Showing a sequence of functions converges to a Gaussian

Question

Suppose we have a thrice continuously differentiable function such that the following hold:

1. $f(0)=1$, $f^\prime(0)=0$, and $f^{\prime\prime}(0)=-1$.
2. There exists $\alpha>0$ such that $\vert f(x)\vert\leq \exp(-\alpha x^2)$

Then,

$$ \lim_{n\rightarrow\infty}\int_{-\infty}^\infty\left(f\left(\frac{x}{\sqrt{n}}\right)\right)^ndx=\int_{-\infty}^\infty e^{-x^2/2}dx $$

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It is easy to show using property 2. that the dominated convergence theorem will apply, and hence we need only show that, at least pointwise a.e,

$$ \lim_{n\rightarrow\infty}\left(f\left(\frac{x}{\sqrt{n}}\right)\right)^n=e^{-x^2/2} $$ The obvious thing to try is a Taylor series, using property $1.$:

$$ f(x)=f(0)+f^\prime(0)x+\frac{f^{\prime\prime}(0)}{2}x^2+\frac{f^{(3)}(\xi)}{6}x^3=1-\frac{x^2}{2}+\frac{f^{(3)}(\xi)}{6}x^3 $$Then, we have

$$ \left(f\left(\frac{x}{\sqrt{n}}\right)\right)^n=\left(1-\frac{x^2}{2n}+\frac{f^{(3)}(\xi)}{6n^{3/2}}x^3\right)^n $$ This is almost clearly converging to $\exp(-x^2/2)$, but how should I go about showing it rigorously? Should I use a different form of the Taylor remainder, or do I need to invoke property 2. again?

Answer 1

I'm not sure how rigorous a proof you're looking for/need to produce, but a quick and dirty argument I came up with is: $$\lim_{n \to \infty}[f(\frac{x}{\sqrt n})]^n= \lim_{n \to \infty}\bigg(1-\frac{x^2}{2n}+\frac{f'''(\xi)}{6n^{3/2}}x^3\bigg)^n$$ As $n \to \infty$, the $\mathcal{O}(n^{-1})$ term dominates the $\mathcal{O(n^{-3/2})}$ term, and so, for large enough n, we can write: $$\lim_{n \to \infty} \bigg(1-\frac{x^2}{2n}\bigg)^n$$ Since $\lim_{n \to \infty} (1+t/n)^n=\exp(t)$, we can see that by substituting $t=-\frac12x^2$ in the above equation, we get this limit. Evaluating it and subbing back in t, the entire expression is equal to $\exp(-\frac12x^2)$, the (un-normalized) Gaussian function we want.