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I want to solve the following problem.

Let $x = (x_n)_{n \geq 1} \in \ell^2$ be a sequence such that $x_n \neq 0$ for all $n \geq 1$. Prove that there exists a sequence $y = (y_n)_{n \geq 1} \in \ell^1$ such that
$ \left(\frac{y_n}{x_n}\right)_{n \geq 1} \not\in \ell^2. $

Here is my attempt: Define the linear operator $S: \ell^1 \to \ell^2$ by
$ S((y_n)) = \left(\frac{y_n}{x_n}\right). $

Linearity follows immediately, and $S$ is injective since for any $y, z \in \ell^1$,
$ S(\alpha y + \beta z) = \alpha \left(\frac{y_n}{x_n}\right) + \beta \left(\frac{z_n}{x_n}\right). $

If $S(y) = S(z)$, then $ \frac{y_n}{x_n} = \frac{z_n}{x_n}, $ which implies $y_n = z_n$, so $y = z$. Hence, $S$ is injective. Now, by the following corollary of the open mapping theorem: Let $E$ and $F$ be two Banach spaces, and let $S$ be a continuous linear operator from $E$ into $F$ that is bijective, i.e., injective (one-to-one) and surjective. Then $S^{-1}$ is also continuous (from $F$ into $E$). If $S$ were also surjective, its inverse would be bounded. However, it is known that there does not exist a bounded linear operator $S^{-1}: \ell^2 \to \ell^1$ (see, e.g., No bounded linear surjectivity from $\ell^2(\mathbb{N})$ to $\ell^1(\mathbb{N})$). This implies that $S$ cannot be surjective. One potential issue with this argument is that $S$ itself may not be bounded. Indeed, if we consider $y_n = 1/n^2$, which belongs to $\ell^1$. Then,
$ S((1/n^2)) = \left(\frac{1/n^2}{1/n^2}\right) = (1,1,1,\dots), $
which does not belong to $\ell^2$, suggesting $S$ is unbounded. Would this reasoning be correct? If so, does the conclusion still hold under an alternative approach?

Hafeez1729
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  • You want to show that for some $y$ the element $Sy$ is not in $l^2$, i.e., you are looking for holes in the domain not in the range. I suggest to study the operator $T(z) = (z_nx_n)$. Then show that this operator is not surjective. – daw Feb 15 '25 at 17:25
  • With $y_n=\frac{x_n}{n}$ you can use Cauchy-Scharz inequality. – Christophe Boilley Feb 15 '25 at 18:00
  • @ChristopheBoilley, with your example, $y_n/x_n = 1/n$ is actually in $\ell^2$. – Hafeez1729 Feb 15 '25 at 18:14

1 Answers1

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We can define the following increasing sequence $(n_k)$: for all $k>0$, $n_k=\min\left\{n>n_{k-1}:|x_n|<\frac{1}{k^2}\right\}$. Then for all $k>0$, define $y_{n_k}=\frac{1}{k^2}$ and the remaining values of the sequence $(y_n)$ are null.

Then by design $(y_n)\in\ell^1$ but the sequence $\displaystyle\left(\frac{y_n}{x_n}\right)$ does not converge to 0 so it is not in $\ell^2$.

Christophe Boilley
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