A series of nonanalytic-smooth functions $f'_n = f_{n+1}$ with finite sum? Good day,
we know, in analytic functions, examples of derivative "ladders", i.e. an infinite set of functions
$$\{f_n\}_{n=-\infty}^{n=+\infty}$$
$$\frac{d}{dx}f_n=f_{n+1}$$
which have finite sums $$\sum_{n=-\infty}^{n=+\infty} f_n(x).$$ Examples:
1) Function $f_0=1$ with condition for its anti-derivatives $f_{-n}(0) = 0$
2) Bessel function $J_0(x)$ with condition for its anti-derivatives $f_{-n}(0) = 0$
Such sums necessarily evaluate into $\alpha\times\exp(x)$ (because are derivative-invariant).
My question: Is it possible to find a set of smooth non-analytic functions $f_n$ with this property?
If yes (my motivation), I think it is interesting to have a non-analytic expansion of an  analytic functions (e.g. $\exp$). All one asks for is a finite sum (and smoothness), what one gets is analyticity.
 A: It is not possible to find such a sequence $f_n.$ This follows from the following theorem from A theorem on analytic functions of a real variable by R. P. Boas, Jr. The proof is an application of the Baire category theorem, similar to a classic problem about functions satisfying $f^{(n(x))}(x)=0$ for each $x$ for some integer $n(x)$.

Let $f(x)$ be a function of class $C^\infty$ on $a\leqq x\leqq b.$ At each point $x$ of $[a, b]$ we form the formal Taylor series of $f(x),$
$$\sum\limits_{k=0}^\infty \frac{f^{(k)}(x)}{k!}(t-x)^k.$$
This series has a definite radius of convergence, $\rho(x),$ zero, finite, or infinite, given by $1/\rho(x)=\overline{\lim}_{k\to\infty}|f^{(k)}(x)/k!|^{1/k}.$ The function $f(x)$ is said to be analytic at the point $x$ if the Taylor development [I believe this just means Taylor series - Dap] of $f(x)$ about $x$ converges to $f(t)$ over a neighborhood $|x-t|<c,$ $c>0,$ of the point; $f(x)$ is analytic in an interval if it is analytic at every point of the interval. [...]
THEOREM A. If there exists a number $\delta>0$ such that $\rho(x)\geqq\delta$ for $a\leqq x\leqq b,$ $f(x)$ is analytic in $[a, b].$

For each $x,$ the convergence of $\sum_{n=-\infty}^\infty f_n(x)$ implies that $f_n\to 0,$ and hence that the Taylor series of $f_0$ has infinite radius of convergence. By Theorem A, this is enough to force $f_0$ to be analytic.
