Let \(\mathcal X\) be a real vector space and let \(p\colon \mathcal X\to\RR\) be sublinear, that is, \(p(x+y) \leq p(x) + p(y)\) and \(p(\lambda x) = \lambda p(x)\) for all \(x,y\in \mathcal X\) and all \(\lambda \geq 0\). Let \(\mathcal M \subseteq \mathcal X\) be a linear subspace and \(f\colon\mathcal M\to\RR\) a linear functional with \(f \leq p\) on \(\mathcal M\). Then there exists a linear functional \(F\colon\mathcal X\to\RR\) with \(\ro{F}{\mathcal M} = f\) and \(F \leq p\) on \(\mathcal X\).
Let \(\mathcal X\) be a complex vector space and let \(p\colon\mathcal X\to\RR\) be a seminorm, that is, \(p(x+y) \leq p(x) + p(y)\) and \(p(\lambda x) = \abs{\lambda} p(x)\) for all \(x,y\in \mathcal X\) and all \(\lambda \in\CC\). Let \(\mathcal M \subseteq \mathcal X\) be a linear subspace and \(f\colon\mathcal M\to\CC\) a linear functional with \(\abs f \leq p\) on \(\mathcal M\). Then there exists a linear functional \(F\colon\mathcal X\to\CC\) with \(\ro{F}{\mathcal M} = f\) and \(\abs F \leq p\) on \(\mathcal X\) [@Fol13].
The proof extends \(f\) one dimension at a time and applies Zorn’s lemma.
The extension \(F\) need not be unique. On \(\mathcal X=\RR^2\) let \(p(x)\coloneqq\abs{x_1}+\abs{x_2}\), \(\mathcal M\coloneqq\RR\times\theset{0}\), and \(f(x_1,0)\coloneqq x_1\). For every \(c\in[-1,1]\), the functional \(F_c(x)\coloneqq x_1+cx_2\) extends \(f\) and satisfies \(F_c\leq p\).
Let \(\mathcal X\) be a normed vector space over \(\RR\) or \(\CC\) with dual space \(\dualof{\mathcal X}\) of continuous linear functionals and dual norm.
If \(\mathcal M \subseteq \mathcal X\) is a closed linear subspace and \(x\in\mathcal X\setminus\mathcal M\), put \(\delta \coloneqq \inf_{y\in \mathcal M}\norm{x-y}\). Then there is \(f \in \dualof{\mathcal X}\) with \(\norm{f} = 1\), \(\ro f {\mathcal M} = 0\), and \(f(x) = \delta\).
For each \(x\in\mathcal X\) with \(x\neq 0\) there is \(f\in \dualof{\mathcal X}\) with \(\norm f = 1\) and \(f(x) = \norm x\). In particular, the functionals in \(\dualof{\mathcal X}\) separate the points of \(\mathcal X\).
The map \(\mathcal X \injects \mathcal X^{\vee\vee}\), \(x \mapsto \hat x\) with \(\hat x(f) \coloneqq f(x)\) for \(f\in\dualof{\mathcal X}\), is a linear isometry.
[@Fol13]