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I read that an orthogonal (Hamel) basis $(e_i)$ of a Hilbert space is always a Schauder basis. I can see why; if $\sum \alpha_i e_i=0$ then taking the inner product with each $e_i$ gives $\alpha_i =0$.

I am confused about the following: take a Hilbert space and an orthogonal Schauder basis $(e_i)$ that is not a basis, e.g. the standard Schauder basis of $\ell^2$. Then $(e_i)$ is in particular linearly independent and thus strictly contained in an orthogonal Hamel basis $(f_i)$, which is a Schauder basis.

But a Schauder basis cannot be strictly contained in a Schauder basis!? - it would contradict the strong linear independence ('strong' meaning we allow countable sums): if $f$ is a member of the larger basis, we can write it as a series in the $e_i$ and this gives a nontrivial series in the $f_i$ that sums to zero.

Bart Michels
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2 Answers2

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Except in the finite-dimensional case there's no such thing as an orthogonal Hamel basis. Your statement ($*$) "and thus strictly contained in an orthogonal Hamel basis" is simply not so.

Indeed, say $H$ is an infinite-dimensional Hilbert space and $B$ is a Hamel basis. Suppose $B$ is orthonormal. Say $e_1,e_2,\dots\in B$. Then the series $x=\sum e_n/n$ converges in $H$. Since $B$ is a Hamel basis, $x$ is also a finite linear combination of elements of $B$, and as you note considering $<x,e_n>$ this shows that there exists $n$ such that $1/n=0$.

David C. Ullrich
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You cannot extend $\{e_i\}$ to a bigger orthogonal basis; it is already a maximal orthonormal set.I don't understand what you mean by 'orthogonal Hamel basis'.

Kavi Rama Murthy
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