# Example of an inner product space with no orthonormal basis

Let $$X$$ be an infinite-dimensional vector space with an inner product $$(\cdot, \cdot)$$. A system of non-zero vectors $$B = \{ x_\alpha \}$$ from $$X$$ is called orthonormal if $$(x_\alpha, x_\beta) = \begin{cases} 0, & \alpha \neq \beta,\\ 1, & \alpha = \beta. \end{cases}$$ The orthonormal system $$B = \{ x_\alpha \}$$ is called an orthonormal basis for $$X$$ if the subspace of $$X$$ generated by finite linear combinations of elements of $$B$$ is dense in $$X$$.

It is well-known that every separable inner product space has an orthonormal basis. What is some good example of non-separable inner product space with no orthonormal basis? I am interested in any concrete example.

P.S.: The definition of orthogonal basis given includes uncountable cases as well.

• A previous answer of mine. $K$ is an incomplete non-separable inner product space with no complete orthonormal system... math.stackexchange.com/a/201149/442 (Dixmier, 1953) Apr 26, 2023 at 17:39
• Thank you! Apparently, this construction is exactly what I need! Apr 27, 2023 at 4:24

Every Hilbert space $$\mathcal{H}$$ has a maximal orthonormal system $$\{e_\alpha\}_{\alpha\in A}.$$ We restrict to the case of infinite dimensional space $$\mathcal{H}.$$ Consider an arbitrary element $$x\in \mathcal{H},$$ which does not belong to the linear span of $$\{e_\alpha\}_{\alpha\in A}.$$ Due to the Bessel inequality for every $$k\in\mathbb{N}$$ the set $$\{\alpha\in A\,:\,|\langle x,e_\alpha\rangle| \ge 2^{-k} \}$$ is finite, therefore the set $$\{\alpha\in A\,:\,\langle x,e_\alpha\rangle \neq 0\}$$ is countable. Let $$\{\alpha\in A\,:\,\langle x,e_\alpha\rangle \neq 0\}=\{\alpha_n\}_{n=1}^\infty$$ By the Parseval identity we get $$\|x\|^2=\sum_{n=1}^\infty |\langle x,e_{\alpha_n}\rangle |^2$$ hence $$\lim_N \sum_{n=1}^N \langle x,e_{\alpha_n}\rangle e_{\alpha_n}\underset{N\to\infty}{\longrightarrow}x$$ Summarizing finite linear combination of the elements of $$\{e_\alpha\}_{\alpha\in A}$$ are dense in $$\mathcal{H}.$$

Remarks

1. Let $$V$$ be an incomplete inner product space and $$\mathcal{H}$$ denote its completion. Assume $$V$$ contains an orthonormal basis of $$\mathcal{H}.$$ Then the reasoning above shows that the linear span of the elements of the basis is dense in $$V.$$

2. Let $$V$$ be a separable inner product space and $$S$$ a countable dense subsets of $$V.$$ Let $$S_0$$ be a maximal linearly independent subset of $$S.$$ By the Gram-Schmidt procedure applied to $$S_0$$ we obtain a countable maximal orthonormal set $$B$$ in $$V.$$ Then $$B$$ is an orthonormal basis in $$\mathcal{H}.$$ Moreover $$B$$ is linearly dense in $$V.$$ Hence we can apply the first remark.

3. An inner product space should be incomplete and not separable in order to satisfy the requirement of OP.

SUMMARY. This is not a strict answer to the question. Here, I show a concrete example of a Hilbert space (that is a complete inner product space) that does not admit countable orthonormal bases.

ANSWER. If an inner product space $$X$$ has a countable orthonormal basis, it must be separable. Indeed, letting $$\{e_n\}_{n\in\mathbb N}$$ denote such basis, the set $$\left\{ \sum_{j=1}^N \lambda_j e_j\ :\ \Re \lambda_j\in\mathbb Q,\ \Im\lambda_j\in\mathbb Q,\ N\in\mathbb N\right\}$$ is dense and countable.

So to answer your question it suffices to exhibit an example of a non-separable inner product space. The standard one that I know of is the following: $$X:=\left\{f\colon\mathbb R\to \mathbb C\text{ measurable }\ :\ \lVert f\rVert:=\sqrt{\lim_{T\to \infty}\frac2T\int_{-T}^T \lvert f(t)\rvert^2\, dt}<\infty\right\}.$$ (This is superficially similar to the usual space $$L^2(\mathbb R)$$, but it is actually rather different). The set $$\mathcal X:=\left\{\chi_\xi(x):=\exp(i\xi x)\ :\ \xi\in\mathbb R\right\}$$ is uncountable and orthonormal in $$X$$. Therefore, $$X$$ is not separable.

Remark. Actually, $$\mathcal X$$ is also complete, hence a basis. Proof: let $$f\in X$$ satisfy $$\langle f, \chi_\xi\rangle=0$$ for all $$\xi\in\mathbb R$$, we claim that $$f=0$$. Note indeed that $$\langle f, \chi_\xi\rangle=\lim_{T\to \infty} \mathcal F\left(f\frac{\mathbf 1_{[-1, 1]}(\frac{\cdot}{T})}{2T}\right)(\xi),$$ where $$\mathcal F(u)=\int_{-\infty}^\infty u(x)e^{-ix\xi}\, dx=\widehat{u}$$ denotes the Fourier transform. By the convolution theorem, $$0=\langle f, \chi_\xi\rangle = \lim_{T\to \infty} \left(\widehat{f}\ast\frac{\sin(T\cdot)}{T\cdot}\right)(\xi)=(\widehat{f}\ast \delta)(\xi)=\widehat{f}(\xi),$$ so we conclude that the (tempered distributional) Fourier transform of $$f$$ vanishes, so $$f$$ vanishes as well, as we wanted to prove.

• Probably by "orthonormal basis" you mean countable orthonormal basis... :) Apr 26, 2023 at 17:40
• @paulgarrett: of course! :-) I checked on Rudin's R&CA and indeed he writes of general orthonormal bases, which then in practice are always countable. I wouldn't know how useful a non-countable basis could be, since even summing on an uncountable set is tricky. But in principle one can perfectly well define bases of any cardinality, as you rightfully remark. Apr 26, 2023 at 17:52
• Indeed, I mostly ignore Hilbert spaces with uncountable bases, except to make the point that it's possible. :) And, to bring up the fun point that a finite-valued infinite sum of positive real numbers can be at most countable. So, even in an uncountably-orthonormal-based Hilbert space, any given vector is just a countably-infinite sum of those basis vectors. :) Apr 26, 2023 at 18:36
• Btw, I would say the standard example of a non-separable Hilbert space is $\ell^2(I)$ for $I$ uncountable. ;) Apr 27, 2023 at 9:46
• I would say that incomplete spaces is where it gets really interesting. For complete spaces, the existence of an ONB is well-known. For incomplete spaces things get tricky. An ONB always exists for separable spaces, but may fail to exist in non-separable spaces. Check out the answer linked to in GEdgar's comment to the question. Apr 27, 2023 at 11:51