I need an help with the following exercise.

Let $(E,\| \cdot \|)$ a n.v.s. and let $f:E\rightarrow \Bbb R$. Show that $H=\{x\in E: f(x)=\alpha\}$ is closed if and only if $f\in E'.$ Actually, I don't know if these hypothesis are enough to conclude, but my exercise states like that.

I know that if $f\in E'$, since $f$ is continuous and $\{\alpha\}$ is closed in $\Bbb R$, then $H$ is close in $E$. But I don't know how to prove the other implication. Any suggestion?

  • $\begingroup$ Is the condition that $H$ be closed for all $\alpha \in \mathbb{R}$? Or just a particular $\alpha$? $\endgroup$ – msteve Aug 4 '14 at 16:53
  • 1
    $\begingroup$ The linearity of $f$ is - apparently tacitly - assumed. For any continuous function $f\colon E\to\mathbb{R}$ the level sets $f^{-1}(\alpha)$ are closed. Since all these hyperplanes are translates of each other (unless $f\equiv0$, in which case one level set is the whole space, all others are empty and none is a hyperplane), the thing to prove is that a linear map $f\colon E\to \mathbb{R}$ is continuous if and only if its kernel is closed. One direction is immediate from general properties of continuous maps. For the other, you could consider the space $E/\ker f$. What do you know about that? $\endgroup$ – Daniel Fischer Aug 4 '14 at 16:53
  • $\begingroup$ @DanielFischer: $E/\ker f$ is isomorphic to $Im f$? $\endgroup$ – batman Aug 5 '14 at 7:32
  • $\begingroup$ @DanielFischer: And also it is not so clear to me when you say that the hyperplanes are the translates of each other. (Sorry for the delay in answering) $\endgroup$ – batman Aug 5 '14 at 7:40
  • $\begingroup$ Let $H_\alpha = f^{-1}(\alpha)$. If $f\not\equiv 0$, then no $H_\alpha$ is empty, and we have $H_\alpha = v_\alpha + H_0$, where $v_\alpha$ is an arbitrary element of $H_\alpha$. So $H_\alpha$ is a translate of $H_0$. And by transitivity, all $H_\alpha$ are translates of each other. $\endgroup$ – Daniel Fischer Aug 5 '14 at 12:00

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.