Let $V$ be an inner product space. Let $W$ be the Hilbert space obtained as the completion of $V$. Is there a complete orthonormal basis of $V$ which is still complete in $W$? This is true if we assume that $V$ is separable (Schumidt's method), but I don't know if this is true or not in general.
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2The answer below answers your question for when it is known that $V$ has an orthonormal basis (orthonormal set with dense linear combinations). Now, it might happen that $V$ doesn't have such a set. Here there is a simple construction of such a $V$, or here an stronger construction without using choice. – Pp.. Feb 07 '15 at 01:13
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What is "Schumidt's method"? – Nov 11 '15 at 15:49
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A complete orthonormal basis is one that cannot be extended to a larger orthonormal basis. A complete orthonormal basis of an inner product space is usually not a Hamel basis (except in the finite-dimensional case), i.e. not every vector in the space is a linear combination of only finitely many members of the basis. If one wants to speak of a linear combination of infinitely many vectors, one must understand what it means for such a series to converge, and for that the inner product is needed. The distance between the $n$th partial sum and the limit is a sequence of real numbers that must approach $0$. Without the inner product one cannot speak of distance. And the idea of convergence is what brings us to the notion of "completion" of the space. The completion contains only points that can be approached from within the space that is getting completed. Therefore one adds no points that cannot be expressed as linear combinations of infinitely many points that were already there, in the space getting completed.
Bottom line: Every complete orthonoral basis of the space getting completed is a complete orthonormal basis of the completion.
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On the other hand, it might be that there is no orthonormal basis of $V$. – Pp.. Feb 07 '15 at 01:03