Example of an inner product space with no orthonormal basis

Question

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.

Answer 1

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.

Answer 2

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.

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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.