# Does every infinite-dimensional inner product space have an orthonormal basis? [duplicate]

For a finite-dimensional inner product space $V$, since it has a finite basis, then we can do the Gram-Schimidt process to produce an orthonormal basis. However, the Gram-Schmidt process does not work with infinitely many vectors. So it is natural to ask, does every infinite-dimensional inner product space have an orthonormal basis? If the answer is yes, how to prove it?

PS: For "basis", I mean the Hamel basis.

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• @Eric in infinite-dimensional settings, there are two common notions of a basis. In a sense, the question amounts to whether you count infinite sums among the "linear combinations" of your basis elements. – Omnomnomnom Sep 12 '17 at 15:14
• @Eric the question is what does "spans $V$" mean to you? Do we include "infinite linear combinations" or not? – Omnomnomnom Sep 12 '17 at 15:19
• Not. I use the definition that the span of a set $S$ means the set of all linear combination of any finitely vectors in $S$. So there're no truly infinite linear combinations. – Eric Sep 12 '17 at 15:21
• @Eric In that case, you're asking about a Hamel basis (as opposed to a "Schauder basis"). – Omnomnomnom Sep 12 '17 at 15:23
• You need to reword your question. Right now it is confusing. Does the "any" in your question mean "one" or "all/every"? The title seems to suggest the former, but the question body seems to suggest the latter. As Chessantor's answer and Omnomnomnom's comment beneath it show, users have already been interpreting your question differently. – user1551 Sep 12 '17 at 17:06

You can define such a space from the ground up: Let $V$ be the vector space of finite (i.e. eventually always zero) linear combinations of a countable set $\{e_i | i \in \mathbb{N}\}$. Then define the inner product as $\langle \sum x_i e_i , \sum y_i e_i \rangle = \sum x_i \bar{y_i}$, where this sum always exists because eventually both the $x_i$ and the $y_i$ become $0$.
In this vector space, the basis vectors $e_i$ are orthonormal.
• I don't see how this answer the question. The question is whether given an arbitrary inner product space $V$, we necessarily have such a basis. – Omnomnomnom Sep 12 '17 at 15:45