Good Kernels: Stein Fourier Analysis Exercise 10 Chapter 5

Question

Let $L_n$ be given by: $$L_n(x)=\frac{(1-x^2)^n}{c_n}\chi_{[-1,1]}(x)$$ Such that $c_n$ is chosen to satisfy $\int L_n(x)dx=1$. Prove $L_n$ is a sequence of good Kernels.

I have proved it is equivalent to calculating a limit, so you may skip to the last statement I wrote if you wish. If you think there is an easier upper bound I would like to see that too.

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We need to verify

1. $\int L_n(x)dx=1$, but this is clear from the definition of $c_n$.

2. $\int |L_n(x)|dx<\infty$, but this is also clear due to the fact $L_n(x)=|L_n(x)|$.

3. For every $\delta>0$, we have $\int_{|x|\geq \delta } |L_n|dx\xrightarrow[]{n} 0$.

So we only need to show the third property. At first I computed $c_n$ explicitly:

$$\int_{-1}^1 (1-x)^n(1+x)^n dx\stackrel{v=(1+x)/2}{=}\int_0^1 2^{2n+1}(1-v)^nv^ndv=2^{2n+1}\mathcal{B}(n+1,n+1)=2^{2n+1}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}$$ $$\therefore c_n=\frac{(2n+1)!}{(n!)^2 2^{2n+1}}$$

With this it is clear that:

$$\int_{|x|\geq \delta}L_n(x)dx=2\int_\delta^1\frac{(1-x^2)^n}{c_n} dx \leq 2\frac{(1-\delta^2)^n(1-\delta)(n!)^22^{2n+1}}{(2n+1)!}$$

I am not really sure how to show this goes to zero in $n$. But it seems like it really does as one can see below.

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Summarizing things we only need to show that for every $0<\delta<1$:

$$2\frac{(1-\delta^2)^n(1-\delta)(n!)^22^{2n+1}}{(2n+1)!}\xrightarrow[]{n\rightarrow \infty} 0$$

Answer 1

We can prove $c_n>\frac{1}{\sqrt{n}}$first:$$ c_n=\int^{1}_{-1}(1-x^2)^ndx\geq2\int^{\frac{1}{\sqrt{n}}}_0(1-x^2)^ndx\geq 2\int^{\frac{1}{\sqrt{n}}}_0(1-nx^2)dx=\frac{4}{3\sqrt{n}}>\frac{1}{\sqrt{n}}.$$ The reason for changing the integral bounds is that we want $(1-nx^2) $ be positive.

Then it's easy to get: $$\int_{\delta\leq|x|\leq1}c_n(1-x^2)^ndx\leq \int_{\delta\leq|x|\leq1}\sqrt{n}(1-x^2)^ndx\leq \int_{\delta\leq|x|\leq1}\sqrt{n}(1-\delta^2)^ndx\leq \int_{|x|\leq1}\sqrt{n}(1-\delta^2)^ndx=\sqrt{n}(1-\delta^2)^n \rightarrow0.$$

Answer 2

As Andrew cleverly remarked, using the asympotic approximation for the exponential yields: $$\lim2\frac{(1-\delta^2)^n(1-\delta)(n!)^22^{2n+1}}{(2n+1)!}=4(1-\delta)\lim\frac{(1-\delta^2)^n (\frac{n}{e})^{2n}2\pi n 2^{2n}}{\left(\frac{2n+1}{e}\right)^{2n+1}\sqrt{2\pi(2n+1)}}=$$ $$4(1-\delta)\lim(1-\delta^2)^n\left(\frac{2n}{2n+1}\right)^{2n}\frac{e}{2n+1}\frac{\sqrt{2\pi}n}{\sqrt{2n+1}}=$$ $$4(1-\delta)e^{-1} e\sqrt{2\pi}\frac{1}{2}\lim \frac{(1-\delta^2)^n}{\sqrt{2n+1}}=0$$

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If this were an exam question, I found a nice mnemonic device for the stirling formula:

$$n!=\int_0^\infty x^n e^{-x} dx=\int_0^\infty e^{n\ln(x)-x}dx$$

Using a Taylor series expansion of order $2$ centered at $x=n$ (for which $n\ln(x)-x$ has a zero derivative) yields $nln(x)-x\approx n\ln(n)-n +\frac{-(x-n)^2}{2\sqrt{n}^2}$. Integrating close to this is approximately equal to integrating everywhere by a lot of hand-waving. Anyways:

$$n!\approx e^{n\ln(n)-n}\int_{-\infty}^\infty e^{\frac{-(x-n)^2}{2\sqrt{n}^2}}=e^{n\ln(n)-n}\sqrt{2\pi n}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sqrt{n}}e^{\frac{-(x-n)^2}{2\sqrt{n}^2}}=\left(\frac{n}{e}\right)^n \sqrt{2\pi n}$$

This last equality follows from recognizing the integrand as a gaussian with mean $\mu=n$ and variance $\sigma^2=n$.