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.

  • $\begingroup$ 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

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 $\ell^2$.


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