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?



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


