Consequence of the Hahn-Banach Theorem
From wikipedia's page on the Hahn-Banach Theorem:
If $V$ is a normed vector space with linear subspace $U$ (not necessarily closed) and if $z$ is an element of $V$ not in the closure of $U$, then there exists a continuous linear map $ψ : V → K$ with $ψ(x) = 0$ for all $x$ in $U$, $ψ(z) = 1$, and $||ψ|| = dist(z, U)^{−1}.$
How is this a consequence of the Hahn-Banach Theorem?
Check that
$$p \colon x \mapsto \operatorname{dist}(x,U)$$
is a sublinear functional. (It's a seminorm, actually.)
Then apply the Hahn-Banach extension theorem to $\tilde{p} = \frac{1}{\operatorname{dist}(z,U)}\cdot p$, and the linear map $t\cdot z \mapsto t$ from the one-dimensional subspace spanned by $z$ to the scalar field.
Alternatively, let $W = z - \overline{U}$, and $C = B_{\operatorname{dist}(z,U)}(0)$. Then $W$ and $C$ are disjoint convex sets, and $C$ is open and balanced. By one of the Hahn-Banach separation theorems, there is a continuous linear functional $\psi_0 \colon V \to \mathbb{R}$ with $\lvert \psi_0(x)\rvert < 1$ for all $x\in C$ and $\psi_0(y) = 1$ for all $y \in W$. These conditions show $\lVert \psi_0\rVert = \operatorname{dist}(z,U)^{-1}$ and $\psi_0(z) = 1$. If the scalar field is $\mathbb{R}$, let $\psi = \psi_0$, if the scalar field is $\mathbb{C}$, let $\psi$ the continuous linear form with real part $\psi_0$, that is, $\psi(x) = \psi_0(x) - i \psi_0(ix)$ for all $x\in V$.
Hints:
- Let $d:= \inf \{||x-z||: x \in U\} (= dist(z,U))$. Then $d>0$.
Let $E$ be the linear span of $ \{z\} \cup U$ and define $h:E \to K$ by
$$h(\alpha z+u)= \alpha d.$$
Then $h(u)=0$ for all $u \in U$ and $h(z)=d$.
Now show that $h$ is continuous and linear and that $||h||=1$
By the the Hahn-Banach Theorem there is a continuous linear map $f:V \to K$ with
$f(u)=0$ for all $u \in U$, $f(z)=d$ and $||f||=1$.
- Now take $ \psi:=\frac{1}{d}f$