Let $I$ be the open interval $(0, 1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.16 Let $E=L^p(I)$ with $1 \leq p<\infty$. Consider the unbounded operator $A: D(A) \subset E \rightarrow E$ defined by $$ D(A)=\left\{u \in W^{1, p}(I) : u(0)=0\right\} \quad \text { and } \quad A u=u^{\prime} . $$
- Check that $D(A)$ is dense in $E$ and that $A$ is closed (i.e., its graph $G(A)$ is closed in $E \times E$).
- Determine $R(A)$ and $N(A)$.
- Compute $A^{*}$. Check that $D\left(A^{*}\right)$ is dense in $E^{*}=L^{p'}(I)$ when $1<p<\infty$, but $D\left(A^{*}\right)$ is not dense in $E^{*}=L^{\infty}(I)$ when $p=1$. Here $p'$ is the Hölder conjugate of $p$.
There are possibly subtle mistakes that I could not recognize in my below attempt of (2). Could you please have a check on it?
Let's prove that $R(A) = E$. Fix $v \in E$ and define $u \in E$ by $u(x) = \int_0^x v(s) \, \mathrm d s$. Clearly, $u(0)=0$. It remains to prove that $u \in W^{1, p}(I)$ with $u' = v$. Indeed, for $\varphi \in C^1_c (I)$ we have $$ \begin{align*} \int_I u \varphi' &= \int_I \left ( \int_0^x v(s) \, \mathrm d s \right ) \varphi' (x) \, \mathrm d s \\ &= \int_I \left ( \int_s^1 \varphi' (x) \, \mathrm d x \right ) v(s) \, \mathrm d s \quad \text{by Fubini's theorem} \\ &= \int_I (\varphi (1) - \varphi (s)) v(s) \, \mathrm d s \\ &= - \int_I \varphi (s) v(s) \, \mathrm d s \quad \text{because} \quad \operatorname{supp} \varphi \subsetneq I \\ &= - \int_I v \varphi. \end{align*} $$
Second, we prove that $N(A) = \{0\}$. Indeed, for $u \in D(A)$ with $u'=0$ we have $u (x) = \int_0^x u'(s) \, \mathrm d s=0$.