Let $p(u,x)=\frac{1}{(4\pi u)^{q/2}}e^{-|x|^2/(4u)},u>0,x \in \mathbb{R}^q.$


1. Prove that for $r\geq 0,c>1$ there exists $C>0$ (depending on $r,c$) such that $$\forall x \in \mathbb{R}^q,u>0,\frac{|x|^{2r}}{u^r}p(u,x) \leq Cp(cu,x).$$
2. Deduce that for $n \in \mathbb{N}^q,k \in \mathbb{N},$ there exists $C'>0$ such that $|\partial\_x^n\partial\_u^k p(u,x)| \leq C' u^{-|n|/2-k}p(cu,x)$ where $\partial^n\_x=\partial x\_1^{n\_1}...\partial x\_q^{n\_q}$


1 follows from $\frac{|x|^{2}}{4ur}(1-\frac{1}{c}) \leq e^{\frac{|x|^2}{4ur}(1-1/c)}$ and we choose $C=c^{q/2} 4^r(r+1)^r/(1-1/c)^r$


How to deduce 2 (for simplicity we can take $q=1)$? *Is it possible to prove the inequality using induction*? If so, how?



Inequality: multivariate normal distribition
$\newcommand\p\partial$By induction,
$$\p\_u^k p(u,x)=\frac1{u^k}\,P\_k\Big(\frac{|x|^2}u\Big)p(u,x)
=\frac K{u^{k+q/2}}\,P\_k\Big(\frac{|x|^2}u\Big)e^{-|x|^2/(4u)}, \tag{1}\label{1}$$
where $K$ is a real number not depending on $u$ or $x$, and $P\_k(z)$ is a polynomial (with coefficients not depending on $u$ or $x$).


Next, for any polynomial $P$ (in $q$ variables) and any $j=1,\dots,q$, there is another polynomial $Q\_j$ (in $q$ variables) such that
$$\p\_{x\_j} \Big(P\Big(\frac{x}{u^{1/2}}\Big)e^{-|x|^2/(4u)}\Big)
=\frac1{u^{1/2}}\,Q\_j\Big(\frac{x}{u^{1/2}}\Big)e^{-|x|^2/(4u)}.$$
So, in view of \eqref{1},for some polynomial $R$ (in $q$ variables) and all real $u>0$,
$$|\p\_x^n\p\_u^k p(u,x)|
=\frac1{u^{|n|/2+k+q/2}}\,\Big|R\Big(\frac{x}{u^{1/2}}\Big)\Big|e^{-|x|^2/(4u)} \\ 
\le \frac C{u^{|n|/2+k+q/2}}\,e^{-|x|^2/(4cu)}
=\frac{C\_1}{u^{k+|n|/2}}\,p(cu,x),$$
where $C$ and $C\_1$ are positive real numbers not depending on $u$ or $x$. $\quad\Box$


