Applications of Stone-Weierstrass

From Density to Useful Approximations

Stone-Weierstrass is more than a statement that a particular collection of polynomials is dense. Its practical value is that a suitable algebra of functions can be dense even when its elements are written in coordinates different from the identity function. This lets us choose expressions that fit a problem: powers of a transformed variable, exponentials, or rational functions with denominators that stay away from zero.

Throughout, let $K$ be a nonempty compact subset of $\mathbb{R}$, and let $C(K)$ carry the supremum norm. We use the Real Stone-Weierstrass Theorem established earlier: a unital subalgebra of $C(K)$ that separates points is dense in $C(K)$. We also use the polynomial approximation corollary for compact subsets of $\mathbb{R}$, and the result that the uniform closure of a unital subalgebra is itself a unital subalgebra.

Approximation in a Transformed Coordinate

Suppose a continuous function $\phi:K\to\mathbb{R}$ assigns a distinct real coordinate to each point of $K$. Instead of using polynomials in $x$, consider polynomials in $\phi(x)$. These functions form an algebra: sums and products of $p(\phi(x))$ and $q(\phi(x))$ are still polynomials in $\phi(x)$, and the constant function $1$ is included.

Theorem (Approximation in an Injective Coordinate): Let $K\subseteq\mathbb{R}$ be nonempty and compact, and let $\phi:K\to\mathbb{R}$ be continuous and injective. For every $f\in C(K)$ and every $\varepsilon>0$, there is a real polynomial $p$ such that $$ \sup_{x\in K}|f(x)-p(\phi(x))|<\varepsilon. $$

Proof. Define

A_\phi=\{p\circ\phi:p\in\mathbb{R}[t]\}\subseteq C(K),

where $t$ denotes the polynomial variable. Since $\phi$ is continuous, each $p\circ\phi$ is continuous. The set $A_\phi$ contains the constant functions. It is closed under addition and scalar multiplication because sums and scalar multiples of polynomials are polynomials. It is closed under multiplication because

\bigl(p(\phi(x))\bigr)\bigl(q(\phi(x))\bigr)=(pq)(\phi(x)).

Thus $A_\phi$ is a unital subalgebra of $C(K)$. If $x,y\in K$ and $x\ne y$, injectivity gives $\phi(x)\ne\phi(y)$. The function $\phi$, which belongs to $A_\phi$ by taking $p(t)=t$, therefore has different values at $x$ and $y$. So $A_\phi$ separates points. The Real Stone-Weierstrass Theorem implies that $A_\phi$ is dense in $C(K)$. Applying density to $f$ and $\varepsilon$ gives the required polynomial $p$. $\square$

The theorem is useful when the desired approximation is naturally expressed in a particular coordinate. Its hypothesis is not that $\phi$ has a simple inverse formula, but that $\phi$ is continuous and injective on the compact set in question. Injectivity ensures that the new coordinate does not identify two points that a target function might need to distinguish.

Worked Examples

Worked Example: Polynomials in a Cube Approximate Continuous Functions

Take $K=[-1,1]$ and $\phi(x)=x^3$. This function is continuous and injective on $K$. The theorem says that for every $f\in C[-1,1]$ and every $\varepsilon>0$, some polynomial $p$ satisfies

\sup_{-1\leq x\leq1}|f(x)-p(x^3)|<\varepsilon.

For the particular target $f(x)=|x|$, define $F(t)=|t|^{1/3}$ on $[-1,1]$. This is continuous, and $F(x^3)=|x|$: indeed, $|x^3|^{1/3}=|x|$. By polynomial approximation on $[-1,1]$, for any chosen $\varepsilon>0$ there is a polynomial $p$ such that

|F(t)-p(t)|<\varepsilon \qquad(-1\leq t\leq1).

Substituting $t=x^3$, which lies in $[-1,1]$, gives

\bigl||x|-p(x^3)\bigr| =|F(x^3)-p(x^3)| <\varepsilon \qquad(-1\leq x\leq1).

Thus polynomials in $x^3$ can approximate this target uniformly, even though $p(x^3)$ has only powers of $x$ whose exponents are multiples of three.

Worked Example: Approximating a Function Using Exponentials

On $K=[0,1]$, let $\phi(x)=e^x$. This function is continuous and strictly increasing, hence injective. The approximation theorem shows that every continuous $f$ on $[0,1]$ can be uniformly approximated by expressions $p(e^x)$, where $p$ is a polynomial.

For example, take $f(x)=x^2$. The range of $\phi$ is $[1,e]$. On this range define $F(t)=(\log t)^2$, which is continuous. For each $x\in[0,1]$,

F(e^x)=(\log(e^x))^2=x^2.

Polynomial approximation on $[1,e]$ gives, for each $\varepsilon>0$, a polynomial $p$ such that $|F(t)-p(t)|<\varepsilon$ for all $t\in[1,e]$. Since $e^x\in[1,e]$ for $x\in[0,1]$, substitution yields

|x^2-p(e^x)|=|F(e^x)-p(e^x)|<\varepsilon \qquad(0\leq x\leq1).

The conclusion is an existence statement; it does not specify the coefficients of $p$. It does show that exponentials can serve as an approximation coordinate just as the identity function can.

Worked Example: Rational Expressions with a Pole Outside the Set

Let $K=[0,1]$ and take $\phi(x)=1/(1+x)$. This is continuous on $K$ because $1+x\geq1$, and it is injective: if $1/(1+x)=1/(1+y)$, multiplication by the positive denominators gives $1+y=1+x$, so $x=y$. Consequently, polynomials in $1/(1+x)$ are dense in $C[0,1]$.

