Prove that the right shift operator $S(x_1,x_2, \dots) = (0,x_1,x_2,\dots)$ is bounded.

Question

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.

Answer 1

$$\|S(x_1,x_2,\dots)\|_1=\|(0,x_1,x_2,\cdots)\|_1=0+|x_1|+|x_2|+\dots = |x_1|+|x_2|+\dots = \|(x_1,x_2,\dots)\|_1.$$

In particular, $$\|S(x_1,x_2,\cdots)\|_1\le 1\cdot \|(x_1,x_2,\cdots)\|_1.$$

This inequality says that $S$ is bounded and $\|S\|\le 1$.

But $S(1,0,0,\dots)=(0,1,0,\dots)$, so $$\|S(1,0,\dots)\|_1 =\|(0,1,0,\dots)\|_1=1=\|(1,0,\dots)\|_1.$$

So $\|S\|=1$

Answer 2

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.