I previously asked a version of this question [on Math.SE](https://math.stackexchange.com/q/554841/822), 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. 



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> **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 dense *subsets* of $X$. Can the $E\_n$ be taken to be *subspaces* of $X$? That is, can a meager subspace always be written as a countable union of nowhere dense subspaces?
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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).



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> **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 countable *increasing* union of nowhere dense subspaces?
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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).




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



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> **Q2.** Let us say a subspace $E \subset X$ **determines weak-\* convergence** if 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? 
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> **Q2a.** If not, what are necessary and sufficient conditions for $E$ to determine weak-\* convergence?
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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](https://mathoverflow.net/questions/3188/are-proper-linear-subspaces-of-banach-spaces-always-meager), 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.




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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 $f\_n(x) \to 0$.



Meager subspaces of a Banach space and weak-* convergence
The answer to Q2 is **No**.


I am grateful to Damian Sobota for drawing my attention to the following paper:



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> Darst, R. B.
>  On a theorem of Nikodym with applications to weak convergence and von Neumann algebras.
>  *Pacific J. Math.* **23** (1967) 473–477. MR0238084
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(Damian mentioned it in the question [Convergence of functionals on compact projections on a separable Hilbert space](https://mathoverflow.net/questions/241935/convergence-of-functionals-on-compact-projections-on-a-separable-hilbert-space?lq=1).) 


Darst's result is as follows (notation slightly adjusted):



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> **Theorem.** Suppose $\{\mu\_n\}$ is a sequence of bounded and finitely additive complex measures on a $\sigma$-algebra $\mathcal{B}$ of subsets of a set $\Omega$ such that for each $B \in \mathcal{B}$ we have $\sup\_n |\mu\_n(B)| < \infty$. Then $\sup\_n \sup\_{B \in \mathcal{B}} |\mu\_n(B)| < \infty$.
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This effectively provides a counterexample to Q2.


To be concrete, let us take $\Omega = \mathbb{N}$, $\mathcal{B} = 2^{\mathbb{N}}$. Then, in the notation of this question, let $X = \ell^\infty(\mathbb{N})$. (For simplicitly let us work over $\mathbb{R}$, and identify $X$ as the space of bounded real-valued functions on $\mathbb{N}$.) Then $X^\*$ can be identified as the set of bounded and finitely additive signed measures on $\mathcal{B} = 2^{\mathbb{N}}$, where the pairing with $X$ is integration, and the dual norm is given by the total variation.


Let $E \subset X$ be the space of simple functions (those with finite range; equivalently, the linear span of the indicator functions). Darst's theorem implies that $E$ determines weak-\* convergence. For suppose $\{\mu\_n\} \subset X^\*$ satisfies $\int x\,d\mu\_n \to 0$ for every $x \in X$. Taking $x = 1\_B$ for arbitrary $B \subset \mathbb{N}$, we see that $\mu\_n(B) \to 0$, and in particular $\sup\_n |\mu\_n(B)| < \infty$. Darst's result then implies that $\sup\_n \sup\_{B \subset \mathbb{N}} |\mu\_n(B)| < \infty$, which means $\sup\_n \|\mu\_n\| < \infty$ (see Claim 1 below). Since $E$ is dense in $X$, it follows by the triangle inequality (see the last part of the footnote to the question) that $\mu\_n \to 0$ weakly-\*.


(In fact, if I am reading Darst's paper correctly, he actually shows more: that $\mu\_n \to 0$ weakly, i.e. in the topology $\sigma(X^\*, X^{\*\*})$.)


On the other hand, $E$ is meager in $X = \ell^\infty$. To see this, just let $E\_k$ be the set of all $x \in \ell^\infty$ whose range contains at most $k$ numbers. Clearly $E\_k$ has empty interior (every ball contains a function with infinite range). And $E\_k$ is closed: suppose $x\_n \in E\_k$ and $x\_n \to x$ in $X$ (i.e. uniformly). If the range of $x$ contains $m$ numbers, $t\_1, \dots, t\_m$, we can choose $\varepsilon$ so small that the balls $B(t\_i, \varepsilon)$ are pairwise disjoint. Yet there is $n$ sufficiently large that $\|x\_n - x\| < \varepsilon$, meaning that each of the $m$ balls must contain a number from the range of $x\_n$. Hence $m \le k$, and we have $x \in E\_k$. So $E\_k$ is nowhere dense, and $E = \bigcup\_k E\_k$.


So $E$ is meager, yet it determines weak-\* convergence.


I would still be interested to know whether there is a counterexample with $X$ separable.




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*Proof of Claim 1.* Suppose $\mu$ is a finitely additive signed measure and $\sup\_{B \subset \mathbb{N}} |\mu(B)| = M$. Let $A\_1, A\_2, \dots, A\_m$ be finitely many disjoint subsets of $\mathbb{N}$. Arrange these sets so that $\mu(A\_i) \ge 0$ for $1 \le i \le k$ and $\mu(A\_i) < 0$ for $k+1 \le i \le m$. Then let $B\_1 = A\_1 \cup \dots \cup A\_k$ and $B\_2 = A\_{k+1} \cup \dots \cup A\_m$. Then
$$\begin{align\*}
\sum\_{i=1}^m |\mu(A\_i)| &= \sum\_{i=1}^k \mu(A\_i) - \sum\_{i=k+1}^m \mu(A\_i) \\
&= \mu(B\_1) - \mu(B\_2) \\
&= |\mu(B\_1)| + |\mu(B\_2)| \le 2M
\end{align\*}$$
Since $\|\mu\|$ is the supremum of the left side over all finite partitions of $\mathbb{N}$, we have $\|\mu\| \le 2M$.


