In Rudin, we are given this corollary, 7.27 to the Stone Weierstrass Theorem:

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where Thm 7.26 is the Stone Weierstrass Theorem: enter image description here

Say instead I replaced $|x|$ in the corollary with a continuous function such that $g(0)=0$, how would the proof change. I honestly don't see any specific changes, although I really feel as though I'm missing something important. Also I'm also confused by Rudin's proof in the first place, by the last line in particular. What does he mean: the polynomials $P_n(x)=P_n^*(x)-P_n^*(0) (n=1,2,3..)$ have the desired properties? Any help would be much appreciated. Thank you.

  • $\begingroup$ That last line is because the corollary puts the additional requirement that each polynomial in the sequence vanishes at $x=0$. Stone-Weierstrass only gives you a sequence that converges, namely $P_n^*$. Then the proof shows that $P_n^*(x)-P_n^*(0)$ also converges and has the additional property that it vanishes at $x=0$. $\endgroup$ – user530511 Feb 11 '18 at 23:50
  • $\begingroup$ As an exercise, see if you can prove the same corollary but with the additional requirement that the polynomials should vanish at $x=0$ and also $P_n(\pm a)=a$. $\endgroup$ – user530511 Feb 11 '18 at 23:53
  • $\begingroup$ So since $P_n^*(0)$ goes to 0 we have $P_n(x)=P_n^*(x)$? Also does the proof change if we replace |x| with a continuous function g(0)=0? $\endgroup$ – kemb Feb 11 '18 at 23:57
  • $\begingroup$ Your $g(0) = 0$ is not a function. Do you mean $g(x) = 0$ for all $x$? $\endgroup$ – user99914 Feb 11 '18 at 23:58
  • $\begingroup$ @kemb $P_n^*(0)$ tends to $0$, which is not the same as being $0$. No, the proof wouldn't change for a continuous function $g$ for which $g(0)=0$. $\endgroup$ – user530511 Feb 12 '18 at 0:01

You are correct, in that the same argument will hold for any $g$ continous such that $g(0)=0.$ The reason Rudin singles out this result, is because he uses it soon afterwards when proving the general Stone Weierstrass theorem (for subalgebras of $C(K)$).

Of course, using the Weierstrass approximation theorem to prove this special case is somewhat overkill. An alternative approach is to consider the Taylor series of $\sqrt{|x|^2+\varepsilon} - \sqrt{\varepsilon}$ and obtain polynomials by taking partial sums while varying $\varepsilon.$

Edit: To answer your second question, let $\varepsilon > 0.$ Then by the properties of $P_n^*,$ there is $N$ such that for all $n > N,$ we have,

$$ | |x| - P_n^*(x)| \leq \varepsilon/2, $$

for all $x \in [-a,a].$ In particular $|P_n^*(0)| < \varepsilon/2,$ so by the triangle inequality,

$$ | |x| - (P_n^*(x)-P_n^*(0))| \leq ||x| - P_n^*(x)| + |P_N^*(0)| < \varepsilon $$

as desired.


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