What is the range of the operator $T$ I mean I want to determine $R(T)$ Given the normed space $\ell^\infty$ of all bounded sequences of (real or complex) numbers with the norm given by $$||x||:= \sup_{j\in Z^+} |\xi_j|,$$ for each $x:=(\xi_j)_{j=1}^\infty$ in $\ell^\infty$, and given the linear operator $T \colon \ell^\infty \to \ell^\infty$ defined as $$T(\xi_j)_{j=1}^\infty  := (\frac{\xi_j}{j})_{j=1}^\infty,$$ 
What  is the range of the operator $T$ I mean I want to determine $R(T)$
 A: Let $\mathbf{x}=(x_n)$ be any sequence in $l^{\infty}$, then it can be easily seen that give any $\epsilon>0$ you can find a natural number $N$ such that $\frac{\|\mathbf{x}\|}{N} < \epsilon$. So, for all $n\geq N$ you can see that $|x_n|\leq\|\mathbf{x}\|$ and $\frac{1}{n}\leq \frac{1}{N}$, so, \begin{equation}|\frac{x_n}{n}| \leq \frac{\|\mathbf{x}\|}{N} < \epsilon.\end{equation}
Hence $T\mathbf{x} \in c_0$. Therefore, it follows that $R(T)\subseteq c_0$. But $R(T)$ is not equal to $c_0$ because $(\frac{1}{\sqrt n}) \in c_0$ but $(\sqrt n) \notin l^{\infty}$. So, the best that can be said about $R(T)$ is that it is a subspace of $c_0$.
A: Let
$$
S = \{(y_n) \in \ell^{\infty} : \exists C > 0 \text{ such that } |y_n| \leq C/n\quad\forall n\in \mathbb{N}\}
$$
If $(x_n) \in \ell^{\infty}$ then $T(x_n) \in S$ with $c = \|(x_n)\|_{\infty}$, and conversely, if $(y_n) \in S$, then the sequence $(x_n)$ defined by
$$
x_n := ny_n
$$
is in $\ell^{\infty}$ and satisfies $T(x_n) = (y_n)$. Hence, $R(T) = S$.
