What is the closure of $\ell_1$ in $\ell_2$?

It is not difficult to show that the sequence space $\ell_1$ is a subspace of $\ell_2$ that is not closed in the $\ell_2$ topology (see Problem 2 in this pdf file for instance)

What is the closure of $\ell_1$ in $\ell_2$?

• – carmichael561 Sep 20 '17 at 5:06

$\ell^1$ is dense in $\ell^2$ (w.r.t. the $\ell^2$ norm topology). The argument is simple: given $\textbf{x} = (x_1, \ldots x_n, \ldots) \in\ell^2$ fix $\epsilon >0$ and take $\textbf{x}_N = (x_1, \ldots x_N, 0, 0, 0 \ldots)$ where $N$ is such that $\displaystyle \left(\sum_{k=N+1}^\infty x_k^2\right)^{1/2} < \epsilon.$
Then $\textbf{x}_N$ is obviously in $\ell^1$ and $\|\textbf{x} - \textbf{x}_N\|_2 < \epsilon$.
Hence the closure of $\ell^1$ in the $\ell^2$ topology is $\ell^2$.
All $e_n$ are in $\ell_1$. What is the closure of their span in $\ell_2$?
Closure of $l_p$ in $l_q$ is $l_q$ for any $p,q$ with $p<q$. Or, $l_p$ is dense in $l_q$ for any $p,q$ with $p<q$. You can prove it either by using the concept/arguement given by Henno Brandsma (above) or by using the technic given by David Bowman (above).