Spanning Criterion of Normed Vector Space

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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 $z \in \vee A$ if and only if:

$\forall \ell \in X^\ast : \ell \restriction_A = 0 \implies \map \ell z = 0$

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


Proof

Necessary condition

Let $z \in \vee A$.

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

Recall, by Definition of $X^\ast$, 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::

\(\ds \map \ell {z - \sum_{i \mathop = 1}^n \alpha_i a_i}\) \(=\) \(\ds \map \ell z - \sum_{i \mathop = 1}^n \alpha_i \map \ell {a_i}\) Definition of Bounded Linear Functional
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \map \ell z\) $\map \ell {a_i} = 0$ by hypothesis


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:

\(\ds \size {\map \ell z}\) \(=\) \(\ds \size {\map \ell {z - \sum_{i \mathop = 1}^n \alpha_i a_i} }\) by $(2)$
\(\ds \) \(\le\) \(\ds C \norm {z - \sum_{i \mathop = 1}^n \alpha_i a_i}\) by $(1)$
\(\ds \) \(<\) \(\ds C \epsilon\) by $(3)$

By $\epsilon \to 0$, we obtain:

$\map \ell z = 0$

$\Box$


Sufficient condition

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

As $\vee A = \map \cl {\map \span A}$, we have:

$\ds d := \inf_{x \in \map \span A} \norm {z - x} > 0$

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 $\alpha \ne 0$:

\(\ds \norm {x + \alpha z}\) \(=\) \(\ds \size \alpha \norm {z - \paren {- \alpha ^{-1} x} }\)
\(\ds \) \(\ge\) \(\ds \size \alpha d\) as $-\alpha ^{-1} x \in \map \span A$
\(\ds \) \(=\) \(\ds \size {\map {\ell_0} {x + \alpha z} } d\)

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:

$\ell \restriction_A = {\ell_0} \restriction_A = 0$
$\map \ell z = \map {\ell_0} z = 1 \ne 0$

$\blacksquare$


Sources

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