I'm struggling on this exercice : Let $E$ be a real pre-Hilbert space, and let $B$ be an open ball in $E$ that does not contain 0. Show that there exists a non-zero linear form $f$ in $E^*$ such that for every $x$ in $B$, $f(x) \geq 0$.
I thought about the Riesz representation theorem but I can't find any way out.
Let $B=B(x_0,r)$. Since $0 \notin B$ we get $\|x_0\|\ge r$. By Hahn-Banach Theorem we can find $f$ such that $\|f\|=1$ and $f(x_0)=\|x_0\|$. If $f(x) <0$ for some $x \in B$ then we can write $x=x_0+y$ with $\|y\|<r$. Now $r+f(y)\le \|x_0\|+f(y)=f(x_0)+f(y)=f(x)<0$, so $f(y)<-r$. But $-f(y)=f(-y)\leq \|f\|\|y\|<r$, a contradiction.
Note: This works in any normed liner space. Pre-Hilbert space is not needed. However, if you have a pre-Hilbert space you can define $f$ explicitly: just take $f(z)=\langle z, \frac x{\|x\|} \rangle$.