Space of Square Summable Mappings is Vector Space

Theorem

Let $\GF$ be a subfield of $\C$.

Let $I$ be a set.

Let $\map {\ell^2} I$ be the space of square summable mappings over $I$.

Then $\map {\ell^2} I$ is a vector space.

Proof

By definition, $\map {\ell^2} I$ is a subset of the vector space $\GF^I$ of all mappings $f: I \to \GF$.

Let us apply the One-Step Vector Subspace Test.

Thus, let $f, g \in \map {\ell^2} I$ and $\lambda \in \GF$.

Then we must show that $f + \lambda g: I \to \GF$ is square summable.

First, note that:

$\set{ i \in I: \map f i + \lambda \map g i \ne 0 } \subseteq \set{ i \in I: \map f i \ne 0 } \cup \set{ i \in I: \map g i \ne 0 }$

By Finite Union of Countable Sets is Countable and Subset of Countable Set is Countable, it follows that:

$\set{ i \in I: \map f i + \lambda \map g i \ne 0 }$

is countable.

Next:

\(\ds \sum_{i \mathop \in I} \cmod{ \map f i + \lambda \map g i }^2\)

\(\le\)

\(\ds \sum_{i \mathop \in I} \cmod{ \map f i }^2 + \cmod{\lambda}^2 \cmod{ \map g i }^2\)

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop \in I} \cmod{ \map f i }^2 + \cmod{\lambda}^2 \sum_{i \mathop \in I} \cmod{ \map g i }^2\)

\(\ds \)

\(<\)

\(\ds \infty\)

so that $f + \lambda g$ is square summable.

Hence the result by the One-Step Vector Subspace Test.

$\blacksquare$