A subset $\( S \)$ of an inner product space $\( V \)$ is called orthonormal provided $\( \mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1 \)$ and $\( x\perp y \)$ for all $\( x \)$ and $\( y \)$ in $\( S \)$.

Let $\( S= \mathopen{}\left\{\, {e}_{1}, \dotsc, {e}_{n}\,\right\}\mathclose{} \)$ be an orthonormal set in an inner product space $\( V \)$. For complex numbers $\( {α}_{1} \)$, …, $\( {α}_{n} \)$ and $\( {β}_{1} \)$, …, $\( {β}_{n} \)$, $$\[ \mathopen{}\left\langle{}\sum_{i=1}^{n}{}{α}_{i}{e}_{i}, \sum_{j=1}^{n}{}{β}_{j}{e}_{j}\right\rangle\mathclose{}= \sum_{i=1}^{n}{}{α}_{i}\overline{{β}_{i}} \text{.} \]$$

Let $\( S= \mathopen{}\left\{\, {e}_{1}, \dotsc, {e}_{n}\,\right\}\mathclose{} \)$ be an orthonormal set in an inner product space $\( V \)$. Let $\( W= \operatorname{span}\mathopen{}\left( S\right)\mathclose{} \)$. Define a linear map $\( Q : V \to W \)$ by $$\[ Q\mathopen{}\left( x\right)\mathclose{}= \sum_{i=1}^{n}{}\mathopen{}\left\langle{}x, {e}_{i}\right\rangle\mathclose{}{e}_{i} \text{.} \]$$ Then, for all $\( x\in V \)$, $\( x-Q\mathopen{}\left( x\right)\mathclose{}\in {W}^{\perp} \)$, $\( \mathopen{}\left\lVert{}Q\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{} \)$, $\( \mathopen{}\left\lVert{}x-Q\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{} \)$, and $\( Q\mathopen{}\left( x\right)\mathclose{} \)$ is the closest point in $\( W \)$ to $\( x \)$.

Let $\( {x}_{1} \)$, $\( {x}_{2} \)$, … be a linearly independent sequence in an inner product space $\( V \)$. Then there exists an orthonormal sequence $\( {e}_{1} \)$, $\( {e}_{2} \)$, …, in $\( V \)$ such that $\( \operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {x}_{1}, {x}_{2}, \dotsc\,\right\}\mathclose{}\right)\mathclose{}= \operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}\right)\mathclose{} \)$.

Let $\( {e}_{1} \)$, $\( {e}_{2} \)$, … be an orthonormal sequence in an inner product space $\( V \)$. Then $$\[ \sum_{i=1}^{\infty}{}{\mathopen{}\left\lvert{}\mathopen{}\left\langle{}x, {e}_{i}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2}\leq {\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2} \]$$ for all $\( x \)$ in $\( V \)$. If $\( V \)$ is complete, then $\( \sum_{i=1}^{\infty}{}\mathopen{}\left\langle{}x, {e}_{i}\right\rangle\mathclose{}{e}_{i} \)$ converges and sums to the orthogonal projection of $\( x \)$ on $\( \overline{\operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}\right)\mathclose{}} \)$.

Let $\( H \)$ be a separable infinite-dimensional Hilbert space. Then there exists an orthonormal set $\( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{} \)$ with $\( \overline{\operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}\right)\mathclose{}}= H \)$. Further, $$\[ x= \sum_{i=1}^{\infty}{}\mathopen{}\left\langle{}x, {e}_{i}\right\rangle\mathclose{}{e}_{i} \]$$ for all $\( x\in H \)$ and $$\[ \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= \sum_{i=1}^{\infty}{}\mathopen{}\left\langle{}x, {e}_{i}\right\rangle\mathclose{}\mathopen{}\left\langle{}{e}_{i}, y\right\rangle\mathclose{} \]$$ for all $\( x\in H \)$ and $\( y\in H \)$.

Let $\( H \)$ be a separable infinite-dimensional Hilbert space. An orthonormal set $\( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{} \)$ such that $\( \overline{\operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}\right)\mathclose{}}= H \)$ is an orthonormal basis of $\( H \)$.

An isomorphism of Hilbert spaces $\( H\simeq K \)$ is an inner-product-preserving linear surjection.

If $\( H \)$ is a separable, infinite dimensional Hilbert space with orthonormal basis $\( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{} \)$, then $\( U : H \to \mathrm{l}^{0} \)$ defined by $\( U\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left(\mathopen{}\left\langle{}x, {e}_{n}\right\rangle\mathclose{}\right)\mathclose{}_{n=1}^{\infty} \)$ is an isomorphism.

