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Is it possible to find a strict infinite dimensional closed subspace of $\ell^1$ and a strict infinite dimensional closed subspace of $\ell^2$ such that they are isomorphic ?

Matey Math
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  • Isomorphic as vector spaces ? Sure the subspace of finite length sequences. Now you probably meant closed subspaces (each closed for a different topology) – reuns Sep 22 '17 at 12:16
  • @reuns yes, i meant closed subspaces, thanks, now i edited the question – Matey Math Sep 22 '17 at 12:24
  • yes if there's a homeomorphism – Matey Math Sep 22 '17 at 12:27
  • So you want a subspace $A \subset l^1$ and a linear map $f: A \to l^2$ such that $b |x|_1 \le |f(x)|_2 \le c |x|_1$ – reuns Sep 22 '17 at 12:30
  • $A \subset \ell^1$ strict subspace infinite dimensional and $B \subset \ell^2$ and linear map $f: A \to B$ homeomorphism as you write – Matey Math Sep 22 '17 at 12:33
  • I don't think so, look at the comments under my answer on one of your previous questions. I don't understand the proof of the argument though. – mechanodroid Sep 22 '17 at 13:02
  • thanks @mechanodroid i already read them i don't understand totally that proof too so i did this question, i think that any infinite dimensional subspace of $\ell^1$ is isomorphic to $\ell^1$ but $\ell^2$ is not isomorphic to $\ell^1$ – Matey Math Sep 22 '17 at 13:11

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It's not possible to embed $l^2$ into $l^1$.

Let $T\colon l^2 \to l^1$ a linear continuous . Since $e_n$ converges weakly to $0$ in $l^2$, $T(e_n)$ must converge weakly to $0$ in $l^1$. However, in $l^1$, if a sequence converges weakly to $0$, it also converges strongly to $0$ ( this is a result of Schur I guess, see reference on this site). Therefore, $T(e_n)$ converges to $0$ in $l^1$. Since $e_n$ does not converge to $0$ in $l^2$, we conclude that $T$ is not an embedding.

orangeskid
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