1
$\begingroup$

I'm wondering about this thing: I have

\begin{equation} T: \ell^2 \rightarrow \ell^2 \end{equation} \begin{equation} Tx_n = \frac{1}{n}x_n \end{equation}

I need to prove that $T$ is injective but not surjective and that $Rg(T)$ is dense in $\ell^2$. For injectivity I have:

\begin{equation} Tx_n = 0 \Leftrightarrow \frac{1}{n}x_n = 0 \Leftrightarrow x_n \equiv 0 \end{equation}

so I think I'm done.

For surjectivity, what came in mind immediately, is that knowing that $x_n, 1/n$ belong to $\ell^2$ their product is in $\ell^1$ (By Holder). So by this quick reasonment, I should conclude that $Rg(T) = \ell^1$ which is dense in $\ell^2$. But my question is:

Can I entirely cover $\ell^1$ by $\frac{1}{n}x_n$ (Both in $\ell^2$)? Probably this is trivial but I was just wondering about it.

$\endgroup$
  • $\begingroup$ Hint for the actual question: you just need to find some $(x_n)\in\ell^2$ such that $(nx_n)\not\in\ell^2$. This shouldn't be especially hard. For your question about $\ell^1$: your question is equivalent to the following: is there a $(y_n)\in\ell^1$ such that $ny_n\not\in\ell^2$. $\endgroup$ – user3482749 Nov 23 '18 at 16:10
  • $\begingroup$ This question already has an answer. $\endgroup$ – Devendra Singh Rana Nov 23 '18 at 16:25
  • $\begingroup$ Where can I find it? Didn't know about $\endgroup$ – James Arten Nov 23 '18 at 16:26
0
$\begingroup$

The range of $T$ is the set of all the sequences $(x_n)_{n\geqslant 1}$ such that the sequence $(nx_n)_{n\geqslant 1}$ belongs to $\ell^2$. Indeed, if $(nx_n)_{n\geqslant 1}$ belongs to $\ell^2$ then let $a_n=nx_n$ and show that $T((a_n)_{n\geqslant 1})=(x_n)_{n\geqslant 1}$ hence $(x_n)_{n\geqslant 1}$ is in the range of $T$. THe converse is not harder.

In particular, the range of $T$ contains the sequences whose only finitely many terms do not vanish and this set is dense in $\ell^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.