proof of Stone-Weierstrass theorem
First, we shall show that, if , then . Since is a continuous function on a compact space must be bounded – there exists constants and such that . By the Weierstrass approximation theorem, for every , there exists a polynomial such that when . (By the way, one does not need the full-blown Weierstrass approximation theorem to show that exists – see the entry “proof of Weierstrass approximation theorem” for an elementary construction of ) Define by . Since is an algebra, . For all , . Since is closed under the uniform convergence topology, this implies that .
A corollary of the fact just proven is that if , then and . The reason for this is that one can write
Second, we shall show that, for every , every , and every , there exists such that and . By condition 1, if , there exists a function such that . Define by , where the constants and have been chosen so that
By condition 2, . (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 . The necessity of condition 2 can be shown by a simple example: Suppose that is the algebra of all continuous functions on which vanish at a point . It is easy to see that this algebra satisfies all the hypotheses of the theorem except condition 2 and the conclusion of the theorem does not hold in this case.)
For every , define the set as
Since and are continuous, is an open set. Because and , is an open cover of . By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset such that . Define
By the corollary of the first part of the proof, . By construction, and .
Third, we shall show that, for every and every , there exists a function such that . This will complete the proof becauase it implies that . For every , define the set as
where is defined as before. Since and are continuous, is an open set. Because , . Hence is an open cover of . By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset such that . Define as
By the corollary of the first part of the proof, . By construction, . Since for every , .
|Title||proof of Stone-Weierstrass theorem|
|Date of creation||2013-03-22 14:35:01|
|Last modified on||2013-03-22 14:35:01|
|Last modified by||rspuzio (6075)|