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 Tamaroff Nov 3 '18 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. – Rebellos Nov 3 '18 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 ? – Rebellos Nov 3 '18 at 15:16
• @Rebellos LEt $(X,\|\cdot\|)$ be a Banach space. $T:X\to X$ is an isometry iff $\|Tx\|=\|x\|$ for all $x\in X$ – Tito Eliatron Nov 3 '18 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 ? – Rebellos Nov 3 '18 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 '18 at 15:21
• @TitoEliatron Oh, nice point ! Understod. – Rebellos Nov 3 '18 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.