# Infinite dimensional Gram-Schmidt

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 $$i,j$$?

• I guess you mean Schauder basis. And yes, it works. Feb 27, 2020 at 23:42
• 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. Feb 27, 2020 at 23:44
• @copper.hat Of course, it can. For example the space of all sequences with only finitely many non-zero terms. Feb 28, 2020 at 8:38
• @Jochen: You are correct, I was thinking F-space which needs completeness. Feb 28, 2020 at 14:25

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 $$e_i: i \leq n$$.