Q:

Is it possible that a closed infinite dimensional subspace of $\ell^2$ is closed in $\ell^1$ too.

Let $V$ a closed infinite dimensional subspace of $\ell^2$
Is it possibile that $V$ is closed in $\ell^1$ too ?
Thanks for any suggestion

A:

It seems that closed in $\ell^2$ implies closed in $\ell^1$ for any subspace of $\ell^1$.
Assume that $M \le \ell^1 \subset \ell^2$ is a closed subspace of $\ell_2$.
We shall prove that $M$ is closed in $\ell^1$ as well.
Take a sequence $(x_n)_{n=1}^\infty$ in $M$ such that it is convergent in $\ell^1$ and set $x_n \xrightarrow{\|\cdot\|_1} x \in \ell^1$.
Since we have $\|\cdot\|_2 \le \|\cdot\|_1$, we get that $x_n \xrightarrow{\|\cdot\|_2} x$ also holds:
$$\|x_n - x\|_2 \le \|x_n - x\|_1 \xrightarrow{n\to\infty} 0$$
Since $M$ is closed in $\ell^2$, we have $x \in M$.
Thus, $M$ is closed in $\ell^1$.
In fact, the situation seems to be the same for any $p < q \in [1, +\infty\rangle$ since:
$$1 = \sum_{n=1}^\infty\frac{|x_n|^q}{\|x\|_q^q} = \sum_{n=1}^\infty\left(\underbrace{\frac{|x_n|}{\|x\|_q}}_{\le 1}\right)^q \le \sum_{n=1}^\infty\left(\frac{|x_n|}{\|x\|_q}\right)^p = \frac{\|x\|_p^p}{\|x\|_q^p} \implies \|x\|_q \le \|x\|_p$$

