Compute operator norm of $l^1$ bounded operator I have a problem with the following exercise:
We have the operator $T: l^1 \to l^1$ given by
$$T(x_1,x_2,x_3,\dots)=\left(\left(1-\frac11\right)x_1, \left(1-\frac12\right)x_2, \dots\right)$$  for $(x_1,x_2,x_3,\dots)$ in $l^1$. Showing that this operator is bounded is easy, but  I am really desperate with showing that the norm $\|T\| = 1$.
I know that for bounded operators the norm is defined as $\|T\|=\sup{\left\{\|T(x)\|: \|x\| \le 1\right\}}$.
I am also wondering if there exists a x in $ l^1$ such that $\|x\|=1 $
and $\|T(x)\|= \|T\|$
Thank you! :)

$$\left|\left| T((x_n)_{n\in\mathbb{N}})\right|\right|_{\ell_1}=\left|\left| \left(\left(1-\frac{1}{n}\right)x_n\right)_{n\in\mathbb{N}}\right|\right|_{\ell_1}=\sum_j \left|\left(1-\frac{1}{j}\right)x_j\right|\leqslant \sum_j |x_j |=||(x_n )_{n\in\mathbb{N}} ||_{\ell_1}$$
hence 
$$||T||\leqslant 1$$
but $$||T||\geqslant \sup_j ||Te_j || =\sup_j \left(1-\frac{1}{j}\right) =1$$
Thus $$||T||=1.$$
