22 Convolution and the Stone–Weierstrass Theorem

25 March 2026 · Denis Auroux

The Stone–Weierstrass theorem

Theorem 22.1 (Weierstrass). Polynomials are dense in , i.e. there exist polynomials such that uniformly on .

Key tool: convolution.

Definition 22.2. The convolution of and is

This is well-defined if e.g. and are piecewise and one of them is compactly supported (i.e. outside some ), to avoid improper integrals.

Principle: “ inherits the best properties of and .” This is because is bounded by (over the relevant intervals, per compact support).

If is , then uniform continuity (over a compact interval) gives (i.e. for all when ), so is continuous.
If is (continuously differentiable) then is and .

Indeed, gives

using the mean value theorem: the integrand involves , which uniformly as by uniform continuity of on the interval of integration.

Hence if is then is !! (Even if isn’t even continuous.)
And… if is a polynomial of degree then so is ! E.g. because , so ; or more directly: gives manifestly a polynomial in .

Approximate identities

Definition 22.3. A sequence of functions approximates the identity if

Theorem 22.4. If is compactly supported and continuous and is an approximate identity, then uniformly.

Proof. Estimate each term as follows. Given , uniform continuity of on its support gives (independent of ) such that . Then (independently of ), while which becomes for sufficiently large, (independent of !). So such that , for all . ∎

Example 22.5. for , elsewhere, where is chosen so that .

Claim: is an approximate identity.

Show solution

For , (the first inequality by convexity), so

For , , so

Theorem 22.6 (Weierstrass). there exist polynomials such that uniformly.

Proof.

By a linear change of variables, we can assume .
Subtracting a degree 1 polynomial from , we can assume . Then extend to by if .
Now let be as above, and . Then is an approximate identity and is and compactly supported, so uniformly.
is a polynomial of degree on , because, given that outside , the formula for doesn’t involve the values of outside , and is polynomial.

∎

Stone’s theorem generalizes this to other families of functions.

Definition 22.7. is an algebra if , , .

separates points if , such that , , , . ( and are arbitrary: this is equivalent to , being surjective for all .)

For complex-valued functions, we further assume is conjugation-invariant, i.e. (equivalently: , ).

Theorem 22.8 (Stone). Let be a compact metric space and an algebra which separates points (and is conjugation-invariant in the case). Then is dense in .

(Weierstrass is the special case , polynomials.)

Proof.

(the uniform closure of ) is an algebra (, , ), so it is enough to show: the assumptions plus closed imply .
Given : since is an algebra and closed, for every polynomial such that . By Weierstrass, is a uniform limit of polynomials on , so . Hence and the same for .
Now: given and , we want to show such that (so ).

Given : for all there is such that and ( separates points). There is an open such that on ; and compact implies such that . Then satisfies

By the same argument, such that satisfies everywhere. (There is an open such that on ; compact implies such that .) ∎

Notes not reached in this lecture, covered at the start of Lecture 23

Fourier series

We consider continuous -periodic functions with complex values, or equivalently functions on , with the inner product The complex exponentials , , satisfy : orthonormality.

Definition 22.9. The Fourier coefficients of are The Fourier series of is

Question 22.10. (Fourier, Dirichlet, Fejér, …) Does the Fourier series accurately represent ? (E.g. does it converge? To ?)

Definition 22.11. Trigonometric polynomials are the vector space of finite linear combinations of the .

Clearly this is an algebra, complex conjugation-invariant, and it separates points of , which is compact. Hence by Stone–Weierstrass, trigonometric polynomials are dense in … hence also in the norm, since

The th Fourier sum is the orthogonal projection of onto for . Indeed: Thus for all , : is the point of closest to for . (This follows from :

Theorem 22.12 (Parseval). Let , its Fourier coefficients, and the partial sums. Then:

in , i.e. as .
(in particular converges, so as ).

Proof. (1) Since the trigonometric polynomials are dense in , such that with . Now for , and is the closest point to , so . This shows in .

(2) Since and , , where by orthonormality, and by the first part. ∎

Corollary 22.13. If have the same Fourier series, then , hence .

The fact that in is the best approximation (in the norm) of by trigonometric polynomials, and that trigonometric polynomials are dense in (so there are trigonometric polynomials uniformly), makes one hope that uniformly, or at least pointwise… alas not!

Fact: there exists such that the Fourier series of does not converge ( unbounded!). (But the example is hard to construct.)

Theorem 22.14 (Dirichlet). If is then uniformly.

The proof uses convolution, redefined for periodic functions as and noting that So where is the Dirichlet kernel. Dirichlet’s proof studies this convolution for to prove uniform convergence.

The fact that convergence can sometimes fail makes it remarkable that for all , can be recovered from the partial sums …

Theorem 22.15 (Fejér). If then converges uniformly to .

The reason is that this process amounts to convolution with the Fejér kernel , which actually approximates the identity (in the sense seen above), unlike .