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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.

Edit 26.04.2022. The problem is now solved; please see my other answer.

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.

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 counter-example, if it exists.

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.

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).

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 counter-example, if it exists.

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 counter-example, if it exists.