# Find the norm of the n-shift map on $l^2$

For every $$n\in\mathbb{N}$$ consider the map $$T_n:l^2\to l^2$$ defined by: $$T_n(x_1,x_2,\ldots)=(\underbrace{0,0,\ldots,0}_\text{n-coordinates},x_1,x_2,\ldots).$$

Show that

1. $$T_n\in B(l^2)$$.
2. Find $$\|T_n\|$$.

Edited 3. Proof that $${T_n\color{red}\rightharpoonup 0}$$.

1. Done.
2. Edited

$$\|T_n(x)\|^2=\displaystyle\sum_{i=1}^{\infty}|x_i|^2=\|x\|^2\Rightarrow \|T_n\|\leq 1$$ take $$x=e_i$$, then $$T_n(e_i)=e_{n+1}$$, so $$\|T_n\|=1$$.

1. Could you give me a hint?

Thank you!

The statement $$T_n \to 0$$ does not make sense by itself since there are several modes of convergence of operators.

My guess is that you are asked to show that $$T_n (x) \to 0$$ weakly for every $$x$$. Since $$\left\langle T_n (x) , y \right\rangle = \sum\limits_{k=1}^{\infty} y_{n+k} x_k,$$ Cauchy Schwarz inequality gives $$\left\lvert \left\langle T_n (x) , y \right\rangle \right\rvert \leq (\sum\limits_{k=n+1}^{\infty} |y_k|^{2})^{1/2}(\sum\limits_{k=1}^{\infty} |x_k|^{2})^{1/2} \to 0$$ since $$\sum \left\lvert y_k \right\rvert^{2}$$ is convergent.

Note that strong convergence is ruled out since $$\|T_nx\|=\|x\|.$$

• you are right, i read wrong! – Framate 2 days ago

You wrote $$T_n(x_1,x_2,\ldots)=(\underbrace{0,0,\ldots,0}_\text{n-coordinates},x_1,x_2,\ldots).$$

But I think it should read

$$T_n(x_1,x_2,\ldots)=(\underbrace{0,0,\ldots,0}_\text{n-coordinates},x_{n+1},x_{n+2},\ldots).$$

Then we have

$$||T_n(x)||^2=||x||^2- \sum_{k=1}^n|x_k|^2.$$

Can you procced ?

• The question title describes $T_n$ as the "n-shift" operator, so I think OP is quite correct in their definition. – postmortes 2 days ago
• But the OP wrote: $\|T_n(x)\|^2=\displaystyle\sum_{i=n+1}^{\infty}|x_i|^2.$ – Fred 2 days ago
• It's probably worth clarifying that with them, but I'd bet it's a typo given the question title and $i$ should start from $1$. I suppose either could be true though, depending on if it's right-shift or left-shift... – postmortes 2 days ago
• @Fred thank you very much, i wrote wrong. The definition is right, i confused on de index of the sum. – Framate 2 days ago

I am assuming that you're using the following definitions:

Let $$\left( X, \| \cdot \|_X \right)$$ and $$\left( Y, \| \cdot \|_Y \right)$$ be normed spaces, either both real or both complex. Then a linear operator $$T \colon X \to Y$$ is said to be bounded if there exists a real number $$\alpha > 0$$ such that $$\| T(x) \|_Y \leq \alpha \|x\|_X \ \mbox{ for all } x \in X.$$ In this case the norm of $$T$$ is defined by $$\lVert T \rVert \colon= \sup \left\{ \ \frac{ \lVert T(x) \rVert_Y }{ \lVert x \rVert_X } \ \colon \ x \in X, x \neq \mathbf{0}_X \ \right\}.$$ It also can be shown that $$\lVert T \rVert = \sup \left\{ \ \lVert T(x) \rVert_Y \ \colon \ x \in X, \lVert x \rVert_X = 1 \ \right\}.$$ And, the set of all the bounded linear operators $$T \colon X \to Y$$ is denoted by $$B(X, Y)$$ and this set is also a normed space; $$B(X, Y)$$ is real or complex according as $$X$$ and $$Y$$ both are real or complex. Thus we can also write $$\lVert T \rVert_{B(X, Y)} \colon= \sup \left\{ \ \frac{ \lVert T(x) \rVert_Y }{ \lVert x \rVert_X } \ \colon \ x \in X, x \neq \mathbf{0}_X \ \right\}.$$ It can be shown that $$\| T \|_{B(X, Y)} = \sup \left\{ \ \| T(x) \|_Y \ \colon \ x \in X, \| x \|_X = 1 \ \right\}.$$ Finally, let $$\left( T_n \right)_{n \in \mathbb{N}}$$ be any sequence in $$B(X, Y)$$, and let $$T \in B(X, Y)$$. Then $$T_n \to T$$ in $$B(X, Y)$$ if, for every real number $$\varepsilon > 0$$, there exists an $$N \in \mathbb{N}$$ (and depending uopn $$\varepsilon$$) such that $$\left\| T_n - T \right\|_{B(X, Y)} < \varepsilon$$ for all $$n \in \mathbb{N}$$ such that $$n > N$$. Of course, by $$0 \in B(X, Y)$$ we mean the bounded linear operator $$\mathbf{0}_{B(X, Y)} \colon X \to Y$$ given by 

