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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 ?

Ryszard Szwarc
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  • See this. And, nice question! – David Mitra Jan 12 '22 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. – Ryszard Szwarc Jan 13 '22 at 13:26
  • See the first answer there. An example is given at the end. – David Mitra Jan 13 '22 at 13:27
  • Thanks again. It looks pretty easy. – Ryszard Szwarc Jan 13 '22 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.

     – Ryszard Szwarc Jan 13 '22 at 14:14

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