Spanning Criterion of Normed Vector Space

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.

Let $X^\ast$ be the vector space of bounded linear functionals on $X$.

Let $A \subseteq X$ be a subset.

Let $\vee A$ be the closed linear span of $A$, i.e. the closure of the linear span of $A$.

Then $z \in \vee A$ :
 * $\forall \ell \in X^\ast : \ell \restriction_A = 0 \implies \map \ell z = 0$

where $\ell \restriction_A$ denotes the restriction of $\ell$ to $A$.

Necessary condition
Let $z \in \vee A$.

Let $\ell \in X^\ast$ such that $\ell \restriction_A = 0$.

Recall, by, there is a $C > 0$ such that:
 * $(1):\quad \forall x \in X : \size {\map \ell x} \le C \norm x$

On the other hand::

Let $\epsilon > 0$.

Then there exist:
 * $n \in \Z_{>0}$
 * $\alpha_1, \ldots, \alpha_n \in \GF$
 * $a_1, \ldots, a_n \in A$

such that:
 * $\ds (3):\quad \norm {z - \sum_{i \mathop = 1}^n \alpha_i a_i} < \epsilon$.

Therefore:

By $\epsilon \to 0$, we obtain:
 * $\map \ell z = 0$

Sufficient condition
Let $z \not \in \vee A$.

Then:
 * $d := \inf_{x \in \vee A} \norm {z - x} > 0$

Consider the linear subspace:
 * $V := \map \span A + \GF z$

Define the linear functional $ \ell_0 : V \to \GF$ by:
 * $\map {\ell_0} {x + \alpha z} = \alpha$

where $x \in \map \span A$ and $\alpha \in \GF$.

$\ell_0$ is well-defined, since if:
 * $x + \alpha z = x' + \alpha' z$

then:
 * $\paren {\alpha - \alpha'} z = x' - x \in \map \span A \cap \GF z = \set 0$

Observe, if $\alpha \ne 0$:

If $\alpha = 0$, then:
 * $\map {\ell_0} {x + \alpha z} = 0$

Thus:
 * $\forall v \in V : \size { \map {\ell_0} v} \le d^{-1} \norm v$

By Hahn-Banach Theorem, there exists an $\ell \in X^\ast$ such that:
 * $\forall x \in X : \size { \map \ell v} \le d^{-1} \norm x$
 * $\ell \restriction_V = \ell_0$

In particular:
 * $\ell \restriction_A = {\ell_0} \restriction_A = 0$
 * $\map \ell z = \map {\ell_0} z = 1 \ne 0$