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 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?

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 countable increasing union 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-* 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?

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 $f_n(x) \to 0$.

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    $\begingroup$ If 1a is not true, you have an interesting situation. Let $L(E)$ be the lattice of subspaces of $E$. Even though $E$ is small in some sense, $L(E)$ is large in that it has "cofinality" greater than $\omega$. (More significantly, some sublattice generated by the nowhere dense subspaces has large cofinality, or otherwise misses E.) I imagine not only AC but some other combinatorial principle is at work here. Perhaps Jonsson lattices might help? Gerhard "Knows Banach Spaces From Nothing" Paseman, 2013.11.12 $\endgroup$ Nov 12, 2013 at 19:06
  • $\begingroup$ Can you provide an example of an explicit space which satisfies this property? It might help to walk through an algorithm for converting meagre subsets to meagre subspaces. Hopefully we could see how to either generalize the argument, or to identify an obstruction to doing so in general. $\endgroup$ Nov 15, 2013 at 18:16
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    $\begingroup$ @TomLaGatta: That's a good question. The only examples where I can see a way to do this explicitly are trivial cases where $E$ has countable dimension (for instance, $X=C([0,1])$ and $E$ is the polynomials). But for example, I don't know whether one can write $E = C([0,1])$ as a countable union of nowhere dense subspaces of $X = L^2([0,1])$. (Nowhere dense subsets is easy: each sup-norm ball is closed and nowhere dense in $L^2([0,1])$.) $\endgroup$ Nov 15, 2013 at 18:18
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    $\begingroup$ @TomLaGatta: Actually, in the case $X = L^2([0,1])$, $E = C([0,1])$, I just realized that there can't be 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. $\endgroup$ Nov 15, 2013 at 18:25
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    $\begingroup$ Concerning Q2: Most probably you only require that $E$ determines weak-$*$ convergence of sequences. If instead $E$ determines convergence of all nets then it determines the topology and this implies $E=X$ because $(X^*,\sigma(X^*,E))=E$. $\endgroup$ Nov 22, 2013 at 7:04

2 Answers 2


Update: Here is a ZFC (probably even ZF+DC) counterexample for Q1. It's from probability and kind of indirect, maybe someone will be able to find something shorter.

Let $X = C_0([0,1])$, the space of continuous functions $\omega$ on $[0,1]$ having $\omega(0)=0$. Fix $0 < \alpha < 1/2$ and let $E = C^{0,\alpha}([0,1]) \cap X$ be the subspace of $\alpha$-Hölder continuous functions. There are several ways to see that $E$ is meager in $X$. For instance, by Arzelà-Ascoli, the balls of $E$ under the Hölder norm are compact in $X$, hence $E$ is a countable union of compact sets, each of which is nowhere dense in $X$ by Riesz's lemma. Or, the argument mentioned in the question: $E$ is Banach in its own norm, hence analytic in $X$, hence has the Baire property in $X$, hence must be meager. (That doesn't work because $E$ is not separable in its own norm, hence not Polish.)

Now let $\mu$ be the Wiener measure on $X$. It is well known that $\mu(E) = 1$, since Brownian motion is almost surely $\alpha$-Hölder continuous for any $\alpha < 1/2$. On the other hand, suppose $F$ is any nowhere dense subspace of $X$, so that its closure is a proper closed subspace of $X$. By Hahn-Banach there is a nonzero continuous linear functional $f$ that vanishes on $F$. Now $\mu$ is a nondegenerate Gaussian measure on $X$, so $f$ has a nondegenerate one-dimensional Gaussian distribution under $\mu$; in particular, $\mu(F) \le \mu(\{f = 0\}) = 0$. Thus every nowhere dense subspace has measure zero, so by countable additivity, $E$ cannot be a countable union of such.

Previous answer. It is consistent with ZF+DC that the answer to Q1 is No.

Consider an example in which $E$ is separable Banach in a stronger norm $\|\cdot\|_E$; for instance, $X = L^2([0,1])$ and $E = C([0,1])$. Then $E$ is analytic in $X$, hence has the Baire property, hence must be meager. If $E$ can be written as a countable union of subspaces $E_n$ then by Baire category one of them must be nonmeager with respect to $\|\cdot\|_E$, hence doesn't have the BP in $E$. But Shelah proved it is consistent with ZF+DC that every subset of a Polish space has the BP.

In our specific example we can see explicitly that $C([0,1])$ is meager in $L^2([0,1])$: the sup-norm balls $B_n = \{f : \|f\|_\infty \le n\}$ are closed in $L^2$ with empty interior.

I don't see whether this helps us with Q2.


The answer to Q2 is No.

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

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

(Damian mentioned it in the question Convergence of functionals on compact projections on a separable Hilbert space.)

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

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 $\sup_n \sup_{B \in \mathcal{B}} |\mu_n(B)| < \infty$.

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.

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 $\|\mu\| \le 2M$.


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