Basis of $L^2(\mathbb R)$ and Fourier transform.

Question

I know that $\{e^{inx}\}_{n\in\mathbb N}$ is a basis of $L^2(\mathbb S^1)$ where $\mathbb S^1=\mathbb R/\mathbb Z$. Using this result, and the fact that $$\left<f ,g \right>=\int_0^1 f(x)e^{-inx}\mathrm d x,$$ is a scalar product over $L^2(\mathbb S^1)$, we can wrrite any function $$f:\mathbb S^1\longrightarrow \mathbb R$$ as $$f(x)=\sum_{n\in\mathbb N}\int_0^1f(x)e^{-iny}\mathrm d ye^{inx},$$ or, as usually denoted, by setting $$c_n=\int_0^1f(x)e^{-inx}\mathrm d x,$$ we write $$f(x)=\sum_{n\in\mathbb Z}c_ne^{inx}=:Sf(x),$$ which is called it's Fourier Series.

Question : Now, does $\{e^{i\alpha x}\}_{\alpha \in \mathbb R}$ a basis of $L^2(\mathbb R)$ ?

If yes, then (without rigor), for $f:\mathbb R\longrightarrow \mathbb R$, we could write $$f(x)=\int_{-\infty }^\infty \int_{-\infty }^\infty f(y)e^{-i \alpha y}\mathrm d ye^{i\alpha x}\mathrm d \alpha ,$$ what is in fact exactly the inversion of Fourier transform, i.e. $$f(x)=\int_{-\infty }^\infty \hat f(\alpha )e^{i\alpha x}\mathrm d \alpha .$$

To me, if my conjecture that $L^2(\mathbb R)=span\{e^{i\alpha x}\}_{\alpha \in\mathbb R}$, then this formula would make totally sense (as replacing $$\sum_{\alpha \in\mathbb R}\hat f(\alpha )e^{i\alpha x}\quad \text{by}\quad \int_{\mathbb R}\hat f(\alpha )e^{i\alpha x}\mathrm d \alpha,$$ since an integral "can be seen" as a continuous sum.) Of coure that to have the existence of the Fourier inverse, we need that $f$ is Schwartz, but as I said, I ask the question without rigor; in other words, we suppose that we have all good conditions for that things exist.

Answer 1

Assuming you're talking about Hilbert-basis : the $e^{inx}$ family is an Hilbert basis of $\mathbf R / 2\pi \mathbf Z$, the $e^{ix\alpha}$ family isn't an Hilbert basis of $L^2(\mathbf R)$. The first reason is that the elements of the family are not in $L^2(\mathbf R)$.

But this is not the only obstruction. The cardinality of an hilbert basis is constant, ie 2 hilbert basis of a same hilbert space must have the same cardinality (which allow one to talk about "hilbert dimension"), which isn't the case here. In particular a Hilbert space with countable hilbert basis is separable, a hilbert space with uncountable basis isn't separable.

Now $L^2(\mathbf R)$ is indeed separable, and the Hermite polynomials are an Hilbert basis, and yes you can do "fourier series" on $L^2(\mathbf R)$ : if $f\in H$ a separable Hilbert space and $(e_n)_{n\in \mathbf N}$ is a hilbert basis then $f=\sum_{n\geq 0} (f|e_n)e_n$.

Now for your idea of using the $e^{i\alpha x}$ as a "basis" to explain fourier transform can be formalized using representation theory.

Last thing to note, exercise 18 chapter 4 of real and complex analysis by Walter Rudin make you construct a Hilbert space with hilbert basis $(e^{i\alpha x})_{\alpha \in \mathbf R}$ (but with a different scalar product), as i said this space is not separable and not isometric to $L^2(\mathbf R)$.

Answer 2

The functions $e^{i\alpha x}$ are not in $L^2$, which means they cannot be a basis. However, there is a general principle associated with Sturm-Liouville theory that can help give you an approximation to what you want. Integrals over any small interval of the parameter $\alpha$ are in the space, and, for such intervals $I$, $J$, you have orthogonality: $$ \left\langle \int_{I} e^{i\alpha x}d\alpha,\int_{J} e^{i\alpha x}d\alpha\right\rangle =0, \;\; \mbox{ if $I\cap J$ is of measure $0$ }. $$ Furthermore, if $|I|$ is the length of the interval $I$, then $$ \left\|\frac{1}{\sqrt{2\pi|I|}}\int_{I}e^{i\alpha x}d\alpha\right\|^2=1. $$ So you can divide up the real axis into very small disjoint intervals $I_n$, and you end up with an orthonormal set $$ \left\{\frac{1}{\sqrt{2\pi|I_n|}}\int_{I_n}e^{i\alpha x}d\alpha\right\}_{n=-\infty}^{\infty} $$ If you let $e_n$ denote the integral over $I_n$, and you attempt to expand in this basis $\{ e_n \}$, you end up with what looks like a type of Riemann integral approximation for the Fourier integral inversion problem. $$ \sum_{n=-\infty}^{\infty}\langle f,e_n\rangle e_n = \frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\left(\frac{1}{|I_n|}\int_{I_n}\hat{f}(\alpha)d\alpha\right)\int_{I_n}e^{i\alpha t}dt $$ The term in parentheses is the integral average of the Fourier transform over $I_n$. So this is very much like a Riemann approximation of the inversion integral applied to the Fourier transform. And it makes Mathematical sense. If the function $f$ is constant on each of intervals $I_n$, then (ignoring values at endpoints) the above is a correct inversion integral to give you back $f$.