proof of Stone-Weierstrass theorem


Let 𝒜¯ denote the closureMathworldPlanetmathPlanetmath of 𝒜 in C0(X,) according to the uniform convergence topologyMathworldPlanetmathPlanetmath. We want to show that, if conditions 1 and 2 are satisfied, then 𝒜¯=C0(X,).

First, we shall show that, if f𝒜¯, then |f|𝒜¯. Since f is a continuous functionMathworldPlanetmathPlanetmath on a compact space f must be boundedPlanetmathPlanetmathPlanetmathPlanetmath – there exists constants a and b such that afb. By the Weierstrass approximation theoremMathworldPlanetmath, for every ϵ>0, there exists a polynomial such that |P(x)-|x||<ϵ when x[a,b]. (By the way, one does not need the full-blown Weierstrass approximation theorem to show that P exists – see the entry “proof of Weierstrass approximation theorem” for an elementary construction of P) Define g:X by g(x)=P(f(x)). Since 𝒜¯ is an algebra, g𝒜¯. For all xX, |g(x)-|f(x)||<ϵ. Since 𝒜¯ is closed under the uniform convergence topology, this implies that |f|𝒜¯.

A corollary of the fact just proven is that if f,g𝒜¯, then max(f,g)𝒜¯ and min(f,g)𝒜¯. The reason for this is that one can write

max(a,b)=12(a+b+|a-b|)
min(a,b)=12(a+b-|a-b|)

Second, we shall show that, for every fC0(X,), every xX, and every ϵ>0, there exists gx𝒜¯ such that gxf+ϵ and gx>f. By condition 1, if yx, there exists a function h~xy𝒜 such that h~xy(x)h~xy(y). Define hxy by hxy(z)=ph~xy(z)+q, where the constants p and q have been chosen so that

hxy(x)=f(x)+ϵ2
hxy(y)=f(y)-ϵ2

By condition 2, hxy𝒜. (Note: This is the only place in the proof where condition 2 is used, but it is crucial since, otherwise, it might not be possible to construct a function which takes arbitrary preassigned values at two distinct points of X. The necessity of condition 2 can be shown by a simple example: Suppose that 𝒜 is the algebra of all continuous functions on f which vanish at a point OX. It is easy to see that this algebra satisfies all the hypotheses of the theorem except condition 2 and the conclusionMathworldPlanetmath of the theorem does not hold in this case.)

For every yx, define the set Uxy as

Uxy={zXhxy(z)<f(z)+ϵ}

Since f and hxy are continuous, Uxy is an open set. Because xUxy and yUxy, {UxyyX{x}} is an open cover of X. By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset {y1,y2,,yn}X such that X=m=0nUxym. Define

gx=min(hxy1,hxy2,,hxyn).

By the corollary of the first part of the proof, gx𝒜¯. By construction, gx(x)=f(x)+ϵ/2 and gx<f+ϵ.

Third, we shall show that, for every fC0(X,) and every ϵ>0, there exists a function g𝒜¯ such that fg<f+ϵ. This will completePlanetmathPlanetmathPlanetmathPlanetmath the proof becauase it implies that 𝒜¯=C0(X,). For every xX, define the set Vx as

Vx={zXgx(z)>f(x)}

where gx is defined as before. Since f and gx are continuous, Vx is an open set. Because gx(x)=f(x)+ϵ/2>f(x), xVx. Hence {Vxx} is an open cover of X. By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset {x1,x2,xn}X such that X=m=0nVxn. Define g as

g(z)=max{gx1(z),gx2(z),,gxn(z)}

By the corollary of the first part of the proof, g𝒜¯. By construction, g>f. Since gx<f+ϵ for every xX, g<f+ϵ.

Title proof of Stone-Weierstrass theorem
Canonical name ProofOfStoneWeierstrassTheorem
Date of creation 2013-03-22 14:35:01
Last modified on 2013-03-22 14:35:01
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 18
Author rspuzio (6075)
Entry type Proof
Classification msc 46E15