operator from $\ell_1 $ to $\ell_1$ and density I want to solve the following problem, but I am stuck...

Let $a=(a_n)_n \in \ell_\infty $. Define the linear operator $
 \displaystyle T : \ell_1 \to \ell_1 $ by $\displaystyle T(x) = a \cdot
 x , \quad \text{i.e.} \quad T(x)= (a_nx_n)_n $, for $\displaystyle
 x=(x_n)_n \in \ell_1 $. Prove that:
(i) If $\displaystyle \text{Ker}(T) =\{0\}$ then $T(\ell_1)$ is dense
  in $\ell_1$
(ii) $\displaystyle T(\ell_1) = \ell_1 $ if and only if $\displaystyle
 \inf_{n \in \mathbb N} |a_n | > 0 $

I have prove that $\displaystyle \text{ker}(T)=\{0\} $ if and only if $ a_n \neq 0 \quad \forall n \in \mathbb N$.
Any solution or hints would be appreciated!
Thank you!
 A: To prove that $\mbox{ker}(T) = \{0\}$ if and only if $a_n \neq 0$ for all $n$, you can  argue as in the finite-dimensional case:


*

*Suppose first $a_k = 0$ for some $k$ and let $e_k = (0,...,0,1,0,...)$ be the sequence whose $k$-th term is $1$ and whose other terms are $0$. Then $Te_k = (0,...,0,a_k,0,...) = 0$ and thus $\mbox{ker}(T) \neq \{0\}$.

*On the other hand, suppose $a_n \neq 0$ for all $n$. Let $x = (x_k), y = (y_k) \in \ell^1$. If $Tx = Ty$, then $a_kx_k = a_ky_k$ for all $k$. But $a_k \neq 0$ so $x_k = y_k$ for all $k$ and hence $x=y$. Thus $T$ is injective and its kernel is trivial.


Knowing that $a_n \neq 0$ for all $n$, we see that $\frac{1}{k}e_k \in \ell^1$ and $T(\frac{1}{k}e_k) = e_k$. Thus the range of $T$ contains the set
$$\mbox{span} \{e_k: k \in \mathbb{N} \}$$
which is a dense set. This proves the first assertion.
For the second, notice that
$$\|Tx\|_1 \geq \inf_{k} |a_k| \|x\|_1$$
so that $T$ is bounded from below if $\inf_{k} |a_k| > 0$. Therefore, it has closed range. Since the range is also dense, it is equal to all of $\ell^1$.
For injective $T$ (which is the case here), the answer linked to above also shows that if the range is closed, then the operator is bounded from below. Thus if $T$ is surjective, there exists a $c > 0$ such that
$$|a_k| = \|a_ke_k\|_1 = |\|Te_k\|_1 \geq c\|e_k\|_1 = c$$
for all $k$. If follows that $\inf_{k} |a_k| \geq c > 0$.
