A linear functional $f$ is continous if and only if... Let be $E$ a topological vector space and a linear functional $f$. How to show that $f$ is continous if and only if exist an oper set $U$ and a real value $t$ such that $t\notin f(U)$.
 A: *

*If $f$ is continuous: Then $f$ is bounded so let $U$ be any non-empty bounded open set. Then $f(U)$ is bounded so let $t=1+\sup \{|f(u)|:u\in U\}.$


*If $f$ is discontinuous: Then $f$ is unbounded so there exists $\{e_n: n\in \Bbb N\}\subset E$ with $\{1\}=\{f(e_n):n\in \Bbb N\}$ and $\lim_{n\to \infty}\|x_n\|=0.$
Let $p\in U\subset E$ where $U$ is open in $E.$ Let $B$ be an open ball in $E$ centered at $p,$ with radius $r>0,$  such that $B\subset U.$
For $t\in \Bbb R$ take $n\in\Bbb N$ large enough that $\|x_n\|\cdot |t-f(p)|<r.$  Then $p+x_n(t-f(p))\in B\subset U$ and $f( p+x_n(t-f(p)))=t.$
A: Here is how it works without a norm:
If $f$ is continuous then $U:=f^{-1}(\,(0,1)\,)$ is open and clearly $f(U)$ is not all of $\Bbb R$ (I'm assuming real vector space, otherwise let $U$ be the pre-image of the complex open unit ball).
If $f$ has the property suppose that $f$ is discontinuous and get a contradiction. Let $U$ be one of the special open sets and by translation assume $0\in U$. Let $z_\alpha\to0$ a net with $f(z_\alpha)\not\to0$. Either $f(z_\alpha)$ is unbounded or admits a convergent (not to $0$) sub-net. Either way we may pass to a sub-net so that $\frac1{f(z_\alpha)}$ converges and is always defined. Now for $\lambda\in \Bbb R$ you have that
$$\frac{\lambda}{f(z_\alpha)}z_\alpha $$
still converges to $0$, hence must eventually lie entirely in $U$. However
$$f\left(\frac{\lambda}{f(z_\alpha)}z_\alpha\right)=\lambda$$
this yields the contradiction, so if $f$ has the property it cannot be discontinuous.
