# orthonormal system which is “complete” but not “closed” [closed]

Let $V$ be an inner product space and $(e_n)_{n=1}^{\infty}$ be an orthonormal system.
We call it complete if $\left \langle v,e_n \right \rangle=0$ for all $n$ implies $v=0$;
and closed if $v=\sum_{n=1}^{\infty}\left \langle v,e_n \right \rangle e_n$ for every $v\in V$.

We've proved in class that a closed system is complete (that's one row), and that in a Hilbert space, a complete system is closed (the contrary holds).

My question is: can you describe an example for a complete, yet not closed system (in a non-complete inner product space)?
I've tried playing with some sequences spaces and functions spaces but nothing seemed to work.

## closed as off-topic by Travis Willse, Xander Henderson, Chris Custer, Isaac Browne, NamasteJul 11 '18 at 23:56

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• What exactly did you try? Please edit the question with the details. – Shaun Jul 9 '18 at 21:34

Let $H$ be a separable Hilbert space with an orthonormal basis $(e_n)_{n=1}^\infty$. Define $e_0 = \sum_{n=1}^\infty \frac1{n}e_n \in H$.
Let $X = \operatorname{span}\{e_0, e_2, e_3, \ldots\}$. Notice that $X$ is not complete.
Show that the orthonormal sequence $(e_n)_{n=2}^\infty$ is complete in $X$, but it is not closed in $X$.