Sequence in $c_0$ (related with $\ell_1$). I need some help with this cuestion:
Let b a sequence $a=\{a(n)\}$ in $\ell_1$. We define, for each $x \in c_0$, the sequence $$T_a: c_0 \longrightarrow c_0, \hspace{0.15cm} T_a(x)(n)=\sum_{k=n}^{\infty}a(k)x(k), n \in \mathbb{N}.$$ Give the properties of the sequences $a \in \ell_1$ for which $T_a$ reach its norm.
I think that I have to give an if-and-only-if property, but I don't know exactly how to do it. I'm trying to prove that maybe $\exists k \in \mathbb{N}$ such that $|a(k)|=||a||_{\infty}$, or something about $||a||_1$, but I don't arrive to any conclusion.
 A: First, let us calculate the norm of $T$. To do this, I suggest extending $T$ to $\ell^\infty$.
Let $S$ be the natural extension of $T$ to $\ell^\infty$, i.e. the same definition, just with sequences from $\ell^\infty$ instead. Let $k \in \ell^\infty$ be the sequence
$$k_n = \begin{cases} \dfrac{|a_n|}{a_n} & \text{if } a_n \neq 0 \\ 1 & \text{if } a_n = 0. \end{cases}$$
Note that $|k_n| = 1$ for all $n$, hence $\|k\|_\infty = 1$. We also have, for any $x \in \ell^\infty$ with $\|x\|_\infty \le 1$,
$$\|S(x)\| = \sup_{n \in \Bbb{N}} \left|\sum_{k=n}^\infty a_k x_k \right| \le \sup_{n \in \Bbb{N}} \sum_{k=n}^\infty |a_k| |x_k| \le \sup_{n \in \Bbb{N}} \sum_{k=n}^\infty |a_k| = \sum_{k=1}^\infty |a_k| = \|S(k)\|.$$
Therefore $S$ achieves its norm on $\ell^\infty$ with this sequence $k$.
Turning our attention back to $T$ and $c_0$, define a sequence $(x^m)_{m \in \Bbb{N}}$ of points in $c_0$ defined by
$$x^m_n = \begin{cases} k_n & \text{if } n \le m \\ 0 & \text{if } n > m. \end{cases}$$
Note that $\|T\| \le \|S\| = \|a\|_1$ and $\|x^m\|_\infty = 1$. Then,
$$\|T(x^m)\| = \max_{1 \le n \le m} \sum_{k = n}^m |a_k| = \sum_{k=1}^m |a_k| \to \sum_{k=1}^\infty |a_k| = \|S(k)\|,$$
thus proving $\|T\| = \|S\| = \|a\|_1$.
Now, we just need to ask, when is there a sequence $x \in c_0$ such that $\|x\|_\infty \le 1$ and $\|T(x)\| = \|a\|_1$? Well, one sufficient condition would be for $a_n$ to eventually become $0$, so that $\|T(x^m)\| = \|a\|_1$ for sufficiently large $m$. Is this necessary too?
Well, yes, it must be. For every $n$ such that $a_n \neq 0$, in order for a point $x \in c_0$ to achieve the norm, we need $|x_n| = 1$. If this happens infinitely often, then this contradicts $x \in c_0$. So, our necessary and sufficient condition is $a \in c_{00}$, the space of sequences which are eventually $0$.
A: Here is a proof that uses the fact that $\ell_1$ is isometrically isomorphic to $c_0^*$ via the map
$\tag1 a\mapsto f:x\mapsto \sum_{n=1}^\infty a_nx_n$
We have
$\tag2 \|T_a\|=\sup\{\|T_a(x)\|:\|x\|\le 1\}$ and
$\tag3 \|T_a(x)\|=\sup \{|T_a(x)(n)|\:n\in \mathbb N\}$
Now, if we regard $T_a(x)(n)=\sum _{k=n}^\infty a_kx_k$ as arising from the linear functional  $f_n:=T(\cdot)(n): x\mapsto \sum _{k=n}^\infty a_kx_k$ then $(1)$ implies that $\|f_n\|=\sum _{k=n}^\infty |a_k|.$ And  since $\sup \{\|f_n(x)\|\:\|x\|\le 1\}\le \sum _{k=1}^\infty |a_k|$ holds for all integers $n$, we have from $(3)$ that $ \|T_a(x)\|\le \sum _{k=1}^\infty |a_k|$, and finally from $(2)$ that
$\tag4 \|T_a\|\le \sum _{k=1}^\infty |a_k|=\|a\|.$
But for each integer $n$ if we define the sequence $x_n\in c_0;\ \|x_n\|=1$ by: $x_k=\operatorname {sgn}\ a_k:k\le n$ and $0:k>n$ then, $\|T_a(x_n)\|=\sum_{k=1}^n|a_k|$ and so $(2)$ implies that $\|T_a\|=\|a\|.$
This argument shows that if $a\in c_{00}$ then the norm is attained because in this case, $(x_n)$ is finite. On the other hand, if $x$ is any other norm one sequence, other than the one defined above, then there is an integer $j$ such that $|x_j||a_j|<|a_j|$  and so $\|T_a(x)(n)\|\le \sum_{k=1}^\infty |a_k||x_k|<\sum_{1=k\neq j}^\infty |a_k||x_k|+|a_j|\le \sum_{k=1}^\infty |a_k|$. It follows then that the norm is not achieved and therefore that if $a\notin c_{00}$ then $x\notin c_0.$ We conclude that the norm of this operator is achieved if and only if $a\in c_{00}.$