Having clearly spelled out the definitions, let us now turn to your particular question.

If $$T_n \to 0$$ were to hold, then we would also have $$\left\| T_n \right\| \to 0,$$ which is clearly false. So $$T_n \to 0$$ does not hold.

However, if for each $$n \in \mathbb{N}$$ your operator $$T_n \colon \ell^2 \to \ell^2$$ were the so-called "lift-shift" operator defined by $$T_n \left( \left( \xi_i \right)_{i \in \mathbb{N} } \right) \colon= \left( \xi_{n+1}, \xi_{n+2}, \ldots \right) \ \mbox{ for all } \left( \xi_i \right)_{i \in \mathbb{N} } \in \ell^2,$$ then we would have $$T_n \to 0$$.

Let $$\varepsilon > 0$$ be given.

Let $$x \colon= \left( \xi_1 \right)_{ i \in \mathbb{N} }$$ be any element of $$\ell^2$$. Then (1) $$\xi_i$$ is a real or complex number for each $$i \in \mathbb{N}$$, and (2) the seties $$\sum \left\lvert \xi_i \right\rvert^2$$ converges, that is, $$\sum_{i = 1}^\infty \left\lvert \xi_i \right\rvert^2 < +\infty,$$ that is, $$\lim_{k \to \infty} \left( \sum_{i = 1}^k \left\lvert \xi_i \right\rvert^2 \right) < +\infty.$$ Then there exists $$N \in \mathbb{N}$$ such that $$\sum_{i = N}^\infty \left\lvert \xi_i \right\rvert^2 < \frac{\varepsilon^2}{4}.$$ And, then for each $$n > N$$, we would have $$\sum_{i = n+1 }^\infty \left\lvert \xi_i \right\rvert^2 < \frac{\varepsilon^2}{4},$$ and so $$\left\lVert T_n(x) - \mathbf{0}_{B\left(\ell^2, \ell^2\right)} (x) \right\rVert_{\ell^2} = \sqrt{\sum_{i = n+1 }^\infty \left\lvert \xi_i \right\rvert^2 } < \frac{\varepsilon}{2},$$ and thus in particular for all $$x \in \ell^2$$ such that $$\| x \|_{\ell^2} = 1$$, we also have $$\left\| T_n(x) - \mathbf{0}_{B \left(\ell^2, \ell^2 \right) } (x) \right\|_{\ell^2} < \frac{\varepsilon}{2},$$ and so $$\left\lVert T_n - \mathbf{0}_{B\left( \ell^2, \ell^2 \right) } \right\rVert_{B\left( \ell^2, \ell^2 \right)} \leq \frac{\varepsilon}{2} < \varepsilon.$$

Thus, given $$\varepsilon > 0$$, there eixsts $$N \in \mathbb{N}$$ such that $$\left\lVert T_n - \mathbf{0}_{B \left( \ell^2, \ell^2 \right) } \right\rVert_{\ell^2} < \varepsilon$$ for every $$n \in \mathbb{N}$$ such that $$n > N$$.

• thank you very much, but I read wrong! It is weak convergence. – Framate 2 days ago