Can a meager linear subspace be written as a countable increasing union of nowhere dense subspaces? Let $X$ be a separable Banach space.  In this question, "subspace" means a linear subspace, not necessarily closed. 

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 $X$.  Without loss of generality, we can also take $E_1 \subset E_2 \subset \cdots$.  Can the $E_n$ be taken to be subspaces of $X$?  That is, can a meager subspace always be written as a countable increasing union of nowhere dense subspaces?

Note that the linear span of a nowhere dense set is not necessarily nowhere dense (consider the unit sphere), nor is the sum of two nowhere dense subspaces (for instance, in $C([0,1])$, consider the mean-zero functions and the constants).
This came up while thinking about this answer: it isn't sufficient to check weak(-*) convergence in $L^2$ on the subspace $C([0,1])$; a sequence of linear functionals may converge pointwise on $C([0,1])$ but diverge somewhere else.  More generally, when is it sufficient to check weak-* convergence on a dense subspace $A \subset X$?  It is sufficient if $A$ is nonmeager, by a version of the uniform boundedness principle and a triangle inequality argument.  But a subspace which is nonmeager lacks the Baire property, so such an $A$ would be a "weird" subspace, something we are not likely to encounter in everyday life.  Most of the examples of dense subspaces I know are either countable dimension (hence meager), or complete in a stronger norm (hence analytic in $X$, hence has the BP, hence meager; as for example $C([0,1]) \subset L^2([0,1])$).
So I was wondering whether one could prove that it is never sufficient to check weak-* convergence on a meager subspace.  If my question above has an affirmative answer, then we can do the following: for any meager subspace $E$, 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) it is not weak-* convergent.  Thus it would not suffice to check weak-* convergence on $E$.
 A: As I posted on MathOverflow some time ago, the answer to this question is No.  I will copy my counterexample here.
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.  Then $E$ is meager in $X$: By Arzelà-Ascoli, the closed balls of  the Hölder norm are compact in $X$, hence by Riesz's lemma they are nowhere dense, and $E$ is their countable union.
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.
The followup question about checking weak-* convergence on a meager subspace remains unanswered as of this writing and is asked more precisely in the MO question linked above.
