(|xp(x)-1|\) on \((0,1]\) is therefore at least \(1\). This holds for every \(p\), so the distance from \(\mathbf{1}\) to \(A\) cannot be made arbitrarily small. The compactness hypothesis in the theorem is what lets the finite construction produce a positive lower bound \(m\) for \(G\); here, a positive function can approach zero without attaining zero.

From Constants to Density

The criterion combines with the Real Stone-Weierstrass Theorem to give a useful density test for algebras that are not initially unital. In this course, “separates points” means that for each distinct pair in \(K\), some member of the algebra takes different values at the two points.

Corollary: Let \(K\) be a nonempty compact subset of \(\mathbb{R}\), and let \(A\) be a subalgebra of \(C(K)\). If \(A\) separates points and has no common zero, then \(A\) is dense in \(C(K)\) in the supremum norm.

Proof. By the theorem, \(\mathbf{1}\in\overline{A}\), and therefore \(\overline{A}\) contains all constants. The closure \(\overline{A}\) is a subalgebra, by the closure result established in Stone-Weierstrass Motivation. Also, \(A\subseteq\overline{A}\), so every separator from \(A\) remains in \(\overline{A}\); thus \(\overline{A}\) separates points. The Real Stone-Weierstrass Theorem applies to the unital subalgebra \(\overline{A}\) and gives \(\overline{A}=C(K)\). Hence \(A\) is dense in \(C(K)\). \(\square\)

The no-common-zero condition is also necessary for density. If every \(f\in A\) vanishes at some fixed \(x_0\in K\), then every uniform limit of members of \(A\) vanishes at \(x_0\), since evaluation at \(x_0\) is continuous in the supremum norm. In particular, such a closure cannot contain \(\mathbf{1}\). The corollary shows that, for a point-separating algebra on compact \(K\), removing the assumption that the algebra itself is unital is possible when it has no common zero.

Keep the Conditions Distinct

Three properties should not be confused. An algebra may contain \(\mathbf{1}\), it may fail to contain \(\mathbf{1}\) but have \(\mathbf{1}\) in its uniform closure, or its closure may fail to contain \(\mathbf{1}\). The theorem distinguishes the last two cases exactly through common zeros. Point separation is a separate condition: the algebra in the second worked example separates points but has a common zero and cannot approximate constants.

Property	What it tells you	What it does not tell you by itself
\(\mathbf{1}\in A\)	The algebra is unital and contains every constant.	That the algebra is dense in \(C(K)\).
\(A\) has no common zero, with \(K\) compact	\(\mathbf{1}\in\overline{A}\), so the closure contains all constants.	That \(A\) separates points.
\(A\) separates points	Every distinct pair can be distinguished by some member.	That constants can be approximated.
\(A\) separates points and has no common zero	The Real Stone-Weierstrass Theorem gives density.	That either condition can be omitted in this nonunital criterion.

Takeaway: For a subalgebra of \(C(K)\) on nonempty compact \(K\), the constant function \(\mathbf{1}\) belongs to the uniform closure exactly when the algebra has no common zero. Combined with point separation, this supplies the hypotheses needed for a nonunital density conclusion.

Check Your Understanding

Use the theorem and examples to answer the following questions.

Why does \(\mathbf{1}\in\overline{A}\) imply that \(A\) has no common zero?
In the proof of the converse, why do the sets \(U_f=\{x:f(x)\ne0\}\) admit a finite subcover?
Why does \(p(t)=tq(t)\) have zero constant term, and why does that ensure \(p(G)\in A\) even if \(A\) is not unital?
Why can the algebra \(x\mathbb{R}[x]\) approximate \(\mathbf{1}\) uniformly on \([1,2]\) but not on \((0,1]\)?
Why does point separation alone not guarantee that \(\mathbf{1}\) belongs to the uniform closure?