When does a real function together with its derivatives form a basis? Consider real functions defined on an interval, let's say $[-\pi,\pi]$. Let $s(x)$ be an analytic, periodic function such that all of its Fourier components are non-zero. We build a set of functions from $s(x)$ and all of its derivatives:
S=$\{s(x),s'(x),s''(x),\dots\}$
What is the condition for $S$ to be a basis?
By "basis" I mean that if I take any $f(x)\in L^2([-\pi,\pi])$ and produce a series of functions $f_N(x)$ which are the projections of $f(x)$ to the subspace spanned by the first $N$ elements of $S$, then $f_N(x)\to f(x)$ in $L^2$ norm. 
What I understood is that if we want that $S$ be a basis, then it is necessary that all Fourier components are not zero. But is this enough or is there some other condition for $s(x)$?
I did not get far with any proofs. All I understand is that if I take the first few members of the set and I project to the space constructed from the first few Fourier components, then we will have a basis there. But I can't go further.

$L^p$ spaces for $1\leq p\leq \infty$ are complete normed vector spaces. Also, $L^2([-\pi,\pi])$ has infinite dimension since $\{1, x, x^2, \ldots\}$ are linearly independent.
Now the key part: no infinite dimensional, complete normed vector space has countably infinite dimension. This follows from the Baire Category theorem. As a rough sketch of the proof, suppose $\{v_1, v_2,\ldots\}$ were a countable basis. Set $X_i=\text{span}\{v_1,\ldots v_i\}$. Then


*

*$X=\bigcup_{i=1}^\infty X_i$

*$X_i$ is closed in $X$ since it is a complete subspace

*$X_i$ has empty interior since it is a proper subspace

*$X$ is a Baire space since it is a complete metric space


These four observations lead to a contradiction. In particular, your countable set $\{s(x), s'(x), s''(x),\ldots\}$ is not a basis for $L^2([-\pi,\pi])$.
