# Question about surjectivity of operator in $\ell^2$

$$$$T: \ell^2 \rightarrow \ell^2$$$$ $$$$Tx_n = \frac{1}{n}x_n$$$$

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

$$$$Tx_n = 0 \Leftrightarrow \frac{1}{n}x_n = 0 \Leftrightarrow x_n \equiv 0$$$$

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.

• 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 $ny_n\not\in\ell^2$. – user3482749 Nov 23 '18 at 16:10
• This question already has an answer. – Devendra Singh Rana Nov 23 '18 at 16:25
• Where can I find it? Didn't know about – 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 $$\ell^2$$.