Let $f : \mathbb{R} \to \mathbb{R}$ be a smooth function which has a smooth inverse and satisfies the estimate
\begin{equation}
\lvert f(x) \rvert \leq \lvert x \rvert.
\end{equation}


Also, let $d\mu$ be the standard normal Gaussian measure on $\mathbb{R}$.


Assume that
\begin{equation}
\int\_{\mathbb{R}} \lvert f(x) -f(y) \rvert^2 d\mu(x) d\mu(y) \leq \int\_{\mathbb{R}} \lvert x-y \rvert^2 d\mu(x) d\mu(y)
\end{equation}
as well.


Now, if we taylor-expand $f(x)$ around some fixed $w \in \mathbb{R}$, we have
\begin{equation}
f(x)-f(w)=f'(w)(x-w)+F(x,w)
\end{equation}


Here, the remainder term $F(x,w)$ satisfies the decay $o(\lvert x-w \rvert)$ when $\lvert x-y \rvert \to 0^+$. However, I wonder if there is any bound over
\begin{equation}
\frac{\int\_{\mathbb{R}}\int\_{\mathbb{R}} \lvert F(x,w)-F(y,w) \rvert^2 d\mu(x)d\mu(y)}{\int\_{\mathbb{R}}\int\_{\mathbb{R}} \lvert x-y \rvert^2 d\mu(x)d\mu(y)}
\end{equation}
with respect to quantities regarding $f$ at $w$


This seems to be related to the behavior of $F(x,w)$ at $x$ far away from $w$, but I cannot figure out such behaviors..


Could anyone please help me?



Estimating the bound of the integral over whole $\mathbb{R}$ of the Taylor remainder term?
$\newcommand\R{\mathbb R}$We want to bound the ratio
\begin{equation\*}
 R:=\frac ND, 
\end{equation\*}
where
\begin{equation\*}
 N:=\iint\_{\R^2}(F(x,w)-F(y,w))^2 \mu(dx)\mu(dy),\quad 
 D:=\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy). 
\end{equation\*}


We have $D=2$. Next,
\begin{equation\*}
 A:=(F(x,w)-F(y,w))^2=(f(x)-f(y)-f'(w)(x-y))^2 \\ 
 \le2(f(x)-f(y))^2+2f'(w)^2(x-y)^2 
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
 R=\frac N2&\le\iint\_{\R^2}(f(x)-f(y))^2 \mu(dx)\mu(dy) \\ 
 &+f'(w)^2\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy) \\ 
 &\le(1+f'(w)^2)\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy)
 =2(1+f'(w)^2). 
\end{aligned}
\tag{1}\label{1}
\end{equation\*}


On the other hand,
\begin{equation\*}
 A\ge\frac12\,f'(w)^2(x-y)^2-(f(x)-f(y))^2 
\end{equation\*}
and hence
\begin{equation\*}
 R=\frac N2\ge\frac14\,f'(w)^2\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy) \\ 
 -\frac12\,\iint\_{\R^2}(x-y)^2 \mu(dx)\mu(dy)
 = \frac12\,f'(w)^2-1. \tag{2}\label{2}
\end{equation\*}


Thus, we have the upper bound on $R$ in \eqref{1} in terms of $|f'(w)|$. However, \eqref{2} shows that, if we cannot control $|f'(w)|$, then there is no finite bound on $R$.


