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Let $X$ be a Banach space.

Suppose there exists a sequence $\{x_n\}$ in $X$ such that

  • for all finite $A\subseteq\mathbb{N}$, and
  • for all functions $\alpha:A\to\{-1,1\}$,

we have that $\|\sum_{n\in A}\alpha_nx_n\|$ equals the number of elements in $A$.

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

This question is a modification to this question, where $\alpha\equiv 1$.

Andrews
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Carucel
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  • you talk about the closed subspace spanned by ${ x_n }$. Shouldn't it be the closure of the subspace spanned by ${ x_n }$ since you dont know if it is already closed? – supinf Feb 26 '19 at 08:59
  • @supinf yes thank you, that is what i meant – Carucel Feb 26 '19 at 09:01
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    @supinf "Closed subspace spanned by" is a perfectly standard synonym for "closure of the subspace spanned by". More briefly, "closed span". – David C. Ullrich Feb 26 '19 at 15:19

1 Answers1

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Yes this is true. It is even possible to show more, namely that the closure of the span is isomtrically isomorphic to $\ell^1$.

The main part of the proof will be to show that $$ \tag{1} \label{eq1} \|\sum_{i=1}^n a_i x_i\| = \sum_{i=1}^n | a_i| $$ is true for all $(a_i)\in \ell^1$ and $n\in\mathbb N$.

We prove \eqref{eq1} by induction in $n$.

We assume that \eqref{eq1} was already shown for $n\in\mathbb N$ and we want to show that it also holds for $n+1$. Wlog we can assume that $a_{n+1}\geq |a_i|$ for $i=1,\ldots n$ (Otherwise we can reorder the indices so that $|a_{n+1}|$ is coefficient with the largest absolute value, and $a_{n+1}\geq0$ can be assumed because otherwise we can multiply everything by $-1$).

We define $\alpha_i:=\operatorname{sgn} a_i$. Then we have $$ \begin{align} \|\sum_{i=1}^{n+1} a_i x_i\| &= \|a_{n+1}\sum_{i=1}^{n+1} \alpha_i x_i - \sum_{i=1}^n (\alpha_i a_{n+1}-a_i)x_i\| \\ &\geq \|a_{n+1}\sum_{i=1}^{n+1} \alpha_i x_i\|-\| \sum_{i=1}^n (\alpha_i a_{n+1}-a_i)x_i\| \\ &= |a_{n+1}|(n+1) - \sum_{i=1}^n |\alpha_i a_{n+1}-a_i| \\ &= |a_{n+1}|(n+1) - \sum_{i=1}^n (|a_{n+1}|-|a_i|) = \sum_{i=1}^{n+1} |a_i|. \end{align} $$ The other inequality follows from the triangle inequality.

Thus \eqref{eq1} holds true for all $n\in\mathbb N, (a_i)\in\ell^1$. By continuity it can also be shown that \eqref{eq1} holds for $n=\infty$.

Let us define the map $$ I:\ell^1\to \overline{\operatorname{span}(x_i)}, e_n\mapsto x_n. $$ Then it follows from the above that $I$ is an isometry. Finally, the property that $I$ is an isometry can be used to show that $I$ is surjective and injective.

supinf
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  • Thank you. Could you explain why we are allowed to take $a_{n+1}\geq |a_i|$ for $i=1,\ldots n$? – Carucel Feb 26 '19 at 12:40
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    We can reorder the indices if necessary. I edited. – supinf Feb 26 '19 at 12:44
  • I think you have made a mistake: why do you think that $| \sum_{i=1}^n (\alpha_i a_{n+1}-a_i)x_i|=\sum_{i=1}^n |\alpha_i a_{n+1}-a_i|$? – Carucel Feb 26 '19 at 12:54
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    that is by induction assumption. We can use the formula (1) for this case. – supinf Feb 26 '19 at 12:58
  • Sorry, you are right. Thank you very much. I found it hard to believe the statement you proved, since it is not true if we only know the assumption for $\alpha\equiv 1$ by this answer. – Carucel Feb 26 '19 at 13:01