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


I have already proved that
$$
\begin{align}
g(t, x+y) &\le 2^{d/2} g(2t, x) e^{|y|^2/(4t)}, \\
\nabla g(t, x) &= \frac{-x}{2t} g(t, x).
\end{align}
$$


I'm trying to prove the inequality (2.3) in [this](https://www.sciencedirect.com/science/article/abs/pii/S0022039620306008) paper, i.e.,



> 
> $$
> |\nabla g(t, x)| \le \frac{2^{d/2}}{\sqrt{t}} g(2t, x).
> $$
> 
> 
> 


Could you explain how to fix my below failed attempt?




---


We have
$$
|\nabla g(t, x)| = \frac{|x|}{2t} g(t, x) \le \frac{2^{d/2} |x|}{2t} g(2t, x).
$$


Then I'm stuck because it's not true that $|x|^2 \le 4t$.



An estimate of the gradient of heat kernel
By your computation, it suffices to show that
$$\frac{|x|}{2\sqrt{t}}\leq 2^{d/2}\frac{g(2t,x)}{g(t,x)}=e^{\frac{|x|^2}{4t}}$$
This follows from the fact that $y\leq e^{y^2}$ for $y\geq 0$.


