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

Then $$ \partial_t g(t, x) = \Delta g(t, x) = \left(\frac{|x|^2-2 d t}{4 t^2}\right) g(t, x). $$

Corollary 1.3 and Theorem 1.2 in the paper Upper Bounds of Derivatives of the Heat Kernel on an Arbitrary Complete Manifold by Alexander Grigor'yan imply there is a constant $C$ such that $$ |\partial_t g| (t, x) \le C\frac{g(2t, x)}{t} \quad \forall t>0, \forall x \in \mathbb R^d. $$

This upper bound is not good enough for my purpose because $\int_0^t \frac{\mathrm d s}{s} = +\infty$ for any $t>0$. I would like to ask if the following improvement is possible, i.e.,

There exist a constant $C \ge 1$ and a measurable function $f:(0, \infty) \to \mathbb R$ such that

  • $\int_0^t f(s) \, \mathrm d s < +\infty$ for all $t>0$.
  • $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$.

Any reference is appreciated. Thank you so much for your help!

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1 Answer 1

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Such a function $f$ does not exist.

Indeed, if the inequality $|\partial_t g|(t,x)\le f(t)g(Ct,x)$ holds for all $t>0$ and $x\in\mathbb R^d$, then it holds for $x=0$, so that $$f(t)\ge f_*(t):=\frac{|\partial_t g|(t,0)}{g(Ct,0)} =C^{d/2}\frac{d }{2 t}$$ for all real $t>0$. So, $\int_0^t f\ge\int_0^t f_*=\infty$ for all $t>0$.

(Instead of choosing $x=0$, here we can more generally let $|x|=O(\sqrt t)$.)

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  • $\begingroup$ Thank you so much for your counter-example! $\endgroup$
    – Analyst
    Commented Jun 9, 2023 at 15:50

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