TITLE: Finding the norm of the shift operator on $l^\infty$, $(x_1,x_2, \dots)\mapsto (x_2,x_3,\dots)$
QUESTION [8 upvotes]: Let $T:l^\infty\rightarrow l^\infty$ be defined by $(x_1,x_2, \dots)\mapsto (x_2,x_3,\dots)$. 

I have seen a claim without justification that $\|T\|=1$, but I am not convinced. I know that 
$\|T\| = \sup_{{\|x\|=1}}\|Tx\|$  . If $\|x\|=1$ then surely $\|Tx\|\le 1$. So 
$$\|T\| = \sup\limits_{\|x\|=1}\|Tx\|\le 1.$$ 

I don't see how it can be claimed that $\|T\| = 1$

Is it true or I am missing something?

REPLY [5 votes]: Clearly $\|Tx\| \le \|x\|$, hence $\|T\| \le 1$. Since $T(0,1,0,...) = (1,0,...)$ we have $\|T\| = 1$.