Exercise 2.22 in Brezis' Functional Analysis

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.

• What is the definition of $A^{\star}$?.... BTW most authors write $E^*,$ not $E^{\star}$, for the dual space of $E$. Commented Jul 21, 2019 at 9:26
• Actually this is not the dual space. $A^{\star}$ is the adjoint operator defined in the context of normed Banach spaces. Commented Jul 21, 2019 at 14:30

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 $$k$$.