Space of Square Summable Mappings is Vector Space
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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.
$\blacksquare$