Kernel of an unbounded linear functional on Banach space

Assume $$f$$ is an unbounded linear functional on Banach space $$X$$. Then $$\ker(f)$$ is a dense linear subspace of $$X.$$ Is $$\ker(f)$$ a set of second category ?

• See this. And, nice question! Jan 12 at 18:52
• Thanks. It seems there exists an unbounded linear functional f such that ker(f) is of first category. I am interested in whether it is possible to find f such that ker(f) is of second category. Let's restrict to Hilbert spaces. Jan 13 at 13:26
• See the first answer there. An example is given at the end. Jan 13 at 13:27
• Thanks again. It looks pretty easy. Jan 13 at 14:02
• My question is related to the following exercise in functional analysis: Let X be a normed linear space and Y a Banach space. Consider a sequence of bounded linear operators T_n : X --> Y, Then the set A={x \in X : lim T_n(x) exists} is either of first category or A=X. The conclusion is not true if we drop the assumption that Y is complete. For this I needed the subspace of second category. Jan 13 at 14:14