Let X be a Banach space, and let Y be a proper non-meager linear subspace of X. If Y is not dense in X, then it is easy to see that the closure of Y has empty interior, contradicting Y being non-meager. So Y must be dense. If Y has the Baire property, then it follows from Pettis Lemma that Y is open and hence closed (since the complement of Y is the union of translates of Y), contradicting Y being proper. Thus, Y must be dense and not have the Baire property. 


My question is: is there a Banach space X with a proper non-meager linear subspace Y? Such a Y must be dense and not have the Baire property. Any such Y must be difficult to construct since all Borel sets and even all continuous images of separable complete metric spaces have the Baire property.


More info:  

1. Meager is just another word for first category, i.e. the countable union of nowhere dense sets.  

2. A set A in a topological space has the Baire property if for some open set V (possibly empty) the set (A-V)U(V-A) is meager.  

3. The collection of sets with the Baire property form a sigma-algebra. All open sets trivially have the Baire property, thus all Borel sets have the Baire property. All analytic sets also have the Baire property.  

4. Pettis Lemma: Let G be a topological group and let A be a non-meager subset of G with the Baire property. Then the set A\*A^{-1} (element-wise multiplication) contains an open neighborhood of the identity. This is an analog to a similar theorem about Lebesgue measure: If A is a Lebesgue measurable subset of the reals with positive Lebesgue measure, then A - A (element-wise subtraction) contains an open set around 0.



Are proper linear subspaces of Banach spaces always meager?
I am afraid that Konstantin's accepted answer is seriously flawed.


In fact, what seems to be proved in his answer is that $\ker f$ is of second category, whenever $f$ is a discontinuous linear functional on a Banach space $X$. This assertion has been known as [Wilansky-Klee conjecture](http://www.ams.org/journals/bull/1966-72-04/S0002-9904-1966-11547-1/S0002-9904-1966-11547-1.pdf)
and has been disproved by [Arias de Reyna](http://dx.doi.org/10.1007/BF01351408) under Martin's axiom (MA). 
He has proved that, under (MA), in any separable Banach space there exists a discontinuous linear functional $f$ such that $\ker f$ is of first category.
There have been some subsequent generalizations, see [Kakol et al.](http://books.google.fr/books?id=k8uYhaIfHuAC&lpg=PA20&ots=yq-Ad-KRHZ&dq=wilansky%20klee%20conjecture&pg=PA20#v=onepage&q=wilansky%20klee%20conjecture&f=false)


*So, where is the gap in the above proof?*


It is implicitly assumed that $\ker f = \bigcup A\_i$. Then $f$ is bounded on $B\_i=A\_i+[-i,i]z$. But in reality, we have only $\ker f \subset \bigcup A\_i$ and we cannot conclude that $f$ is bounded on $B\_i$.


*And finally, what is the answer to the OP's question?*


It should not be surprising (remember the conjecture of Klee and Wilansky) that the answer is:
in every infinite dimensional Banach space $X$ there **exists** a discontinuous linear form $f$ such that $\ker f$ is of second category.


Indeed, let $(e\_\gamma)\_{\gamma \in \Gamma}$ be a normalized [Hamel basis](http://en.wikipedia.org/wiki/Hamel_basis#Related_notions)
 of $X$. 
Let us split $\Gamma$ into countably many pairwise disjoint sets $\Gamma =\bigcup\_{n=1}^\infty \Gamma\_n$, each of them infinite. We put $X\_n=span\{e\_\gamma: \gamma \in \bigcup\_{i=1}^n \Gamma\_i\}$. It is clear (from the definition of Hamel basis) that $X=\bigcup X\_n$. Therefore there exists $n$ such that $X\_n$ is of second category. Finally, we define $f(e\_\gamma)=0$ for every $\gamma \in \bigcup\_{i=1}^n\Gamma\_i$ and $f(e\_{\gamma\_k})=k$ for some sequence $(\gamma\_k) \subset \Gamma\_{n+1}$. We extend $f$ to be a linear functional on $X$. It is clearly unbounded, $f\neq 0$, and $X\_n \subset \ker f$. Hence $\ker f\neq X$ is dense in $X$ and of second category in $X$.


