$\int_{\mathbb{R}}|p(v-r,x)-p(u-r,x)|\,dx \leq C\frac{v-u}{u-r}$

Question

Consider $p(u,x)=(4\pi u)^{-d/2}e^{-\frac{|x|^2}{4u}},u>0,x\in \mathbb{R}^d.$

Prove that there exists $C>0$ such that for all $0<u\leq v,r\in[0,u[,$ $$\int_{\mathbb{R}^d}|p(v-r,x)-p(u-r,x)|\, dx \leq C\frac{v-u}{u-r}$$

Any ideas how to prove it?

Answer 1

$\newcommand\R{\mathbb R}$Letting $s:=u-r$ and $t:=v-r$, rewrite the inequality in question as \begin{equation*} \int_{\R^d}dx\,|p(t,x)-p(s,x)| \le C\Big(\frac ts-1\Big) \tag{0}\label{0} \end{equation*} given that $0<s\le t<\infty$.

Note that \begin{equation*} |p(t,x)-p(s,x)|\le\int_s^t dw\,|D_w p(w,x)|, \end{equation*} where $D_w$ is the operator of partial differentiation with respect to $w$. So, \begin{equation*} \int_{\R^d}dx\,|p(t,x)-p(s,x)| \le\int_s^t dw\, \int_{\R^d}dx\,|D_w p(w,x)|. \tag{1}\label{1} \end{equation*} Next, \begin{equation*} \int_{\R^d}dx\,D_w p(w,x)=D_w\int_{\R^d}dx\, p(w,x)=D_w1=0 \end{equation*} and \begin{equation*} D_w p(w,x)=p(w,x)\Big(\frac{|x|^2}{4w^2}-\frac d{2w}\Big). \end{equation*} So, with $z_+:=\max(0,z)$ for real $z$, \begin{equation*} \int_{\R^d}dx\,|D_w p(w,x)| =2\int_{\R^d}dx\,p(w,x)\Big(\frac d{2w}-\frac{|x|^2}{4w^2}\Big)_+ \le2\int_{\R^d}dx\,p(w,x)\frac d{2w}=\frac dw. \end{equation*}

Thus, by \eqref{1}, \begin{equation*} \int_{\R^d}dx\,|p(t,x)-p(s,x)| \le d\,\ln\frac ts, \end{equation*} whence \eqref{0} follows, with $C=d$.