Let $X$ be a locally convex space and $A\subseteq X$. The polar of $A$ is a subset $A^\circ$ of the dual $X^\ast$ defined by $A^\circ\equiv \{x^\ast\in X^\ast:|\langle a,x^\ast\rangle|\leq 1,\, \forall a\in A\}$.
Now an exercise in Conway's book on functional analysis states that a subset $A$ is bounded (in the sense of a topological vector space) in the weak topology if and only if $A^\circ$ is absorbing.
Now I wonder how to prove this and also if this is even true?
My skepticism comes from the fact that if $x^\ast\in A^\circ$ satisfies $|\langle a,x^\ast\rangle| = 1$ for some $a\in A$ then by mutliplying with a unimodular constant we can assume that $\langle a,x^\ast\rangle = 1$. However then $A^\circ$ is not absorbing at $x^\ast$ since there is no $r>0$ such that $x^\ast+rx^\ast\in A^\circ$. (In that case $\langle a,x^\ast+rx^\ast\rangle = 1+r>1$!)
So if I'm correct this implies that $|\langle a,x^\ast\rangle|<1$ for every $a\in A$ and $x^\ast\in A^\circ$ in the event that $A$ is bounded? But this surely fails for normed spaces: If $X$ is normed and $B$ is the closed unit ball in $X$ then $B^\circ$ would be the closed unit ball in $X^\ast$. If $\|x\| = 1$ for some $x\in X$ then by Hahn-Banach we can find a continuous linear functional $x^\ast$ with norm $1$ such that $\langle x,x^\ast\rangle = 1$.