Approximation Theory
Stone–Weierstrass Theorem
You should know: equicontinuity and arzela ascoli, uniform convergence, banach spaces
Overview
The Stone-Weierstrass theorem is a vast generalisation of Weierstrass's classical result that every continuous function on [a,b] can be uniformly approximated by polynomials. Marshall Stone's theorem (1948) identifies exactly which subalgebras of C(K) are dense: a subalgebra A of C(K) (K compact Hausdorff) is dense in the sup-norm if and only if it separates points and vanishes nowhere (or, for the real version, if A is a real subalgebra that separates points and contains constants). This single result unifies the density of polynomials, trigonometric polynomials, and many other approximating families.
Intuition
The Weierstrass theorem says polynomials are 'enough' to approximate any continuous function on a closed interval. Stone asked: what is the essential feature of polynomials that makes this work? His answer: they form an algebra (closed under addition and multiplication), they separate points (p(x) \neq p(y) if x \neq y for some polynomial p), and they include the constants. Any family of continuous functions with these three properties is dense. This gives a powerful abstract framework: to check that your favourite family of functions is dense, just verify separation and the algebraic closure conditions.
Formal Definition
Let K be a compact Hausdorff space and A \subseteq C(K, \mathbb{R}) a subalgebra (closed under addition, scalar multiplication, and pointwise multiplication).
Worked Examples
Let K = [a,b] \subset \mathbb{R} and A = \{\text{polynomials on } [a,b]\}.
A is a subalgebra of C([a,b]): sums, scalar multiples, and products of polynomials are polynomials.
A separates points: p(x) = x separates any two distinct real numbers.
A contains the constant 1, so it vanishes nowhere.
By Stone-Weierstrass, A is dense in C([a,b]) with the sup-norm, i.e., every continuous function on [a,b] is a uniform limit of polynomials.
Answer: Polynomials form a point-separating subalgebra containing constants, so Stone-Weierstrass implies they are dense in C([a,b]).
Practice Problems
Does Stone-Weierstrass apply to A = \{f \in C([0,1]) : f(0) = 0\}? Is A dense in C([0,1])?
Prove that if A is a self-adjoint subalgebra of C(K, \mathbb{C}) that separates points and contains the constants, then A is dense in C(K, \mathbb{C}).
Show that there is no polynomial p(x,y) in two variables that uniformly approximates |x-y| on [0,1]^2 to within any \varepsilon > 0 using only polynomials of one variable.
Common Mistakes
Any subalgebra of C(K) is dense.
Only subalgebras that separate points and vanish nowhere are dense. For example, functions vanishing at a fixed point form a proper closed subalgebra.
The complex version of Stone-Weierstrass follows immediately from the real version.
The complex version requires the additional hypothesis that A is self-adjoint (closed under complex conjugation). Without it, the conclusion fails: polynomials in z alone on the unit disk are not dense in C(\partial D, \mathbb{C}).
Quiz
Summary
- Stone-Weierstrass: a subalgebra of C(K, \mathbb{R}) that separates points and contains a nonzero constant is dense in C(K).
- The classical Weierstrass approximation theorem is the special case K = [a,b], A = polynomials.
- Trigonometric polynomials are dense in C(\mathbb{T}) by the complex self-adjoint version.
- The hypotheses (separation of points, vanishing nowhere) are both necessary and sufficient.
- Stone-Weierstrass has applications in approximation theory, probability (moments determine distributions), and operator algebras.
References
- BookRudin, W. — Real and Complex Analysis (3rd ed.), McGraw-Hill, 1987, Chapter 5.
Mathematics