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! :)

Hint: Define $X^n = (\underbrace{\frac{1}{n}, \dots ,\frac{1}{n}}_{n}, 0, 0, ...)$. Show that for every $\epsilon >0$, there exists $N$ such that $||X^N||_{l^1} > 1-\epsilon$, thus proving the statement.
