Let $D$ be a dense subspace of a separable Hilbert space $H$. Let $x \in D$ be a unit vector. Is it possible to construct an orthonormal basis of $H$ containing $x$ and contained in $D$?
Some thoughts: Let $\mathcal{A}$ be the set of all orthonormal sets contained in $D$ and contains $x$. ie if $A \in \mathcal A$ then $A \subset D$, $x \in A$ and $A$ is an orthonormal set. Give the partial order of set inclusion for $\mathcal A$. Then it is clear that every chain has an upper bound and thus Zorn's lemma guarantees the existence of a maximal element (say, $M$) in $\mathcal A$.
I would like to claim that $M$ is an orthonormal basis for $H$ which satisfies the conditions, but I am unable to prove this. Any thoughts are highly appreciated. Thanks.
Let $X = \{x_n \}$ be a countable dense subset of $D$, the existence of which is guaranteed by the separability of $H$. Now we apply the Gram-Schmidt orthogonalisation process, meaning we define $$y_0 = x_0 \\ y_1 = x_1 - P_{y_0}(x_1) \\ \dots \\ y_n = x_n - \sum_{k=0}^{n-1}P_{y_k}(x_n) $$ where $P_y$ is the projection onto $z \in H$ defined by $P_y(x) = \frac{\langle x, y \rangle}{\langle y, y \rangle} y$. Then we normalise $z_n = \frac{y_n}{\|y_n\|}$ if $y_n \neq 0$. Now it's pretty clear that $ X \subset \text{span}( \{ z_n \}), $ so the closure of $\text{span}(\{z_n\})$ will be equal to $H$ because $X$ is dense. It's also an orthonormal set, which isn't too hard to verify, so it will be an orthonormal base.