Representation of weak*-continuous linear functionals on a subspace which is not necessarily weak* closed

Question

Let $X$ be a Banach space and let $X^{*}$ be its dual space. Let $M$ a subspace of $X^{*}$ on which a weak*-continuous linear functional $L$ is given. In the case where $M$ is weak*-closed, we know that there exists an element $x\in X/{M}_\perp$ which represents $L,$ i.e., $$L(m)=m(x)\qquad (m\in M),$$ where $M_\perp$ denotes the pre-annihilaor of $M$ in $X.$ I have the following question;

Is the conclusion above true even when $M$ is not necessarily weak*-closed?

Answer 1

You can extend $L$ uniquely (by weak* continuity) to the weak*-closure of $M$. And the preannihilator of $M$ coincides with the preannihilator of its weak* closure.