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

Let $0 < t_1 < t_2 <\infty$ and $1\le\lambda<\infty$ such that $\frac{t_2}{t_1} \le \lambda$. I would like to ask if there are constants $C_1, C_2>0$ (depending only on $\lambda$) such that $$ \frac{g(t_1, x)}{t_1} \le C_1 \frac{g(C_2 t_2, x)}{\sqrt{t_2}} \quad \forall x \in \mathbb R^d. $$

Thank you so much for your elaboration!

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    $\begingroup$ If I didn't make mistake, taking $x = 0$, $t_2 = 2t_1$ and sending $t_1$ to zero gives a counterexample. $\endgroup$
    – Aleksei Kulikov
    Commented Jun 5, 2023 at 16:10
  • $\begingroup$ Also, $\lambda$defined this way depends on $t_{1}$ and $t_{2}$. So, it does not make sense to let $C_{1}$ or $C_{2}$ depend on $\lambda$. $\endgroup$
    – Medo
    Commented Jun 5, 2023 at 16:42

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The answer is no. E.g., let $t_1\sim t_2\downarrow 0$ and $|x|\sim\sqrt{t_2}$.

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