Exercise 2.22 in Brezis' Functional Analysis

Question

I am trying to solve the excercise 2.22 from Brezis' Functional Analysis.

$2.22$ The purpose of this exercise is to construct an unbounded operator $A: D(A) \subset$ $E \rightarrow E$ that is densely defined, closed, and such that $\overline{D\left(A^{\star}\right)} \neq E^{\star} .$
Let $E=\ell^{1}$, so that $E^{\star}=\ell^{\infty}$. Consider the operator $A: D(A) \subset E \rightarrow E$ defined by $$ D(A)=\left\{u=\left(u_{n}\right) \in \ell^{1} ;\left(n u_{n}\right) \in \ell^{1}\right\} \text { and } A u=\left(n u_{n}\right) $$

1. Check that $A$ is densely defined and closed
2. Determine $D\left(A^{\star}\right), A^{\star}$, and $\overline{D\left(A^{\star}\right)}$.

The fist part is easy but I have problems with the second part. Brezis solution claims that:

$$ \begin{aligned} D\left(A^{\star}\right) &=\left\{v=\left(v_{n}\right) \in \ell^{\infty} ;\left(n v_{n}\right) \in \ell^{\infty}\right\}, \\ A^{\star} v &=\left(n v_{n}\right) \text { and } \overline{D\left(A^{\star}\right)}=c_{0} . \end{aligned} $$

But I can't prove that. In order to find $D(A^{\star})$, I concluded that if $(y_{n}) \in D(A^{\star})$, then $(y_{n}) \in l^{\infty}$ and $\left | \sum_{n\in\mathbb{N}}(n y_{n} u_{n}) \right | \leq \sum_{n \in \mathbb{N}} \left | u_{n} \right |$ for all $u_{n} \in D(A^{\star})$. My attempt is take in particular the secuences $(u_{n})=(\frac{1}{n^{r}})$ with $r > 2$ (note that $(u_{n})$ thus defined is in $D(A^{\star})$ ). So we have that: $$\left | \sum_{n \in \mathbb{N}} n y_{n} \frac{1}{n^{r}} \right | \leq \sum_{n \in \mathbb{N}} \left | \frac{1}{n^{r}} \right |, \hspace{1 cm } \forall r > 2 $$ And it is suffices to prove that $(ny_{n})$ is a bounded secuence in $\mathbb{R}$, but I can't get that.

Answer 1

You should write $|\sum nu_ny_n| \leq C \sum|u_n|$ for $(u_n) \in D(A)$. [There are two mistakes in your statement]. Just take $u_n=1$ for $n=k$ and $u_n=0$ for $ n \neq k$ where $k$ is fixed. You get $|ky_k| \leq C$ for all $k$.