I was working with my thesis and I needed to use the existence of a countable orthonormal basis of $L^2(\mathbb{R}).$
It should be true that the Hilbert Space $L^2(\mathbb{R})$ is separable, this because the measure space $(\mathbb{R},\lambda),$ where $\lambda$ is the Lebesgue measure, is $\sigma-$finite.
Now, this implies that there exists a countable orthonormal basis, but this comes from an abstract type of reasoning, i.e. the Zorn's Lemma for the existence of an orthonormal basis and the use of separability to say that it is countable.
The question that came up to me is: is there an explicit representation of this basis?
The answer should be no, otherwise the use of the continuous Fourier transform doesn't make much sense, but who knows, perhaps there are some good reasons to prefer the continuous transform even if there is a countable basis.
My idea was to take inspiration from the system $\{sin(\frac{n x}{m})\}_{n,m\in\mathbb{N}},$ but I don't know how to continue.
Best Answer
As in the comments, the Hermite functions will work. You can think of them as being obtained by applying Gram-Schmidt to the functions $x^n e^{- \frac{x^2}{2} }$ (up to some factors of $2$ depending on your conventions).
The point of the Fourier basis of $L^2(S^1)$ is not just that it's a countable orthonormal basis, it's also that it diagonalizes translation, or said another way, that it diagonalizes differentiation; this is why it's useful for solving differential equations, which were the original motivation for Fourier series (Fourier used them to solve the heat equation). The Hermite basis is great but it doesn't do this.