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.

  • 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

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.

  • $\begingroup$ 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

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