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From Terence Tao's Analysis II:

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I'm having problems with 14.8.2.(c), since it seems like the choice of $N$ depends on $c$ (and vice versa). The only proof I have relies on the convergence of the sequence $\sqrt{n} (1-\delta^2)^n$ to zero, but I don't know if that's even true.

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For $|x|>1$ let $p_n(x)=0.$ For $|x|\le 1$ let $p_n=c_n(1-x^2)^n$ where $c_n\cdot\int_{-1}^1(1-x^2)^ndx=1.$ By 14.8.2.(b) we have $c_n\le \sqrt n\,.$ The idea is to show that $p_n\to 0$ uniformly on $[-1,-\delta]\cup [\delta,1]$ for any fixed $\delta \in (0,1).$ So, given any $\epsilon^*$ in $(0,1)$ we find some "small" $x_n$ in $(0,1)$ such that $x_n\le |x|\le 1\implies p_n(x)\le\epsilon^*,$ and hope that $x_n\to 0$ as $n\to \infty.$

Since $c_n\le \sqrt n ,$ it would suffice that $\sqrt n\,(1-x_n^2)^n=\epsilon^*,$ equivalently $$x_n^2= 1-(\epsilon^*/\sqrt n)^{1/n}.$$

We have $x_n\to 0 \iff x_n^2\to 0 \iff (\epsilon^*/\sqrt n)^{1/n}\to 1.$

(i). Let $\epsilon^*=(1+k)^{-1}$. Then $k>0.$ Let $(1+k)^{1/n}=1+k_n.$ Then $k_n> 0.$ By the Binomial Theorem $1+k=(1+k_n)^n\ge 1+nk_n,$ so $k/n\ge k_n>0,$ so $k_n\to 0.$

So $(\epsilon^*)^{1/n}\to 1.$

(ii). Let $n^{1/2n}=1+y_n.$ For $n\ge 2$ we have $y_n>0$ and by the Binomial Theorem $$n=(1+y_n)^{2n}\ge 1+\binom {2n}{1}y_n+\binom {2n}{2}y_n^2> 1+\binom {2n}{2}y_n^2>1+2(n-1)^2y_n,$$ so $n-1>2(n-1)^2y_n^2,$ so $1/\sqrt {2(n-1)}>y_n>0$, so $y_n\to 0.$

So $(\sqrt n )^{1/n}=n^{1/2n}\to 1.$

(iii). Therefore $x_n\to 0.$

In summary, given $\epsilon>0$ and given $0<\delta<1,$ let $\epsilon^*=\min(1/2,\epsilon).$ Now for all $n$ large enough that $x_n$ (as defined above) is less than $\delta,$ we have $$1\ge |x|\ge \delta \implies 1\ge |x|>x_n\implies 0\le p_n(x)<p_n(x_n)\le \epsilon^*\le \epsilon.$$

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  • $\begingroup$ To get $x_n\to 0$ we want $(c_n)^{1/n}\to 1,$ and $c_n$ is small enough that it works. $\endgroup$ – DanielWainfleet Nov 9 '19 at 9:48

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