TITLE: What's so special about the Orthonormal base $\{e_n\}$ of $L^2[0,1]$, where $e_n(x)=e^{2\pi i nx }$?
QUESTION [7 upvotes]: Let $f \in L^2([0,1])$ . Then Carleson's Theorem states that
$$\lim_{N\to \infty} \sum_{|n|<N} \langle f,e_n\rangle e_n(x)=f(x),\quad\text{a.e. } x\in[0,1],$$
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  $$\lim_{N\to \infty} \sum_{|n|<N} \langle f,w_n\rangle w_n(x)=f(x),\quad\text{a.e. } x\in[0,1]?$$ 
And what if we ask the same question with $L^2$ replaced everywhere by $C[0,1]$ ?

REPLY [5 votes]: 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.