0
$\begingroup$

I've been reading the proof of the Stone-Weierstrass Theorem in Rudin and I'm stuck on this inequality.

$$\sqrt{n}(1-\delta^2)^n<\varepsilon \text{ for } n\geq N \text{ where } 0<\delta\leq1$$

Why is this true?

I can't find the solution by searching here and I've tried looking at the other Stone-Weierstrass questions.

$\endgroup$
  • 2
    $\begingroup$ It's certainly not true if $\delta =0.$ But basically it just boils down to $\sqrt n a^n \to 0$ if $0\le a <1.$ Remember, exponential decay always wins vs polynomial growth, so surely it beats $\sqrt n.$ $\endgroup$ – zhw. Dec 14 '16 at 21:29
  • $\begingroup$ Yes it should've said $\delta>0$. With regards to your comment I seem to have forgotten this fact, would you remind me how to show this? $\endgroup$ – Strange Brew Dec 14 '16 at 21:33
  • $\begingroup$ It's right there in Rudin in the earlier chapters. $\endgroup$ – zhw. Dec 14 '16 at 21:36
  • $\begingroup$ In Theorem 3.20 it says $ \lim_{n\to\infty}(n)^{1/n} =1 $ but doesn't mention $lim_{n\to\infty }\sqrt{n}$. But maybe you meant another Theorem. $\endgroup$ – Strange Brew Dec 14 '16 at 21:43
2
$\begingroup$

The statement is equivalent to $\displaystyle \lim_{n \to \infty} \sqrt{n(1-\delta^2)^{2n}} = 0$. Since $0 \le 1 - \delta^2 < 1$, put $\dfrac{1}{1+a} = \left(1- \delta^2\right)^2 \implies a >0\implies 0 \le n(1-\delta^2)^{2n} = \dfrac{n}{(1+a)^n}< \dfrac{n}{1+na+\dfrac{a^2n^2}{2}}< \dfrac{2}{a^2n}$. By squeeze theorem $n(1-\delta^2)^{2n} \to 0$ as $n \to \infty$, and the result follows.

$\endgroup$
  • $\begingroup$ Very nice! I was thinking it might even be possible to apply the result of Theorem 3.20 d): for $p > 0 $ and $\alpha \in \mathbb{R}$, then $$lim_{n\to\infty}\frac{n^\alpha}{(1+p)^n}=0 $$ directly since $\sqrt{n(1-\delta^2)^{2n}}\leq n(1-\delta^2)^{2n}$. Given that the expression is always positive. $\endgroup$ – Strange Brew Dec 14 '16 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.