For the target $f(x)=\sqrt{x}$, the range of $\phi$ is $[1/2,1]$. Define

F(t)=\sqrt{\frac{1}{t}-1} \qquad\left(\frac12\leq t\leq1\right).

The expression under the square root is nonnegative on this interval, and $F$ is continuous there. Direct substitution verifies the target identity:

F\left(\frac{1}{1+x}\right) =\sqrt{\frac{1}{1/(1+x)}-1} =\sqrt{1+x-1} =\sqrt{x}.

Given $\varepsilon>0$, choose a polynomial $p$ that approximates $F$ within $\varepsilon$ on $[1/2,1]$. Then

\left|\sqrt{x}-p\left(\frac{1}{1+x}\right)\right|<\varepsilon \qquad(0\leq x\leq1).

Each approximant is a rational expression whose denominator is a power of $1+x$, so it has no pole on $[0,1]$. This illustrates how a coordinate with a pole outside the domain can generate a dense approximation algebra.

Reciprocals in Closed Algebras

A second useful application concerns division. An algebra is closed under sums and products, but it need not contain the reciprocal of each of its nonzero elements. Uniform closure changes this: a closed unital subalgebra of $C(K)$ contains the reciprocal of each of its functions that never vanish.

Theorem (Reciprocals in a Closed Unital Subalgebra): Let $B$ be a closed unital subalgebra of $C(K)$. If $f\in B$ and $f(x)\ne0$ for every $x\in K$, then $1/f\in B$.

Proof. Since $f$ is continuous and $K$ is compact, $f(K)$ is compact. The assumption that $f$ never vanishes means $0\notin f(K)$. The function $t\mapsto1/t$ is continuous on $f(K)$, so the polynomial approximation corollary gives polynomials $q_n$ such that

\sup_{t\in f(K)}\left|q_n(t)-\frac1t\right|\longrightarrow0.

Because $B$ is a unital algebra and $f\in B$, each $q_n(f)$ belongs to $B$: if $q_n(t)=\sum_{j=0}^{m}c_jt^j$, then $q_n(f)=\sum_{j=0}^{m}c_j f^j$, with the constant term represented using the unit of $B$. For every $x\in K$, $f(x)\in f(K)$, and hence

\left|q_n(f(x))-\frac1{f(x)}\right| \leq \sup_{t\in f(K)}\left|q_n(t)-\frac1t\right|.

Taking the supremum over $x\in K$ shows that $q_n(f)$ converges uniformly to $1/f$. Since $B$ is closed in the supremum norm and each $q_n(f)$ belongs to $B$, its limit $1/f$ also belongs to $B$. $\square$

Worked Example: Polynomial Approximation of a Reciprocal

On $[0,1]$, the function $f(x)=2+x$ is a polynomial and is bounded away from zero, since $2+x\geq2$. Its reciprocal can be approximated directly by polynomials. Rewrite it as

\frac1{2+x} =\frac13\frac1{1-\frac{1-x}{3}}.

This identity holds because $1-(1-x)/3=(2+x)/3$. For each integer $N\geq0$, define the polynomial

p_N(x)=\frac13\sum_{n=0}^{N}\left(\frac{1-x}{3}\right)^n.

The finite geometric-sum identity, with $r=(1-x)/3$, gives

\frac1{2+x}-p_N(x) =\frac13\frac{r^{N+1}}{1-r} =\frac{r^{N+1}}{2+x}.

For $x\in[0,1]$, $0\leq r\leq1/3$ and $2+x\geq2$. Therefore

\left|\frac1{2+x}-p_N(x)\right| \leq\frac{(1/3)^{N+1}}{2} =\frac{1}{2\cdot3^{N+1}}.

This explicit sequence not only confirms that the reciprocal lies in the uniform closure of the polynomial algebra, but also gives a concrete error bound for this particular function.

What These Applications Do—and Do Not—Give

The transformed-coordinate theorem provides a simple test for density: if a continuous coordinate is injective, its polynomial algebra separates points, so Stone-Weierstrass applies. This can make an approximation family easier to use or better suited to the structure of a problem. The reciprocal theorem gives a related structural consequence: uniform limits in a closed unital algebra can support division whenever the denominator has no zeros on the compact domain.

A common pitfall is to confuse density with an error estimate. Stone-Weierstrass guarantees that an approximant exists for each positive tolerance; by itself, it does not give a degree bound or a formula for the polynomial. The geometric-series example supplies such a bound because its special algebraic structure allows a direct calculation. In other situations, rates require additional information about the target function and a separate approximation theorem.

The no-zero condition in the reciprocal result is essential. If $f(x_0)=0$ at some point $x_0$, then $1/f$ is not a real-valued continuous function on all of $K$, so it cannot belong to $C(K)$. Compactness helps ensure that a continuous function with no zeros stays a positive distance from zero, which is precisely what makes reciprocal approximation possible on its range.

Takeaway: Stone-Weierstrass applies to any unital algebra that separates points, including polynomial algebras in injective coordinates. In addition, closed unital subalgebras of $C(K)$ contain reciprocals of their nowhere-zero functions.

Check Your Understanding

Use the results and examples in this tutorial to answer the following questions.

Why does a continuous injective function $\phi:K\to\mathbb{R}$ ensure that the algebra of polynomials in $\phi$ separates points?
For the coordinate $\phi(x)=x^3$ on $[-1,1]$, what continuous function of $t$ corresponds to the target $f(x)=|x|$?
Why does $1/(1+x)$ define a valid approximation coordinate on $[0,1]$, even though its formula has a potential pole at $x=-1$?
In the reciprocal theorem, where is compactness used to approximate $1/t$ by polynomials?
Why must the function $f$ in that theorem have no zeros on $K$?
What additional information does the geometric-series example provide that Stone-Weierstrass alone does not?