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Let $b:\RR\_+ \times \RR^d \to \RR^d$ and $\sigma:\RR\_+ \times \RR^d \to \mathcal M^{\text{sym}}\_{d \times d} (\RR)$ be measurable. Let $a(t, x) := \sigma \sigma^\top (t, x)$. We assume that


1. There is $\lambda >0$ such that
$$
\lambda^{-1} |\xi|^2 \le \xi^\top a(t, x) \xi \le \lambda |\xi|^2
\quad \forall (t, x, \xi) \in {\RR}\_+ \times {\RR}^d \times {\RR}^d.
$$
2. There is $\eta, L>0$ such that
$$
\sup\_{(t, x) \in \RR\_+ \times \RR^d} |b(t, x)| + \sup\_{(t, x) \in \RR\_+ \times \RR^d} |\sigma(t, x)| + \sup\_{\substack{(x, y) \in \RR^d \times \RR^d \\ t \in \RR\_+, x \neq y}} \frac{|a(t, x) - a(t, y)}{|x-y|^\eta} \le L.
$$


We define Gaussian-like densities
$$
\begin{align}
p\_c (t, x, x') &:= \left ( \frac{c}{2 \pi t} \right )^{d/2} \exp \left ( - \frac{c |x'-x|^2}{2t} \right )
&& \forall c, t >0, \forall x,x' \in {\RR}^d, \\
G(y) &:= (2\pi)^{-d/2} \exp (-|y|^2/2) && \forall y \in {\RR}^d.
\end{align}
$$


Let $x,x' \in \RR^d$ and $h, t>0$. We are interested in bounding the absolute value of the quantity
$$
\alpha := \frac{1}{h} \left ( \frac{G ((\sqrt h \sigma (t, x))^{-1}(x'-x-b(t, x)h)}{\sqrt{h^d \det a(t, x)}} - \frac{G ((\sqrt h \sigma (t, x'))^{-1}(x'-x-b(t, x')h)}{\sqrt{h^d \det a(t, x')}} \right ).
$$


At the end of page 21 of the paper [On Some non Asymptotic Bounds for the Euler Scheme](https://arxiv.org/pdf/1001.1347.pdf), the authors said that



> 
> ...tedious but elementary computations involving the mean value theorem yield that $\exists c>0, \exists C \ge 1$ s.t.
> $$
> |\alpha| \le C h^{-1 + \eta/2} p\_c (h, x, x').
> $$
> 
> 
> 


The authors did not provide these "elementary computations", so I'm trying to derive above inequality.


**My attempt:** Let
$$
\begin{align}
y\_1 &:= (\sqrt h \sigma (t, x))^{-1}(x'-x-b(t, x)h \\
y\_2 &:= (\sqrt h \sigma (t, x'))^{-1}(x'-x-b(t, x')h \\
z\_1 &:= \sqrt{h^d \det a(t, x)} \\
z\_2 &:= \sqrt{h^d \det a(t, x')}.
\end{align}
$$


Then
$$
\begin{align}
|\alpha| &= \frac{1}{h} \left | \frac{G(y\_1)}{z\_1} - \frac{G(y\_2)}{z\_2} \right | \\
&= \frac{1}{h} \frac{|z\_2 G(y\_1) - z\_1 G(y\_2)|}{z\_1 z\_2}.
\end{align}
$$


By assumption (1.), we can lower bound $z\_1 z\_2$.



> 
> Could you elaborate on how to upper bound $|z\_2 G(y\_1) - z\_1 G(y\_2)|$?
> 
> 
> 


Thank you so much for your help!



How to upper bound the difference between these two Gaussian-like densities?
$\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand\la\lambda$We have
\begin{equation\*}
 |\al|=h^{-1-d/2}|g(1)-g(0)|\le h^{-1-d/2}\sup\_{s\in[0,1]}|g'(s)|,
\end{equation\*}
where
\begin{equation\*}
 g(s):=\frac{G((\sqrt h\si(s))^{-1}y(s))}{\sqrt{\det a(s)}}, 
\end{equation\*}
$y(s):=u-hb(s)$, $u:=x'-x$,
$\si(s):=(1-s)\si(t,x)+s\si(t,x')$, $b(s):=(1-s)b(t,x)+sb(t,x')$, $a(s):=(1-s)a(t,x)+sa(t,x')$.
For $s\in[0,1]$,  

\begin{equation\*}
 \frac{g'(s)}{g(s)}=g\_1+g\_2+g\_3, 
\end{equation\*}
where
\begin{equation\*}
 g\_1:=-\frac{\det a(1)-\det a(0)}{2a(s)}, \quad
 g\_2:=-(b(1)-b(0))^\top a(s)^{-1}y(s), 
\end{equation\*}
\begin{equation\*}
 g\_3:=\frac1h\, y(s)^\top a(s)^{-1}(a(1)-a(0))a(s)^{-1}y(s). 
\end{equation\*}


