I am working on an exercise in Brezis' functional analysis book (ex 2.23) and I'm having trouble with a certain step:
Let $E=l^1$ and consider $T: E\to E$ with $Tu = (u_n/n)_{n\geq 1}$. Determine $N(T), N(T)^\bot, T^*, R(T^*), \overline{R(T^*)}$.
I got that $N(T) = \{0\}$, hence $N(T)^\bot = l^{\infty}$, the adjoint looks identical to $T$ but operates on the dual space $E^* = l^\infty$. The last two objects are interesting, though:
$$R(T^*) = \{(v_n/n)_n, v\in l^\infty\} = \bigcap_{p>1}l^p$$
because $\|(v_n/n)_n\|_{l^p} < \infty$ for any $v\in l^\infty$.
Now I don't know what $$\overline{\bigcap_{p>1}l^p}^{l^\infty}$$ is. Is that just $l^\infty$?
The closure in $l^{\infty}$ of $\{(v_n/n),v\in l^{\infty}\}$ is $c_0$, the space of sequences that converge to $0$. If $a_n \to 0$ and $n$ is a positive integer then $(a_1,a_2,...,a_N,0,0,...)$ belongs to $\{(v_n/n),v\in l^{\infty}\}$ (just take $v_n=na_n$ for $n\leq N$ and $0$ for other $n$). Also $||(a_n)-(a_1,a_2,...,a_N,0,0,...)||{\infty} \to 0$ as $N\to \infty$. Hence $(a_n)$ is in the closure of $\{(v_n/n),v\in l^{\infty}\}$. Conversely, suppose $(a_n)$ is in the closure of $\{(v_n/n),v\in l^{\infty}\}$. Given $\epsilon >0$ let $(v_n) \in l^{\infty}$ and $||(a_n)-(v_n/n)||{\infty} <\epsilon$. Then $|a_n| \leq \epsilon +|v_n| /n \leq \epsilon +M/n$ where $M$ is the norm of $(v_n)$ in $l^{\infty}$. It is clear from this that $a_n \to 0$.