When does a real function together with its derivatives form a basis?

It is indeed enough to assume that all Fourier coefficients of $s$ are non-zero.

The problem actually seems to become quite simple after a Fourier transform: Since $s$ is assumed to be holomorphic on a neighborhood of $[-\pi,\pi]$, we have that $\widehat{s}(k) = O(e^{-2\delta |k|})$.

In terms of the Fourier transforms, our task is to show that the vectors $(k^n \widehat{s}(k))_{k\in\mathbb Z}$, for $n=0,1,2, \ldots$, have dense linear span in $\ell^2(\mathbb Z)$. This is the case: I'm going to use the fact that we can do weighted uniform polynomial approximation on $\mathbb R$ with respect to an exponential weight $w(x)=e^{\delta |x|}$ (this is perhaps an overkill; the result is discussed in Koosis, The logarithmic integral I). In particular, for any finitely supported sequence $b_k$, we can find a polynomial $p(k)$ so that $$ \sup_{k\in\mathbb Z} \left| \frac{b_k}{\widehat{s}(k)} - p(k) \right| e^{-\delta |k|}< \epsilon . $$ In other words, $$ \left| b_k - p(k)\widehat{s}(k) \right| < \epsilon\, |\widehat{s}(k)|e^{\delta |k|} , $$ and this means that I can make $\|b-p\widehat{s}\|_2$ arbitrarily small: first take a ridiculously small $\epsilon$ to keep things under control as long as $\widehat{s}e^{\delta |k|}$ is not small, but eventually this term will take over and make everything exponentially small.

Suppose that $s(z)$ has period $T$ and is analytic on the strip $|\mathrm{Im}(z)| < a$ for some $a>\pi T$. Then I claim that

• The derivatives $s^k(z)$ are dense in the space of all $T$-periodic functions BUT

• Said derivatives are not linearly independent.

I can not currently figure out whether the condition that $a>\pi T$ is necessary or an artifact of the proof method. By rescaling $T$, we may (and do) assume that $T=1$ for this rest of this answer, so we are assuming $a>\pi$.

The linear relations between the derivatives:

I claim that $$\sum_{k=0}^{\infty} \frac{1}{2^{2k+1} (2k+1)!} s^{(2k+1)}(z)=0.$$ I remark that this is only one of many relations, and in particular the simpler relation $\sum_{k=0}^{\infty} \frac{1}{(2k+1)!} s^{(2k+1)}(z)=0$ also holds if $s$ extends to a wide enough strip. But that simpler relation requires (or at least my proof of it requires) $a>2 \pi$ rather than $a> \pi$.

We first compute, then justify. Let the Fourier series of $s$ be $$s(z) = \sum_{m=-\infty}^{\infty} c_k e^{(2 \pi i) mz}.$$ Then $$\sum_{k=0}^{\infty} \frac{s^{(2k+1)}(z)}{2^{2k+1} (2k+1)!} = \sum_{k=0}^{\infty} \sum_{m=-\infty}^{\infty} \frac{ c_m (m \pi i)^{2k+1}}{(2k+1)!} e^{(2 \pi i) mz} = $$ $$\sum_{m=-\infty}^{\infty} c_m e^{(2 \pi i) m z} \frac{e^{m \pi i} - e^{-m \pi i}}{2} = \sum_{m=-\infty}^{\infty} 0.$$

These formal manipulations are justified if the double sum at the end of the first line is absolutely convergent. The first few steps of our bound are easy: $$\sum_{k=0}^{\infty} \sum_{m=-\infty}^{\infty} \left| \frac{ c_m (m \pi i)^{2k+1}}{(2k+1)!} e^{(2 \pi i) mz} \right| = \sum_{m=-\infty}^{\infty} |c_m| \sum_{k=0}^{\infty} \frac{(\pi |m|)^{2k+1}}{(2k+1)!} \leq \sum_{m=-\infty}^{\infty} |c_m| e^{\pi |m|}.$$

Now, $c_m = \int_0^1 s(z) e^{-m(2 \pi i)z} dz$. Since $s(z)$ is periodic and analytic on the strip $|\mathrm{Im}(z)| < a$, we may slide this integral up to a line segment at height $b$ for some $b \in (\pi, a)$ and deduce $c_m = O(e^{-m b})$. Sliding down instead, we get $c_m = O(e^{-|m| b})$. So $\sum_{m=-\infty}^{\infty} |c_m| e^{\pi |m|}$ is bounded by a multiple of the geometric series $\sum_{m=-\infty}^{\infty} e^{(\pi-b) |m|}$ and we are done. $\square$.

Spanning To show that the derivatives of $s$ span the periodic functions, it is enough to show that their span contains $e^{r (2\pi i) z}$, for any integer $r$. Define $$g(u) = (-1)^r \frac{e^{u/2} - e^{-u/2}}{u-r(2 \pi i)}$$ and let $$g(u) = \sum_{k=0}^{\infty} b_k u^k.$$ Then the same formal manipulations as above show that $$\sum_{k=0}^{\infty} b_k s^{(k)}(z) = \sum_{m=\infty}^{\infty} c_m g(m(2 \pi i)) e^{m (2 \pi i) z}.$$ But $g(m(2 \pi i))$ is $1$ for $m=r$ and $0$ otherwise, so we have shown that $e^{r (2 \pi i)z}$ is in the span of the derivatives.

To justify the formal manipuluations, one needs a bound like $|b_k| = O(1/(2^k k!))$. You should be able to get this out of a contour integral on a square of side length $2k$, but it is a bit of work so I'm going to hold off writing it up until I know whether you actually need it for something. (And until I see whether someone comes up with a better method which allows any positive $a$, not just $a > \pi$.)

$L^p$ 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 $L^2([-\pi,\pi])$.