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Given an infinite dimensional inner product space $(V,\langle \cdot,\cdot \rangle)$, with a countable Hamel basis, is it always possible to perform the Gram--Schmidt process and produce an orthonormal basis for $(V,\langle \cdot,\cdot \rangle)$? (To be precise, by orthonormal basis I mean a Hamel basis $\{e_i\}_{i \in \mathbb{N}}$ such that $\langle e_i,e_j \rangle) = \delta_{ij}$, for all $i,j$?

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  • $\begingroup$ I guess you mean Schauder basis. And yes, it works. $\endgroup$
    – Berci
    Feb 27, 2020 at 23:42
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    $\begingroup$ I think the OP means Hamel basis (i.e., basis in the sense of linear algebra - a linearly independent spanning set). And yes it works: just keep extending an orthonormal bsis for the span of the first $n$ elements of the Hamel basis to an orthonormal basis for the span of the first $n+1$ elements. $\endgroup$
    – Rob Arthan
    Feb 27, 2020 at 23:44
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    $\begingroup$ @copper.hat Of course, it can. For example the space of all sequences with only finitely many non-zero terms. $\endgroup$
    – Jochen
    Feb 28, 2020 at 8:38
  • $\begingroup$ @Jochen: You are correct, I was thinking F-space which needs completeness. $\endgroup$
    – copper.hat
    Feb 28, 2020 at 14:25

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Yes and this is quite straightforward. Let $(v_n)$ be a Hamel basis. Having found an orthonormal basis $e_1,e_2,...,e_n$ for the span of $v_1,v_2,..,v_n$ we can always find coefficients $c_j$ such that $e_{n+1}=\sum c_je_j+c_{n+1} v_{n+1}$ has norm $1$ and is orthogonal to $e_i: i \leq n$.

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  • $\begingroup$ You are correct, I was thinking F-space which needs completeness. $\endgroup$
    – copper.hat
    Feb 28, 2020 at 14:25

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