$X=C_0(\mathbb R^n)$ is the closure of Schwartz function space $\mathcal S(\mathbb R^n)$ under the $L^{\infty}(\mathbb R^n)$ norm. Define $$ T_tu=\left\{ \begin{aligned} &\frac{1}{t^{\frac n2}}G\left(\frac{x}{\sqrt t}\right)*u,\,t>0,\\ &u, t=0, \end{aligned} \right. $$ where $G$ is the Heat-kernel $p_l$ for $l=1$: $$G(x)=\frac{1}{({4 \pi})^{\frac{n}{2}}}e^{-\frac{|x|^2}{4}}.$$ Then $T_t$ is a strong continous contraction semigroup.
Show that the generator of $T_t$ is $A=\Delta$, $\mathcal D(A)=\{ u\in C_0:\Delta u\in C_0\}$.
I proved that $T_t$ is a strong continous contraction semigroup. But I cannot get its generator.
I denote by $$p_t(x)=\frac{1}{(2\pi t)^{n/2}}e^{-\frac{|x|^2}{4t}}.$$ the Heat-kernel. Since $u\in \mathcal S(\mathbb R^n)$, your claim is equivalent to prove that $$\lim_{h\to 0}\frac{1}{(2\pi)^{n/2}}\frac{\hat p_h \hat u-\hat u}{h}=\widehat{\frac{1}{2}\Delta u},$$ where $$\hat f(\xi):=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb R^d}e^{-i\xi\cdot x}f(x)\,\mathrm d x\quad \text{and}\quad \widehat{\Delta f}(\xi)=-\xi^2\hat f(\xi).$$
Therefore, the claim is equivalent to $$\lim_{h\to 0}\frac{e^{h\psi}g-g}{h}=\psi g,\quad g\in \mathcal S(\mathbb R^n),$$ where $\psi(\xi)=-\frac{\xi^2}{2}$. To see the statement, consider $$\Phi(z):=\frac{e^z-1}{z}=\sum_{n\geq 2}\frac{z^{n-1}}{n!}.$$ Then, for $g\in \mathcal S(\mathbb R^n)$, $$\left\|\frac{e^{\psi h}g-g}{h}-\psi g\right\|_{L^2}^2=\|\Phi\circ (h\psi)\cdot \psi g\|_{L^2}^2=\int_{\mathbb R^n}\left|\Phi\left(-\frac{h\xi^2}{2}\right)\right|^2\left|\frac{\xi^2}{2}g(\xi)\right|^2\,\mathrm d \xi\underset{h\to 0}{\longrightarrow }0,$$ where the last equality comes from the dominated convergence theorem, and the claim follows.