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View $C([0,1])$ inside the complex Lp space $L^{\infty }((0,1))$. Fix $x_0\in [0,1]$. If a unit function $\psi$ satisfies $\Vert g+\psi {\Vert }_{\infty }\ge |g(x_0)|+1$ for every continuous function $g$, then the possible values at $\psi$ of norm-one extensions of point evaluation at $x_0$ form exactly the closed complex unit disc.

The estimate first excludes $\psi \in C([0,1])$. On the direct sum $C([0,1])\oplus ⟨\psi ⟩$, prescribe $L_a(g+c\psi )=g(x_0)+ca$ for $|a|\le 1$. Scaling the estimate gives $\Vert g+c\psi {\Vert }_{\infty }\ge |g(x_0)|+|c|\ge |L_a(g+c\psi )|$, and $L_a(1)=1$ proves operator norm one. The complex Hahn-Banach theorem extends $L_a$ to the whole Lp space. No value outside the disc is possible because $\Vert \psi {\Vert }_{\infty }=1$.

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.

Write $\psi ={\psi }_{x_0}$; its value at $x_0$, if that point is in $(0,1)$, may be assigned arbitrarily because Lp spaces identify functions equal almost everywhere. Its L-infinity norm is one.

The supplied estimate shows that $\psi$ cannot agree almost everywhere with any continuous function $q$: choosing $g=-q$ would give $0\ge |q(x_0)|+1$. Henceis a direct sum, so every element has a unique expression $g+c\psi$.

For any complex $a$ with $|a|\le 1$, define the linear functionalIf $c\ne 0$, the supplied estimate, applied to $g/c$, and absolute homogeneity of a norm giveFor $c=0$, the same bound follows from point evaluation. Thus $\Vert L_a\Vert \le 1$, and $L_a(1)=1$ makes its operator norm exactly one. The complex Hahn-Banach theorem extends $L_a$ to a norm-one linear functional on all of $X$, with value $a$ at $\psi$.

Conversely every norm-one extension satisfies $|{\ell }_{x_0}(\psi )|\le \Vert \psi {\Vert }_{\infty }=1$. We have therefore determined all possible values:In particular, taking $a=0$ and $a=1$ proves nonuniqueness of norm-preserving point evaluation extensions.

For a continuous function $g$ on $[0,1]$, the L-infinity norm of its restriction to $(0,1)$ equals its supremum norm on $[0,1]$. Indeed, values arbitrarily close to a maximum occur on an interval of positive Lebesgue measure, even if the maximum lies at an endpoint. Also, continuous functions equal almost everywhere on $(0,1)$ are equal everywhere on $[0,1]$. Thus restriction embeds $C([0,1])$ isometrically as a linear subspace of $X$.

The point evaluation functional $E_{x_0}(g)=g(x_0)$ is complex-linear and satisfies $|E_{x_0}(g)|\le \Vert g{\Vert }_{\infty }$. The constant function one attains equality, so $\Vert E_{x_0}\Vert =1$. The complex Hahn-Banach theorem extends this bounded linear functional from that linear subspace to $X$, preserving its operator norm. Therefore