# Is there $f$ such that $\int_0^t f(s)\,\mathrm d s<\infty$ and $|\partial_t g| (t, x) \le f(t)g(Ct, x)$ for all $t>0$ and $x \in \mathbb R^d$?

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!

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 $$|x|=O(\sqrt t)$$.)