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Suppose $(\alpha_k)$ is a sequence in $\mathbb{C}$ and let $$ M:=\{ (x_k) \in \ell^2 | (\alpha_k x_k) \in \ell^2 \} $$ I want to show that $M$ is dense in the sequence space $\ell^2$ (equipped with the ususal norm), so $\overline{M}=\ell^2$. This is the case if for each $x \in \ell^2$ there exists a sequence $(x_k) \in M$ such that $x_k \rightarrow x$ in $\ell^2$. However, I don't know how to prove this.

Also, I would like to know if we can make certain restrictions on $(\alpha_k)$ such that $M=\ell^2$. I thought this would be the case for $\underset{0\leq i \leq k}{\sup}|\alpha_i|^2 \leq C$ (where $C$ is a positive constant). Is this indeed correct? Or do we need other restrictions?

Cathematics
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2 Answers2

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If $(x_n) \in \ell^{2}$ then $(x_1,0,0,...), (x_1,x_2,0,0...),(x_1,x_2,x_3,0...),....$ are all elements of $M$ and these vectors converge to $(x_n)$ in the norm of $\ell^{2}$.

If $(\alpha_k)$ is bounded then $M=\ell^{2}$. Suppose $(\alpha_k)$ is not bounded. There exists a sequence $n_k$ increasing to $\infty$ such that $|\alpha_{n_k}|> k$. Define $x_n=0$ if $n$ is not one of the $n_k$'s and $x_{n_k}=\frac 1 {\alpha_{n_k}}$. Then $(x_n) \in \ell^{2}$ but $(x_n) \notin M$.

Kavi Rama Murthy
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The set $M$ contains the set of all the elements of $\ell^2$ having only finitely many non-zero coordinates, and this set is dense in $\ell^2$.

Suppose that $M=\ell^2$, then $M=\bigcup_{N\geqslant 1}F_N$, where $$ F_N=\bigcap_{R\geqslant 1}\left\{x\in \ell^2\mid \sum_{k=1}^R\lvert \alpha_k\rvert^2\lvert x_k\rvert^2\leqslant N\right\}. $$ hence by Baire's theorem, one of the $F_N$ has a non-empty interior. We derive that there exists a constant $C$ such that for all $x\in \ell^2$, $$ \sum_{k=1}^\infty\lvert \alpha_k\rvert^2\lvert x_k\rvert^2\leqslant C\sum_{k=1}^\infty \lvert x_k\rvert^2 $$ hence taking $x_k=e_k$, the sequence $(\alpha_k)_{k\geqslant 1}$ is bounded.

Davide Giraudo
  • 172,925