Let $X$ be a TVS and $X'$ denotes the space of all continuous linear functionals on $X$. Let us denote the $weak^*$-topology on $X'$ by $\sigma(X',X).$
My question is this. Why does every equicontinuous subset $M$ of $X'$ is bounded with respect to $\sigma(X',X)?$
So far, I can show that if $M$ is an equicontinuous subset of $X'$ then $M\subseteq V^o$ for some open neighborhood of $0$ in $X$. Here, $V^o$ refers to the polar set of $V$. I don't have any idea on how to proceed...In fact I cannot imagine how the basic elements look like in $X'$ (in terms of polar sets, at least.) Thanks in advance.
So you have $M \subset V^\circ =\{ S\in X'; |S(v)| \leq 1 \forall v\in V\}$. Fix $x\in X$. You need to show the existence of a constant $c>0$ with $$ |S(x)| <= c \operatorname{ for all }S\in M.$$ But as $V$ is absorbing, $cx\in V$ for some $c>0$, so $|S(cx)| \leq 1$ for all $S\in M$. Hence you can conclude.