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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.

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  • $\begingroup$ You might be interested in the Lagrange–Bürmann formula. $\endgroup$ Oct 23, 2017 at 12:20
  • $\begingroup$ I don't see why the resulting function need be derivative-invariant, or even continuous, without some extra assumptions about the type of convergence. $\endgroup$
    – Dap
    Oct 31, 2017 at 7:49
  • $\begingroup$ 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) ? $\endgroup$
    – F. Jatpil
    Nov 3, 2017 at 9:00

1 Answer 1

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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.

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  • $\begingroup$ Could you maybe refine on implication " $f_n \rightarrow 0$, and hence that the Taylor series of $f_0$ has infinite radius of convergence"? $\endgroup$
    – F. Jatpil
    Dec 15, 2017 at 9:16
  • $\begingroup$ @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. $\rho(x)=\infty.$ $\endgroup$
    – Dap
    Dec 15, 2017 at 9:26
  • $\begingroup$ In the last comment, $f=f_0.$ $\endgroup$
    – Dap
    Dec 15, 2017 at 9:34

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