Let $I$ be the open interval $(0, 1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.17 Let $H=L^2(I)$ and $A: D(A) \subset H \rightarrow H$ be the unbounded operator defined by $A u=u^{\prime \prime}$, whose domain $D(A)$ will be made precise below. Determine $A^{*}, D\left(A^{*}\right), N(A)$, and $N\left(A^{*}\right)$ in the following cases:
- $D(A)=\left\{u \in H^2 (I) : u(0)=u(1)=0\right\}$.
- $D(A)=H^2(I)$.
- $D(A)=\left\{u \in H^2(I) : u(0)=u(1)=u'(0)=u'(1)=0\right\}$.
- $D(A)=\left\{u \in H^2(I) : u(0)=u(1)\right\}$.
There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on my computation of $D(A^*)$?
Let $A^* : D(A^*) \subset H \to H$ be the adjoint of $A$. We have the duality $$ \int_I v (Au) = \int_I (A^*v)u, \quad v \in D(A^*), u \in D(A), $$ which is equivalent to $$ \int_I v u'' = \int_I (A^*v)u, \quad v \in D(A^*), u \in D(A). \quad (*) $$
We have $(*)$ and the fact $C^\infty_c (I) \subset D(A)$ imply $D(A^*) \subset H^2(I)$ and $A^*v = v''$. It is clear that $C^\infty_c (I) \subset D(A^*)$.
We have $(*)$ and integration by parts (IbP) imply $$ D(A^*) = \{ v \in H^2 (I) : v(1) u'(1) - v(0) u'(0) - [v'(1) u(1) - v'(0) u(0)] = 0 \quad \forall u \in D(A) \}. \quad (**) $$
We have $(**)$ implies $$ \begin{align} D(A^*) &= \{ v \in H^2 (I) : v(1) u'(1) - v(0) u'(0) = 0 \quad \forall u \in D(A) \} \\ &= \{ v \in H^2 (I) : v(1) = v(0) = 0 \}. \end{align} $$
We have $(**)$ implies $$ \begin{align} D(A^*) &= \{ v \in H^2 (I) : v(1) u'(1) - v(0) u'(0) - (v'(1) u(1) - v'(0) u(0)) = 0 \quad \forall u \in H^2 (I) \} \\ &= \{ v \in H^2 (I) : v(1) = v(0) = v'(1) = v'(0) = 0 \}. \end{align} $$
Clearly, $D(A^*) = H^2 (I)$.
We have $(**)$ implies $$ \begin{align} D(A^*) &= \{ v \in H^2 (I) : v(1) u'(1) - v(0) u'(0) - [v'(1) - v'(0)] u(0) = 0 \quad \forall u \in D(A) \} \\ &= \{ v \in H^2 (I) : v(1) = v(0) = v'(1) - v'(0) = 0 \}. \end{align} $$