TITLE: Prove that the right shift operator $S(x_1,x_2, \dots) = (0,x_1,x_2,\dots)$ is bounded.
QUESTION [5 upvotes]: Exercise :

Let $S: l^1 \to l^1$ be the right-shift operator :
  $$S(x_1,x_2,\dots) = (0,x_1,x_2,\dots)$$
  Prove that $S$ is bounded and find its norm.

Attempt :
The space $l^1$ is : $$l^1 = \{(x=(x_n) : \sum_{i=1}^\infty x_n < + \infty\}$$
To show that the operator $S$ is bounded, I must show that :
$$\exists \; M>0 : \|Sx\|\leq M\|x\|$$
But I can't really see how to proceed on this particular proof without having any knowledge of the norm that should be used. Shall the norm of $l^1$ space be used ? If so, what does  "find the norm of the operator S" ? 
I would really appreciate any tips/solutions or clarifications regarding this particular exercise.
Note : I have NOT been introduced to isometries in my Functional Analysis class yet, so I am looking for an elementary bounded operator approach.

REPLY [2 votes]: From scratch: 
$\vec x=(x_1,x_2,\cdots )$ and  $S(\vec x)=(0,x_1,x_2,\cdots)$ so 
$\|\vec x\|=\sum^{\infty}_{i=1}|x_i|$ and $\|S(\vec x)\|=\sum^{\infty}_{i=1}|x_i|=\|\vec x\|.$
Now, $\|S|\|=\sup_x\frac{\|S(\vec x)\|}{\|\vec x\|}=\frac{\|\vec x\|}{\|\vec x\|}=1$.
Therefore, $\|S\|=1$ and, in particular $S$ is bounded.