In Example 7.20 p.99 of enter link description here
It was stated $D(A)=W^{2,p}$.
Could anyone give a proof of above claim? Here, \begin{equation*} D(A) :=\left\{v\in L^p(\mathbb{R}) : \lim_{t\to 0}\frac{(S(t)-I)v}{t}\,\,\text{exists}\right\}\subset L^p(\mathbb{R}) \end{equation*}
Recall some basic facts. The Sobolev space $W^{k,p}$ is the space of all locally summable functions $u:\Omega\mapsto\mathbb{R}$ such that, for every multi-index $\alpha$ with $|\alpha|\le k$, the weak derivative $D^{\alpha}u$ exists and belongs to $L^p(\Omega)$. In other words, \begin{align*} W^{k,p}=\{f\in L^p(\Omega),D^{\alpha}f\,\,\text{exists,}\quad D^{\alpha}f\in L^p(\Omega),|\alpha|\le k\} \end{align*} At $k=1$, we have $W^{1,p}$ which implies there exists weak derivatives in $L^p(\Omega)$, written $$g=D^{\alpha}f,$$ provided \begin{align*} \int_{\Omega}f D^{\alpha}\phi\,dx&=(-1)^{\alpha}\int_{\Omega} g\phi\,dx\quad\forall\phi\in C_0^{\infty}(\Omega) \end{align*} Now if $u\in W^{k,p}(\Omega),$ we define its norm to be \begin{align*} \|u\|_{k,p}(\mathbb{R})&:= \begin{cases} \left(\sum_{|\alpha|\le k}\int_{\mathbb{R}}|D^{\alpha}u|^pdx\right)^{1/p}, &\text{ if }1\le p<\infty\\ \sum_{|\alpha|\le k}\text{esssup}_{\mathbb{R}}|D^{\alpha}u|, &\text{ if }p=\infty \end{cases} \end{align*} Hence, \begin{align*} \|u\|_{2,p}(\mathbb{R})&=\left(\sum_{|\alpha|\le 2}\int_{\mathbb{R}}|D^{\alpha}u|^pdx\right)^{1/p}=\left(\int_{\mathbb{R}}|u|^p+|Du|^p+|D^2u|^p dx\right)^{1/p} \end{align*}
The Heat semigroup has an explicit form as convolution with respect to a Gaussian distribution $G(t,x)$. That is, $H(t)f=\int G(t,x-y)f(y)dy$. This kernel is positive and $\int G(t,y)dy = 1$ for all $t > 0$. It follows that $H(t)$ is a contractive semigroup on every $L^{p}(\mathbb{R}^{n})$ for $1 \le p \le \infty$, but it's only $C^0$ for $1 \le p < \infty$. Let $A_p$ be the generator $H$ on $L^{p}$. The domain $\mathcal{D}(A_p)$ of the generator consists of all $f \in L^p$ for which there exists $g \in L^p$ such that $$ \lim_{t\downarrow 0}\|\frac{1}{t}\{T(t)f-f\}-g\|_{L^p}=0. $$ If such a $g$ exists, then it is unique and one defines $A_pf=g$. The generation theorem for $C^0$ semigroups guarantees that $A_p$ is a densely-defined, closed linear operator on the Banach space $X$ (in this case $X=L^p$.) The existence of the limit in $L^p$ of the time derivative is a very strong condition. For example, if $f \in \mathcal{D}(A_p)$, then \begin{align} & \int_{\mathbb{R}^n}G(s,x-y)(A_pf)(y) dy \\ & = \lim_{t\downarrow 0} \int_{\mathbb{R}^n} G(s,x-y)(\frac{1}{t}\{T(t)-I\}f)(y) dy \\ & = \lim_{t\downarrow 0}T(s)\frac{1}{t}\{T(t)-I\}f \\ & = \lim_{t\downarrow 0}\frac{1}{t}\{T(t)-I\}T(s)f \\ & = A_p T(s)f = T(s)A_pf. \end{align} The inner integral limit occurs pointwise in $y$ because the derivative limit occurs in $L^p$. And the limit occurs in $L^p$ in the $x$ variable because $T(s)$ is a bounded linear operator on $L^p$. The operator $T(s)$ has a spatial smoothing effect for any $s > 0$ because $G(s,x)$ is $C^{\infty}$ in the spatial variable for $s > 0$ and all orders of spatial derivatives of $G(s,x)$ are in every $L^p$ space for $1 \le p \le \infty$. And $G(s,x)$ is rapidly decaying in $x$ for $s > 0$.
Suppose $f \in W^{2,p}$. Normally one would require a compactly supported test function, but because $G(s,x)$ is so nice in $x$, this is easily extended to obtain the same using using $G(s,x)$ as a test function instead. Therefore, if $\nabla_w^2$ is the weak Laplacian, then $$ \int_{\mathbb{R}^n}G(t,x-y)\nabla^2_wf(y)dy = \int_{\mathbb{R}^n}f(y)\nabla_y^{2}G(t,x-y)dy. $$ The right side switches to actual derivatives in the weak equation. And, $$ \nabla_y^2G(t,x-y)dy = \frac{\partial}{\partial t}G(t,x-y), $$ which is a property of the heat kernel. Therefore, $$ T(t)\nabla_w^2f = \int_{\mathbb{R}^n}\frac{\partial}{\partial t}G(t,x-y)f(y)dy $$ Because of how $G$ is rapidly decaying, you can interchange the derivative with respect to $t$ with the spatial integral in order to obtain the pointwise identity $$ T(t)\nabla_w^2f = \frac{\partial}{\partial t}\int_{\mathbb{R}^n}G(t,x-y)f(y)dy $$ The derivative on the right also exists in an $L^p$ sense for $t > 0$. Therefore, it follows that $T(t)f \in \mathcal{D}(A_p)$ for all $t > 0$, $f \in L^p$, and $$ T(t)\nabla_w^2f = A_p T(t)f $$ As $t \downarrow 0$, $T(t)\nabla_w^2f$ converges in $L^p$ to $\nabla_w^f$ and $T(t)f$ converges in $L^p$ to $f$. Because $A_p$ is closed, that means that $f \in \mathcal{D}(A_p)$ and $\nabla_w^2 f = A_pf$. In other words, $W^{2,p}\subseteq \mathcal{D}(A_p)$.