Is the Fourier Basis a complete basis of $L^2$ if we consider functions that differ on a set of Lebesgue-measure 0 to be distinct?

Question

I am currently working on replicating some results from the following paper:

https://arxiv.org/abs/1803.00798

https://onlinelibrary.wiley.com/doi/abs/10.1002/jae.2846

In it, the authors define the square-integrable functions somewhat unconventional in that they consider two functions that differ on a set of Lebesgue-measure zero to be distinct. So effectively, we are not looking at the equivalence classes but the functions themselves.

If I understand the paper correctly, the authors later make use of the fact that the Fourier basis is a complete orthogonal basis of the space of square-integrable functions when approximating a functional integral over $L^2$. My question is, whether this way of defining $L^2$ creates a problem with the desired properties of the Fourier basis. Sadly, my theoretical knowledge in this area is somewhat lacking and I would be glad to have any advice.

[I originally asked this question on CrossValidated, but I was told that this is probably the more appropriate place.]

Answer 1

The notion of a basis doesn't really work in the context that you are proposing. To understand why, I'll have to explain a few things first.

When we say the Fourier basis is a basis (putting aside orthogonality for now), what we really mean is a Schauder basis. This is a term used in the study of normed vector spaces (I don't like to assume any mathematical knowledge, so if you're not familiar with normed vector spaces, I'd recommend you read about them first). Specifically, here is the definition of a Schauder basis.

Definition: Let $(V, \|\cdot\|)$ be a normed vector space. A sequence $(e_n)_{n \in \mathbb{N}}$ of vectors in $V$ is called a Schauder basis if for any vector $v \in V$, there exists a unique sequence of scalars $(\alpha_n)_{n \in \mathbb{N}}$ such that $$ v = \sum \limits_{n=1}^{\infty} \alpha_n e_n. $$ More specifically, what this sum means is that $$ \lim \limits_{N \rightarrow \infty} \left\| v - \sum \limits_{n=1}^N \alpha_n e_n \right\| = 0. $$

So, when we say that the Fourier basis is a basis for $L^2[a,b]$ (where $[a,b]$ is an interval), what we mean is that it is a Schauder basis. This makes sense because $L^2[a,b]$ is a normed vector space. In particular, the norm of an element $f \in L^2[a,b]$ is given by: $$ \|f\|_2 = \left( \int_a^b f(x)^2 dx \right)^{1/2} $$ Here's the problem. If we're considering the elements of $L^2[a,b]$ to be functions and not equivalence classes (i.e. functions which differ on a measure-zero set are distinct), then this is no longer a norm! Specifically, it doesn't satisfy the condition that $\|f\| = 0$ implies $f=0$ (in fact, this makes it a pseudonorm). This is precisely why we define $L^2[a,b]$ to consist of equivalence classes of functions: so that we can consider it to be a normed vector space.

If we're not working with a normed vector space, then the notion of a Schauder basis no longer makes sense. Or at least, it doesn't have the same meaning anymore. You may still be able to get the properties you want out of the Fourier basis, but that will depend on what you're using it for.

Answer 2

The authors don't seem to explain what topology they give their space $L_2(\mathcal{I})$, and what they mean by ``orthonormal basis." If they give it the topology induced by the obvious seminorm, then their space is not Hausdorff. In particular, their function $Z(t)$ is not well-defined if they intend the infinite sum to be a limit defined with the seminorm topology. The function $Z(t)$ does make sense, of course, as an element of the usual Hilbert space $L^2(I)$. (This is, by the way, apparently not an uncommon problem for some physicists to run into.)

I believe there's a consensus that one should not discuss correctness of current research on this site, and you should probably consult with your advisor before you do anything else, but I suspect this will be cleared up only by communicating directly with the authors.