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Let $X$ be a Banach space.

Suppose there exists a sequence $(x_n)$ in $X$ such that for all finite $A\subseteq\mathbb{N}$ we have that $\|\sum_{n\in A}x_n\|$ equals the number of elements in $A$.

Does this imply that the subspace spanned by $\{x_n\}$ is isomorphic to $\ell^1$?

Carucel
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  • If $\sum_{n\in A} x_n$ equals the number of elements of $A$, the $x_n$ must be numbers and in fact $x_n=1$ (use the condition for singletons $A={n}$). – Jochen Feb 21 '19 at 12:21
  • @Jochen. Notice the norm now around $\sum_{n\in A}x_n$. Without this norm, this must be a number and indeed $x_n=1$ follows. – Carucel Feb 21 '19 at 12:25
  • My comment was before your edit. – Jochen Feb 21 '19 at 12:27

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In order that the subspace spanned by $(x_n)$ is isomorphic to $l^1$ it has to be closed. Hence, we have to talk about the closure of the span of $(x_n)$.

If $X$ is reflexive, then this subspace is also reflexive, and cannot be isomorphic to $l^1$.

If $X$ is a Hilbert space (or strictly convex space) then no such sequence can exist.

Also the example by @Jochen ($X=\mathbb R^1$, $x_n=1$) shows that span of $(x_n)$ needs not to be infinite-dimensional.

daw
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    But it is not clear that such a sequence exists in reflexive spaces. – Jochen Feb 21 '19 at 12:29
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    @Carucel As you accepted the answer so quickly: What do I miss? Does, e.g., $\ell^2$ admit such a sequence? – Jochen Feb 21 '19 at 12:32
  • Thank you for your interesting proof. Thank you also @Jochen for your very clever remark. The reason I accepted so quickly is that I was completely not expecting such a proof and was too easily convinced. – Carucel Feb 21 '19 at 12:33
  • For any $x\in X$ with norm $1$, the constant sequence $x_n=x$ satisfies the requirement. This example, however, is cheated. It would be more interesting to have a linear independent sequence. – Jochen Feb 21 '19 at 12:37
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    @Jochen $X=l^\infty$, $x_n= (1,...,1,0,0,...)$ with $n$ non-zero entries. Is the closure of the span equal to $c$? – daw Feb 21 '19 at 12:39
  • Nice example @daw. The closed span of ${x_n:n\in\mathbb N}$ is equal to the closed spaan of ${e_n: n\in\mathbb N}$ (with the unit vectors $e_n=x_n-x_{n-1}$) and this is the space $c_0$ of sequences converging to $0$. – Jochen Feb 21 '19 at 12:53
  • @daw: without considering the closure of its span, your example also shows that $|\sum_{n=1}^N(-1)^nx_n|{\ell^\infty}=1$, which makes it impossible for ${x_n}$ to be isomorphic to $\ell^1$, since $|\sum{n=1}^N(-1)^ne_n|_{\ell^1}=N$. – Carucel Feb 26 '19 at 08:29