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The uniform closure of a real function algebra is a vector lattice

Statement

Let X be a compact Hausdorff space and let AC(X,R) be a real function algebra, not necessarily unital. Let A consist of the continuous functions that can be approximated uniformly by members of A. Then A is a real function algebra and a real vector sublattice of C(X,R).

Facts & Assumptions

Given: A compact Hausdorff space X, a real function algebra AC(X,R), and its uniform closure A.

[L1]

For ab, every continuous real function on [a,b] is a uniform limit of polynomials (Polynomials are uniformly dense in C([a,b],R) for every closed interval).

[L2]

If for every ε>0 a function f:XY has a continuous approximant g with d(f(x),g(x))<ε for every x, then f is continuous (A uniform limit of continuous functions is continuous, so C(X,Y) is closed in YX under the uniform metric, clause 1).

[L4]

A real function algebra is a real vector subspace of C(X,R) closed under pointwise multiplication (Unital, point-separating, and nowhere-vanishing real function algebras on a compact Hausdorff space).

Proof

technique · direct
1.1

If X=, then C(X,R) consists of the unique empty function, which is the zero element of the vector subspace A; hence A=A=C(X,R) and the claim is immediate.

L4
1.2

Assume X. By the definition of uniform closure, every fA has, for every positive error, a continuous approximant from A, so [L2] confirms that all such uniform limits remain continuous.

L2L4
1.3

The set A is a real vector subspace: approximants to f and g add to an approximant to f+g, scalar multiples approximate scalar multiples, and the zero function belongs to A.

L4algebra
2.1

The set A is closed under multiplication. Indeed, for f,gA, [L3] gives finite bounds Mf:=maxXf and Mg:=maxXg. Given η>0, choose a,bA with af<min{1,η/(2(Mg+1))},bg<η/(2(Mf+1)). Then aMf+1 and abfgabg+gaf<η pointwise. Thus products of members of A again lie in A.

L3L4step 1.3choosealgebra
2.2

Fix fA and ε>0. By [L3], f has a maximum M0; if M=0 then f=0 and f=0A.

L3step 1.3
3.1

If M>0, apply [L1] on [M,M] to choose a polynomial q with q(t)t<ε/2 there, and put p(t):=q(t)q(0). Then p(0)=0, p(t)t<ε on [M,M], and steps 1.3 and 2.1 give p(f)A. Hence f is uniformly approximable by members of A, and therefore belongs to the closed set A. Together with the M=0 case in step 2.2, this proves fA for every fA.

L1step 1.3step 2.1step 2.2choosealgebra
4.1

For f,gA, the pointwise identities fg=(f+g+fg)/2 and fg=(f+gfg)/2, together with steps 1.3 and 3.1, put both functions in A; hence A is a real vector sublattice.

step 1.3step 3.1algebra

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