Let us denote $T$ by the unit circle. Let $\{e\_n\}$ be an orthonormal basis for $L^2(T)$, with respect to Lebesgue measure. 


We say $\{e\_n\}$ is ***smooth*** if it satisfies the following property:


$$f(t){=}\lim\_{N\to \infty}\sum\_{-N}^{N}\langle f,e\_n\rangle e\_n(t)~~~~~(\forall f\in C(T))$$ 


Q. Does there exists any smooth orthonormal basis for $L^2(T)$?



Point-wisely dense orthonormal basis
Yes, you just need to find a Schauder basis of $C(T)$ that is an orthonormal basis of $L^2(T)$ at the same time. On the unit interval this can be done using the Franklin system, which has a periodic version, suitable for the unit circle. If you want the basis indexed by $\mathbb{Z}$ instead of $\mathbb{N}$, you can just reindex it. The coefficients of the decomposition in $C(T)$ are given by the inner products, since convergence in $C(T)$ implies convergence in $L^2(T)$, thus it has to be the usual orthonormal decomposition.


I must say I do not understand why you decided to call such a basis smooth.


