# Can a Hamel basis for an infinite-dimensional Hilbert space be orthonormal?

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.)

• Note that if you replace "Hilbert space" with "inner product space" the answer changes. E.g., consider $\mathbb R^{(\mathbb N)}$, the space of all sequences of real numbers with finitely many nonzero entries, with the infinite dot product. Then $\{e_n \,|\, n \in \mathbb N\}$ is a Hamel basis and orthonormal. Commented Jun 10 at 19:40

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.

• Another remark. An orthonormal basis for Hilbert space $L^2[0,1]$ has the cardinal $\aleph_0$, but a Hamel basis for that space has the larger cardinal $2^{\aleph_0}$. Commented Aug 14, 2018 at 16:21
• @GEdgar I think you have misunderstood my question, which has nothing to do with orthonormal bases. Commented Aug 14, 2018 at 19:12
• OK, improve my remark to: any orthonormal set is countable, but any Hamel basis is uncountable. Therefore, no Hamel basis is orthonormal. Commented Aug 14, 2018 at 21:11
• @GEdgar Although your comment is from several years ago, I just want to clarify one point: The comment refers specifically to $L^2[0,1]$, correct? In general, orthonormal sets can be uncountable (in non-separable, infinite dimensional Hilbert spaces) and Hamel bases can be countable and even orthonormal (in inner product spaces of countable algebraic dimension). Robert's answer, on the other hand, shows that no Hamel basis of any Hilbert space can be orthonormal. Commented Jun 10 at 19:33

No, and this follows from the answer to the question you linked.

Indeed, the proof of 'No orthonormal basis can be a Hamel basis' presented there actually shows 'No orthonormal set can be a Hamel basis of its completion' and hence 'No Hamel basis [of a complete space] can be an orthonormal set'.

An orthonormal set can be uncountable. Consider the space A.P. of almost periodic functions on $\mathbb{R}$. Its orthonormal basis is the set $\{\exp(ixt)\}$ with $x$ in $\mathbb{R}$, inner product $$(f,g) = \lim_{T\rightarrow \infty} \frac{1}{2T}\int_{-T}^{+T} (f(t)\cdot g*(t))\ dt$$, where g*(t) is the complex conjugate of g(t). The concept of a Hamel basis of an infinitely-dimensional space is improper, even though popularly used. My email: [email protected]

• "i" was missing in the definition of the base. I added it. The base is {exp(ixt)} Commented Aug 26, 2018 at 12:37