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 ?

  • $\begingroup$ See this. And, nice question! $\endgroup$ Jan 12 at 18:52
  • $\begingroup$ 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. $\endgroup$ Jan 13 at 13:26
  • $\begingroup$ See the first answer there. An example is given at the end. $\endgroup$ Jan 13 at 13:27
  • $\begingroup$ Thanks again. It looks pretty easy. $\endgroup$ Jan 13 at 14:02
  • $\begingroup$ 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. $\endgroup$ Jan 13 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.