Let \begin{align} \frac{\partial}{\partial_t}u(t,x)&=\Delta u(t,x),\quad t>0,\, x\in\mathbb{R}^n\\ u(0,x)&=f(x) \end{align}
where $f\in L^2(\mathbb{R}^n)$ (Heat's equation)
The heat kernel is the function $p_t(x-y)=\int_{\mathbb{R}^n}\mathrm{e}^{i(x-y)\cdot \xi}\mathrm{e}^{-t|\xi|^2}\,d\xi$. This kernel satisfies that $\partial_t p_t=\Delta p_t$ and $\lim_{t\to 0}p_t*u=u$ in the $L^2$-norm (I just checked).
Now I would like to verify, from these properties, that $\Delta$ is the infinitesimal generator of the semigroup $T(t)$ given by $(T(t)u)(x)=\int_{\mathbb{R}^n}P_t(x-y)u(y)\,dy,\quad u\in L^2$.
By definition, I must verify that $\lim_{t\to 0^+} \frac{T(t)u-u}{t}=Au$ for all $u\in D(A)=\left\{u\in L^2: \Delta u\in L^2\right\}$
Question 1. This limit is in $L^2$? or not?
Question 2. How proves that $\Delta$ is the infinitesimal generator of the semigroup $T(t)$?
Actualization. In question 1, the answer is affirmative. See In what sense is the limit in the definition of infinitesimal generator?
Question 2. I still don't know how to proves it. Only i have this: \begin{align} (\partial_t-\Delta) (S(t)u)(x)&=(\partial_t-\Delta)\int p_t(x-y)u(y)\,dy\\ &=\int (\partial_t-\Delta)p_t(x-y)u(y)\,dy\\ &=0 \end{align} but I don't know how to justify the step baoj the intergal sign
Actualization 2.
\begin{align} \partial_t (S(t)u)&=\partial_t (p_t*u)\\ &=(\partial_t p_t)*u\\ &=(\Delta p_t*u)\\ &=p_t*(\Delta u)\\ &\to \Delta u \end{align}