Convergence of a sequence of elements of $L(l^1,l^\infty)$ I am trying to solve this interesting problem:

I think I solved (a):
for any $x\in l^1$,$(T_nx)_n$ converges to $x$. In fact, letting $\epsilon>0$, $\parallel (T_nx)_n-x) \parallel_\infty=\sup_{n\in\mathbb{N}}|T_nx-x_n|=\sup_{n\in\mathbb{N}}0=0<\epsilon$.  
I do not see how to solve (b), though. I know that $L(l^1,l^\infty)$ is a banach space...
 A: To avoid confusion, I denote $x(n)$ to be the $n-th$ term of a sequence $x$ when $x$ is an element of either $\ell^1$ or $\ell^2$.
Part a). You are right that $T_nx\to x$ in $\ell^\infty$, but your reasoning is incorrect. Here's the approach: Denote $y_n:=T_nx-x$ and $y_n(k)$ the $k-$th term of the sequence $y_n\in\ell^\infty$. Then we have
$$y_n(k)=\begin{cases} 0, &k\leq n\\
x(n)-x(k), &k>n\end{cases}.$$
Since $x\in\ell^1$, this means $(x(k))_{k=1}^\infty$ is a Cauchy sequence in the scalar field, thus for every $\epsilon>0$,there exists $N\in\mathbb{N}$ such that
$$|x(n)-x(k)|<\epsilon,\quad\forall n,k>N,$$
and this is equivalent to say that
$$\|y_n\|_\infty=\sup_{k>n}|x(n)-x(k)|<\epsilon,\quad\forall n>N.$$
Thus we have
$$\|T_nx-x\|_\infty=\|y_n\|_\infty\to 0.$$
Part b). It suffices to show that $(T_n)_n$ is not a Cauchy sequence in $L(\ell^1,\ell^\infty)$ with respect to the operator norm. For any $x\in\ell^1$ note that
$$T_{n+1}x-T_nx=(0,0,\dots,x(n+1)-x(n),x(n+1)-x(n),\dots),$$
i.e. the first $n$ terms of $T_{n+1}x-T_nx$ all equal to $0$, and starting from the $n+1-$th term all term equal to $x(n+1)-x(n)$.
Then we let $e_n\in\ell^1$ be defined as
$$e_n(k)=\begin{cases} 0, &k\neq n\\
1, &k=n\end{cases}$$
It follows that
$$T_{n+1}e_n-T_ne_n=(0,0,\dots,-1,\dots,-1).$$
Now we see that
$$\|T_{n+1}e_n-T_ne_n\|_\infty=1=\|e_n\|_1.$$
This means
$$\|T_{n+1}-T_n\|\geq 1,$$
and therefore $(T_n)_n$ cannot be a Cauchy sequence in $L(\ell^1,\ell^\infty).$
