For a finite-dimensional Hilbert space, any orthonormal basis is trivially a Hamel basis (because there's only one natural notion of "basis" in finite dimensions). But for an infinite-dimensional Hilbert space, no orthonormal basis is a Hamel basis, as proven here.
But is it possible for a Hamel basis $B$ for an infinite-dimensional Hilbert space to form an orthonormal set, i.e. $\forall x, y \in B, \langle x, y\rangle = \delta_{xy}$ (the Kronecker delta)?
(There are some linguistic issues here, because unfortunately "a basis that is orthonormal" is not necessarily the same thing as "an orthonormal basis". In the title of my question, I'm refering to an orthonormal set that is also a Hamel basis, not to an orthonormal basis.)
Best Answer
No, a Hamel basis of an infinite-dimensional Hilbert space can't be orthonormal. Consider an infinite sequence of basis elements $\beta_1, \beta_2, \ldots$ that are orthonormal. Take a convergent series in these with nonzero coefficients, say $v = \sum_{n=1}^\infty 2^{-n} \beta_n$. This must be a finite linear combination of basis elements, say $\sum_{j=1}^k c_j \alpha_j$. But if $\beta_n$ is not one of the $\alpha_j$, we have $2^{-n} = (v, \beta_n) = \sum_{j=1}^k c_j (\alpha_j, \beta_n)$, so some $\alpha_j$ and $\beta_n$ are distinct basis elements that are not orthogonal.