Definitions
Given inner product space ${ (X,\langle \cdot,\cdot \rangle) }$,
orthonormal system is a subset ${S }$ of ${ X }$ s.t.
- (orthogonality) ${ \langle x,y \rangle = 0 }$ for all distinct ${ x,y \in S }$
- (normality) ${ \langle x,x \rangle =1 }$ for all ${ x \in S }$
maximal orthonormal system (MONS) is a orthonormal system of maximal with inclusion ${ \subseteq }$
orthonormal basis (ONB) is a orthonormal system s.t.
\[x = \sum_{y \in S} c_{y} y\]for some sequence ${ \left\{ c_{y} \right\} }$ of scalars of ${ X }$.
Gram-Schmidt process
Let
\[\left \{ u_{1}, u_{2}, u_{3} \dots \right \}\]be at most countable linearly independent susbset of inner product space ${ X }$. Define
\[\begin{align} w_{1} &= u_{1} \\ w_{n+1} &= u_{n+1}-\sum_{j=1}^{n} \left( u_{n+1}, \frac{w_{j}}{\lVert w_{j} \rVert} \right) \frac{w_{j}}{\lVert w_{j} \rVert} \end{align}\]and define
\[v_{j}:= \frac{w_{j}}{\lVert w_{j} \rVert}\], then the set ${ \left\{ v_{j} \right\} }$ is orthornomal system such that
\[\left\{ v_{j} \right\}_{j=1}^{n} \mbox{ spans the same subspace as } \left\{ u_{j} \right\}_{j=1}^{n}\]for all ${ n \in \mathbb{N} }$.
Separable Hilbert space
We can prove every separable Hilbert space has an ONB without AC.
Theorem If ${ X }$ is a separable Hilbert space, then ${ X }$ has a ONB.
proof) Since ${ X }$ is separable, there exists a countable subset
\[S = \left\{ x_{1},x_{2}, x_{3},\dots \right\}\]such that ${ S }$ is dense in ${ X }$. Let ${ S’ =S \setminus D }$ where
\[D = \left\{ x_{k} \in S : x_{k} \mbox{ is a linear combination of } x_{i} \mbox{ for } i < k \right\}\]Then ${ S’ }$ is a linearly independent set of ${ X }$.
If apply Gram-Schmidt process on ${ S’ }$, we get orthonormal system ${ S’’ }$. Since ${ S’’ }$ is dense in ${ X }$, ${ S’’ }$ is the desired ONB.
Hilbert space
Theorem Any Hilbert space has an ONB. Any MONS in a Hilbert space is an ONB.
pf) By the Zorn’s lemma, any inner product space has at least one MONS.
Suppose ${ { x_{i} }_{i \in I} }$ is an arbitrary MONS in a Hilbert space ${ H }$. By the Bessel’s inequality, there is at most a countable subset ${ I_{o} }$ of ${ I }$ s.t.
\[\langle x_{i},x\rangle =0 \mbox{ for any } i \in I\setminus I_{0}\]The completeness of ${ H }$ entails that
\(x_{o} = \sum_{i \in I} \langle x,x_{i} \rangle x_{i} = \sum_{i \in I_{o} } \langle x,x_{i}\rangle x_{i}\) On the other hand, ${ x=x_{o} }$ since
\[\langle x_{o}-x,x_{i} \rangle = \langle x_{o},x_{i} \rangle-\langle x,x_{i} \rangle =0\]for any ${ i \in I }$ and the maximality of ${ \left\{ x_{i} \right\} }$ gives ${ x_{o} -x = 0}$