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?



On statistical bases in Banach spaces
**Edit 26.04.2022**. The problem is now solved; please see [my other answer](https://mathoverflow.net/a/359434/15129).




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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](https://gowers.wordpress.com/2017/09/19/two-infinities-that-are-surprisingly-equal/#more-6359) on T. Gowers' blog for more details).



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> T. Kochanek, [ℱ-bases with brackets and with individual brackets in Banach spaces](https://arxiv.org/abs/1207.3097), *Studia Math.* **211** (2012), 259-268.
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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.


