Existence of Orthonormal Bases for Inf. Dim. Separable Hilbert Spaces

# Existence of Orthonormal Bases for Infinite Dimensional Separable Hilbert Spaces

Recall from the Orthonormal Bases page that if $H$ is a Hilbert Space then an orthonormal basis for $H$ is an orthonormal sequence $(e_n)$ of $H$ such that every $h \in H$ can be uniquely written as:

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

The following theorem tells us that an orthonormal basis always exist when we're looking at infinite dimensional separable Hilbert spaces.

 Theorem 1: Every infinite dimensional separable Hilbert space has an orthonormal basis.
• Proof: Let $\mathcal F$ be the set of all orthonormal subsets of $H$. Then elements of $\mathcal F$ can be ordered by inclusion.
• Let $\mathcal E \subseteq \mathcal F$ be a chain in $\mathcal F$, that is, if $E, E' \in \mathcal E$ then either $E \subseteq E'$ or $E \supseteq E'$. We will show that $\mathcal E$ has an upperbound. Let:
(2)
\begin{align} \quad \tilde{E} = \bigcup_{E \in \mathcal E} E \end{align}
• Observe that $\tilde{E}$ is an orthonormal set. To see this, let $e, e' \in \tilde{E}$. Then there exists an $E_0 \in E$ such that $e, e' \in E_0$. So $\langle e, e' \rangle = 0$. Furthermore, $\| e \| = 1$ and $\| e' \| = 1$. So $\tilde{E}$ is an orthonormal subset of $H$ and clearly, $\tilde{E}$ is an upperbound for $\mathcal E$.
• So every chain in $\mathcal F$ has an upper bound. By Zorn's Lemma, $\mathcal F$ has a maximal element, call it $E^*$.
• Now since $H$ is separable, from the theorem on the Orthonormal Sets in Separable Inner Product Spaces page we have that $E^*$ is countable. Let $E^* = (e_n)$ be any enumeration of $E^*$. Then $(e_n)$ is such that for every $h \in H$ we have that $\displaystyle{h - \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n}$ is orthogonal to each $e_n$.
• So $\displaystyle{h - \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n = 0}$, otherwise we could create a larger orthonormal set consisting of $E^*$ with $\displaystyle{\frac{h - \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n}{\| h - \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n \|}}$. Thus, for each $h \in H$:
(3)
\begin{align} \quad h = \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n \end{align}
• So $E^* = (e_n)$ is an orthonormal basis for $H$. $\blacksquare$