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?


1 Answer 1


$\newcommand\R{\mathbb R}$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, $$\frac{I(a)}{a^2}\to F'(0)^2f'(0)^2.$$

  • $\begingroup$ Oh sorry. I need to reformulate my question regarding the mean. Sorry for deleting the comment. $\endgroup$
    – Isaac
    Commented May 26, 2023 at 18:10
  • $\begingroup$ 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. $\endgroup$
    – Isaac
    Commented May 26, 2023 at 18:40
  • 1
    $\begingroup$ @Isaac : I suggest you state your further question fully, with all details, in another post. $\endgroup$ Commented May 26, 2023 at 18:54

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