Let $E = L^p(0, 1)$ with $1 \le p < \infty$. Consider the unbounded operator $A: D(A) \subset E \to E$ defined by$$D(A) = \{u \in W^{1, p}(0, 1),\text{ }u(0) = 0\} \text{ and }Au = u'.$$I have two questions.
On
A different approach for 2:
Take $(u_n)$ in $D(A)$ such that $u_n\to u$ and $Au_n\to v$, both convergences in $E$.
Fact 1. $u\in W^{1,p}(0,1)$ with $u'=v$.
Proof: Take $\varphi\in C_c^\infty(0,1)$. Since $L^p(0,1)$ is continuously embedded into $L^1(0,1)$, there exists a constant $c_0$ such that
$$0\leq \left|\int_0^1u_n\varphi'-\int_0^1u\varphi'\right|\leq (\max |\varphi'| )\int_0^1|u_n-u|\leq c_0(\max |\varphi'| )\|u_n-u\|_{L^p},\quad\forall \ n \in\mathbb{N}.$$
and
$$0\leq \left|\int_0^1Au_n\varphi-\int_0^1v\varphi\right|\leq (\max |\varphi |)\int_0^1|Au_n-v|\leq c_0(\max \varphi )\|Au_n-v\|_{L^p},\quad\forall \ n \in\mathbb{N}.$$
Thus
$$\int_0^1u\varphi'=\lim_{n\to\infty}\int_0^1u_n\varphi'=\lim_{n\to\infty}\left(-\int_0^1Au_n\varphi\right)=-\int_0^1v\varphi.$$
As $\varphi$ is arbitrary, we get the desired fact.
Fact 2. $u(0)=0$.
Proof: Since $W^{1,p}(0,1)$ is continuously embedded into $C([0,1])$, there exists a constant $c_1$ such that
$$0\leq |u(0)|=|u_n(0)-u(0)|\leq \sup_{x\in[0,1]}|u_n(x)-u(x)|\leq c_1\|u_n-u\|_{W^{1,p}},\quad\forall\ n\in\mathbb N.$$
Taking the limit as $n\to\infty$, we get the desired fact.
Conclusion: $u\in D(A)$ and $v=Au$. So, $G(A)$ is closed in $E$ (see Theorem 4.13-3 in Kreyszig's book or this Wikipedia's page).
For $(1)$ $D(A)$ is dense, because it contains all smooth compactly supported functions $C_0^\infty$ which are dense in $L^p(0,1)$.
For $(2)$, you have the inverse operator $(A^{-1}y)(t)=\int\limits_{0}^{t}{y(s)ds}$ for each $y\in L^p(0,1)$, because $\int\limits_{0}^{t}{y(s)ds}$ is absolutely continuous and $(A^{-1}y)(0)=0$. Also $A^{-1}: L^p(0,1) \to L^p(0,1)$ is bounded: $$\|A^{-1}y\|_{L^p}^p=\int\limits_{0}^{1}{\left|\int\limits_{0}^{t}{y(s)ds}\right|^pdt}\leq 1\cdot\|y\|_{L^p}^p$$ Therefore $A$ is closed. The last is a special case of a more general
Proposition: Let $A: D(A)\subset X\to Y$, where $X,Y$ are Banach and $A$ is linear. If $A^{-1}: Y\to X$ exists and is bounded, then $A$ is closed.
Proof: Let $u_n\subset D(A)$ and $u_n\to u$, and $Au_n\to y$. Then, because $Y$ is closed, $y\in Y$. Also, because $A^{-1}$ is bounded $\Rightarrow A^{-1}(Au_n)=u_n\to A^{-1}y$ in $X$. But $u_n\to u$ in $X$ and limits are unique $\Rightarrow u=A^{-1}y$ and therefore $u\in D(A)$ and $Au=y$.