There exists a sequence $(y_n) \in \ell_2$ such that $(x_n / y_n) \in \ell_2$ for every $ (x_n) \in \ell_1$? I'm trying to figure out if there exists a sequence $(y_n) \in \ell_2$ such that $(x_n / y_n) \in \ell_2$ for every $(x_n) \in \ell_1$. There some obvious reductions to this problem: we have to consider $y_n \neq 0$ for every $n \in \mathbb{N}$. Moreover, $(y_n) \notin \ell_1$, because it would imply $(1)_n = (y_n/y_n)_n \in \ell_2.$ I think it is necessary that $(y_n) \notin \ell_p$ for any $p<2,$ but I don't know if this condition is sufficient.
Anyone can help me? Thank you.
 A: This may be a long-winded, but I think it works: Suppose such a sequence $y = (y_n)$ exists, then we may define $T : \ell^1\to \ell^2$ by
$$
T((x_n)) = (x_n/y_n)
$$
We claim that $T$ is continuous: Suppose $x^k \to x$ in $\ell^1$ and $T(x^k) \to z$ in $\ell^2$, then for any $n\in \mathbb{N}$,
$$
\lim_{k\to \infty} |z_n - \frac{x^k_n}{y_n}|^2 \leq \lim_{k\to \infty} \|z - T(x^k)\|_2^2 = 0
$$
$$
\Rightarrow \lim_{k\to \infty} |y_nz_n - x^k_n|^2 = 0
$$
But
$$
\lim_{k\to \infty} |x_n - x^k_n| \leq \lim_{k\to \infty} \|x - x^k\|_1 = 0
$$
So
$$
z_n = \frac{x_n}{y_n} \quad\forall n\in \mathbb{N}
$$
Hence, $T(x) = z$, so $T$ is bounded by the closed graph theorem. Hence, there is $M> 0$ such that
$$
\|T(x)\|_2 \leq M\|x\|_1 \quad\forall x\in \ell^1
$$
Applying this to the `standard basis' $e_n \in \ell^1$, we conclude that
$$
\frac{1}{|y_n|^2} \leq M \quad\forall n\in \mathbb{N}
$$
Since $M > 0$, this implies that
$$
|y_n|^2 \geq \frac{1}{M} \quad\forall n\in \mathbb{N}
$$
contradicting the assumption that $(y_n) \in \ell^2$. Thus, no such sequence $(y_n)$ can exist.