Let's look at the sine series in $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$. Consider $\( {e}_{n}\mathopen{}\left( t\right)\mathclose{}= \sqrt{2}\sin\mathopen{}\left( n\mathrm{\pi}t\right)\mathclose{} \)$ for $\( n\in \mathbb{N} \)$. From calculus its easy to see that $\( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{} \)$ is orthonormal. We claim that it is in fact dense in $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$. Take $\( f\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ such that $\( \mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}= 0 \)$ for all $\( n \)$. Extend $\( f \)$ to $\( \mathopen{}\left[{-}1, 1\right]\mathclose{} \)$ by $\( f\mathopen{}\left( {-}t\right)\mathclose{}= {-}f\mathopen{}\left( t\right)\mathclose{} \)$ for $\( 0\lt t\lt 1 \)$. Consider also the function $\( g \)$ defined (almost everywhere) on the unit circle $\( \mathbb{T}= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, \mathopen{}\left\lvert{}z\right\rvert\mathclose{}= 1\,\right\}\mathclose{} \)$ by $\( g\mathopen{}\left( {\mathrm{e}}^{\mathrm{\pi}\mathrm{i}θ}\right)\mathclose{}= f\mathopen{}\left( θ\right)\mathclose{} \)$. Notice $$\[ \int _{\mathbb{T}}{}g\mathopen{}\left( z\right)\mathclose{}{z}^{n}\,\mathrm{d}z= \int _{{-}1}^{1}{}f\mathopen{}\left( θ\right)\mathclose{}{\mathrm{e}}^{\mathrm{\pi}\mathrm{i}nθ}\,\mathrm{d}θ= \int _{{-}1}^{1}{} f\mathopen{}\left( θ\right)\mathclose{}\mathopen{}\left(\cos\mathopen{}\left( n\mathrm{\pi}θ\right)\mathclose{}+\mathrm{i}\sin\mathopen{}\left( n\mathrm{\pi}θ\right)\mathclose{}\right)\mathclose{} \,\mathrm{d}θ= 0 \]$$ because $\( f \)$ is an odd function and $\( \mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}= 0 \)$ for all $\( n \)$.

So we have that $\( \int _{\mathbb{T}}{}g\mathopen{}\left( z\right)\mathclose{}φ\mathopen{}\left( z\right)\mathclose{}\,\mathrm{d}z= 0 \)$ for all $\( φ\in \mathrm{C}\mathopen{}\left( \mathbb{T}\right)\mathclose{} \)$ because polynomials in $\( z \)$ and $\( {z}^{-1} \)$ are uniformly dense in $\( \mathrm{C}\mathopen{}\left( \mathbb{T}\right)\mathclose{} \)$ by the Stone-Weierstrass Theorem and $\( g= g\mathbf{1} \)$ in $\( \mathrm{L}^{\mathrm{1}}\mathopen{}\left( \mathbb{T}\right)\mathclose{} \)$. If a sequence $\( \mathopen{}\left({φ}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ converges to $\( φ \)$ uniformly, then $$\[ \mathopen{}\left\lvert{}\int _{\mathbb{T}}{}g\mathopen{}\left( z\right)\mathclose{}{φ}_{n}\mathopen{}\left( z\right)\mathclose{}\,\mathrm{d}z-\int _{\mathbb{T}}{}g\mathopen{}\left( z\right)\mathclose{}φ\mathopen{}\left( z\right)\mathclose{}\,\mathrm{d}z\right\rvert\mathclose{}\leq g\mathopen{}\left\lVert{}{φ}_{n}-φ\right\rVert_\infty\mathclose{} \text{.} \]$$

For any arc $\( A \)$ in $\( \mathbb{T} \)$, $\( \chi_{A} \)$ is the $\( \mathrm{L}^{\mathrm{2}} \)$-limit of a sequence of continuous functions on $\( \mathbb{T} \)$, so $\( \int _{\mathbb{T}}{}g\mathopen{}\left( z\right)\mathclose{}\chi_{A}\mathopen{}\left( z\right)\mathclose{}\,\mathrm{d}z= 0 \)$. For every measurable subset $\( S\subseteq \mathbb{T} \)$ we get a sequence $\( \mathopen{}\left({U}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ of countable unions of open arcs with $\( A \)$ and the measure of $\( {U}_{n}\setminus S \)$ converges to zero by the regularity of Lebesque measure. So $\( \int _{{U}_{n}}{}g\mathopen{}\left( z\right)\mathclose{}\,\mathrm{d}z= 0 \)$ for all $\( n \)$ implies $\( \int _{S}{}g\mathopen{}\left( z\right)\mathclose{}\,\mathrm{d}z= 0 \)$. Thus if $\( g \)$ is zero almost everywhere, then $\( f \)$ is zero almost everywhere.

We have shown that $\( f= \sum_{n=1}^{\infty}{}\mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}{e}_{n} \)$, and thus $\( {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}= \sum_{n=1}^{\infty}{}{\mathopen{}\left\lvert{}\mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \)$ for all $\( f\in \mathrm{L}^{\mathrm{2}} \)$. For any choice of $\( f \)$ that is readily integrable against each $\( {e}_{n} \)$, we get an amusing series summation. For instance, $\( f\equiv \mathbf{1} \)$ gives $$\[ \mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}= \sqrt{2}\int _{0}^{1}{}\sin\mathopen{}\left( n\mathrm{\pi}t\right)\mathclose{}\,\mathrm{d}t= \begin{cases}\frac{2\sqrt{2}}{\mathrm{\pi}n}, & n|\text{odd}; \\ 0, & n|\text{even.}\end{cases} \]$$ Then $\( {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}= \sum_{n=1}^{\infty}{}{\mathopen{}\left\lvert{}\mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \)$, so $$\[ 1= \frac{8}{{\mathrm{\pi}}^{2}}\sum_{n|\text{odd}}{}\frac{1}{{n}^{2}} \]$$ implies $$\[ \sum_{n|\text{odd}}{}\frac{1}{{n}^{2}} = \frac{{\mathrm{\pi}}^{2}}{8} \]$$and $$\[ \sum_{n=1}^{\infty}{}\frac{1}{{n}^{2}} = \frac{4}{3}\mathopen{}\left(\frac{{\mathrm{\pi}}^{2}}{8}\right)\mathclose{}= \frac{{\mathrm{\pi}}^{2}}{6} \text{.} \]$$