I am studying the concept of semigroup with the example of the heat equation and I have a concern. Once you have the semigroup generated by the Laplace operator (as in the case of the heat equation). How can I explicitly show that $\left.\frac{d}{dt}\right|_{t=0}T(t)=\Delta$?.
I will give a bit of context to understand what I would like to demonstrate.
Let the Heat equation \begin{align} u_t(t,x)&=\Delta u(t,x)\\ u(0,x)&=f(x) \end{align} By Fourier transform, \begin{align} \widehat{u}_t(t,\xi)&=-|\xi|^2\widehat{u}(t,\xi)\\ \widehat{u}(0,\xi)&=\widehat{f}(\xi) \end{align} This equation has as a solution \begin{align} \widehat{u}(t,\xi)=\mathrm{e}^{-t|\xi|^2}\widehat{f}(\xi) \end{align}
If $P_t(x)=\mathcal{F}^{-1}(\mathcal{e}^{-t|\xi|^2})(x)$ is the heat kernel, then
\begin{align} \widehat{u}(t,\xi)=\widehat{P_t}(\xi)\widehat{f}(\xi) \end{align} now, by Convolution theorem, \begin{align} u(t,x)=(P_t*f)(x) \end{align} Let $(T(t))_{t\geq 0}$ the semigroup generated by the operator $\Delta$ defined by $T(t)u=(P_t*u)$
Question 1. How can I show that $\left.\frac{d}{dt}\right|_{t=0}T(t)=\Delta$?
My attempt (formally):\begin{align} \left.\frac{d}{dt}\right|_{t=0}(P_t*u)(x)=\left.\frac{d}{dt}\right|_{t=0}\int P_t(x-y)u(y)\,dy\\ =\int \left.\frac{d}{dt}\right|_{t=0}P_t(x-y)u(y)\,dy \end{align} and \begin{align} \left.\frac{d}{dt}\right|_{t=0} P_t(x-y)&=\left.\frac{d}{dt}\right|_{t=0} \int \mathrm{e}^{i(x-y)\cdot \xi} \mathrm{e}^{-t|\xi|^2}\,d\xi\\ &=\int \mathrm{e}^{i(x-y)\cdot \xi}\left.\frac{d}{dt}\right|_{t=0}\mathrm{e}^{-t|\xi|^2}\,d\xi\\ &=\int \mathrm{e}^{i(x-y)\cdot \xi}\cdot -|\xi|^2\,d\xi \end{align}
Then, by Fubini's theorem, \begin{align} \int \left.\frac{d}{dt}\right|_{t=0}P_t(x-y)u(y)\,dy&=\int \left(\int \mathrm{e}^{i(x-y)\cdot \xi}\cdot -|\xi|^2\,d\xi\right)u(y)\,dy\\ &=\int\mathrm{e}^{ix\cdot\xi}\cdot -|\xi|^2\left(\int\mathrm{e}^{-iy\cdot \xi}u(y)\,dy\right)\,d\xi\\ &=\int\mathrm{e}^{ix\cdot \xi}\cdot -|\xi|^2\widehat{u}(\xi)\,d\xi\\ &=\mathcal{F}^{-1}(-|\xi|^2\widehat{u}(\xi))(x)\\ &=\Delta{u}(x) \end{align}
Therefore, $\left.\frac{d}{dt}\right|_{t=0}T(t)=\Delta$
Question: Why can the derivative with respect to $t$ enter under the integral sign?
