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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.

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

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis Willse, Xander Henderson, Chris Custer, Isaac Browne, Namaste
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  • $\begingroup$ What exactly did you try? Please edit the question with the details. $\endgroup$ – Shaun Jul 9 '18 at 21:34
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Hint:

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 $X$.

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