Let $X$ be a Hausdorff topological vector space over $\mathbb{K}$. Suppose $W$ is a convex subset of its topological dual $X'$. How to prove that if for any $x\in X\setminus\{0\}$ set $\{f(x):f\in W\}$ is dense in $\mathbb{K}$ then $\overline{W}^{\sigma(X',X)}=X'$ ?

  • $\begingroup$ You need $x\neq 0$, for $x = 0$ we have $\{f(x) : f\in W\} = \{0\}$. Use (one of) the Hahn-Banach separation theorem(s). $\endgroup$ – Daniel Fischer Feb 6 '16 at 22:03
  • $\begingroup$ So, suppose there is $f_0\notin \overline{W}^{\sigma(X',X)}$. Then there exists $\xi\in X''$ s.th. $\xi\left(\overline{W}^{\sigma(X',X)}\right)<\xi(f_0)$. But $\{f(x):f\in W\}\subset \xi(W)$, hence $\{f(x):f\in W\}$ cannot be dense in $\mathbb{K}$. Is it correct ? $\endgroup$ – mikis Feb 6 '16 at 22:47
  • $\begingroup$ You need $\xi = x$. But you're looking at $(X',\sigma(X',X))$, and $(X',\sigma(X',X))' = X$. $\endgroup$ – Daniel Fischer Feb 6 '16 at 22:50

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.