# Polynomials can approximate the identity (proof)

From Terence Tao's Analysis II:

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.

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)

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