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

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.

• The shift operator is an isometry, so in particular it is bounded.
– Pedro
Nov 3, 2018 at 15:08
• @PedroTamaroff We haven't been introduced to isometries yet in our Functional Analysis course, so I assume we should seek a more elementary approach. Nov 3, 2018 at 15:09

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

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 $$\|S\|=1$$

• I cannot really understand what an isometry is, since we haven't been introduced to them yet. Also, what does "find the norm of the operator S" mean ? Nov 3, 2018 at 15:16
• @Rebellos LEt $(X,\|\cdot\|)$ be a Banach space. $T:X\to X$ is an isometry iff $\|Tx\|=\|x\|$ for all $x\in X$ Nov 3, 2018 at 15:18
• As I said again, I am really not interested in an isometry approach or terminology. I must solve the exercise using an elementary bounded operator approach, as we have not been introduced to isometries yet. Also, the second part of my question is not answered, what does find the norm of the operator S mean ? Nov 3, 2018 at 15:19
• @Rebellos the fact that it is an isometry is icing on the cake. This answer shows that the operator is bounded via the definition you give.
– Dave
Nov 3, 2018 at 15:21
• @TitoEliatron Oh, nice point ! Understod. Nov 3, 2018 at 15:30

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.