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Polynomial ridge functions are dense on compact sets

Statement

Let d≥1 and let K⊆Rd be compact. Let Rd be the real linear span of the functions x↦(w⋅x)k with w∈Rd and k∈N0, using the convention s0=1 for every real s. Then the restrictions Rd∣K are uniformly dense in C(K,R).

Facts & Assumptions

Given: An integer d≥1 and a compact set K⊆Rd.

[F1]

Every unital point-separating real function algebra on a compact Hausdorff space is uniformly dense in the continuous real-valued functions (Real Stone–Weierstrass theorem for compact Hausdorff spaces).

Proof

1.1givenalgebra

Fix k≥1 and linear forms ℓj(x)=vj⋅x for 1≤j≤k. Expanding the powers and summing signs gives the polarization identity ∏j=1kℓj(x)=12kk!∑ϵ∈{−1,1}k(∏j=1kϵj)(∑j=1kϵjℓj(x))k. Indeed, in the multinomial expansion, a monomial ∏jℓjrj has sign sum ∏j∑ϵj=±1ϵjrj+1, which is zero unless every rj is odd. Since ∑jrj=k, the only surviving tuple is r1=⋯=rk=1, with coefficient 2kk!. Thus every product of k linear forms lies in Rd.

2.1step 1.1algebra

Every coordinate monomial x1α1⋯xdαd of positive total degree k is a product of k coordinate linear forms, with xi repeated αi times. Step 1.1 puts it in Rd. Constants lie in Rd by k=0, so every real polynomial restricts to an element of Rd∣K.

3.1step 2.1F1algebra∎

The restrictions to K of real polynomials form a unital real function algebra: sums, scalar multiples, and products are polynomials, and the constant-one polynomial is a unit. If x,y∈K are distinct, some coordinate differs, so the corresponding coordinate polynomial separates them. By [F1] these polynomial restrictions are uniformly dense in C(K,R). Since step 2.1 places them inside Rd∣K, the latter is dense as well.

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