I got stuck with the following problem while going through the proof of Lemma $1.9$(i) from the book 'Lectures on von Neumann Algebras' by Strătilă and Zsidó.
Let $\mathscr{B}$ be a Banach space and $\mathscr{B}^*$ be the dual of $\mathscr{B}$. Let $\mathscr{B}_*\subseteq \mathscr{B}^*$ be a norm-closed vector subspace such that $\mathscr{B}=(\mathscr{B}_*)^*$ (i.e. isometrically isomorphic) through the canonical bilinear form on $\mathscr{B}\times \mathscr{B}_*$. Consider the $\sigma (\mathscr{B};\mathscr{B}_*)$-topology on $\mathscr{B}$ which is defined by the family of semi-norms $\{p_{\varphi}:\varphi\in \mathscr{B}_*\}$, where $p_{\varphi}(x):=|\varphi (x)|$ for $x\in \mathscr{B}$. Let $\mathscr{M}\subseteq \mathscr{B}$ be a $\sigma (\mathscr{B};\mathscr{B}_*)$-closed vector subspace.
Problem: If $\mathscr{M}^{\circ}:=\{\varphi \in \mathscr{B}_*;\varphi |_{\mathscr{M}}=0\}$, then prove that $\mathscr{M}=\{x\in\mathscr{B};\varphi (x)=0 \text{ for any } \varphi\in\mathscr{M}^{\circ}\}$.
The authors call $\mathscr{M}^{\circ}$ the polar of $\mathscr{M}$ and then says that the conclusion follows from the bipolar theorem. But I did not find any reference for that 'bipolar theorem'. So could someone help me with a proper reference for that 'bipolar theorem' along with the solution. Thanks in advance.
Let $$ \mathscr{M}^{oo}=\{x\in\mathscr{B};\varphi (x)=0 \text{ for any } \varphi\in\mathscr{M}^{\circ}\}. $$ We have $\mathscr M\subset \mathscr M^{oo}$ by definition of $\mathscr M^o$. Suppose that there exists $z\in \mathscr M^{oo}\setminus \mathscr M$. As $\mathscr M$ is convex and $\sigma (\mathscr{B};\mathscr{B}_*)$-closed, by Hahn-Banach there exists $\varphi$, linear and $\sigma (\mathscr{B};\mathscr{B}_*)$-continuous, with $\varphi(z)=1$ and $\varphi|_{\mathscr M}=0$. The $\sigma (\mathscr{B};\mathscr{B}_*)$-continuous functionals on the dual of the locally compact space are precisely those in the predual (see, for instance, Theorem V.1.3 in Conway's A Course in Functional Analysis); so $\varphi\in\mathscr B_*$. Then $\varphi\in\mathscr M^o$; as $z\in\mathscr M^{oo}$, we would have $\varphi(z)=0$, a contradiction. So $\mathscr M^{oo}=\mathscr M$.