I previously asked a version of this question on Math.SE, but didn't receive an answer. (But there is a bounty there if you want to claim it!)

Let $X$ be a Banach space. (If it helps, feel free to assume that $X$ is separable.) In this question, "subspace" means a linear subspace of $X$, not necessarily closed.

Q1.Suppose $E \subset X$ is a subspace which is meager, so that we can write $E = \bigcup_n E_n$, where the $E_n$ are nowhere densesubsetsof $X$. Can the $E_n$ be taken to besubspacesof $X$? That is, can a meager subspace always be written as a countable union of nowhere dense subspaces?

Note that the trivial approach of replacing each $E_n$ with its linear span does not work, since the linear span of a nowhere dense set is not necessarily nowhere dense (consider the unit sphere).

Q1a. If the answer to Q1 is yes, can the subspaces $E_n$ be taken to be increasing, i.e $E_1 \subset E_2 \subset \cdots$? Can we write $E$ as a a countableincreasingunion of nowhere dense subspaces?

The trivial approach of replacing $E_n$ by $E_1 + \dots + E_n$ does not work, since the sum of two nowhere dense subspaces may not be nowhere dense (for instance, in $C([0,1])$, consider the mean-zero functions and the constants).

This came up in the context of thinking about the following related question.

Q2.Let us say a subspace $E \subset X$determines weak-* convergenceif for every sequence $\{f_n\} \subset X^*$ such that $f_n(x) \to 0$ for every $x \in E$, we have $f_n(x) \to 0$ for every $x \in X$. Is it true that $E$ determines weak-* convergence if and only if $E$ is nonmeager?

Q2a.If not, what are necessary and sufficient conditions for $E$ to determine weak-* convergence?

Clearly it is necessary that $E$ be dense. A rookie mistake is to guess that density is also sufficient, but this is false and it is easy to find counterexamples.

By a version of the uniform boundedness principle and the triangle inequality, every nonmeager subspace determines weak-* convergence (see below for a sketch). This is a very strong condition, since as discussed in this question, a nonmeager proper subspace of $X$ lacks the Baire property and so must be rather pathological, not a space we are likely to encounter in everyday life. For practical purposes, requiring that $E$ be nonmeager is essentially as strong as requiring that $E = X$.

I was wondering whether "nonmeager" is in fact necessary.

Suppose we could write $E = \bigcup_n E_n$ where $E_n$ are increasing nowhere dense subspaces. Since $E_n$ is nowhere dense, it is not dense, so by Hahn-Banach we may find $f_n \in X^*$ with $f_n(E_n) = 0$ and $\|f_n\| = n$. Then $f_n(x) \to 0$ for every $x \in E$, but $\{f_n\}$ is unbounded so by the uniform boundedness principle, there exists $x \in X$ with $\{f_n(x)\}$ unbounded. So such an $E$ does not determine weak-* convergence.

Therefore, if Q1a has an affirmative answer, so does Q2.

Footnote: Here's a sketch of the argument that nonmeager subspaces determine weak-* convergence. Suppose $E$ is nonmeager and $\{f_n\} \subset W^*$ with $f_n(x) \to 0$ for $x \in E$. Let $A_k = \bigcap_n \{x : |f_n(x)| \le K\}$. $A_k$ is closed, and since $\{f_n(x)\}$ is bounded for $x \in E$, we have $E \subset \bigcup_k A_k$. By Baire, some $A_k$ has nonempty interior, and a little rearrangement shows that $M := \sup_n \|f_n\| < \infty$, as in the usual uniform boundedness principle. Now since $E$ is nonmeager, it must be dense, since non-dense subspaces are nowhere dense. For any $x \in X$, fix $\epsilon$ and choose $x' \in E$ with $\|x-x'\| < \epsilon$. Then $|f_n(x)| \le |f_n(x')| + M \epsilon$; letting $n \to \infty$ we get $\limsup_n |f_n(x)| \le M \epsilon$, but $\epsilon$ was arbitrary so $f_n(x) \to 0$.

can'tbe a way to do it explicitly. Since $E$ is Banach in its own norm, one of the subspaces $E_n$ must be nonmeager in the sup norm, and hence won't have the BP. So we won't be able to construct it without a fair amount of choice. – Nate Eldredge Nov 15 '13 at 18:25sequences. If instead $E$ determines convergence of allnetsthen it determines the topology and this implies $E=X$ because $(X^*,\sigma(X^*,E))=E$. – Jochen Wengenroth Nov 22 '13 at 7:04