Sufficient and necessary conditions for a given operator $T$ to be $T \in \mathcal{B}$, $T \in \mathcal{K}$ or $T \in \mathcal{F}$ I was rewiewing for an upcoming exam and found this problem. It is quite basic and straightforward but still causing me some confusion mainly since I would have to justify my answer.
Let a linear operator $T$ on Banach space $\ell^1$ be given with respect to the normal baiss $(e_n)_ {n=1}^\infty$ s.t
$Te_n := \alpha_n (e_n +e_{n+1} + ... + e_{2n})$  for all  $n \in \mathbb{N}$.
Here $e_1 = (1,0,0,...)$, $e_2 = (0,1,0,0,...)$ etc.
Give sufficient and necessary conditions for the sequence $(\alpha_n)_{n=1}^\infty$ ($\alpha_n \in \mathbb{C}$) so that
a) $T \in \mathcal{B}(\ell^1)$
b) $T \in \mathcal{K}(\ell^1)$
c) $T \in \mathcal{F}(\ell^1)$
The a) part seems quite clear, I just need to check when $T$ is bounded i.e. when $||T|| < \infty$. As result I got $\sup_j |\alpha_j| < \infty$.
The other two parts, however, are a bit more confusing.
 A: a) If $T$ is bounded, then the sequence $\left(\left\lVert Te_n\right\rVert_1\right)_{n\geqslant 1}$ should be bounded. Since $\left\lVert Te_n\right\rVert_1=(n+1)\left\lvert \alpha_n\right\rvert$, the sequence $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ should be bounded. 
Conversely, if $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ is bounded, then $\sup_{n\geqslant 1}\left\lVert Te_n\right\rVert_1$ is finite which is sufficient to ensure that $T$ is bounded.
b) Assume that $T$ is compact. Then the set $\left\{Te_n,n\geqslant 1\right\}$ should have a compact closure hence 
$$\lim_{N\to+\infty}\sup_{n\geqslant 1}\sum_{i\geqslant N}\left\lvert \left(Te_n\right)_i\right\rvert=0,$$
where $\left(v\right)_i$ denotes the $i$-th coordinate of $v\in\ell^1$. Since 
$$
\sup_{n\geqslant 1}\sum_{i\geqslant N}\left\lvert \left(Te_n\right)_i\right\rvert\geqslant \sum_{i\geqslant N}\left\lvert \left(Te_N\right)_i\right\rvert =(N+1)\left\lvert \alpha_N\right\rvert
$$
we should have that $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ converges to $0$. 
Conversely, if $\left(n\left\lvert \alpha_n\right\rvert \right)_{n\geqslant 1}$ converges to $0$, define the linear operator $T_N$ in the following way: if $1\leqslant n\leqslant N$, then $T_N(e_n)=T(e_n)$ and if $n\gt N$, then $T_N(e_n)=0$. The operator $T_N$ has a finite rank and the operator norm of $T-T_N$ can be bounded by 
$\sup_{i\geqslant N}(i+1)\left\lvert \alpha_i\right\rvert $.
c) If $a_n=0$ for $n$ large enough, then $T$ has a finite rank. If  $a_n\neq 0$ for infinitely many $n$'s, let $n_i$ be integers such that $a_{n_i}\neq 0$ and $n_{i+1}\gt 3n_i$. Then $\left\{Te_{n_i}\right\}$ is linearly independent and infinite.
