Equivalence of Definitions of Closed Linear Span

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Theorem

Let $H$ be a Hilbert space over $\Bbb F \in \set {\R, \C}$, and let $A \subseteq H$ be a subset.

The following definitions of the concept of closed linear span of $A$ are equivalent:

$(1): \quad \ds \vee A = \bigcap \Bbb M$, where $\Bbb M$ consists of all closed linear subspaces $M$ of $H$ with $A \subseteq M$
$(2): \quad \vee A$ is the smallest closed linear subspace $M$ of $H$ with $A \subseteq M$
$(3): \quad \ds \vee A = \map \cl {\set {\sum_{k \mathop = 1}^n \alpha_k f_k: n \in \N_{\ge 1}, \alpha_i \in \Bbb F, f_i \in A} }$, where $\cl$ denotes closure


Proof

Let the proposition $(1)$ hold:

Assume the closed linear subspace $M'$ contains the set $A$.

Then because $M' \in \Bbb M$:

$\vee A \subseteq M'$

The intersection of arbitrary family of subspaces is a subspace.

For suppose $\CC$ is a family of subspaces.

Denote $\bigcap \CC = \set {f\in H: \forall V \in \CC: \exists f \in V}$

If $f \in \bigcap \CC$, then for any $V \in \CC$, $f \in V$ there exists $a f \in V$ for $a \in \Bbb F$.

If $f, g \in \bigcap \CC$, we have for any $V \in \CC$:

$f + g \in V$

Therefore, $\vee A$ is a subspace.

It is closed, as intersection of arbitrary family of closed sets is closed.

The choice of $M'$ is arbitrary.

Hence, $(2)$ holds.

This is the proof $(1) \implies (2)$.


Next, assume $(2)$ holds.

Since $A \subseteq \vee A$:

$\vee A \in \Bbb M$

$\vee A$ is the smallest one in $\Bbb M$.

Hence:

$\ds \vee A = \bigcap \Bbb M$

Thus we have established the equivalence between $(1)$ and $(2)$.


Finally we come to $(3)$:

We prove that $\map \cl {\map \span A}$ is a subspace.

Let $f \in \map \cl {\map \span A}$.

$H$, the Hilbert space, is a metric space, which satisfies the first countability axiom.

For reason then one shall consider the collection of open balls which all are centred at a point and all have their length of radii be rational.

We then need the Sequence Lemma, which is:

Let $A$ be a subset of a topological space $X$.

If there is a sequence of points of $A$ converging to $x$, then $x \in \bar A$.

The converse holds if $X$ is first-countable.

$\Box$


So there is a sequence $\sequence {f_i}$ in $\map \span A$ that its limit is $f$.

By the continuity of the function of the multiplication of numbers of the field $\Bbb F$ and points in $H$, the sequence $\sequence {af_i}$ with $a\in \Bbb F$ converges to $af$.

Because all the terms of the sequence $\sequence {af_i}$ are points in $\map \span A$, we have $af\in \map \cl {\map \span A}$.


The proof regards the addition of two vectors is by a similar manner.

If $f \in \map \cl {\map \span A}$ and $g\in \map \cl {\map \span A}$, then a sequence $\sequence {f_i}$, converges to $f$, and a sequence $\sequence {g_i}$, converges to $g$, of $\map \span A$ are given.

By the continuity of the function of the addition of points in $H$, $\sequence {f_i + g_i}$ converges to $f+g$.

Because each terms of the sequence $f_i + g_i$ is in $\map \span A$, we have $f + g \in \map \cl {\map \span A}$.

Notice that in general, if $H'$ is a subspace, then $\map \cl {H'}$ is a subspace.


Finally we come to the proof of equivalence of $(2)$ and $(3)$:

$\map \cl {\map \span A}$, the closed linear subspace, contains $\map \span A$ and thus contains $A$.

For any closed linear subspace $M$ which contains $A$, $\map \span A \subseteq M$, since the linear span of $A$ is the smallest subspace that contains $A$.

Because $M$ is closed, $\map \cl {\map \span A} \subseteq M$.

$\map \cl {\map \span A}$ is the the smallest closed linear subspace $M$ of $H$ with $A \subseteq M$.

Because arbitrary intersection of closed sets is closed and arbitrary intersection of subspaces is a subspace, the smallestness is unique.

Hence $\vee A = \map \cl {\map \span A}$.

$\blacksquare$


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