What's so special about the Orthonormal base $\{e_n\}$ of $L^2[0,1]$, where $e_n(x)=e^{2\pi i nx }$?

Let $$f \in L^2([0,1])$$ . Then Carleson's Theorem states that

$$\lim_{N\to \infty} \sum_{|n| where $$\{e_n\}$$ is the Orthonormal basis of $$L^2([0,1])$$ defined by $$e_n(x)=e^{2\pi in x}$$ and $$\langle\,\cdot\,,\,\cdot\,\rangle$$ is the usual inner product of $$L^2[0,1]$$ defined as $$\langle f,g\rangle:=\int_0^1 f(x)\overline {g(x)} dx.$$

Now my question is: what makes the particular orthonormal base $$\{e^{2\pi i nx}\}$$ so special?

For which Orthonormal basis $$\{w_n\}$$ of $$L^2[0,1]$$, can we say that for every $$f \in L^2[0,1]$$, that $$\lim_{N\to \infty} \sum_{|n|

And what if we ask the same question with $$L^2$$ replaced everywhere by $$C[0,1]$$ ?

• At first it is a property of the convolution with the Dirichlet kernel – reuns Dec 16 '18 at 8:49
• It is not clear what is meant by the second question, about replacing $L^2[0,1]$ with $C[0,1]$. Perhaps an example could be given to explain what the question is there. – Josiah Park Dec 16 '18 at 9:44
• Remark: the functions from $L^2$ do not have well-defined values at specific points, like $f(0)$ – Fedor Petrov Dec 16 '18 at 10:04
• @FedorPetrov: Sure, but I do the convergence to hold for such functions with $f(0)=f(1)$ ... I am nowhere saying that any such $L^2$ function has that property ... – user521337 Dec 16 '18 at 11:25
• I want to say that this property just has no meaning. – Fedor Petrov Dec 16 '18 at 11:42

Yes, there are other systems that have the Carleson convergence property. Notably, Billard proved in 1967 the Walsh Paley case of Carleson's theorem. Often Carleson theorem results are phrased on the real line because one can dilate there. In this setting, for some wavelet packet series $$\{b_{n}\}$$, it is possible to get bounds on the Carleson operator $$Lf(x)=\sup\limits_{N\geq 0} \sum\limits_{n=0}^{N-1} \langle f, b_{n} \rangle b_{n}$$ thus proving convergence a.e. of expansions. Wickerhauser's paper explains how the support and size of the wavelet packets play into getting such bounds.