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
\begin{equation}
d\mu\_a(x):=\frac{1}{\sqrt{2\pi}f(a)} e^{-\frac{1}{2}\frac{x^2}{[f(a)]^2}}
\end{equation}
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
\begin{equation}
\int\_{\mathbb{R}}F(x)d\mu\_a(x) \to F(0)
\end{equation}
as $a \to 0$.


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


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?



Convergence of Gaussian measures $\{ d\mu_a \}$ whose variances depend smoothly on the index $a$
$\newcommand\R{\mathbb R}$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,
$$\frac{I(a)}{a^2}\to F'(0)^2f'(0)^2.$$


