Let $f \in C_c^{\infty}(\mathbb{R}^d)$ and $K_t = (4\pi t)^{-\frac{d}{2}}e^{-\frac{x^2}{4t}} $ for every $t>0$.
I want to prove this :
\begin{align*} \frac{d}{dt}(K_t*f)(x)&=\frac{d}{dt}\int_{\mathbb{R}^d} K_t(x-y) f(y)dy = \lim_{h \to 0} \int_{\mathbb{R}^d} \frac{K_{t+h}-K_t}{h}(x-y)f(y) dy\\ &= \int_{\mathbb{R}^d} \lim_{h \to 0} \frac{K_{t+h}-K_t}{h}(x-y)f(y) dy = \int_{\mathbb{R}^d} \Delta K_t(x-y)f(y)dy=((\Delta K_t)*f)(x) \end{align*}
In order to prove the third equality, I thought I could use the dominated convegrence theorem since $||\Delta K_t * f||_1$$ \leq ||{\Delta K_t}||_1 ||{f}||_1< \infty.$
$$\frac{\partial}{\partial t} K_t(x-y) f(y) = \lim_{h\to 0} \frac{K_{t+h}-K_t}{h}(x-y)f(y) = \Delta K_t(x-y)f(y) .$$ Now, by definition of the limit, for every $\epsilon > 0 $, there exists $\delta>0 $ such that for every $h\leq \delta$, $$\left| \frac{K_{t+h}-K_t}{h}(x-y)f(y) \right| \leq \Delta K_t(x-y)f(y) + \epsilon$$ taking $\epsilon = \Delta K_t(x-y)f(y)$, there exists $T$ such that for every $h \leq T$ $$\left| \frac{K_{t+h}-K_t}{h}(x-y)f(y) \right| \leq 2\Delta K_t(x-y)f(y) .$$
clearly I have a dependance of T to y which mean I can't use this, I can't find a better bound, does someone have a good idea ?
wlog let $d=1$. The heat kernel satisfies the heat equation, so $\partial_tK_t(x)=\Delta K_t(x)$. Therefore $(\partial_tK_t*f)(x)=(\Delta K_t*f)(x)$. We indicate the Fourier transform with $\hat{g}(\xi)=\int_{\mathbb{R}}g(x)e^{-i\xi x}dx$. We have that $(\widehat{\Delta K_t*f})(\xi)=(-\xi^2\hat{K}_t(\xi))\hat{f}(\xi)=\hat{K}_t(\xi)(-\xi^2\hat{f}(\xi))=(\widehat{K_t*\Delta f})(\xi)$. This implies $(\Delta K_t*f)(x)=(K_t*\Delta f)(x)$ a.e. Now note that $$(K_t*\Delta f)(x)=E^x[\Delta f(\sqrt{2}W_t)]$$ where $W=(W_t)_{t \geq 0}$ is a Brownian motion and $E^x[.]$ the expectation wrt the law of the process started at $x$ i.e. $E^x[f(\sqrt{2}W_t)]=E[f(x+\sqrt{2}W_t)]$. Dynkin's formula yields $E^x[\Delta f(\sqrt{2}W_t)]=\partial_tE^x[f(\sqrt{2}W_t)]$. Finally note that $(K_t*f)(x)=E^x[f(\sqrt{2}W_t)]$. In summary, the following holds for a.a. $x$: $$\begin{aligned}&(\partial_tK_t*f)(x)=(\Delta K_t*f)(x)=(K_t*\Delta f)(x)=\\ &=E^x[\Delta f(\sqrt{2}W_t)]=\partial_tE^x[f(\sqrt{2}W_t)]=\partial_t(K_t*f)(x)\end{aligned}$$