Is the fourier orthonormal system special? If I chose some arbitary orthonormal system, could I hope for pointwise convergence for nice functions? As the title says, I would like to understand and get intution for fourier expanstion converging pointwise for nice functions.
In particular, I might start by asking how often (of course this isn't well defined), when I pick an orthonormal system with respect to the usual integral inner product, does it give pointwise convergence for nice functions.
Let's define nice functions as analytic (of course I'll be much happer if things hold for just diffrentiable functions for instance).
I'm not sure how to define an arbtiary orthonormal system, so I leave this open to the answerer that is more knowledgeable than me.
Thanks
 A: This is a very interesting question. First, it is not at all obvious how (or whether it is possible) to extend the relatively elementary Fourier-Dirichlet result on pointwise convergence to general orthonormal systems (let's say even for (the simplest, "non-singular") Sturm-Liouville problems with smooth coefficients). Yes, of course, the $L^2$ convergence is the main point, and for utilitarian purposes not so many people ask further. And, yes, everyone imagines that for very smooth functions, the spectral expansion in terms of the eigenfunctions should converge pointwise, etc.
It cannot be quite this simple, unfortunately/interestingly enough. For example, the eigenfunctions for the Dirichlet problem $u''=\lambda u$ and $u(0)=0=u(2\pi)$ are $\sin(nx/2)$. So of course things like the constant function $1$ are in their $L^2$ span. But that $L^2$ expansion cannot possibly converge pointwise to $1$, because $1$ is just $1$, at the endpoints, while all the eigenfunctions are $0$ there. 
Yes, this failure can be viewed as essentially irrelevant. And, indeed, for functions in the corresponding $H^1$ Sobolev space (attached to the operator + boundary conditions), since we have Sobolev imbedding $H^1\subset C^o$, convergence is in $C^o$. 
But, still, the eigenfunctions themselves are only in $H^1$, not $H^\infty$. So there will inevitably be troubles... understandably, at the endpoints, but such troubles propagate in spectral expansions (for the same reasons that Fourier transforms interchange local smoothness and global decay).
A: An orthogonal set $\mathbf A  $ in an inner product space is one such that for all distinct $ x,y \in \mathbf A $, we have $\langle\,x,y\rangle = 0$. An orthogonal set $\mathbf A $ is called orthonormal whenever we have $\lVert x\rVert = 1$ for all $x \in A$. Furthermore, an othonormal set is called complete (or an orthonormal basis) whenever it is not contained in another orthonormal set. By using Zorn's Lemma, we can extend any orthonormal set to a complete orthonormal set. In any Hilbert space, such as that of the square integrable functions $L^2[\pi,-\pi]$ over a closed interval in which we do ordinary Fourier analysis, we can give an expansion of any point in terms of any complete orthonormal set. In particular, if $A=\{a_1, a_2, a_3...\}$ is a complete orthonormal set and if $x$ is any point we have $x =\sum_{i=1}^\infty \langle x,a_i \rangle a_i $.
