Space of Square Summable Mappings is L-2 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 equal to the $L^2$ space $\map {L^2} {I, \powerset I, \mu}$, where $\mu$ is the counting measure on $I$.

Proof

First, let us unpack the definition of the Lebesgue space $\map {\LL^2} {I, \powerset I, \mu}$:

$\ds \map {\LL^2} {I, \powerset I, \mu} = \set{ f: I \to \GF: f \in \map \MM { \powerset I }, \int \size f^2 \rd \mu < \infty}$

By Function Measurable with respect to Power Set, all mappings $f: I \to \GF$ are measurable.

By Integral over Counting Measure is Sum of Values, $\ds \int \size f^2 \rd \mu = \sum_{i \mathop \in I} \cmod{ \map f i }^2$.

Hence:

$\ds \map {\LL^2} {I, \powerset I, \mu} = \set{ f: I \to \GF: \sum_{i \mathop \in I} \cmod{ \map f i }^2 < \infty}$

which is exactly the definition of $\map {\ell^2} I$.

Next, the definition of $\map {L^2} {I, \powerset I, \mu}$ is:

$\map {L^2} \mu = \map {\LL^2} {\mu} / \sim_\mu$

where $\sim_\mu$ is the $\mu$-almost-everywhere equality relation.

Now let $f, g \in \map {\ell^2} I$ such that $\map \mu { \set{ f \ne g } } = 0$.

Then since $\mu$ is the counting measure, it follows that $\set{ f \ne g } = \varnothing$.

That is, $f = g$.

Therefore, $\sim_\mu$ is the trivial equivalence and $\map {\ell^2} I = \map {L^2} {I, \powerset I, \mu}$.

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: Example $1.7$