We have $\la^{-1}I\le a(s)\le\la I$ and hence
\begin{equation\*}
 \la^{-1}\le |a(s)|\le\la, \quad \det a(s)\ge\la^{-d},
\end{equation\*}
and, in view of [this answer](https://mathoverflow.net/a/447138/36721), $|\det a(1)-\det a(0)|\le\la^{d-1}d\,|a(1)-a(0)|\le\la^{d-1}d\,L|u|^\eta$. So,
\begin{equation\*}
 |g\_1|\ll|u|^\eta;
\end{equation\*}
here and in what follows, $A\ll B$ and $B\gg A$ mean that $A\le CB$ for some real $C>0$ depending only on $d,\la,L,T$ -- given that $h\in(0,T]$. Next,
\begin{equation\*}
 |g\_2|\le2L\la(|u|+hL)\ll|u|+h,
\end{equation\*}
\begin{equation\*}
 |g\_3|\le\frac1h\,\la^2 L|u|^\eta (|u|+hL)^2\ll\frac{|u|^{\eta+2}}h+|u|^\eta, 
\end{equation\*}
since $h\in(0,T]$.


So,
\begin{equation\*}
 \frac{h^{1+d/2}|\al|}{g(s)}\ll |u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h. 
\end{equation\*}


Next,
\begin{equation\*}
 g(s)\le\la^{d/2}\exp\Big(-\frac{|u|^2/2-(Lh)^2}{2\la h}\Big)
 \ll\exp\Big(-\frac{|u|^2}{4\la h}\Big)
 \ll h^{d/2}e^{-c|u|^2/h}p\_c(h,x,x') 
\end{equation\*}
for $c:=\frac1{6\la}$. So,
\begin{equation\*}
 \frac{|\al|}{h^{-1+\eta/2} p\_c (h,x,x')}
 \ll \Big(|u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h\Big)h^{-\eta/2}e^{-c|u|^2/h}. \tag{10}\label{10}
\end{equation\*}
Further,
\begin{equation\*}
 |u|^\eta h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2}e^{-c|u|^2/h}
 \le\max\_{w\ge0}w^{\eta/2}e^{-cw}\ll1; \tag{20}\label{20}
\end{equation\*}
\begin{equation\*}
 |u| h^{-\eta/2}e^{-c|u|^2/h}=h^{(1-\eta)/2}(|u|^2/h)^{1/2}e^{-c|u|^2/h}
 \le T^{(1-\eta)/2}\max\_{w\ge0}w^{1/2}e^{-cw}\ll1 \tag{30}\label{30}
\end{equation\*}
provided that
\begin{equation\*}
 \eta\le1; \tag{$\*$}\label{\*}
\end{equation\*}
\begin{equation\*}
 h\,h^{-\eta/2}e^{-c|u|^2/h}\le h^{1-\eta/2}\ll1,
\end{equation\*}
again provided \eqref{\*};
\begin{equation\*}
 \frac{|u|^{\eta+2}}h h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2+1}e^{-c|u|^2/h}
 \le\max\_{w\ge0}w^{\eta/2+1}e^{-cw}\ll1. \tag{40}\label{40}
\end{equation\*}


Now the desired inequality
\begin{equation\*}
 |\al|\ll h^{-1+\eta/2} p\_c (h,x,x') \tag{50}\label{50}
\end{equation\*}
follows from \eqref{10}, \eqref{20}, \eqref{30}, and \eqref{40}, provided \eqref{\*}. $\quad\Box$




---


Without condition \eqref{\*}, inequality \eqref{50} will fail to hold in this generality.  

E.g., suppose that for all $t,x$ we have $\si(t,x)=I$ and $|b(t,x)|\in[0,1]$, so that $a(t,x)=I$ and your conditions 1 and 2 hold with $\la=1$, $L=2$, and any real $\eta>0$. Suppose also that $x=0$, $|x'|=\sqrt h$, $b(t,0)=0$, and $b(t,x')=-x'/|x'|$. Then
\begin{equation}
 \frac{|\al|}{h^{-1+\eta/2} p\_c (h,x,x')}\asymp h^{1/2-\eta/2}\to\infty
\end{equation}
as $h\downarrow0$ if $\eta>1$. $\quad\Box$


