 Solution

The complex Hahn-Banach theorem extends every bounded complex-linear functional on a subspace to the whole normed space with exactly the same norm. Let $d=\mathrm{dist}(x_0,M)>0$ and define on $M+\mathbb{C}{x}_0$The representation is unique. The distance inequality $\Vert y+\lambda x_0\Vert \ge |\lambda |d$ gives $\Vert g_0\Vert \le 1$. Taking $y=-m$ with $\Vert x_0-m\Vert \downarrow d$ proves the reverse norm bound. Extend by the Hahn-Banach theorem to $g\in X^{\prime }$; then $g{|}_M=0$, $\Vert g\Vert =1$, and $g(x_0)=d\ne 0$. The labels (a)-(c) specify these simultaneous properties of this one functional.

Let $Y=\overline{\mathrm{span}F}$. Every functional vanishing on $F$ vanishes on $Y$ by linearity and continuity, so $Y\subseteq (F^{\circ }{)}_{\circ }$. If $x\notin Y$, the separation just proved gives a bounded functional zero on $Y$ and nonzero at $x$. It belongs to $F^{\circ }$ and excludes $x$ from its preannihilator. ThereforeFor the dual-side counterexample take $X={\ell }^1$, $X^{\prime }={\ell }^{\infty }$, and $G=c_0$. It is norm closed, since a uniform limit of sequences tending to zero still tends to zero. Every coordinate unit vector belongs to $G$, so the only $x\in {\ell }^1$ annihilated by every element of $G$ has all coordinates zero. Thus $G_{\circ }=\{0\}$ andThe dual-side double annihilator corresponds to weak-star closure, which need not equal norm closure.