# Riesz Representation Theorem (Hilbert Spaces)/Examples/Space of Square Summable Mappings

## Example of Use of Riesz Representation Theorem (Hilbert Spaces)

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

Let $N \in \N$.

Let $L_N: \map {\ell^2} \N \to \GF$ be defined by:

$\map {L_N} {\sequence{ a_n } } := a_N$

Let $\delta_N \in \map {\ell^2} \N$ be given by:

$\forall n \in \N: \paren{ \delta_N }_n = \begin{cases} 1 & n = N \\ 0 & n \ne N \end{cases}$

Then for all $a \in \map {\ell^2} \N$:

$\map {L_N} a = \innerprod a {\delta_N}$

## Proof

By Space of Square Summable Mappings is Hilbert Space, $\map {\ell^2} \N$ is a Hilbert space.

Since for all $a \in \map {\ell^2} \N$:

 $\ds \cmod{ \map {L_n} a }$ $=$ $\ds \cmod{ a_N }$ $\ds$ $=$ $\ds \sqrt{ \cmod{ a_N }^2 }$ $\ds$ $\le$ $\ds \sqrt{ \sum_{n \mathop \in \N} \cmod{ a_n }^2 }$ $\ds$ $=$ $\ds \norm a$

it follows that $L_N$ is a bounded linear functional.

Hence the Riesz Representation Theorem (Hilbert Spaces) applies, so that there exists a unique $b \in \map {\ell^2} \N$ such that for all $a \in \map {\ell^2} \N$:

$\map {L_N} a = \innerprod a b$

Let us check that $\delta_N$ fulfils the claim:

 $\ds \innerprod a {\delta_N}$ $=$ $\ds \sum_{n \mathop \in \N} a_n \paren{ \delta_N }_n$ $\ds$ $=$ $\ds a_N$ since $\paren {\delta_N}_n = 0$ for $n \ne N$

The result follows.

$\blacksquare$