0

I must construct an example of a non separable Euclidean space which does not have an orthonormal basis. In particular, we know that such a space must be (uncountably)infinite-dimensional. Otherwise, by the Gram–Schmidt process, we could make the $n$ linearly independent vectors that can be found in the space into an orthonormal system.

So by the mere fact that this is an actual linear space, we definetively must have some set of elements $\{\phi_{\alpha}\}$ which are linearly independent. Hence, we could use the Gram-Schmidt process to make them orthogonal (at least for some subset of them).

My actual question is: when you are required to prove the non-existence of an orthogonal basis, does it suffice to show that the space has uncountably many linearly independent elements? Becuase then you can't really use the Gram-Schmidt process? Or are there Euclidean spaces with an uncountably infinite dimension, which indeed have an orthonormal base?

Happy Sunday!

  • 2
    No. $\ell_2$, for example, has uncountable (Hamel) dimension (in fact all infinite dimensional complete spaces do). This MO post seems relevant. – David Mitra Jul 25 '21 at 14:37
  • 1
    "Orthonormal basis" may be uncountable. Your example must be non-separable, must be non-complete. – GEdgar Jul 25 '21 at 14:44
  • 1
    My post describing Dixmier's example: https://math.stackexchange.com/a/201149/442 (A non-separable inner product space, where any orthonormal system is countable, and thus there is no orthonormal basis) – GEdgar Jul 25 '21 at 14:48
  • I am finding it hard to understand the example in your post @GEdgar. Do you know maybe any simpler example of an Euclidean space without an orthonormal basis? I recalled now that the elements obtained as an infinite sum of these elements also belong to the space, which is beginning to cause some confusion. After all, they are linearly independent elements in a vector space. And I get even more confused when I think of the "uncountable" sum of these elements, when this set is made up of uncountably many elements. –  Jul 27 '21 at 13:43

1 Answers1

0

It's not enough. For instance, consider the space of functions $\mathbb R\to\mathbb R$ which vanish at all except finitely many points. The functions which are $0$ everywhere except at a single point, where they are one, form a basis of this space, and there are uncountably many of these (one for every point on the real number line). Now define the inner product

$$\langle f,g\rangle:=\sum_{x\in\mathbb R}f(x)g(x).$$

You'll see that the basis given above is orthonormal with respect to this inner product.

Vercassivelaunos
  • 13,226
  • 2
  • 13
  • 41
  • And to be a basis, shouldn't every element of the space be represented by a \textbf{finite} linear combination of these elements? The function that takes the value 1 at each integer and 0 otherwise, can not be represented by any finite linear combination of these functions. –  Jul 25 '21 at 15:16
  • 3
    @MathiasBarreto The function is not an element of the space, as it has positive value at an infinite number of points. – Xander Henderson Jul 25 '21 at 15:30