Let $M\subset X$ be a subspace and define the polar set $M^{\circ}:=\{x^{\ast}\in X^{\ast}:|\langle u,x^{\ast}\rangle|\le 1\forall u\in M\}$ and $M^{\perp}:=\{x^{\ast}\in X^{\ast}:\langle u,x^{\ast}\rangle =0\forall u\in M\}$
I want to show $M^{\circ}=M^{\perp}$.
The problem is that in every piece of literature I have found, this is stated as a trivial result and so the proof is usually omitted.
But I am essentially trying to show $|\langle u,x^{\ast}\rangle|\le 1\iff\langle u,x^{\ast}\rangle=0$, for all $u\in U$.
i.e. $|x^{\ast}(u)|\le 1\iff x^{\ast}(u)=0$.
And unless I am missing something of the operations of the inner product within the dual space, it seems rather nonsensical (at first glance, at least).
I write only the forward direction. The inverse is trivially true. Assume that $|x^*(u)|=\alpha\in (0,1)$ for some $u\in M$. Because $M$ is a subspace, then $\lambda u\in M$ for each $\lambda\in\mathbb R$. Therefore you must have $|x^*(\lambda u)|=|\lambda|\cdot|x^*(u)|=|\lambda|\cdot\alpha<1$ for all $\lambda\in\mathbb R$. But this is impossible, unless $x^*(u)=0,\forall u\in M$