Hilbert Bases (Orthonormal Bases) for Hilbert Spaces

# Hilbert Bases (Orthonormal Bases) for Hilbert Spaces

Recall from the [[[Convergence Criterion for Series in Hilbert Spaces page that if $H$ is a Hilbert space and $(x_n)_{n=1}^{\infty}$ is an orthonormal sequence of points in $H$ then for every $y \in H$ the series:

(1)\begin{align} \quad \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n \end{align}

converges to some $z \in H$ and:

(2)\begin{align} \quad z - y \perp \{ x_1, x_2, ... \} \end{align}

We will now define a new type of basis and use the result above to give us a criterion for when such an orthonormal sequence of points in a Hilbert space is that particular type of basis.

Definition: Let $H$ be a Hilbert space. A Hilbert Basis or simply, Orthonormal Basis for $H$ is an orthonormal sequence $(x_n)_{n=1}^{\infty}$ of points in $H$ such that for every $y \in H$, $\displaystyle{y = \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$. |

*Observe that in general, a Hilbert basis need not be a Hamel basis.*

The following theorem gives us a criterion for determining when an orthonormal sequence of points in a Hilbert space is a Hilbert basis.

Theorem 1: Let $H$ be a Hilbert space and let $(x_n)_{n=1}^{\infty}$ be an orthonormal sequence of points in $H$. Then $(x_n)_{n=1}^{\infty}$ is a Hilbert basis of $H$ if and only if $\{ x_1, x_2, ... \}^{\perp} = \{ 0 \}$. |

**Proof:**$\Rightarrow$ Let $(x_n)_{n=1}^{\infty}$ be a Hilbert basis of $H$. Then for every $y \in H$ we have that:

\begin{align} \quad y = \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n \end{align}

- We want to show that the only vector in $H$ orthogonal to every $x_1, x_2, ...$ is $0$. So, let $y \in H$ and suppose that $\langle y, x_m \rangle = 0$ for every $m \in \mathbb{N}$. Then:

\begin{align} \quad y = \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n = \sum_{n=1}^{\infty} 0x_n = 0 \end{align}

- So $y = 0$ and hence:

\begin{align} \quad \{ x_1, x_2, ... \}^{\perp} = \{ 0 \} \quad \blacksquare \end{align}

- $\Leftarrow$ Suppose that $\{ x_1, x_2, ... \}^{\perp} = \{ 0 \}$. Let $y \in H$. Then by the theorem referenced on the top of this page the series $\displaystyle{\sum_{n=1}^{\infty} \langle y, x_n \rangle x_n}$ converges in $H$, and:

\begin{align} \quad y - \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n \perp \{ x_1, x_2, ... \} \end{align}

- But then by hypothesis we must have that:

\begin{align} \quad y - \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n = 0 \quad \Leftrightarrow \quad y = \sum_{n=1}^{\infty} \langle y, x_n \rangle x_n \end{align}

- Since this holds for every $y \in H$ we conclude that $(x_n)_{n=1}^{\infty}$ is a Hilbert basis for $H$. $\blacksquare$