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 $$\|T_nx\|=\|x\|. $$

  • $\begingroup$ you are right, i read wrong! $\endgroup$ – Framate Feb 14 at 17:06

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


Then we have

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

Can you procced ?

  • $\begingroup$ The question title describes $T_n$ as the "n-shift" operator, so I think OP is quite correct in their definition. $\endgroup$ – postmortes Feb 14 at 10:09
  • $\begingroup$ But the OP wrote: $\|T_n(x)\|^2=\displaystyle\sum_{i=n+1}^{\infty}|x_i|^2.$ $\endgroup$ – Fred Feb 14 at 10:13
  • $\begingroup$ 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... $\endgroup$ – postmortes Feb 14 at 10:14
  • $\begingroup$ @Fred thank you very much, i wrote wrong. The definition is right, i confused on de index of the sum. $\endgroup$ – Framate Feb 14 at 17:10

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 $n > N$.

  • $\begingroup$ thank you very much, but I read wrong! It is weak convergence. $\endgroup$ – Framate Feb 14 at 17:07

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