Here is the question:
Let be $X$ a real normed vector space and $ \varphi : X \longrightarrow \mathbb{R} $ linear. Show that if $\varphi$ is unbounded then for all $\delta$, $ \varphi (B(0, \delta)) = \mathbb{R}$ ( the image of the open ball with center zero and radius delta ).
I tried to answer for contraposition, i.e, I wanna show that $\varphi$ is bounded. Here is my draft:
I supposed that it exists a $\delta$ such that $ \varphi (B(0, \delta)) \neq \mathbb{R}.$ Thus, exists an $a \in \mathbb{R}$ such that for all $\xi \in B(o, \delta) $ one have $ \varphi (\xi) \neq a $. Then $ \varphi (\xi) > a $ or $ \varphi (\xi) < a .$
But I can't finish this question. Can Someone help me? I will be very grateful.
You are basically done. If there exist $\xi,\eta\in B(0,\delta)$ with $\varphi(\xi)>a$ and $\varphi(\eta)<a$, then there exists $t\in(0,1)$ with $a=t\varphi(\xi)+(1-t)\varphi(\eta)$ (that is, $a$ is in the segment that joins $\varphi(\xi)$ and $\varphi(\eta)$. Then $\nu=t\xi+(1-t)\eta\in B(0,\delta)$ and $$ \varphi(\nu)=a, $$ a contradiction. So either $\varphi(\xi)>a$ for all $\xi\in B(0,\delta)$, or $\varphi(\xi)<a$ for all $\xi$. In the first case, we may rewrite the inequality as $\varphi(-\xi)<-a$, and since $0\in B(0,\delta)$ we get $0<-a$. Thus $$ |\varphi(\xi)|<-a$$ for all $\xi\in B(0,\delta)$. In the second case, we get $0<a$ and $|\varphi(\xi)|<a$ for all $\xi\in B(0,\delta)$.