# Stone-Weierstrass Theorem (Rudin)

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.

• 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 $\sqrt n.$ – zhw. Dec 14 '16 at 21:29
• 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? – Strange Brew Dec 14 '16 at 21:33
• It's right there in Rudin in the earlier chapters. – zhw. Dec 14 '16 at 21:36
• 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. – 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.
• 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. – Strange Brew Dec 14 '16 at 22:04