- Corollary 1.8. Let $F \subset E$ be a linear subspace such that $\bar{F} \neq E$. Then there exists some $f \in E^{\star}, f \not \equiv 0$, such that $$ \langle f, x\rangle=0 \quad \forall x \in F . $$ Proof. Let $x_0 \in E$ with $x_0 \notin \bar{F}$. Using Theorem 1.7 with $A=\bar{F}$ and $B=\left\{x_0\right\}$, we find a closed hyperplane $[f=\alpha]$ that strictly separates $\bar{F}$ and $\left\{x_0\right\}$. Thus, we have $$ \langle f, x\rangle<\alpha<\left\langle f, x_0\right\rangle \quad \forall x \in F . $$ It follows that $\langle f, x\rangle=0 \quad \forall x \in F$, since $\lambda\langle f, x\rangle<\alpha$ for every $\lambda \in \mathbb{R}$.
2026-05-17 12:17:50.1779020270