Question about Rudin (Principles of Math Analysis) theorem 7.26. Why does $Q_n \to 0$ uniformly?

Question

In the proof of the Stone-Weierstrass theorem (7.26), Rudin claims $Q_n \to 0$ uniformly. Can someone explain why this is the case? I don't see how that immediately follows from the bound.

Answer 1

Asserting that a sequence $(f_n)_{n\in\Bbb N}$ of functions from a set $D$ into $\Bbb R$ converges uniformly to a function $f$ is the same thing as asserting that the functions $f-f_n$ are bounded (at least if $n$ is large enough) and that $\lim_{n\to\infty}\sup_{x\in D}|f(x)-f_n|(x)=0$. So, asserting that $(Q_n)_{n\in\Bbb N}$ converges uniformly to $0$ is the same thing as asserting that each $Q_n$ is bounded (again, at least if $n$ is large enough) and that$$\lim_{n\to\infty}\sup_{x\in D}|Q_n(x)|=0.$$But$$\sup_{x\in D}Q_n\leqslant\sqrt n\left(1-\delta^2\right)^n\text{ and }\lim_{n\to\infty}\sqrt n\left(1-\delta^2\right)^n=0,$$since $0\leqslant1-\delta^2<1$.

Answer 2

$0<\delta <1$, then $0<1-\delta^2 <1;$

$Q_n(x)\le √n(1-\delta^2)^n;$

Set: $(1-\delta^2)=:\frac{1}{1+\kappa}$, where $0<\kappa$.

Binomial expansion :

$(1-\delta^2)^n=\dfrac{1}{(1+\kappa)^n}=$

$\dfrac{1}{1+n\kappa+...}< \dfrac{1}{n\kappa};$

$0< Q_n(x) <√n\dfrac{1}{n\kappa}=\dfrac{1}{\kappa√n};$

Thus

$\lim_{n \rightarrow \infty}\sup_{x}Q_(x)=0,$

uniformly convergent.

What happens if $\delta=1?$