A functional analysis text book contained the following exercise.
Let $V$ be an incomplete inner product space. Show there exists a maximal orthonormal system that is not complete. Hint: Use a closed hyperspace $F$ which has trivial orthogonal complement.
Here maximal refers to the inability to add more elements and complete refers to the linear span being dense.
I manage to prove the hint as follows.
Let $H$ be a completion of $V$, so $V$ is a dense subspace of $H$ which is complete. Let $x\in H\setminus V$. Then $x^{\perp\perp}=<x>$ in $H$, so if we let $F=x^\perp|_V$ then since $F$ is dense in $x^\perp$ we find $F$ has trivial orthogonal complement in $V$.
I tried to use this to solve the exercise as follows.
If $F$ has a complete orthonormal system, then it is maximal in $V$ as the orthogonal complement of $F$ is trivial. Since $x$ can be approximated in $V$ but not in $F$, we find it can not be a complete orthonormal system in $V$.
So I have two questions. The main question is how do I finish the proof? What if $F$ does not have a complete orthonormal system? I found this which shows that separable inner product spaces always have a complete orthonormal system, but inseparable inner product spaces may not. So at least I solved the exercise for separable inner product spaces, but the inseparable case remains unsolved.
The other question I have is less important, but the book has not mentioned the completion of normed or inner product spaces, so my solution for the hint feels like cheating. I wonder if someone can come up with a solution that does not require the use of a completion.
