unilateral shift operator on $l^1(\mathbb{N})$

Question

We consider the unilateral shift operator $T$ on $l^1(\mathbb{N})$ defined as follows: for all $x\in l^1(\mathbb{N})$

$$T(x_0,x_1,..)=(0,x_0,x_1,...)$$ I want to prove that for all $x\in l^1(\mathbb{N})$, $||T^kx||_1\rightarrow0 $ when, $k\rightarrow \infty.$

Well let $x\in l^1(\mathbb{N})$, then $x=\sum_{n\geq0}x_n.e_n$ with $e_n=(0,0,..,1,0,0,..)$

and also $\sum_{n\geq 0}|x_n|< \infty$, then for all $\epsilon>0$, exists $N\in \mathbb{N}$, such that for all $n>N$ we have $\sum_{n>N}|x_n|<\epsilon$

then we have that $$||T^kx||_1\leq \epsilon+||\sum_{n>N}x_n.e_{n+k}||_1$$ but I'm stuck with proving that the second term converges to $0$, when $k\rightarrow\infty.$

Answer 1

$T$ in this case is an isometry, $||x||=||T^k x||$ for all $x\in l^1$.