# 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.

• You might be interested in the Lagrange–Bürmann formula. Oct 23, 2017 at 12:20
• I don't see why the resulting function need be derivative-invariant, or even continuous, without some extra assumptions about the type of convergence.
– Dap
Oct 31, 2017 at 7:49
• You might be fully right, yet, if the derivative exists then presumably you can differentiate term-by-term (proof needed :) ). Anyway the question points to a different direction, and I apologize if I asked it incorrectly: Can one find such "n-a-s-m" functions $f_n$ that can be summed (<$\infty$) and the sum can be differentiated term by term (now as assumption) ? Nov 3, 2017 at 9:00

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>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.

• Could you maybe refine on implication " $f_n \rightarrow 0$, and hence that the Taylor series of $f_0$ has infinite radius of convergence"? Dec 15, 2017 at 9:16
• @F.Jatpil: fixing $x,$ the sum $\sum_{n=-\infty}^\infty f_n(x)$ can only converge if $f_k(x)\to 0$ as $k\to +\infty$ (necessary but not sufficient). Plugging this into the quoted formula $1/\rho(x)=\limsup_{k\to\infty}|f^{(k)}(x)/k!|^{1/k}$ gives $1/\rho(x)=\limsup_{k\to\infty}|f^{(k)}(x)|^{1/k}|1/k!|^{1/k}=0,$ i.e. $\rho(x)=\infty.$
– Dap
Dec 15, 2017 at 9:26
• In the last comment, $f=f_0.$
– Dap
Dec 15, 2017 at 9:34