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Let $E$ be a infinite dimensional normed vector space.

1 - How to define a "not continuous" linear functional $f$ in $E$ such that the set $\ker(f)=\{x\in E:\ f(x)=0\}$ is dense in $E$ but $\ker(f)\neq E$?

2 - If $f$ is the functional defined above, can we find $y\in E\setminus\ker(f)$ such that the set $\{\lambda y:\ \lambda\in\mathbb{R}\}$ is not contained in $\ker(f)$?

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For part 1., see here. In particular, this section (note a discontinuous linear functional will have a dense kernal and the kernal will not be all of $E$). –  David Mitra Dec 8 '12 at 17:13
    
You are right @DavideGiraudo, the answer for 2 wass easy. I forgot that $Ker f$ is a linear space –  Tomás Dec 8 '12 at 17:19

1 Answer 1

up vote 2 down vote accepted
  1. Not continuous and the property you want is redundant, as the kernel of a non-zero linear functional is either closed or dense (closed when the linear form is continuous, and dense otherwise). So you just have to find a discontinuous linear functional, which can be done using Hamel basis.

  2. Any $y$ such that $f(y)\neq 0$ will do the job, as $1\cdot y\notin \ker f$.

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