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Let $K$ be a subset of the positive integers $\mathbb{N}$. For each $n\in \mathbb{N}$, $K_{n}$ denotes the set $\{k\in K: k\leq n\}$ and $|K_{n}|$ denotes the number of the elements in $K_{n}$. The natural density of $K$ is defined by $$\delta(K)=\lim_{n\rightarrow \infty}\frac{|K_{n}|}{n}.$$ A sequence $(x_{k})_{k}$ in a Banach space $X$ is said to be norm statistically convergent to $x\in X$ if $\delta(\{k\in \mathbb{N}:\|x_{k}-x\|\geq \epsilon\})=0$ for every $\epsilon>0$.

It is natural to define a series $\sum_{k=1}^{\infty}x_{k}$ to be norm statistically convergent to $x\in X$ by $(s_{n})_{n}=(\sum_{k=1}^{n}x_{k})_{n}$ to be norm statistically convergent to $x$.

We say that a sequence $(x_{k})_{k}$ in a Banach space $X$ is a statistical basis for $X$ if, for each $x\in X$, there exists a unique $(a_{k})_{k}$ such that the series $\sum_{k=1}^{\infty}a_{k}x_{k}$ norm statistically convergent to $x$.

Clearly, statistical basis is a generalization of the notion of Schauder basis.

Question: For each $n$, we can define a linear functional $f_{n}$ in $X$ by $f_{n}(x)=a_{n}$. If $(x_{k})_{k}$ is a Schauder basis, we know that $f_{n}$ is continuous. However, if $(x_{k})_{k}$ is a statistical basis for $X$, is $f_{n}$ continuous?

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    $\begingroup$ what is an example of a statistical basis which is not a Schauder basis? $\endgroup$ Oct 31 '17 at 13:00
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    $\begingroup$ Example 1 on page 254 in bit.ly/2yil4fa gives a statistical M-basis, which is not a Schauder basis.However, the coordinate functionals in that basis are continuous. $\endgroup$ Oct 31 '17 at 17:56
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This is an open problem due to Vladimir Kadets. I would not expect an easy answer here. The problem with statistical convergence is that this filter is not countably generated (in which case these functionals would be continuous).

A more general question is also open. Suppose that $\mathcal{F}$ is a filter on $\mathbb N$. Say that a sequence $(e_n)$ in a Banach space $X$ is an $\mathcal F$-basis if every element $x\in X$ has a unique expansion as $$\mathcal{F}-\lim_{n\to\infty} \sum_{k=1}^n e_k^*(x)e_k.$$ Are the linear functionals $(e_n^*)$ continuous?

The only positive result so far is a theorem of Kochanek which says that the answer is yes when $\mathcal F$ is generated by less than $\mathfrak p$ elements (here $\mathfrak p$ denotes the pseudo-intersection number; see this entry on T. Gowers' blog for more details).

T. Kochanek, ℱ-bases with brackets and with individual brackets in Banach spaces, Studia Math. 211 (2012), 259-268.


For the statistical convergence, one may suggest a meta-proof. The definition of statistical convergence does not involve (full) Axiom of Choice. There are models of ZF+DC where all linear functionals on Banach spaces are continuous hence you cannot have a constructible counterexample, should it exist.

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  • $\begingroup$ You say that this is an open problem due to Vladimir Kadets. Could you tell me in which paper Vladimir Kadets posed this problem? $\endgroup$ Nov 1 '17 at 13:42
  • $\begingroup$ Are there any papers about statistical bases in Banach spaces? Could you tell me any reference about statistical bases? I am interested in it. $\endgroup$ Nov 1 '17 at 13:46
  • $\begingroup$ @DongyangChen all relevant information is contained on the first page of the above-linked paper by Kochanek. $\endgroup$ Nov 2 '17 at 10:30
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In a recent note with J. Swaczyna (arXiv:2005.04873), we proved that assuming the existence of certain large cardinals (for example, the existence of a super-compact cardinal) that filter bases whose underlying filter is a projective subset of the Cantor set, have continuous coordinate functionals. The role of large cardinals is to make the heuristic proof I outlined in my other answer work.

The filter of statistical convergence is actually $F_{\sigma \delta}$ (hence Borel, hence projective), so in a theory stronger than ZFC, the answer to your question is affirmative. We still expect that, at least for Borel filters, the question about continuity of coordinate functionals should have a positive answer in ZFC. However, our proof method does not give a chance to invoke Schoenfield's absoluteness theorem in that context.

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  • $\begingroup$ Do you use some determinacy? $\endgroup$ May 5 '20 at 13:55
  • $\begingroup$ @MonroeEskew, yes, in fact, we wanted to prove this under PD, but got lost in some technicalities. $\endgroup$ May 5 '20 at 14:07

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