# Convergence of Gaussian measures $\{ d\mu_a \}$ whose variances depend smoothly on the index $a$

Let $$f: \mathbb{R} \to \mathbb{R}$$ be a smooth function such that $$f(x)$$ is positive in a small punctured neighborhood of $$x=0$$ but $$f(0)=0$$.

Now, define a collection of centered Gaussian measures on $$\mathbb{R}$$ as $$$$d\mu_a(x):=\frac{1}{\sqrt{2\pi}f(a)} e^{-\frac{1}{2}\frac{x^2}{[f(a)]^2}}$$$$ where $$a$$ is any sufficiently small nonzero real number.

Then, it is well-known that $$d\mu_a \to \delta(0)$$ in the sense of probability measures as $$a \to 0$$.

Theerfore, for any smooth function $$F : \mathbb{R} \to \mathbb{R}$$, we have $$$$\int_{\mathbb{R}}F(x)d\mu_a(x) \to F(0)$$$$ as $$a \to 0$$.

However, I wonder what will happen to the following limit: $$$$\frac{1}{a^2}\int_{\mathbb{R}}[F(x)-F(0)]^2d\mu_a(x)$$$$ as $$a \to \infty$$. Since $$\frac{f(a)}{a} \to f'(0)$$, it is clear that $$$$\frac{1}{a^2}\int_{\mathbb{R}}x^2d\mu_a(x)= \Bigl( \frac{f(a)}{a} \Bigr)^2 \to [f'(0)]^2$$$$

However, is it possible to generalize to any smooth $$F$$ as in the above? This kind of issue is completely new to me and I can't quite see how to handle..

Could anyone please provide some insights?

$$\newcommand\R{\mathbb R}$$You need some restriction on the rate of growth of the smooth function $$F$$. Otherwise, if e.g. $$F(x)=e^{|x|^p}$$ for $$p>2$$, then $$\int_{\R}F(x)\mu_a(dx)=\infty\not\to F(0)$$ (as $$a \to 0$$) and similarly $$\int_\R(F(x)-F(0))^2\mu_a(dx)=\infty$$.
So, assume that $$|F(x)|\le Ce^{Cx^2} \tag{1}\label{1}$$ for some real $$C>0$$ and all real $$x$$. Then $$I(a):=\int_\R(F(x)-F(0))^2\mu_a(dx)=\int_\R(F(f(a)x)-F(0))^2g(x)\,dx =I_1(a)+I_2(a),$$ where $$g$$ is the standard normal p.d.f., $$I_1(a):=\int_{|x|\le1/a}(F(f(a)x)-F(0))^2g(x)\,dx,\quad I_2(a):=\int_{|x|>1/a}(F(f(a)x)-F(0))^2g(x)\,dx.$$ Next, by \eqref{1} and because $$f$$ is smooth with $$f(0)=0$$, for all $$a$$ close enough to $$0$$ we have $$0\le I_2(a)\le2\int_{|x|>1/a}(C^2e^{2Cf(a)^2x^2}+F(0)^2)g(x)\,dx \\ \le2\int_{|x|>1/a}(C^2e^{x^2/4}+F(0)^2)g(x)\,dx=o(a^2).$$ Further, because $$F$$ and $$f$$ are smooth and $$f(0)=0$$, $$I_1(a)=\int_{|x|\le1/a}(F'(0)+o(1))^2f(a)^2x^2g(x)\,dx \\ =(F'(0)+o(1))^2(f'(0)+o(1))^2a^2\int_{|x|\le1/a}x^2g(x)\,dx \\ =(F'(0)+o(1))^2(f'(0)+o(1))^2a^2(1+o(1)).$$ Thus, $$\frac{I(a)}{a^2}\to F'(0)^2f'(0)^2.$$
• Ok, suppose that the center of $d\mu_a$ is some $g(a)$ which is smooth w.r.t to $a$ and $g(0)=0$. And we change nothing else. Then, I am trying to figure out what changes. Commented May 26, 2023 at 18:40