Brezis' exercise 8.16.3: check that $D\left(A^{*}\right)$ is not dense in $E^{*}=L^{\infty}(I)$ when $p=1$

Question

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 $$ D(A)=\left\{u \in W^{1, p}(I) : u(0)=0\right\} \quad \text {and} \quad A u=u^{\prime} . $$

1. Check that $D(A)$ is dense in $E$ and that $A$ is closed (i.e., its graph $G(A)$ is closed in $E \times E$).
2. Determine $R(A)$ and $N(A)$.
3. 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 $p$.
4. Same questions for the operator $\widetilde{A}$ defined by $$ D(\widetilde{A})=W_0^{1, p}(I) \quad \text {and} \quad \widetilde{A} u=u'. $$

There are possibly subtle mistakes that I could not recognize in my below attempt of (3). Could you please have a check on it?

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Let $A^* : D(A^*) \subset E^* \to E^*$ be the adjoint of $A$. We have the duality $$ \int_I v (Au) = \int_I (A^*v)u, \quad v \in D(A^*), u \in D(A), \quad (*) $$ which implies $$ \int_I v u' = \int_I (A^*v)u, \quad v \in D(A^*), u \in D(A), \quad (**) $$

We have $(**)$ and the fact $C^1_c (I) \subset D(A)$ imply $D(A^*) \subset W^{1, p'}(I)$ and $A^*v = v'$. In particular, $D(A^*) \subset C(\bar I)$. For $\varphi \in C^1_c (I)$ we have $$ \left | \int_I \varphi (Au) \right | = \left | \int_I \varphi' u \right | \le \|\varphi'\|_{L^{p'}} \|u\|_{L^{p}}, \quad \forall u \in D(A), $$ which implies $\varphi \in D(A^*)$. Hence $C^1_c (I) \subset D(A^*)$. For $v \in D(A^*)$ the integration by parts implies $$ u(1) v(1) = u(0) v(0) = 0 \quad u \in D(A), $$ and thus $v(1)=0$.

If $p \in (1, \infty)$ then $p' \in (1, \infty)$. In this case, $C^1_c (I)$ and thus $D(A^*)$ are dense in $E^*$.

Now we consider the case $p=1$. Then a strictly positive constant function in $L^\infty(I)$ could not be approximated by any sequence $(v_n)$ in $D(A^*)$ because $v_n(1)=0$ for all $n$.

Answer 1

Let $p \in [1, \infty)$. You have proved that $$ D(A^*) \subset \left\{v \in W^{1, p'}(I) : v(1)=0\right\}. $$

The reverese inclusion actually holds. Fix $v \in W^{1, p'}(I)$ with $v(1)=0$. By integration by parts, we have $$ \left | \int_I v (Au) \right | = \left | \int_I v u' \right | = \left | \int_I v' u \right | \le \|v'\|_{L^{p'}} \|u\|_{L^{p}}, \quad \forall u \in D(A). $$

The second equality above is due to $v(1)=u(0)=0$. The claim then follows.