Let $E = L^p(0, 1)$ with $1 \le p < \infty$. Consider the unbounded operator $A: D(A) \subset E \to E$ defined by$$D(A) = \{u \in W^{1, p}(0, 1),\text{ }u(0) = 0\} \text{ and }Au = u'.$$My two questions are as follows.
- What is $A^*$ here?
- For what $1 \le p < \infty$ is $D(A^*)$ dense in $E^* = L^{p'}(0, 1)$?
For $1 < p < \infty$, $L^{p}(0,1)^{\star}=L^{p/(p-1)}(0,1)$. And $L^{1}(0,1)^{\star}=L^{\infty}(0,1)$. Therefore, the domain of $A_p^{\star}$ is the set of functionals $\Phi_{g} \in L^{p}(0,1)^{\star}$ such that $$ \langle A_p f, \Phi_{g}\rangle = \langle f,A_p^\star\Phi_g\rangle, \;\;\; f \in \mathcal{D}(A_p). $$ That is, $\Phi_g \in \mathcal{D}(A_p^{\star})$ iff there exists $h\in L^p(0,1)^{\star}$ such that $$ \int_{0}^{1}f'gdx=\int_{0}^{1}fhdx,\;\;\; f\in \mathcal{D}(A_p). $$ Because $\mathcal{D}(A_p)$ includes compactly supported test functions, then it is necessary that $g$ be absolutely continuous with $g'=-h\in L^p(0,1)^{\star}$. Therefore, $$ \mathcal{D}(A_p^{\star})= \{ \Phi_g : g \in (L^{p})^{\star} \mbox{ is abs. cont. with } g'\in (L^{p})^{\star}\},\\ A_p^{\star}\Phi_g = -\Phi_{g'}, \\ g(1)=0. $$ Absolutely continuous functions are dense in $L^p(0,1)$ for $1 \le p < \infty$, but not for $p=\infty$ because the closure of continuous functions in $L^{\infty}(0,1)$ consists of continuous functions.