Estimative for Hessian of Heat Kernel in $\mathbb{R}^d$

Question

We consider the heat kernel $$ g :\mathbb R_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ). $$

The inequality (2.3) in this paper, shows that $$ |\nabla g(t, x)| \le \frac{2^{d/2}}{\sqrt{t}} g(2t, x). $$ So I imagine this is also true for the Hessian of the Heat Kernel $$ |\nabla^2 g(t, x)| \le C(t) g(2t, x). $$ Can you confirm if this happens? In fact, I was unsure of what was done to remove the x-dependent term in this estimate, even in the case of the gradient, since $\nabla g(t,x) = - \dfrac{x}{2t}g(t,x)$, looks like he used that $|x/2t|e^{-|x|^2/4t} \leq 1.$ I appreciate any comments.

Answer 1

We have $$\nabla^2 g(t,x)=g(t,x)\Big(\frac{xx^\top}{(2t)^2}-\frac{I_d}{2t}\Big),$$ where $I_d\in\mathbb R^{d\times d}$ is the identity matrix and $\mathbb R^d$ is identified with $\mathbb R^{d\times1}$.

So, $$\|\nabla^2 g(t,x)\|\le g(t,x)\Big(\frac{|x|^2}{(2t)^2}+\frac1{2t}\Big)\le \frac{C_d}t\,g(2t,x),$$ where $\|\cdot\|$ is the spectral norm of the matrix and $C_d$ is a positive real number depending only on $d$; here we used (with $u=|x|^2/(8t)$) the inequality $ue^{-u}\le e^{-1}$ for all real $u$. (One may take $C_d=(2/e+1/2)2^{d/2}$.)